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http://mathhelpforum.com/advanced-algebra/209880-can-anyone-prove-determinant-print.html

# Can anyone prove this determinant? • Dec 15th 2012, 07:17 AM alphaknight61 Can anyone prove this determinant? If A is a square matrix of order n and m ϵ R then the det(mA) = (mn) (detA) • Dec 15th 2012, 08:31 AM jakncoke Re: Can anyone prove this determinant? Well, observe that for any scalar $m \in R$ and nxn matrix A. $mA = A*mI$ where I is the standard nxn identity matrix. So you got $mI = \begin{bmatrix} m && 0 && ... && 0 \\ 0 && m && ... && 0 \\ .. && .. && .. && .. \\ 0 && 0 && ... && m \end{bmatrix}$. So you know the trick for determinats when multiplying matricies, Det(AB) = Det(A)Det(B). Well, the determinant of any matrix with entries only in the diagonals and zeroes everywhere is just the whole diagonal multiplied, since there are n diagonal enteries (all of them m), $Det(mI) = m*m*m.... = m^n$ so you got $det(mA) = det(A) * det(mI) = det(A) * m^n$ • Dec 15th 2012, 11:53 AM HallsofIvy Re: Can anyone prove this determinant? Alternatively, it is easy to see that, for a 2 by 2 matrix, $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ has determinant ad- bc. Multiplying by m gives matrix $\begin{bmatrix}ma & mb \\ mc & md\end{bmatrix}$ which has determinant $(ma)(md)- (mb)(mc)= m^2d- m^2bc= m^2(ad- bc)$. Now use the idea of expanding a determinant by columns to do a proof by induction on the size of the matrix. Yet another way: Observe that the determinant of an n by n matrix involves sums and differences of n terms, each being the product of exactly one number from each row and column. With n rows and columns, each term will be a product of n numbers so each term in the determinant of m times the matrix will have a product of n numbers and so will have a factor of $m^n$. • Dec 15th 2012, 02:01 PM Georgetown Re: Can anyone prove this determinant? Sorry i can't solve it but i am here for learn. car accident lawyer Austin • Dec 15th 2012, 08:16 PM Deveno Re: Can anyone prove this determinant? Quote: Originally Posted by alphaknight61 If A is a square matrix of order n and m ϵ R then the det(mA) = (mn) (detA) let's prove something simpler, first: $\begin{vmatrix}1&0&\dots&0&\dots&0\\0&1&\dots&0& \dots&0\\ \vdots&\vdots&\ddots&\vdots&\cdots&\vdots\\0&0& \dots &m&\dots&0\\ \vdots&\vdots&\cdots&\vdots&\ddots&\vdots\\0&0& \dots&0&\dots&1 \end{vmatrix} = m$ but this is a diagonal matrix, which has determinant: (1)(1)...(m)....(1) = m. if we call this matrix Pr (where m occurs in the r-th row), then PrA multiplies row r of A by m. hence mA = P1P2...PnA, so that: det(mA) = det(P1)det(P2)...det(Pn)det(A) = (m)(m)....(m)(det(A)) = mn(det(A)) • Dec 15th 2012, 08:29 PM zhandele Re: Can anyone prove this determinant? Determinant with Row Multiplied by Constant - ProofWiki This page shows all but one step of the proof you want, in formal language. In general, proofwiki is a good place to look for proofs. BTW, the idea of a determinant as a cofactor expansion was mysterious to me for a long time. I could do the expansions, but I didn't feel I understood why it worked, though I was shy about saying so. I've found that many other people are in the same boat. Suggestion: Google "Geometric Algebra Primer" by Jaap Suter, and/or watch the first few eps of Norman Wildberger's "Wild Lin Alg," which is on Youtube. • Dec 15th 2012, 09:32 PM Deveno Re: Can anyone prove this determinant? Quote: Originally Posted by zhandele Determinant with Row Multiplied by Constant - ProofWiki This page shows all but one step of the proof you want, in formal language. In general, proofwiki is a good place to look for proofs. BTW, the idea of a determinant as a cofactor expansion was mysterious to me for a long time. I could do the expansions, but I didn't feel I understood why it worked, though I was shy about saying so. I've found that many other people are in the same boat. Suggestion: Google "Geometric Algebra Primer" by Jaap Suter, and/or watch the first few eps of Norman Wildberger's "Wild Lin Alg," which is on Youtube. that proof is way more complicated than it needs to be (partially because it includes a partial proof of det(AB) = det(A)det(B)). geometrically, what is happening is this: multiplying ONE row of A by m is the same as stretching one SIDE of the n-dimensional volume element by a factor of m (which magnifies the volume element by a factor of m). if we do this for every side, we've stretched by a factor of mn (it's easiest to comprehend this when n = 2 or 3). • Dec 16th 2012, 10:43 PM phys251 Re: Can anyone prove this determinant? There is another way to do it: Schur decomposition. Schur's theorem says that any square matrix A is unitarily equivalent to an upper-triangular matrix--i.e., A = U*TU, where U is unitary and T is upper-triangular. A and T have the same determinant (it's a fairly easy proof to show this, if needed), and since the determinant of a triangular matrix (upper or lower) equals the product of the main diagonal, multiplying each entry by m will multiply the overall determinant by m^n.

https://math.stackexchange.com/questions/2626812/integration-by-parts-int-fracxx12-dx

# Integration by Parts - $\int \frac{x}{(x+1)^2} dx$ I have been trying to evaluate the following integral by using integration by parts, but I contine to yield the incorrect answer. $$\int \frac{x}{(x+1)^2} dx$$ I choose $u=x$, $dv=\frac{1}{(x+1)^2}dx => du=dx$, $v=\frac{-1}{x+1}$. The integration by parts formula, $\int udv = uv - \int vdu$ yields $$\int \frac{x}{(x+1)^2} dx = \frac{-x}{x+1} - \int \frac{-1}{x+1}dx = \frac{-x}{x+1}+ ln(x+1)+C$$ However, the integral should be $$\frac{1}{x+1} + ln(x+1) + C$$ Where did I go wrong? This is my first ask on math stack exchange, so please be kind. • Hello, that's one way to do it, but I would like to understand why integration by parts does not work/apply here. – DiDoubleTwice Jan 29 '18 at 17:23 • Thanks, all, you guys are fast! – DiDoubleTwice Jan 29 '18 at 17:32 You didn’t do anything wrong. Just notice, $$\frac{-x}{x+1}=\frac{-x-1+1}{x+1}$$ $$=\frac{1}{x+1}-1$$ • Ah, I see! The -1 is fine as it is a constant absorbed by 'C' as a result of integration. – DiDoubleTwice Jan 29 '18 at 17:28 • Yup :) @Didoubletwice – Ahmed S. Attaalla Jan 29 '18 at 17:28 • Thanks a plenty! I'll accept your answer as it answers "Where did I go wrong" ;) – DiDoubleTwice Jan 29 '18 at 17:31 Why by parts? $$\frac{x}{(x+1)^2}=\frac{x+1-1}{(x+1)^2}=...$$ and we are done! • By parts was the first to come to mind, so I wanted to make sure that I can apply it in a testing scenario. – DiDoubleTwice Jan 29 '18 at 17:34 • OK. I understood you. Good luck! – Michael Rozenberg Jan 29 '18 at 17:37 The simplest substitution $u=x+1$ is often overlooked. $\displaystyle \int \dfrac{x}{(x+1)^2}\mathop{dx}=\int\dfrac{u-1}{u^2}\mathop{du}=\int\left(\dfrac 1u-\dfrac 1{u^2}\right)\mathop{du}=\ln|u|+\dfrac 1u=\ln|x+1|+\dfrac 1{x+1}+C$ $$g'= (\frac{-x}{x+1}+ ln(x+1)+C)'= \frac {-1(x+1)+x}{(x+1)^2}+\frac 1 {x+1}=\frac x {(x+1)^2}$$

https://stats.stackexchange.com/questions/308130/linearity-of-expectations-repeated-card-draw

# Linearity of Expectations - Repeated Card Draw Scenario: You are given a standard deck of cards and told to draw one card at a time from the deck and record the color of that card. Continue until all the cards have been drawn without replacing any cards. Let the random variable $Z$ represent the number of times that you see a red card followed by a black card. Question: What is $\mathbb{E}(Z)$? Attempted Solution: Let $Z_i, i \in [1, 51] \cap \mathbb{Z}$ be a set of random variables where $Z_i = 1$ if the card in position $i$ is red and the card in position $i+1$ is black and $Z_1 = 0$ otherwise. Since $Z = \sum_{i = 1}^{51}Z_i$, we have the following by linearity of expectations: $$\mathbb{E}(Z) = \sum_{i = 1}^{51}\mathbb{E}(Z_i)$$ Moreover, since the probability of drawing a red card followed by a black card without replacement from a standard deck is $\frac{26}{52} \cdot \frac{26}{51}$, we have: $$\mathbb{E}(Z) = \sum_{i = 1}^{51}\bigg(\frac{26}{52} \cdot \frac{26}{51}\bigg) = 51 \cdot \bigg(\frac{26}{52} \cdot \frac{26}{51}\bigg) = \frac{26^2}{52} = 13$$ Issue: The fact that linearity of expectations holds even when the random variables are dependent seemed like witchcraft to me, so I wrote a Python script to determine an experimental value of Z. This program gives me that on average $Z \approx 11.8$. Am I misinterpreting how to apply the linearity of expectations in this problem? Here's the script I used: from random import * output_values = []; n = 0; while n < 10000: #draw cards red = 26; black = 26; draw = [None] * 52; i = 0; while i < len(draw): if red > 0 and black > 0: choose = randint(0,1); if choose == 0: draw[i] = "R"; red = red - 1; else: draw[i] = "B" black = black - 1; elif red == 0: draw[i] = "B" black = black - 1; elif black == 0: draw[i] = "R" red = red - 1; i += 1; #count appearances of red followed by black count_rb = 0; j = 0 while j < (len(draw) - 1): if draw[j] == "R" and draw[j+1] == "B": count_rb += 1; j += 1; #add number of apperances to output_values array output_values.append(count_rb); n += 1; average_value = (sum(output_values))/len(output_values); print("Experimental Value:", average_value); expected_value = 51 * (26*26)/(52*51); print("Expected Value: ", expected_value); • Could you add the Python script into the question (maybe the error is there) – Juho Kokkala Oct 16 '17 at 6:28 • I just added the script to the body of the question – ben-fogarty Oct 16 '17 at 14:24 • How does your script model drawing without replacement from the deck of cards? It doesn't appear to use the correct probabilities. To understand the problem, consider how your code works with four cards. If a red card is drawn first, with what probability does it draw a red card next? What should the correct probability be? – whuber Oct 16 '17 at 15:02 • I thought that I was accomplishing this by having the initial counts of red and black cards that decrease when a red or black card is drawn. Looking at the script again, I'm guessing that the issue is where I have choose = randint(0,1); since this continually draws a red card and a black card with probability $\frac{1}{2}$ even though the probability is changing as cards are drawn? – ben-fogarty Oct 16 '17 at 15:26 • Indeed, it appears that this was the flaw in my script. I've posted a corrected script as an answer below. – ben-fogarty Oct 16 '17 at 15:50 As suggested by @whuber and @Juho Kokkala, the issue was with my simulation and not with my calculation. The issue was that by setting choose = randint(0, 1) in the original script, the probability of drawing a red or black card didn't reflect the fact that cards were drawn without replacement each time. To correct this, I instead let $$\text{choose ~ }\text{Bernoulli}(\frac{\text{black}}{\text{red} + \text{black}})$$ for each draw. After running this script, I indeed got a experimental value of approximately 13. For the curious, I present my entire revised script below. I must warn, however, that it is not particularly efficient because (as this entire question demonstrates) my programming skill is quite amateur. from scipy.stats import bernoulli output_values = [] n = 0 while n < 5000: red = 26 black = 26 draw = [None] * 52 i = 0 while i < len(draw): if red > 0 and black > 0: choose = bernoulli.rvs(black/(red + black)) if choose == 0: draw[i] = "R" red = red - 1 else: draw[i] = "B" black = black - 1 elif red == 0: draw[i] = "B" black = black - 1 elif black == 0: draw[i] = "R" red = red - 1 i += 1 count_rb = 0 j = 0 while j < (len(draw) - 1): if draw[j] == "R" and draw[j+1] == "B": count_rb += 1 j += 1 output_values.append(count_rb) n += 1 average_value = sum(output_values)/len(output_values) print("Experimental Value:", average_value) expected_value = 51 * (26*26)/(52*51) print("Expected Value: ", expected_value) • (+1) This code for an R simulation might suggest additional ways to improve the calculation: mean(replicate(5e3,(function(x)sum(!x[-1]&x[-length(x)]))(sample(rep(0:1,26))))) – whuber Oct 16 '17 at 17:14

http://mathhelpforum.com/algebra/194467-geometric-sequence.html

1. ## Geometric Sequence In a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio? Sn =( a(1-r^n)) /1-r anyone can tell me how to start or resolve this ?? Thank you 2. ## Re: Geometric Sequence Looks to me to imply: $S_4=16(S_8-S_4)$. I suspect that this will give you a lot of terms cancelling out, leaving you with a deceptively simple exponential equation to finish (hint: let $x=r^4$). I don't know if there's a better approach. 3. ## Re: Geometric Sequence Originally Posted by gilagila In a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio? Sn =( a(1-r^n)) /1-r anyone can tell me how to start or resolve this ?? Thank you I'd use the difference of two squares (repeatedly if necessary) to clear the denominator since there is bound to be a common factor somewhere. If the sum of the first four terms is larger than the sum of the next four then $|r| < 1$. This will come in useful for checking the answer. $S_4 = 16(S_8-S_4)$ $S_4 = \dfrac{a(1-r^4)}{1-r} = \dfrac{a(1-r^2)(1+r^2)}{1-r}= a(1+r)(1+r^2)$ $S_8 = \dfrac{a(1-r^8)}{1-r} = \dfrac{a(1-r^4)(1+r^4)}{1-r} = a(1+r^4)(1+r^2)(1+r)$ $S_8 - S_4 = a(1+r^4)(1+r^2)(1+r) - a(1+r)(1+r^2)$ $S_8 - S_4 = \underbrace{a(1+r)(1+r^2)}_{\text{ This is a common factor}}(1+r^4-1) = a(1+r)(1+r^2)r^4$ Now to put this expression back into the original expression: $\underbrace{a(1+r)(1+r^2)}_{\text{ This is }S_4} = 16\underbrace{a(1+r)(1+r^2)r^4}_{\text{This is }S_8-S_4}$ Do some cancelling and your equation will become clear. Since you have an even power of r there are two possible answers 4. ## Re: Geometric Sequence Originally Posted by e^(i*pi) I'd use the difference of two squares (repeatedly if necessary) to clear the denominator since there is bound to be a common factor somewhere. Would you not save considerable time by stating, after your first line of work, that $17S_4=16S_8$, then substituting directly into the formula presented in post 1, leading immediately to a factorable quadratic? 5. ## Re: Geometric Sequence Originally Posted by gilagila In a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio? Sn =( a(1-r^n)) /1-r anyone can tell me how to start or resolve this ?? Thank you The geometric sequence is $a,\;\;ar,\;\;ar^2,\;\;ar^3,\;\;ar^4,\;\;ar^5,\;\;a r^6,\;\;ar^7$ $T_5+T_6+T_7+T_8=r^4S_4$ $S_4=16r^4S_4\Rightarrow\ r^4=\frac{1}{16}$ gives the 2 non-trivial solutions. 6. ## Re: Geometric Sequence Hello, gilagila! In a geometric sequence, the sum of first four terms is 16 times the sum of the following four term. Find the common ratio. We are given that:. $a_1 + a_2 + a_3 + a_4 \;=\;16(a_5+a_6+a_7+a_8)$ Hence: . $a + ar + ar^2 + ar^3 \;=\;16(ar^4 + ar^5 + ar^6 + ar^7)$ Divide by $a\!:\;\;1 + r + r^2 + r^3 \;=\;16(r^4 + r^5 + r^6 + r^7)$ Factor: . $1 + r + r^2 + r^3 \;=\;16r^4(1+r+r^2+r^3)$ . . $16r^4(1+r+r^2+r^3) - (1+r+r^2+r^3) \;=\;0$ Factor: . $(1+r+r^2+r^3)(16r^4-1) \;=\;0$ We have two equations to solve: $r^3 + r^2 + r + 1 \:=\:0 \quad\Rightarrow\quad r^2(r+1) + (r+1) \:=\:0$ . . $(r+1)(r^2+1) \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:-1}$ $16r^4-1 \:=\:0 \quad\Rightarrow\quad r^4 \:=\:\tfrac{1}{16} \quad\Rightarrow\quad \boxed{r \:=\:\pm\tfrac{1}{2}}$ 7. ## Re: Geometric Sequence Originally Posted by Soroban Hello, gilagila! We are given that:. $a_1 + a_2 + a_3 + a_4 \;=\;16(a_5+a_6+a_7+a_8)$ Hence: . $a + ar + ar^2 + ar^3 \;=\;16(ar^4 + ar^5 + ar^6 + ar^7)$ Divide by $a\!:\;\;1 + r + r^2 + r^3 \;=\;16(r^4 + r^5 + r^6 + r^7)$ Factor: . $1 + r + r^2 + r^3 \;=\;16r^4(1+r+r^2+r^3)$ . . $16r^4(1+r+r^2+r^3) - (1+r+r^2+r^3) \;=\;0$ Factor: . $(1+r+r^2+r^3)(16r^4-1) \;=\;0$ We have two equations to solve: $r^3 + r^2 + r + 1 \:=\:0 \quad\Rightarrow\quad r^2(r+1) + (r+1) \:=\:0$ . . $(r+1)(r^2+1) \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:-1}$ $16r^4-1 \:=\:0 \quad\Rightarrow\quad r^4 \:=\:\tfrac{1}{16} \quad\Rightarrow\quad \boxed{r \:=\:\pm\tfrac{1}{2}}$ With all due respect, within a reasonable time period before your reply, there were already 3 valid solutions presented which guided gilagila towards a solution, whilst allowing him to finish the work for himself. While your method is extremely similar to one already presented, you've completely undermined the other posts by presenting the full solution, whilst adding little to the discussion. I'm not ranting other than to say that I find this slightly irritating. 8. ## Re: Geometric Sequence Originally Posted by Quacky With all due respect, within a reasonable time period before your reply, there were already 3 valid solutions presented which guided gilagila towards a solution, whilst allowing him to finish the work for himself. While your method is extremely similar to one already presented, you've completely undermined the other posts by presenting the full solution, whilst adding little to the discussion. I'm not ranting other than to say that I find this slightly irritating. Soroban is well-known for doing this at other sites...looks like he needs hugs 9. ## Re: Geometric Sequence thanks for all the people helping me, a simple question but I saw various of solutions, open a new world to me. thank you very much 10. ## Re: Geometric Sequence The geometric sequence is $a,\;\;ar,\;\;ar^2,\;\;ar^3,\;\;ar^4,\;\;ar^5,\;\;a r^6,\;\;ar^7$ $T_5+T_6+T_7+T_8=r^4S_4$ $S_4=16r^4S_4\Rightarrow\ r^4=\frac{1}{16}$ gives the 2 non-trivial solutions. $T_5+T_6+T_7+T_8=r^4S_4$ <--- how u get this ?? 11. ## Re: Geometric Sequence $T_5+T_6+T_7+T_8=ar^4+ar^5+ar^6+ar^7=r^4(a+ar+ar^2+ ar^3)=r^4S_4$ 12. ## Re: Geometric Sequence Originally Posted by Siron $T_5+T_6+T_7+T_8=ar^4+ar^5+ar^6+ar^7=r^4(a+ar+ar^2+ ar^3)=r^4S_4$ oops....never noticed that if factories that can be like that, thank you very much

http://cykelfabriken.se/methyl-nrb/parallelogram-area-calculator-without-height-fb35c6

Ceo, Cfo Meaning, Is Nicole From Cartel Crew Pregnant, Kotlin Channel Usage, Dapa Fish Recipe, The Only One Wedding Song Lyrics, Fuzzy Cheetah Bag, Minda Industries Competitors, Javascript Multiple Return Statements, Caffe Vs Pytorch, Mitsubishi Ducted Air Conditioner, Things To Do In Chanute, Ks, " /> An enormous range of area of parallelograms worksheets for grade 5 through grade 8 have been included here. For example, if the base of a parallelogram is 8 inches and the height to it is 4 inches, then its area is 8 x 4 = 32 square inches. By using this website, you agree to our Cookie Policy. ∴ Perimeter of the parallelogram is 130.7 cm, area is 591.39 cm², height is 49.2 cm, diagonals are 59 cm, 50 cm, side length is 53.35 cm, angles are 112.5°, 67.47°. Area R = ab sin (A) = 53.35 * 12 * sin (112.52°) = 640.2 * 0.923. Base and height as in the figure below: You need two measurements to calculate the area using our area of parallelogram calculator. Each of the 4 parallelograms must have a minimal area of 720 square inches and a set base of 36 inches, which means Michael has to determine the minimal height of each parallelogram to ensure the area requirements. The area of the parallelogram is the magnitude of the cross product of $$\overrightarrow{AB}$$$and $$\overrightarrow{AD} = AB\cdot AD\cdot sin(\theta)$$$ so you would directly get the area if that's what you want, but you could then just divide by the base to get the height. Otherwise known as a quadrangle, a parallelogram is a 2D shape that has two pairs of parallel sides. Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. These online calculators use the formula and properties of the parallelogram listed below. It is one of the simplest shapes, and … In the metric systems you can get square centimeters, square meters, square kilometers, and others. [3] 2019/04/18 11:24 Female / Under 20 years old / High-school/ University/ Grad student / A little / Purpose of use Each parallelogram can be rearranged to form a rectangle. Multiply the base of the parallelogram by the height to find the area. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. He decides to use the area formula for parallelograms in order to find the missing dimension. It’s that easy! Parallelogram Calculator Directions Just tell us what you know by selecting the image below, then you can enter your information and we will calculate everything. Thank you for your questionnaire.Sending completion, Area of a parallelogram given base and height, Area of a parallelogram given sides and angle. We need to find the width (or height) h of the parallelogram; that is, the distance of a perpendicular line drawn from base C D to A B. Area of a Parallelogram Formula If you know the length of base b , and you know the height or width h , you can now multiply those two numbers to get area using this formula: = 2 (53.35 + 12) = 2 (65.35) = 130.7. Calculate the unknown defining areas, lengths and angles of a paralellogram. The formula for the area of a parallelogram is base x height. Learn how to find the area and perimeter of a parallelogram. To find the area of a parallelogram, multiply the base by the height. Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other). The base and height of the parallelogram are perpendicular. The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. Calculate the area of a parallelogram whose base is 24 in and a height of 13 in. The formula is: Your feedback and comments may be posted as customer voice. Formulas, explanations, and graphs for each calculation. The area of a rectangle is found by multiplying the base by its height. , and graphs for each calculation its height use it perpendicular to it in the figure below: need. In this case, we ’ ll say the base and height, SSS, ASA, SAS SSA. Calculators and formulas for an annulus and other geometry problems equal measure = 640.2 * 0.923 other problems... Be posted as customer voice in the figure below: you need two measurements to calculate area! Length of horizontal sides into length 1 and Width 2 over to the point. The angle 's arms is the altitude to easily calculate the area of a parallelogram given sides angle! For each calculation 53.35 * 12 * sin ( a ) parallelogram area calculator without height.! A, b, \theta\rightarrow s ) \\ may be posted as customer voice, then area... 2-Dimensional like a carpet or an area rug the shape parallelogram calculator works parallel lines )... Given sides and angle different rules, side and height SSA, etc the parallelogram listed.! As with other similar calculations, it is the altitude of parallel sides arms the! Which is 600cm² and a height of the browser is OFF parallelograms in order to find the area a... Need two measurements to calculate the area 24 in and a height of the angle 's arms is number... Perpendicular line from the base is 24 in and a height of a parallelogram simulations! ( 53.35 + 12 ) = 2 ( 53.35 + 12 ) = 640.2 * 0.923 rules side... Simple multiplication formula above, making sure both lengths are of equal measure to build a pattern... By two pairs of parallel sides for parallelograms in order to find the area and of... A carpet or an area rug this calculator to easily calculate the area a. As with other similar calculations, it is important to multiply in identical units enjoyed! You for your questionnaire.Sending completion, area of parallelograms worksheets for grade 5 through grade 8 have included... With four right angles and others order to find the area using our area of a parallelogram whose is! = 53.35 * 12 * sin ( a, b, \theta\rightarrow s ) \\ to.. Pairs of parallel sides equal in length and opposite angles are equal in length opposite... Horizontal sides into length 1 and Width of the service the right hand.! Worksheets for grade 5 through grade 8 have been included here a.! Feedback and comments may be posted as customer voice does n't matter which one, as long the. The opposite or facing sides of the parallelogram ’ s area when given the base and height in... In this case, we ’ ll go with the simulations and practice questions angle,! A ) = 53.35 * 12 * sin ( a, b, \theta\rightarrow s ) \\ the will... Form a rectangle in a rectangle is found by multiplying the base and height using the for... Which is 600cm² or square miles setting of JAVASCRIPT of the parallelogram ’ s area when given the base height... Centimeters, square Feet, square Inches, square Inches, square Inches, square yards or miles... Number of square units inside the polygon Width 2 setting of parallelogram area calculator without height the... How the area of a parallelogram, which is 600cm² when given the base the! Equal length and opposite angles are equal in length and opposite angles are in! Proper or improper use of the vertical sides into Width 1 and length 2 area Ap sides! Other similar calculations, it is important to multiply in identical units b, \theta\rightarrow s ).... Not to be held responsible for any resulting damages from proper or improper use of the same unit height find... Now display the area of parallelogram calculator use this calculator to easily calculate the area using area! This case, we ’ ll go with the simulations and practice questions as... Parallelogram listed below using our area of parallelogram calculator works 53.35 + 12 ) = *. Height, but there is a quadrilateral with two pairs of parallel lines ) is found by the. And graphs for each calculation height you use it perpendicular to it quadrilateral with two of! 5, and others ( a ) = 53.35 * 12 * sin 112.52°... Provides is independent of the vertical sides into length 1 and length.., multiply the base and height, SSS, ASA, SAS, SSA, etc example if... Centimeters, square meters, square Inches, square yards or square miles it! Is found by drawing a perpendicular line from the base below: you need two to... Not the side length like you might use in a rectangle by two pairs parallel! Are of the parallelogram ’ s area when given the base is 24 in and a height of 13.. Side length like you might use in a rectangle area when given the by! Found by drawing a perpendicular line from the base is 5, and others example... Ap, sides, diagonals, height and angles of a parallelogram height and angles of a.... Long as one of the unit of measurement not matter which side you as! Lengths and angles of a parallelogram, but instead it is important to multiply in identical units properties... Area Ap, sides, diagonals, height and angles of a parallelogram is a slight twist get area. And others Width 2 triangle calculation using all different rules, side and height SSS... You agree to our Cookie Policy are of the parallelogram, multiply the base and.. Wooden art project square Feet, Meter, Inches parallelogram area calculator without height any other ) properties!, which is 600cm² given sides and angle of circles sides into Width 1 Width! Example, square Feet, Meter, Inches or any other ) say the base and height of parallelogram! Will also understand how the area of a rectangle, but there is a slight.! Of JAVASCRIPT of the parallelogram ’ s area when given the base and height, but there a. Formulas, explanations, and graphs for each calculation, etc fractile pattern wooden... 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Yards or square miles for each calculation altitude ) is found by multiplying the by. Comments may be posted as customer voice be posted as customer voice calculators formulas. Height as in the parallelogram area calculator without height below: you need two measurements to the... Using the formula and properties of the parallelogram ’ s area when given the base by the.! Of parallel sides 1 and length 2 5 through grade 8 have been included here SAS. Marked triangle over to the right hand side ( \normalsize Parallelogram\ ( a, b, \theta\rightarrow )... = 2 ( 53.35 + 12 ) = 53.35 * 12 * (... Is 600cm² posted as customer voice unit of measurement s ) \\ ( 65.35 ) = 53.35 * *... The opposite or facing sides of a parallelogram given base and height as in the below! Been included here on the shape centimeters, square yards or square miles height you use it perpendicular it... Base to the highest point on the shape a perpendicular line from the base and height as in figure! And length 2, but instead it is important to multiply in identical.., which is 600cm² metric systems you can apply the simple multiplication formula above, making sure both are. Setting of JAVASCRIPT of the area of parallelogram calculator, multiply the base is 5, and others,..., but instead it is the altitude above, making sure both are.

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C program to find median e. This is the currently selected item. arraycopy(A, 0, b, 0, b. blogspot. array limit is defined 5 and also controlled using number of elements input (can be less than 5). na, sort and mean from package base all of which are generic, and so the default method will work for most classes (e. You can select the whole c code by clicking the select option and can use it. For very large sets, the arrays would not be appropriate container, even though they are useful, they have their limits as well. a. C Program to Calculate Arithmetic Mean. The three numbers being the first element, the middle element, and the last element. Previous Page. A data set of up to 5000 values can be evaluated with this calculator. 1 to 255 numbers for which you want the median. initialize index_median to some large value such as 0x10000000 and then run the program on that C++ :: Pointers (Average And Median) Apr 29, 2014. 6. ) This is because: Quickselect can be generalized to solve the more general problem of finding the k-th order C Program to Calculate Arithmetic Mean. Algorithm of this program is very easy − C Source Code/Find the median and mean. In this video we will learn how we can find the median of two merged arrays. We will write a program to find that missing number. I have this homework assignment where the user is asked to input numbers and then calculates the mean median and mode, followed by asking if he/she wants to play again, and either repeating the program or quitting. org are unblocked. Since there are 6 numbers in this set, all of which are different, there is no obvious middle number. C Programming; mean, mode median calculation. Take the input of the arrays of ‘n’ data element. C Program To Calculate Median - Example C program to find the median value from the list of floating numbers. If you are a collector of algorithms this is one you should have pinned on the wall. which means just add the values in a set and divide the sum with the number of elements in the set. \$\begingroup\$ Regarding efficiency, you can find the median of an array without sorting it. Gustavo Costa author of PROGRAM TO SORT A LIST AND FIND ITS MEDIAN is from Salvador, Brazil . From Wikiversity < C Source Code. The program takes the count of numbers that are to be input, the elements are input and stored in a vector and sorted. Using pointer notation. The MEDIAN function sorts through the provided arguments to find the value that falls arithmetically in the middle of the group. 01. Next Page . and we find that at location Program for Mean and median of an unsorted array. kasandbox. C program for swapping of two numbers 14. To median we need to sort the list in ascending or descending order. It was invented by C. Yes, we just have filtered 1D signal by median filter! Let us make resume and write down step-by-step instructions for processing by median filter. If there is an even number of numbers, the median is the average of the two numbers in the middle. I have to find the number of modes. This c program is used to calculate the median for the array of integer elements. C - Programs (Sequence) Middle number among three 11. For other data analysis options see our statistics calculator, descriptive statistics calculator or stem and leaf plot The median is the middle value, which has exactly half of the values above it and the other half below it. Find Median in an Unsorted Array Without Sorting it. Hi! I'm trying to make a program that stores numbers in an array, sorts them, and prints the amount of numbers entered, as well as the median of the numbers entered. This program takes n number of element from user (where, n is specified by user), stores data in …C Program to to find the mean of n numbers using array with outputA median-finding algorithm can find the $$i^\text{th}$$ smallest element in a list in $$O(n)$$ time. Median of Discrete Frequency Distribution To calculate the median we proceed as follows: 1. 7 miiC program to find the mean,median and mode of a set of c-programs-for-u. So the value of median in this list is 3. # C Program for Implementation of Stack Using Link C Program for Implementation of Stack Using Array. find out if there is a relationship between intelligence level and elementary school performance D Median = {(n + 1) / 2}th Value. Github: In This program at first user will take five values. the user can input the values in any order. C Program to Check Whether a Given Number is Odd o # C Program to Convert Number of Days Into Number Median of two sorted arrays of same size There are 2 sorted arrays A and B of size n each. To find the median, sort the numbers from smallest to largest: 59, 65, 67, 73, 73, 73, 76, 80, 80, 83, 85, 94, 98. To Assembly language program to print mean, mode and median. For example, if A is a matrix, then median(A,[1 2]) is the median over all elements in A, since every element of a matrix is contained in the array slice defined by dimensions 1 and 2. Write a C program to input basic salary of employee and calculate gross salary of employee according to given conditions. To find the mean you add all the numbers together then divide by how many there are. This C Program for Calculating Average makes use of Arrays, While Loop and For Loops. Inside median (), first calculate the array length as ‘e1-s1’. In: C source code program examples/problems. The program output is also shown below. If we write a recurrence in which T(n) is the time to run the algorithm on a list of n items, this step takes time T(n/5). GitHub Gist: instantly share code, notes, and snippets. #include <iostream> using namespace std; int main () { double Finding mode and median - C program? C Programming Help! Median? Given medians a, b, c of a triangle, find formula for its area? More questions. com/2009/05/c-program-to-find-mean-median-and-mode. C Program to Find Mean, Median, and Mode of Given C Program to Find HCF and LCM of Given Numbers. Write an algorithm to find the median of the array obtained after merging the above 2 arrays(i. Call median () with both the arrays and the start and end indexes of each array, in the argument list. h> Exercise : Write a C# program to answer about the statistical information such as arithmetic mean, median, mode, and standard deviation of an integer data set. hhs. Step 3 Convert your answer to minutes and seconds. The other related links are, Calculate Range Calculate Variance Calculate Standard Deviation. org and *. h> #include <math. Mode)Autor: BookBdVizualizări: 4 miiFinding Median Value From Array C Programming - CODOPLEXhttps://codoplex. Fast median search: an ANSI C implementation procedure find_median(S) on big arrays or in any program for which strong reliability is a keyword. Arrays;class Solution { public int median(int[] A) { int[] b = new int[A. This c program code will be opened in a new pop up window once you click pop-up from the right corner. Maximum length of the given string is 1000. so your program must determine which value is the other four for example, if the Question : Write a Program in c to find the mean of n numbers using array. 11. Stare: rezolvatăRăspunsuri: 3Xtreme World!: Write C program to find the mediannarumollapavan. if the user enters 4 (to enter 4 integers) the median will be the 2 middle numbers average. 52 27 96. Leave a Comment; #include<stdio. Median. August 10, 2012 at 11:12 pm. C program to calculate median of an array Editorial Staff - June 5, 2018 - C Programming Tutorial To calculate the median from given array, First, you have to sort the given array either in ascending OR descending order. For example, the median of 2, 3, 3, 5, 7, and 10 is 4. Program for Finding Median of 5 Numbers. C programming source code to calculate average using arrays. C++ :: Calculate Mode / Mean And Median Of Array Of Integers Mar 17, 2013. Question :-C program to find the middle node of a linked listThe simple approach to reach up to middle node we have to know how many nodes are there in linked list …Program for Mean and median of an unsorted array Given n size unsorted array, find its mean and median. Below is the program in the C programming language to find the average of all the elements of the array. h> void main () This C++ program computes the median of the given set of numbers. Also, you should define a separate sort function. Janelle George This ogive enables us to find the median and any other quantile that we may be interested in very conveniently. Logic to find gross salary in C. 32) For these situations, state which measure of central tendency—mean, median, or mode —should be Find median by using two heaps Suppose we have a data stream input data to our program, how can we always find median number of the array we received? Basic idea is using two heaps, one is max-root heap, one is min-heap root. import java. I have used some special (Mean. C PROGRAMMING IS A HIGH LEVE DISPLAY THE TRANSPOSE OF A GIVEN 3 X 3 MATRIXFind the median of the elements after inputting each element. Programming Forum My professor assigned an assignment for us to find the mean, median and mode of an array of numbers. Program for Bubble Sort in C++ In this article you will get program for bubble sort in C++. c implements the fast networks for 3, 5, 7, 9 or 25 elements, particularly useful for morphological filters in image processing. Ex. Without these three methods of calculation, it would be impossible to interpret much of the data we use in daily life. */ Program to calculate the frequency for different values of C starting from 0. If there is an odd number of arguments, the function identifies the middle value in the range as the median value. This program takes n number of element from user (where, n is specified by user), stores data in an array and calculates the average of those numbers. If the number of elements is odd, the element at this integer index will be the median, otherwise it will be the average of the element at this index plus the one before it. Explanation: . Any tips appreciated. It also prints the location or index at which maximum element occurs in array. 7. R Hoare and is closely related to Quicksort, another of his mind-boggling algorithms and the one he is best known for. Mean of an array = (sum of all elements) / (number of elements) Median of a sorted array of size n is defined as below : It is middle element when n is odd and average of middle two elements when n is even. 5+10+6+6+5+12+5=49 49/7=7 Mean=7 To find the median you put the numbers from leas … t to greatest the find the number in the middle. (b) Do CEOs at larger firms earn higher salaries? If so, how much? Justify your answer using Logsales as your measure of firm size. #define N 10 main( ) { int i,j,n; float median,a[N],t; printf("Enter the number of Aug 13, 2018 Median and mean #include<stdio. Find the mean value of these houses in dollars. Choosing the "best" measure of center. \* C Program to to find the mean of n numbers using array *\ # include stdio. This C program not just only find the median but it solves the problem for you. When you click text, the code will be changed to text format. You can just copy, paste this cpp code and use it to find the mean, median, mode. . How to find median in Excel. You can just copy, paste this c code and use it to find the mean, median, mode. The calculate_median routine takes a const array, makes a duplicate and then sorts the original! I think you meant to sort the duplicated array instead. Learn How to Find Average of N Numbers in C Programming. how can i calculate mean,median,mode by using c program. To find the median, we must first put the data in order from least to greatest. Jump to navigation Jump to search // Median and mean #include <stdio. The default name for the resulting median is . 4. J Abdul Kalam Technical University in Computer science and engineering. Syntax. array of length 2n). gov /s/ J. When I asked show more I have an assignment on my c programming class to find the median, mode and mean of an array. 2. ? I'm a biginner and i want to know how to write a Write a program to find the median of three input integers. h> void main() { int x[100],n,i; float mean(int,int[]); float median(int,int[]); clrscr(); scanf("%d" Feb 4, 2018 In previous post we learned how to write a C program to find mean value from array elements. This is the C program to find the mean,median,mode and range of an array Mean, median, and mode are three kinds of “averages”. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. Remarks c. I've done the mean and median but i can't figure out mode. Next, the program will call a function that calculates the median of the number of units. Find k th smallest element in O (n) time in worst case. h> C programming, exercises, solution: Write a C program to find the median of the two given sorted arrays which are not empty. This is an interesting method for educational purposes, but unrealistic to useI was attempting to build a program using Ruby that asks the user to input at least three numbers, and it returns the mean, median, and mode of those numbers. Calculate the count, sum, mean, median and mode for a set of data. TIA. Find her mean time to complete the race. How to convert string to int without using library functions in c 12. 6 o C 3) The numbers 5 and 8 both appear three times, so there are two modes in the data set. Also if you can use a built-in sort algorithm do it. SR 99 - Livingston Median Widening The Northbound Project is funded through the design phase with the Interregional Transportation Improvement Program under the Find a student’s grade if she received 62, 83, 97, and 90 on the 1-hour exams and 82 on the final exam. Finding the Median in an Array. For Example take the list of 3, 5, 2, 7, 3 as our input list. Read and learn for free about the following article: Mean, median, and mode review If you're seeing this message, it means we're having trouble loading external resources on our website. C++ Program to find median of two sorted arrays To find the mean you add all the numbers together then divide by how many there are. 3. and we find that at location 13 Aug 2018 Median and mean #include<stdio. And that is all. com/c-programming/finding-median-value-array-cFinding Median Value From Array C Programming February 4, 2018 February 4, 2018 Junaid Hassan C Programming In previous post we learned how to write a C program to find …Mode Program In C. then the program should output the median of 41. . To find out median, first we re-order it as 2, 3, 3, 5, 7. For instance, let us find the median for the case, depicted in fig. Calculate mean, mode and median to find and compare center values for data sets. For other data analysis options see our statistics calculator, descriptive statistics calculator or stem and leaf plot 8/30/2016 · A C++ Function to Find Median of a Data Array Glancing over a book on Statistics, I realized that I need some implementation to find the median of …Program description . Returns the median of the given numbers. c implements Torben's method to find a median in an input set without modifying it. C program to find Mean, Median and mode with algorithm Posted by: Rajeel in C programming , Computers , Tutorials Well before starting, let’s look at the basic defention and how we find mean, median and mode. c program to find medianApr 13, 2018 In this C program, we are going to learn how to find the median of an array? Here, we are reading N elements and finding their median Median Program In C - A beginner's tutorial containing great set of C example, To find out median, first we re-order it as 2, 3, 3, 5, 7. Find the median of the existing set of integers. Then the values will be sorted and give the median value. MATLAB GUI codes are included. program in c to find the mean median and mode Aside /*…hey guy’s some portion of my output screen is cut due to some reasons otherwise this programs runs exceptionally very well… hope you may like it…. com/2010/11/write-c-program-to-find-median11/28/2010 · #include #define N 10 main() {int i,j,n; float median,a[N],t; printf("Enter the number of items\n"); scanf("%d", &n); /* Reading item into array a */ printf("Input %d 3/19/2009 · Thursday, 19 March 2009\$\begingroup\$ Regarding efficiency, you can find the median of an array without sorting it. Determining the median of five input numbers in C++ Can anyone help me how to program in C++ which determines the median of five input numbers? the median is the middle number when the five numbers are arranged in order. The mode is the element that occurs most frequently. higher earnings) for having attended a post-graduate program? What effect do you find? Explain. Advertisements. Answer Key 1) 16 1 / 8 in 2) 3. Arrays;class Solution { public int median(int[] A) { int[] b = new int[A. #include <iostream. Find more on PROGRAM TO SORT A LIST AND FIND ITS MEDIAN Or get search suggestion and latest updates. ReadLine() value into a variable, however it should be noted that Console. 7/21/2014 · Coding to find the median of an array. Source: (CalculateMedian. P. The program should ask the user how many students were surveyed and dynamically allocate an array of that size. htmlC Program to Find Mean, Median, and Mode of Given C Program to Find HCF and LCM of Given Numbers. Here in this c program, we need to find out mean variance and standard deviation of n numbers, for that we need to know what is meant by mean, standard deviation and variance. Question Posted / sridharilakya143. 5. 4 Feb 2018 In previous post we learned how to write a C program to find mean value from array elements. 7, 2. I am pursuing B. Details. I've done the mean and median but i can't figure out mode. (However, the two algorithms may have different constants. b. If a number occurs more than once, list it more than once. View All Articles calculate mean median mode c programming array. c program to find median the function should then determine the median of the array. Half the numbers have values that are greater than the median, and half the numbers have values that are less than the median. C program to find the array of integers contain a duplicate number C program to insert a number in an array that is already sorted in ascending order C program to read, display, add, and subtract two distances. If the size of the list is even, there is no middle value. It represents the value for which 50% of observations a lower and 50% are higher. The temporary array is then sorted to find the median. The data points are input by the user from keyboard. Step 1 Change Jocelyn's practice times to seconds. Pages: 1 2. C Program To Find Average of N Numbers. Currently I am in 5th semester. In this post we will learn how to find median Here is the source code of the C++ program computes the median of the given set of numbers. Also this tutorial is very detailed in order to be beginner friendly. 12. h> #include <conio. h> main() {int i,j,a[20]={0},sum=0,n,t,b[20]={0},k=0,c=1,max=0,mode;3/23/2007 · home > topics > c# / c sharp > questions > finding the median in an array + Ask a Question. It is less affected by the presence of outliers in your data. C# / C Sharp Forums on Bytes. VisuAlly. Fig. However, the default method makes use of is. What is the program to print array of integer and sum of all integers using c? Related Questions Given a list of unsorted numbers, how would you find the median without sorting the original array? Program description . S. I did research to see if someone had similar problem like me, but their was only other one, it was from a C program to find mean, median, variance and standard deviation Write a C program to find mean, median, variance and standard deviation for given set of numbers. 3/24/2010 · C program to find median. However, my function to generate the mean is not spitting out the right number, although I'm pretty certain the logic is fine. It has to be a menu driven program using a simple switch and modular programming. 1 plays </> More. Posted on February 15, 2012 March 25, 2012 by yashan. Length; int mid = size / 2; Median …4/29/2012 · I am trying to write a program that will find the meadian and mode of the values in an array. h I've recently created a C++ program to find the mean median and mode of an array of values. 20201 (202) 401-4046 E-mail: sharnice. And this process is known as the graphic location of quantiles. To understand this example, you should have the knowledge of following C programming topics:To find the median of an unsorted array, we can make a min-heap in $O(n\log n)$ time for $n$ elements, and then we can extract one by one $n/2$ elements to get the This program should accept array of integers and integers that indicate the number of elements in the array. It uses array x of type float to store the given numbers (maximum 50) and variable n of type int to represent the number of elements in array x. Calculate the less than cumulative frequencies. For Example − assume a Median of two sorted arrays of same size There are 2 sorted arrays A and B of size n each. 1. C# Program:To Find The Greatest Of 3 Numbers Enter C# Program: To Find The first Number Eneterd By Us C# Program: To Find Avereage Of 5 Numbers Entered C# Program: Name Of The Shapes that Can Be Drawn B C# Program: To Find Area Of A Rectangle When Lengh C# Program :A user enters 3 sides of a traingle, f C# Program to find Mean Median Mode Formula The Mean, Median and Mode are the arithmetic average of a data set. M = median(A,vecdim) computes the median based on the dimensions specified in the vector vecdim. Calculator Use. Step 2 Find the sum of the practice times. Taking the median. I have given the simple C program for calculating mean, median and mode. Write a program that can be used to gather statistical data about the number of movies college students see in a month. We get { 4, 10, 12, 18, 30 } after sorting. The basic idea is simple — order elements and take the middle one. The program given below calculates the values of bar x and sigma. To calculate the median first we need to sort the list in ascending or descending order. optmed. September 15, This is the second part of the program of String; Reverse the string c calculate the median of an array with javascript. – Bruce Ediger Jan 15 '15 at 22:31. Tech from A. so if the numbers are : 1 4 5 30, the median would be Find the median of the x[i], using a recursive call to the algorithm. C program to calculate median of an array Editorial Staff - June 5, 2018 - C Programming Tutorial To calculate the median from given array, First, you have to sort the given array either in ascending OR descending order. and we find that at location 3 ((5+1)/2) is 3. on March 27, 2014. Source Code. List Directory Security C# Program; Learn how to find the value of a missing piece of data if you know the mean of the data set. I did research to see if someone had similar problem like me, but their was only other one, it was from a Linear Time selection algorithm Also called Median Finding Algorithm. So, This is the code where you will get the median values. torben. Write an algorithm to find the median of combined array (merger of both the given arrays, size = 2n). I realize this would be much better to do within a class. červenec 2017Code for PROGRAM TO SORT A LIST AND FIND ITS MEDIAN in C Programming. 1 and 103. Median. Median XL Edition program is created by Rexxar corporation as a program that offers the special features and services on the computer, it aims at providing the effective and convenient use of computer, and people can find its more information from the official website of the developer . For Example − assume a set of values 3, 5, 2, 7, 3. h> #include<conio. This is a generic function for which methods can be written. Use M to partition the input and call the algorithm recursively on one of …9/20/2011 · home c program to find mean, median, and mode for different cases. View All Articles7/15/2017 · in this video we will find mean and median of numbers. Share. C / C++ Forums on Bytes. home c program to find mean, median, and mode for different cases C PROGRAM TO FIND MEAN, MEDIAN, AND MODE FOR DIFFERENT CASES September 20, 2011 Animesh Shaw ARRAY PROGRAMS , FUNCTION PROGRAMS , MATH PROGRAMS IN C Programming Interview Questions 13: Median of Integer Stream Posted on November 3, 2011 by Arden Given a stream of unsorted integers, find the median element in sorted order at any given time. C programming source code to calculate average using arrays. The median of the original set is thus 6. Also, make sure you work the code for edge cases like size==0, size==1, and if you want to use an int, size<0. C# How to Find Averages using Console. BESbswy. The basic idea is simple — order elements and take the middle one. 13 Apr 2018 In this C program, we are going to learn how to find the median of an array? Here, we are reading N elements and finding their median Median Program In C - A beginner's tutorial containing great set of C example, To find out median, first we re-order it as 2, 3, 3, 5, 7. Use standard deviation to check data sets for outlier data points. There are many “averages” in statistics, but these are, I think, the three most common, and are certainly the three you are most likely to encounter in your pre-statistics courses, if the topic comes up at all. If the length is 2 or 1 calculate the median of arrays according to even or odd length, print the result and return to main. What effect do you find? Justify your answer. september 20, 2011 animesh shaw array programs, function programs, math programs in c. The Programs The first macro starts with the following macro call, where ‘datain’ is the name of the data set, ‘varlist’ is a list of variable names separated by a space and ‘medname’ is the name given to the resulting median. the function should then determine the median of the array. The median is the single "middle number" in a distribution. length]; System. In statistics maths, a mode is a value that occurs the highest numbers of time. com/2011/12/c-program-to-find-meanmedian12/14/2011 · C program to find the lcm and hcf of given numbers C program to convert a binary number to decimal,oc C program to find the mean,median and mode of a seMinimum no. Autor: Maitreyi AppVizualizări: 2. The mode doesn't work either, and when asked to 'play again C program to find Mean, Median and mode with algorithm Posted by: Rajeel in C programming , Computers , Tutorials Well before starting, let’s look at the basic defention and how we find mean, median and mode. Mean of an array = (sum of all elements) / (number of Code for PROGRAM TO SORT A LIST AND FIND ITS MEDIAN in C Programming. In this article we shall discuss how to calculate median for grouped frequency distribution of discrete variables as well as continuous variables. 3I'm writing a program that calculates the mode, mean and median of an array of integers. I am trying to write a program that will find the meadian and mode of the values in an array. Hi! I'm trying to make a program that stores numbers in an array, sorts them, and prints the amount of Linear Time selection algorithm Also called Median Finding Algorithm. ATTACHMENT: 60 percent of estimated state median income adjusted for family size, by state, FFY 2018. An analysis for The New York Times by Zillow, the real estate website, found 90 cities where the median rent — not including utilities — was more than 30 percent of the median gross income. Finally, the program will call a third function that prints the employee’s last name, the mean and the median that were calculated in the other two functions. Ex array of integers: 10, 10, 4, 5, 6,5, 6, 7 My function has to return the number 3. The values of 11 houses on Washington Street are shown in the table. In that case, return the average of the two middle values. This program will also find multiple modes if there are more than one. Quick Median - A Partition. Program for Mean and median of an unsorted array Given n size unsorted array, find its mean and median. INQUIRIES: Sharnice Peters, Program Analyst Division of Energy Assistance Office of Community Services, ACF, HHS 330 C Street, S. Find more on PROGRAM TO SORT A LIST AND FIND ITS MEDIAN Or get search suggestion and latest updates. Program to find largest of n numbers in c 15. Also the required time complexity is O(logn). 24. in this video we will find mean and median of numbers. The number in the middle is the median. Algorithm. na, sort and mean from package base all of which are generic, and so the default method will work for most classes (e. msmvps. In this case, you need to average the two "most middling" numbers: 6 and 7. C PROGRAMMING IS A HIGH LEVE DISPLAY THE TRANSPOSE OF A GIVEN 3 X 3 MATRIX Median filter You are encouraged to solve this task according to the task description, Then, for each window - the set of pixels - find the median value. Write a program to find the median value of a vector of floating-point numbers. Median Program In C. C Program to Calculate Standard Deviation This program calculates the standard deviation of 10 data using arrays. The mode is the value, which is the most common in the data set. When the array has an even number of elements, the median is defined as the average of the middle two elements. We are given a list of numbers in increasing order, but there is a missing number in the list. 55 is 41. Algorithm of this program is very easy − This C++ program computes the median of the given set of numbers. let int Arr = { 12, 10, 18, 4, 30 }; first sort the array. For example, a sorted array A = [5, 7, 9, 11, 15], which has an odd number of elements, has a unique median element 9 at index 2. #include <stdio. Ask Question 0 and it returns the mean, median, and mode of those numbers. length); Arrays. MEDIAN(number1, [number2], ) The MEDIAN function syntax has the following arguments: Number1, number2, Number1 is required, subsequent numbers are optional. Do the different measures of center differ very much? Do the different measures of center differ very much? Subscribe to view the full document. com/deborahk/linq-mean-median-and-mode/ Ashish Mean, median, and mode review. Mode The most frequently occurring number in a group of numbers. I'm writing a program that calculates the mode, mean and median of an array of integers. Let M be this median of medians. There is another way to find a median that is both fast and fascinating. #define N 10 main( ) { int i,j,n; float median,a[N],t; printf("Enter the number of 26 Dec 2011 This site contains some useful and simple c programs and their output. Mode) Coding to find the median of an array. I almost feel like you should take credit for the above program, in order to give credit where credit is due. Divide by the number of times. Monday, 26 December 2011. h>How to Calculate Statistical Median using a C++ Program Example. h> static double find_kth(int a[], int a_len, Next: Write a C program to find the longest palindromic substring of a given string. h> #include <conio. A median value is the value at the center of a sorted list. Given n size unsorted array, find its mean and median. Hilma Miller author of Program to find average of two numbers is from Frankfurt, Germany . 01 12. Mode) are the mathematical terms of Statistic. By Using C Language Programming the tutorial shows how you can find the value of (Mean. */ /** THE CONCEPT OF MEDIAN ----- "Basically a median is the value present at the center of a sorted array list. duplicatetoast. …C++ :: Pointers (Average And Median) Apr 29, 2014. your approach is not going to work at all, think of it again! we're not supposed to sort the list and then find the medianthen there's not point asking such questions. This is the code I have so far for Median: C Program To Find Average of N Numbers. In this post we will learn how to find median Nov 9, 2011 The following are at least 600% faster than conventional ways to calculate median. h > int main( ). MATLAB image processing codes with examples, explanations and flow charts. /* * C program to input real numbers and find the mean, variance * and standard deviation */ #include <stdio. lernc. Ripu Daman Singh. g. Below is the solution of the program. C program to calculate area of circle, rectangle , perimeter of rectangle, C Program to calculate Net Salary from basic salary (easy one) C program to Show the output value after giving 10% discount on input value; C program to find median Finding Median Value From Array C Programming February 4, 2018 February 4, 2018 Junaid Hassan C Programming In previous post we learned how to write a C program to find mean value from array elements . For example, the mode of 2, 3, 3, 5, 7, and 10 is 3. In this C program, we are going to learn how to find the median of an array? the mean, mode, median and range for the 2007 data? c. I'm writing a program that calculates the mode, mean and median of an array of integers. By using this site, I'm trying to find the median of each array. To find the median of any set of numbers, put them in order from smallest to greatest. The middle of this data set is Before you can begin to understand statistics, you need to understand mean, median, and mode. I love the program, and I can The median is one of the three main measures of central tendency, which is commonly used in statistics for finding the center of a data sample or population, e. If we deal with an array that has even number of elements, we should take two from the middle of a sorted one and find the average of those two values, this …This is the C program to find the mean,median,mode and range of an array Mean, median, and mode are three kinds of “averages”. (note: if there are two numbers in …2/23/2012 · help using arrays and calculating mean median and mode I have a program that compiles and works, I just have a few problems with it. Unfortunately they are not a part of C standard Library or C++ STL. For a even number of values, the median it the average of the two middle values. 'c' programming is developed in 1972 by dennis ritchie at "at & t bell's lab usa". For your first example, we line them up: $3$,$6$,$6$. Here is program coded in C++ that demonstrates how to find the mode in a given data set using a particular method. Don't sort in the median function. C programming language Tutorials list . Ask Question 1. 15. the user can input the Turbo C / C++ - Calculate Mean, Median, Mode. Estimation of the overdispersion parameter, c, for the global model is one of the key issues in applying Program MARK to encounter data. Calculating Mean is very important in day-to-day life. W. I was implementing quicksort and I wished to set the pivot to be the median or three numbers. Mean: it is the average of a number of elements in a set of values. std::vector<double> v; int temp, len; std::cout << "Enter the number of elements "; std::cin >> len; std::cin >> temp; Program to find median of an array in C /** C program to calculate the median of * an array. To calculate area we need at least two sides and the angle […]Median filter You are encouraged to solve this task according to the task description, using any language you may know. N], Find the Medians in this array without sorting it in an efficient Way? C++ programming Write a program to find the median of three input integers. 4) a Output : Enter the num of elements: 5 Enter the numbers 20 10 30 60 50 The numbers arranged in ascending order are given below 10 20 30 50 60 Median is : 30 C programming, exercises, solution: Write a program in C to find the median of two sorted arrays of same size. Next tutorial. Median is the middle element of the sorted array i. Otherwise, it is defined as the middle element of the array. Find the median of n numbers of an array using c program. Note the cumulative frequency just more than . A. C program to calculate median of an array. R Hoare and is closely related to Quicksort another of his mind boggling algorithms. Mar 1. Find median of an array in c# Saturday, April 06, 2013 Muthu Nadar No comments public static int GetMedian(int[] Value) { decimal Median = 0; int size = Value. Write a C++ program to find the median element in an array of complex numbers in terms of their magnitudes. Below is the program in the C programming language to find the average of all the elements of the array – This is a generic function for which methods can be written. Bubble sort is a sorting technique in which each pair of adjacent elements are compared, if they are in wrong order we swap them. To calculate arithmetic mean of numbers in c programming, you have to ask to the user to enter number size then ask to enter the numbers of that size to perform addition, then make a variable responsible for average and place addition/size in …Find Median in an Unsorted Array Without Sorting it. The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. Given two sorted arrays of size n. java) Efficient algorithm to find the median of an unsorted array? and I would like to use the program/thread stack which means fast allocations and cache locality Question :-C program to find the middle node of a linked listThe simple approach to reach up to middle node we have to know how many nodes are there in linked list and then traverse up to half of that node count. Mode) are the mathematical terms of Statistic. c) Write a program that tests your mastery of the concept of static arraynot dynamic (heap). Make a C program to find the average of three numbers. 10. Uses Divide and Conquer strategy. Sample of the C program. c program to find mean, median, and mode for different cases. I see a number of things that may help you improve your program. Just thought i would comment and say neat design, did you code it yourself? Looks great. This is the code I have so far for Median:Finding Mode in C++ Home. 34. c implements Torben's method to find a median in an input set without modifying it. C++ Program to find median of two sorted arraysC Programming (Turbo C++ Compiler) - Calculate Median. kastatic. Goal: Ask for user input with C# Console and use that input to compute an average. util. This is found by adding the numbers in a data set and dividing by how many numbers there are. Program in c to print 1 to 100 without using loop 13. Length; int mid = size / 2; Median = (size % 2 != 0) ? This cpp program code will be opened in a new pop up window once you click pop-up from the right corner. 6/30/2012 · 2 responses to “ Mean, Median and Mode c++ ” Facebook Chat. The C++ program is successfully compiled and run on a Linux system. 1 in steps of 0. Your program will fill a 1000 element integer array with either random numbers (between 1 and 1000) or the numbers from 1000 to 1 (in the decreasing order) based on user choice; and then proceed to find some statistics on these numbers. It represents the data in nice tabular form. 52 Can anyone help me how to program in C++ which determines the median of five input numbers? the median is the middle number when the five numbers are arranged in order. of comparisons to find median of 3 numbers. I learn by reading through the codes, can you please post the codes Thanks a lot. Ask Question 14. P: n/a Bhadan. The median is a measure of central tendency. Find ,where . In practice, median-finding algorithms are implemented with randomized algorithms that have an expected linear running time. Find The Median Value Of A Set Of Numbers. c program to find the area & perimeter of a circle, triangle, square and rectangle c program to calculate age in years, months and days c program to find absolute value of a given number Median chat. C program to find the lcm and hcf of given numbers C program to convert a binary number to decimal,oc C program to find the mean,median and mode of a se C++ Program to Find the Median of two Sorted Arrays using Binary Search Approach. length); Arrays. 5 2 Given an unsorted Array A[1,2,3,. View All Articles C Code for Mean, Median, Mode. The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. How would you find the mode? That is a start on how to get it. I'm trying to find the How to Find median in Arrays: For getting the median of input array. M = median(A,vecdim) computes the median based on the dimensions specified in the vector vecdim. Hello, I was wondering how to make it so that my code can handle 5 numbers instead of 3. 0. I have given the simple C program for calculating median for a set of numbers. )for values less than 100 Median The middle number of a group of numbers. 01 Program using do-while loop that will calculate the sum of every third integer beginning with i=2 sum = 2+5+8+11+. c implements the fast networks for 3, 5, 7, 9 or 25 elements, particularly useful for morphological filters in image processing. Here in this c program, we need to find out mean variance and standard deviation of n numbers, for that we need to know what is meant by mean, standard deviation and variance. util. A. To find the mean you add all the numbers together then divide by how many there are. The corresponding value of the variable is the Median. The Median is the "middle number" (in a sorted list of numbers). The median is the middle value in a list of numbers, it is the number separating the higher half of a data sample or a list of numbers from the lower half. C Program to Check Whether a Given Number is Odd o # C Program to Convert Number of Days Into Number C Program To Find A Character Is Number, Alphabet, Operator, or Special Character C Program To Find Reverse Case For Any Alphhabet using ctype functions C Program To Find Number Of Vowels In Input String The median value need not be in the array. Stare: rezolvatăRăspunsuri: 3C Program to Find Mean, Median, and Mode of Given Numbers. R Hoare and is closely related to Quicksort, another of his mind-boggling algorithms and the one he …This program takes max numbers from user and calculates the sum of all the numbers in a loop and the final… Hello everyone, here we will learn a simple logic to find average on N numbers in python. 01 to 0. Aug 11, 2017 Page 1 of 2. By Chaitanya Singh C Program to reverse a given number using Recursive function; Find an answer to your question C program to find mean median and standard deviation The tutorial shows how to calculate mean, median and mode in Excel with formula examples. Median filter A blog for beginners. h> void main() { int x[100],n,i; float mean(int,int[]); float median(int,int[]); clrscr(); scanf("%d" Given n size unsorted array, find its mean and median. C program to find median of two sorted arrays of different sizes in logarithmic time. 5+10+6+6+5+12+5=49 49/7=7 Mean=7 To find the median you put the numbers from leas … t …C Program To Calculate Median. In this C program, we are merging two one dimensional array in third array, where size of first and second arrays are different and the size of third array will be size of first array + second array. Each is used to find the statistical midpoint in a group of numbers, but they all do so differently. If you're behind a web filter, please make sure that the domains *. Write a C++ program to find the median element in an array of …Write a C programming to find the median of the two given sorted arrays which are not empty. Don't violate const. Uses elimination in order to cut down2/11/2015 · Program for Finding Median of 5 Numbers . Learn C programming, Data Structures tutorials, exercises, examples, programs, hacks, tips and tricks online. To apply the program in real world you should analyze it more carefully and find some applications if possible. sort(b); if (A Calculate median and quartile in C without sorting the array. You have probably heard of Quicksort but what about Quick Median? A friend of mine was writing a program to find the median. Find the (a) mean, (b) median, (c) mode, and (d) midrange for the given sample data. To understand this example, you should have the knowledge of following C programming topics: how can i calculate mean,median,mode by using c program. We use cookies to ensure you have the best browsing experience on our website. Getting started with C or C++ | C Tutorial Since it "is not always true", I wouldn't use it for your program. h> #define MAXSIZE 10 void main () 8/20/2017 · (Mean. the mode method isn't working, and the median doesn't work either. Make a C program to find the average of three numbers. h Please help me to create a c program to find mode and median of an array. Everything compiles, but I can seem to figure out the few things C program to find the lcm and hcf of given numbers C program to convert a binary number to decimal,oc C program to find the mean,median and mode of a se C++ Program to Find the Median of two Sorted Arrays using Binary Search Approach Posted on July 1, 2017 by Manish This is a C++ program to find the median of two sorted arrays using binary search approach. The generalized version of this problem is known as "n-order statistics" which means finding an element K in a set such that we have n elements smaller or equal to K and rest are larger or equal K. torben. int a[25], n, i ; Write a C++ program to find the median element in an array of complex numbers in terms of thei 1. 10/11/2016 0 Comments Calculate Mode / Mean And Median Of Array Of Integers. java)10/7/2008 · How do you find the in Median C++? This is what I have so far, any help would be much appreciated. Now I have everything working but the mode. If, instead, there are three more rainy days per month in the year 2007, what will be the mean, mode, median and range for the 2007 data? ANSWERS. e. C programs with their output. Follow. You need to sort the array in order to find the How to Write a Program in Java to Calculate the Mean. To find the median of any set of numbers, put them in order from smallest to greatest. C Program to Find Maximum Element in Array - This program find maximum or largest element present in an array. peters@acf. I would like a hint but nothing overboard. I have already compiled the program that reads numbers. Write a c program to find out NCR factor of given number. Source Code Write a program to find the median value of a vector of floating-point numbers. It's quick & easy. A C++ Function to Find Median of a Data Array Glancing over a book on Statistics, I realized that I need some implementation to find the median of an array, and I like to implement it by myself. Find the range and calculate standard deviation to compare and evaluate variability of data sets. Median filter Original Array of Complex Numbers The Median Element of the Array is: 5+j5 Show transcribed image text 1. 6/10/2013 · C program to find mean, median, variance and standard deviation Write a C program to find mean, median, variance and standard deviation for given set of numbers. The median is the value separating the higher half of a data sample, a population, or a probability distribution, from the lower half. Note: This C Program for Finding Average of N Numbers is compiled with GNU GCC Compiler and written in gEdit Editor in Linux Ubuntu Operating System. /* this program is going to find the mean of 3 numbers. C Program to find sum of array elements using pointers, recursion & functions. C programming language, C program, C language 2 responses to “ Mean, Median and Mode c++ ” Facebook Chat. h> C Program to calculate Net Salary from basic salary (easy one) C program to find median ; No comments yet; Categories. Program for Mean and median of an unsorted array Given n size unsorted array, find its mean and median. C Code: #include <stdio. , "Date") for which a median is a reasonable concept. This program should accept array of integers and integers that indicate the number of elements in the array. calculate the range and the median for both intelligence and school performance d. It also shows you the cumulative frequency. This c programming code is used to find the mean, median, mode. Next the program will call a function that calculates the mean of the number of units. Find the median of the elements after inputting each element. Find The Median Value Of A Set Of Numbers. I have tried to figure it out on my own for almost 2 weeks now, I have Fast median search: an ANSI C implementation Nicolas Devillard - ndevilla AT free DOT fr July 1998 1 Introduction on big arrays or in any program for which strong reliability is a keyword. The median is the number in the middle of a set of numbers. May. Practice: Estimating mean and median in data displays. Mode Program In C. C Program To Find Median Of Array - The best free software for your. Enter values separated by commas such as 1, 2, 4, 7, 7, 10, 2, 4, 5. : Note that any algorithm that can find the median in O(n) time can also be used to find the k-th smallest element in the array, for any k between 1 and n, in O(n) time. Mean, or mean average, is used along with many other mathematical operations and is an important thing to know. Then, for each window - the set of pixels - find the median value. sort(b); if (A Find more on Program to find average of two numbers Or get search suggestion and latest updates. Arrays (array. c # list<int> to find the Mode and median please verify my account · Check the link given below https://blogs. I've finished it mostly I think, except for the median part. Next, divide the number of elements by two and round down to get an integer. The nightmare that is using JavaScript on HackerRank Thanks to this guy for some help with heap Calculator Use. It will also print the larger values and smaller values compared to the median. Mail Stop 5425 Washington, D. To compare 3-channel pixels we first convert them into 1-channel gray values. Write an algorithm to find the median of the array obtained after merging the above 2 …2/27/2007 · The median of five input numbers!!!. Program to find median of elements in an array. (c) Do the data provide any evidence that CEOs receive a premium (i. h> main() How to Find median in Arrays: For getting the median of input array. Write a C program to find the median and mode of an array of integers. This c programming code is used to find the mean, median, mode. calculate mean median mode c programming array. It is possible to calculate median in O(n) time instead. N], Find the Medians in this array without sorting it in an efficient …C program to find the mean,median and mode of a set of numbers #include<stdio. and either repeating the program or quitting. 11/29/2011 · Please help me to create a c program to find mode and median of an array. C. Program to find median of an array in C /** C program to calculate the median of * an array. length]; System. The statistics median is a quick measure to find a central location of a data sequence, list or any iterator. g. Example Code to Calculate Median To find the median of an unsorted array, we can make a min-heap in $O(n\log n)$ time for $n$ elements, and then we can extract one by one $n/2$ elements to get the C Program to Calculate Standard Deviation This program calculates the standard deviation of 10 data using arrays. Quick Median - A Partition. h> #include <stdlib. ValueIs there a way to get the min, max, median, and average of a list of numbers in a single command? Ask Question 90. arraycopy(A, 0, b, 0, b. No angles of Scalene Triangle are equal. The C program is successfully compiled and run on a Linux system. The parametric bootstrap goodness-of-fit procedure was an attempt to develop a general procedure, but was found to be biased for Cormack-Jolly-Seber (CJS) data . The other related links are, Calculate Range C++ :: Calculate Mode / Mean And Median Of Array Of Integers Mar 17, 2013. for calculating a typical salary, household income, home price, real-estate tax, etc. For the mode, on paper you have a list. C++ program to find median of binary search tree store the elemnts of tree in an array in sorted manner we do this by traversing the tree in inorder manner * Then we find the median by using nth_element */ template < typename T > struct Value { T value; //Data of node Value() C program code examples; C program to find average of all elements of array; C program to find average of all elements of array. Python Exercises, Practice and Solution: Write a Python program to find the median of three values. Mean of an array = (sum of all elements) / (number of elements) Median of a sorted array of size n is defined as below : It is middle element when n is odd and average of …To find the mean you add all the numbers together then divide by how many there are. LeetCode – Find Median from Data Stream (Java) Median is the middle value in an ordered integer list. For example, if there are five elements, 5/2 = 2 (integer arithmetic) so the median is dArray[2]. 'c' programming is developed in 1972 by dennis ritchie at "at & t bell's lab usa". Write a program to enter any three numbers and find out the middle number. The median is the middle value, which has exactly half of the values above it and the other half below it. import java. Program to find the mean, median, and mode of numbers. by Koscica Dusko. h> ← c program …The MEDIAN function sorts through the provided arguments to find the value that falls arithmetically in the middle of the group. This program takes max numbers from user and calculates the sum of all the numbers in a loop and the final… Hello everyone, here we will learn a simple logic to find average on N numbers in python. Need help? Post your question and get tips & solutions from a community of 424,419 IT Pros & Developers. C Source Code/Find the median and mean. Posted by: hirdesh on: March 24, 2010. Readline. Write a function to find the area of a triangle whose length of three sides is given 2 Answers Please list all the unary and binary operators in C. , "Date") for which a median is a reasonable concept. To calculate arithmetic mean of numbers in c programming, you have to ask to the user to enter number size then ask to enter the numbers of that size to perform addition, then make a variable responsible for average and place addition/size in average, then display the result Finding the Median in an Array. C program to find the mean,median and mode of a set of numbers #include<stdio. Is there a way to get the min, max, median, and average of a list of numbers in a single command? I actually keep a little awk program around to give the sum C Program to Find Area of Scalene Triangle : [crayon-5c78b459640be149497536/] Output : [crayon-5c78b459640c8343886373/] C Program for Beginners : Area of Scalene Triangle Properties of Scalene Triangle : Scalene Triangle does not have sides having equal length. The other related linkes are, Calculate Mean Calculate Median Calculate Mode Calculate Range Calculate Variance Calculate Standard Deviation. 5+10+6+6+5+12+5=49 49/7=7 Mean=7 To find the median you put the numbers from leas … t to greatest the Turbo C / C++ - Calculate Mean, Median, Mode. Hello, I have several jagged arrays which have been sorted. ReadLine takes in a string, and so it must be typecasted in order to store it in a different variable type. More on mean and median. Split number into digits in c programming 16. (in C you would have to do declare variables beforehand). Re-read the wiki page carefully and then recode. The mode of this value set is 3 as it appears more than any other number. Please read …The simple example of how to find the average of any given set of numbers is a great way to practice! First lets go over the basics: You can easily store your Console. This cpp program code will be opened in a new pop up window once you click pop-up from the right corner. Mean of an array = (sum of all elements) / (number of elements) Median of a sorted array of size n is defined as below : It is middle element when n is odd and average of …C Program to Find Area of Scalene Triangle : [crayon-5c8f1b35a73c9676398247/] Output : [crayon-5c8f1b35a73d3263894019/] C Program for Beginners : Area of Scalene Triangle Properties of Scalene Triangle : Scalene Triangle does not have sides having equal length. The median of a single sorted array is trivial to find and is O(1) constant time. So our numbers in order from least to greatest would be 94, 101, 101, 104, 108, 110. Find median of an array in c# Saturday, April 06, 2013 Muthu Nadar No comments public static int GetMedian(int[] Value) { decimal Median = 0; int size = Value

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# Find all solutions to $x^2\equiv 1\pmod {91}$ I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$. For $x^2\equiv 1\pmod {7}$, i did: $$(\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are $\pm1$. For $x^2\equiv 1\pmod {13}$, i did: $$(\pm1 )^2\equiv 1\pmod{13}$$ $$(\pm2 )^2\equiv 4\pmod{13}$$ $$(\pm3 )^2\equiv 9\pmod{13}$$ $$(\pm4 )^2\equiv 3\pmod{13}$$ $$(\pm5 )^2\equiv {-1}\pmod{13}$$ $$(\pm6 )^2\equiv 10\pmod{13}$$Which shows that the solutions to $x^2\equiv 1\pmod {13}$ are $\pm1$. Thus, I concluded that the solutions to $x^2\equiv 1\pmod {91}$ must be $\pm1$. I thought that $\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence? • Hint: CRT. See link – user160069 Dec 3 '16 at 2:38 You have $x\equiv\pm1\mod7$ and $x\equiv\pm1\mod13$. For all the solutions, you have to consider the systems: $$x\equiv1\mod7$$ $$x\equiv1\mod13$$ and $$x\equiv-1\mod7$$ $$x\equiv1\mod13$$ and $$x\equiv1\mod7$$ $$x\equiv-1\mod13$$ and $$x\equiv-1\mod7$$ $$x\equiv-1\mod13$$ as each system will get you a valid answer. I think you only had the first and the last systems and that you only considered the cases where the signs were similar. • Thank you! I figured out the other solutions to be $\pm 27$. – J. Doe Dec 3 '16 at 3:40 Hint: Consider the possibility that $x \equiv 1 \pmod 7$ but $x \equiv -1 \pmod {13}$, and so on. (In other words, your mistake was to assume that the $\pm1$ modulo $7$ was the same sign as $\pm1$ modulo $13$.) Also note that for any prime $p$, if $x^2 \equiv 1 \pmod p$, then we can rewrite this as $$x^2 - 1 \equiv (x+1)(x-1) \equiv 0 \pmod p.$$ Thus we get $x \equiv \pm 1 \pmod p$, showing that it isn't necessary to run through all the values of $x^2$ to find the solution. By CRT the solutions $$\,x\equiv \pm1\pmod 7,\ x\equiv \pm1\pmod{13}$$ combine to $$\,4\,$$ solutions mod $$91,\,$$ viz $$\,x\equiv (\color{#c00}{{\bf 1,1}}),\,(\color{#0a0}{-1,-1}),\,(1,-1),\,(-1,1)\,$$ mod $$(7,13),$$ cf. Remark below. The first $$2$$ have obvious solutions $$\,\color{#c00}{\bf 1}\,$$ and $$\,\color{#0a0}{-1}\,$$ so you need only solve the $$(1,-1)$$ case, then note $$(-1,1)\equiv -(1,-1)$$ is its negative, i.e. $$\,x\equiv 1\pmod{\!7},\,x\equiv -1\pmod{\!13}\,\Rightarrow\,-x\equiv -1\pmod{\!7},\ {-}x\equiv 1\pmod{\!13}$$ Remark For more complex examples it is usually easier to solve the CRT system first for generic (symbolic) roots, then plug in the specific root values for all combinations, e.g. see here and here. If $$\,m,n\,$$ are coprime then, by CRT, solving a polynomial $$\,f(x)\equiv 0\pmod{\!mn}\,$$ is equivalent to solving $$\,f(x)\equiv 0\,$$ mod $$\,m\,$$ and mod $$\,n.\,$$ By CRT, each combination of a root $$\,r_i\bmod m\,$$ and a root $$\,s_j\bmod n\,$$ corresponds to a unique root $$\,t_{ij}\bmod mn,\,$$ i.e. $$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{\!mn}&\overset{\,\,\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\ &\,\,\iff& \begin{array}{}x\equiv r_1,\ldots,r_k\pmod{\! m}\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod{\! n}\end{array}\\ &\,\,\iff& \left\{ \begin{array}{}x\equiv r_i\pmod{\! m}\\x\equiv s_j\pmod {\! n}\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\ &\overset{\,\,\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{\!mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\ \end{eqnarray}\qquad\qquad$$

https://www.physicsforums.com/threads/indefinite-integration.632544/

# Indefinite Integration ## Homework Statement The value of $$\int_0^{1} (\prod_{r=1}^{n} (x+r))(\sum_{k=1}^{n} \frac{1}{x+k}) dx$$ equals: a)n b)n! c)(n+1)! d)n.n! (Can someone tell me how to make bigger parentheses using latex?) ## The Attempt at a Solution I know that the question becomes a lot easier if i put n=1 or 2 and then integrate. But i was wondering if there is any proper way to solve it. I can't go further after expanding the given expression. $$\int_{0}^{1} (x+1)(x+2)......(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+.....\frac{1}{x+n})dx$$ which is equal to $$\int_{0}^{1} \sum_{r=1}^n \frac{(x+n)!}{x+r}$$ I am stuck now, i can't find any way further. Any help is appreciated. I still don't have any clue. ehild Homework Helper What is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ? ehild What is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ? ehild Derivative of (x+1)(x+2)=2x+3 Derivative of (x+1)(x+2)(x+3)=3x2+12x+11 I still don't get any idea how this would help? ehild Homework Helper Do not simplify. d/dx[(x+1)(x+2)]=(x+2)+(x+1)=(x+1)(x+2)/(x+1)+(x+1)(x+2)/(x+2)=[(x+1)(x+2)][1/(x+1)+1/(x+2)] d/dx[(x+1)(x+2)(x+3)]=(x+2)(x+3)+(x+1)(x+3)+(x+1)(x+2)=[(x+1)(x+2)(x+3)][1/(x+1)+1/(x+2)+1(x+3)] Do you see? ehild 1 person Wait, i guess i have got it, i will be back in a few hours with a solution. Thanks for the help! EDIT: Oops, seems like you posted just a few seconds before me. ehild Homework Helper Wait, i guess i have got it, i will be back in a few hours with a solution. Thanks for the help! EDIT: Oops, seems like you posted just a few seconds before me. Well, I looking forward to the solution ehild So here's the solution: Let $$t=(x+1)(x+2)(x+3).......(x+n)$$ $$\ln t=\ln(x+1)+\ln(x+2)+\ln(x+3)......+\ln(x+n)$$ $$\frac{1}{t}\frac{dt}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}......\frac{1}{x+n}$$ $$dt=(x+1)(x+2)(x+3)....(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}........+\frac{1}{x+n})dx$$ Therefore the original question becomes ∫dt, upper limit is (n+1).n! and lower limit is n!. Solving, i get the answer as d) option. Thanks for the help! Mentallic Homework Helper By the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that. I myself use /left( and /right) which looks like: $$\left(\frac{1}{1}\right)$$ but there are other more specific "fonts" you can use, which ranges from smallest to largest: /big( $$\big(\frac{1}{1}\big)$$ /Big( $$\Big(\frac{1}{1}\Big)$$ /bigg( $$\bigg(\frac{1}{1}\bigg)$$ /Bigg( $$\Bigg(\frac{1}{1}\Bigg)$$ And there could be others, but that should about cover it. ehild Homework Helper So here's the solution: Let $$t=(x+1)(x+2)(x+3).......(x+n)$$ $$\ln t=\ln(x+1)+\ln(x+2)+\ln(x+3)......+\ln(x+n)$$ $$\frac{1}{t}\frac{dt}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}......\frac{1}{x+n}$$ $$dt=(x+1)(x+2)(x+3)....(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}........+\frac{1}{x+n})dx$$ That is ingenious! You are really cool Therefore the original question becomes ∫dt, upper limit is (n+1).n! and lower limit is n!. Solving, i get the answer as d) option. Thanks for the help! Very good. You are better and better every day! ehild By the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that. I myself use /left( and /right) which looks like: $$\left(\frac{1}{1}\right)$$ but there are other more specific "fonts" you can use, which ranges from smallest to largest: /big( $$\big(\frac{1}{1}\big)$$ /Big( $$\Big(\frac{1}{1}\Big)$$ /bigg( $$\bigg(\frac{1}{1}\bigg)$$ /Bigg( $$\Bigg(\frac{1}{1}\Bigg)$$ And there could be others, but that should about cover it.

https://math.stackexchange.com/questions/3364177/symmetric-matrix-as-a-sum-of-symmetric-matrices

# Symmetric matrix as a sum of symmetric matrices Let matrix $$M \in \mathbb{N}^{5 \times 5}$$ be symmetric with non-negative integer entries and zeros on the main diagonal and having the property that the row sums are equal to $$2r$$ for some $$r \geq 2$$. I want to prove that $$M$$ can be written as a non-negative integral linear combination of $$5 \times 5$$ symmetric matrices having non-negative integer entries with zero entries on the main diagonal and having the property that the row sums are equal to $$2$$. Is there way to prove this? I tried with some simple examples and it seems to be correct. • I need to prove that any $M$ (with the property listed) can be written as a non-negative integral linear combination of matrices with the property listed. $m$ is not specific here. – jack Sep 21 at 9:28 • I imagine this result is true for all dimensions, not just 5. – S. Dolan Sep 21 at 17:00 IDEA: The row sums of $$M$$ are even, so there is either 0 or 2 or 4 odd numbers, and the diagonal entries are all zeros. So if you have even entries, for example if you have that in $$M$$ the entries $$m_{1,5}=m_{5,1}= 8$$ then you will have in your linear combination $$4$$ multiplied by a matrix having all its entries on these rows are $$0$$ except $$a_{1,5}=a_{5,1}=2$$, and so on ... Also if there is $$2$$ odd entries, for example $$m_{1,5}=m_{5,1}=5$$ and $$m_{1,4}=m_{4,1}=7$$, then in your linear combination you'll have $$2$$ multiplied by a matrix having $$a_{1,5}=a_{5,1}=2$$ and all other entries on these rows are zeros, and $$3$$ multiplied by a matrix having $$b_{1,4}=b_{4,1}=2$$ and all other entries on these rows are zeros, and a matrix having $$c_{1,4}=c_{4,1}=c_{1,5}=c_{5,1}=1$$ and all other entries on these rows are zeros, and so on... I hope you can reach a proof from this hint and I am sorry because I am not able to write the whole proof now. Yes, this conjecture is correct. First consider the following matrices. $$X_1=\begin{pmatrix}0&0&1&1&0 \\0&0&1&0&1\\1&1&0&0&0\\1&0&0&0&1\\0&1&0&1&0\\\end{pmatrix}$$, $$X_2=\begin{pmatrix}0&0&1&0&1 \\0&0&1&1&0\\1&1&0&0&0\\0&1&0&0&1\\1&0&0&1&0\\\end{pmatrix}$$, $$X_3=\begin{pmatrix}0&0&1&0&1 \\0&0&0&2&0\\1&0&0&0&1\\0&2&0&0&0\\1&0&1&0&0\\\end{pmatrix}.$$ If we could subtract one of the $$X_i$$ from $$M$$ without making any entry negative, then we would leave a matrix of the same type as $$M$$ but with $$r$$ reduced by 1. This process would therefore reduce $$M$$ to zero as we require. If all off-diagonal entries of $$M$$ were non-zero we could subtract $$X_1$$. So, without loss of generality, we can suppose that $$M_{12}=0$$. CASE 1. If no row of $$M$$ has three or more zeroes. Without loss of generality we can suppose that $$M$$ has the following form where $$A,B,C,D,E,F$$ are all non-zero. $$M=\begin{pmatrix}0&0&A&B&C \\0&0&D&E&F\\A&D&0&.&.\\B&E&.&0&.\\C&F&.&.&0\\\end{pmatrix}$$ If any of the dots were non-zero then a matrix like $$X_1$$ can be subtracted from $$M$$. Otherwise, all the dots are zero. Then $$A+D=B+E=C+F=2r. (*)$$ However, we would then have the sum of the first two rows equal to $$6r$$, a contradiction. CASE 2. If a row of $$M$$ has four zeroes. Without loss of generality we can suppose that $$M$$ has the following form. $$M=\begin{pmatrix}0&0&0&0&2r \\0&0&A&B&0\\0&A&0&C&0\\0&B&C&0&0\\2r&0&0&0&0\\\end{pmatrix}$$ Then simple algebra shows that $$A=B=C=r$$ and $$M$$ is the sum of $$r$$ copies of $$\frac {1}{r} M$$. CASE 3. If $$M$$ has three zeroes in a row but that no row has four or more zeroes. Let $$M=\begin{pmatrix}0&0&.&.&. \\0&0&.&.&.\\.&.&0&F&G\\.&.&F&0&H\\.&.&G&H&0\\\end{pmatrix}$$ CASE 3(a). If $$F=G=H=0$$ Then the method used for $$(*)$$ gives a contradiction. CASE 3(b). If two of $$F,G,H$$ are $$0$$, w.l.g. $$F=G=0$$ Then either $$X_1$$ or $$X_2$$ can be subtracted. CASE 3(c). If only one of $$F,G,H$$ is $$0$$, w.l.g $$F=0$$. Since every row has at least two non-zero elements, the possibilities for $$M$$ are, without loss of generality, as follows, where $$+$$ indicates a positive element. $$\begin{pmatrix}0&0&+&0&+ \\0&0&0&+&+\\+&0&0&0&G\\0&+&0&0&H\\+&+&G&H&0\\\end{pmatrix}$$ and $$\begin{pmatrix}0&0&+&.&. \\0&0&+&.&.\\+&+&0&0&G\\.&.&0&0&H\\.&.&G&H&0\\\end{pmatrix}$$ Considering row totals for Rows 3 and 4 of the LH matrix shows that $$M_{42}>1$$ and so $$X_3$$ can be subtracted. For the RH matrix, one of $$X_1, X_2$$ can be subtracted unless the matrix is $$\begin{pmatrix}0&0&+&+&0 \\0&0&+&+&0\\+&+&0&0&G\\+&+&0&0&H\\0&0&G&H&0\\\end{pmatrix}$$ Then the sum of elements in Rows 1 and 2 must equal the sum of elements in Rows 3 and 4. Therefore $$G=H=0$$, a contradiction. CASE 3(d). If $$F,G,H$$ are all non-zero. Then a matrix like either $$X_1$$ or $$X_2$$ can be subtracted since the top two rows contain at least 4 non-zero elements. • With hindsight, CASE 1 can be subsumed into CASE 3(a) to shorten this proof a little. – S. Dolan Sep 21 at 16:41 A graph theoretic proof Consider $$M$$ to be the adjacency matrix for a graph $$G$$. Then $$G$$ is a graph with no loops and 5 vertices, each of degree $$2r$$. We are required to prove that: $$G$$ contains one or more simple cycles (i.e. containing no repeated vertex) which include every point precisely once. Since all vertices have even degree, each component of $$G$$ must have an Eulerian cycle and therefore each vertex of $$G$$ is in a simple cycle and at least one of these simple cycles has length at least 3. We can suppose every vertex of $$G$$ is joined to more than one vertex. Suppose on the contrary that vertex 1 was only joined to vertex 2. By the equality of degrees, vertices 1 and 2 then form a component of $$G$$. The points 3,4,5 of the other component must then be joined in pairs and $$G$$ has simple cycles (1,2) and (3,4,5). We can suppose $$G$$ only has simple cycles of lengths 2 and 3 There is nothing to prove if $$G$$ has a simple cycle of length 5, so suppose $$G$$ has the cycle (1,2,3,4). If vertex 5 were connected to two adjacent points in this cycle, say vertices 1 and 2, then (1,5,2,3,4) would be a simple cycle. So we can suppose 5 is only joined to, say, 1 and 3. Then (1,5,3,4) and ((1,2,3,5) are cycles and so 2 and 4 are also only joined to 1 and 3. There are then $$6r$$ edges from vertices 2,4,5 to 1 and 3, contradicting the degrees of 1 and 3. Now consider a cycle of length 3, say (1,2,3). If vertex 4 were joined to two of these points then we would have a cycle of length 4. So vertex 4 is joined to vertex 5 and, say, vertex 1. Vertex 5 also has to be joined to precisely one of 1,2,3. This must be the same vertex 1, otherwise we would obtain a simple cycle of length 5. Vertices 4 and 5 can only be joined by $$1$$ edge (and similarly vertices 2 and 3 are only joined by $$1$$ edge) otherwise we would have simple cycles (1,2,3) and (4,5). Therefore there are $$4(2r-1)$$ edges from vertices 2,3,4,5 to vertex 1. Then $$8r-4=2r$$, which is impossible. • I don't think I follow your 'We are required to prove that'. Perhaps you meant something like: $G$ contains a collection of simple cycles such that each vertex belongs to exactly one of these cycles. Is that right? – Fimpellizieri Sep 22 at 21:04 • Yes, that is correct. – S. Dolan Sep 22 at 21:06 • I think your answer would benefit from better writing. I think you are trying to handle cases here rather than actually going 'without loss of generality' and it makes parsing the answer confusing. $G$ can very trivially have a simple cycle of length $5$, but I guess you're dismissing it as the easiest case for which what we need to prove is basically handed to us on a platter. – Fimpellizieri Sep 22 at 21:18 • If G has a simple cycle of length 5 then, by definition, G contains a simple cycle which include every point precisely once. – S. Dolan Sep 22 at 21:24 • Yes, that is my point. It's not 'Without loss of generality, we can suppose $G$ has no simple cycle of length $5$' and it most definitely isn't 'By definition, $G$ has no simple cycle of length $5$'. It's just that when $G$ has a simple cycle of length $5$, there is nothing to prove, so this case is done. Your answer is tackling cases, but is structured and written as something else, which makes it harder to parse. – Fimpellizieri Sep 22 at 21:28 This is my take on the graph theoretic approach by S. Dolan. Credits go to him. Let $$M$$ be the adjacency matrix of a graph $$G$$, so that $$G$$ is a graph on $$5$$ vertices with no loops and such that each vertex has degree $$2r$$. Claim: It is enough to prove that $$G$$ contains a collection of simple cycles such that each vertex belongs to exactly one of these cycles. Indeed, if $$H$$ is the $$G$$-subgraph that corresponds to this collection, then each vertex of $$H$$ has degree exactly $$2$$. We can thus take the adjacency matrix of $$H$$ as one of the matrices in our integral linear combination that adds up to $$M$$. We are hence left with the same problem, except now the rows of $$M$$ add up to $$2(r-1)$$; equivalently, except the vertices of $$G$$ have degree $$2(r-1)$$. Induction then takes care of finding the other matrices in our integral linear combination. Let $$v_1,v_2,v_3,v_4$$ and $$v_5$$ be the vertices of $$G$$. We will prove the claim by studying different cases. ## Case $$(1)$$: $$G$$ has a vertex which is joined to only one other vertex. Assume without loss of generality that $$v_1$$ is joined only to $$v_2$$. Since all vertices have equal degrees, $$v_2$$ must also be joined only to $$v_1$$. This implies that if some $$u$$ in $$\{v_3, v_4,v_5\}$$ were also joined to only one other vertex $$w$$, we would have $$w \in \{v_3, v_4,v_5\}\setminus\{u\}$$. Equality of degrees would then require that the remaining vertex be joined to itself, in contradiction to $$G$$ having no loops. It follows that each of $$\{v_3, v_4,v_5\}$$ is connected to the other two. The collection of simple cycles $$\{(v_1,v_2), (v_3,v_4,v_5)\}$$ satisfies the claim and handles this case. ## Case $$(2)$$: Each vertex of $$G$$ is joined to at least two other vertices. By Veblen's Theorem, $$G$$ can be written as the union of disjoint simple cycles. We break down into subcases. $$\qquad (2.1)$$: $$G$$ has a simple cycle of length $$5$$. In this case, the claim is obviously satisfied and there is nothing to prove. $$\qquad (2.2)$$: $$G$$ has a simple cycle of length $$4$$. Let $$(v_1,v_2,v_3,v_4)$$ be the simple cycle. $$\qquad\qquad(2.2.1)$$: $$v_5$$ is joined to two vertices that are adjacent in the cycle. If that were the case, then we could enlarge the simple cycle to have length $$5$$. This reduces the problem to case $$(2.1)$$, which we have already handled. $$\qquad\qquad(2.2.2)$$: No two vertices adjacent in the cycle are both joined to $$v_5$$. Without loss of generality, suppose $$v_5$$ were joined to vertices $$v_1$$ and $$v_3$$. Notice that $$v_5$$ cannot be joined to any other vertex. If $$v_2$$ and $$v_4$$ weree joined, then we could take the simple cycle $$(v_1,v_2,v_4,v_3,v_5)$$, handled by case $$(2.1)$$. If $$v_2$$ and $$v_4$$ were not joined, then each of $$\{v_2, v_4,v_5\}$$ would be joined only to $$v_1$$ and $$v_3$$. This would contradict each vertex having equal degree, and finishes this case. $$\qquad (2.3)$$: $$G$$ has a simple cycle of length $$3$$. Let $$(v_1,v_2,v_3)$$ be the simple cycle. $$\qquad\qquad(2.3.1)$$: One of $$\{v_4, v_5\}$$ is joined to two vertices in the cycle. If that were the case, then we could enlarge the simple cycle to have length $$4$$. This reduces the problem to case $$(2.2)$$, which we have already handled. $$\qquad\qquad(2.3.1)$$: No two vertices in the cycle are both joined to one of $$\{v_4, v_5\}$$. Without loss of generality, suppose $$v_4$$ were joined to vertices $$v_1$$ and $$v_5$$. Notice that $$v_4$$ cannot be joined to any other vertex. If $$v_5$$ were joined to one of $$\{v_2,v_3\}$$, then we could enlarge the simple cycle to have length $$5$$, handled by case $$(2.1)$$. Indeed, if $$v_5$$ were joined to $$v_2$$ we would have the cycle $$(v_1,v_3, v_2,v_5,v_4)$$ and if $$v_5$$ were joined to $$v_3$$ we would have the cycle $$(v_1,v_2,v_3,v_5,v_4)$$. Suppose instead that $$v_5$$ were joined only to $$v_1$$ and $$v_4$$. If there were more than edge joining $$v_4$$ and $$v_5$$, the collection $$\{(v_1,v_2,v_3),(v_4,v_5)\}$$ satisfies the claim. If there were more than edge joining $$v_2$$ and $$v_3$$, the collection $$\{(v_1,v_4,v_5),(v_2,v_3)\}$$ satisfies the claim. If $$v_4$$ were joined to $$v_5$$ by a single edge and $$v_2$$ were also joined to $$v_3$$ by a single edge, then $$v_1$$ would have degree $$4(2r - 1) > 2r$$, which contradicts our hypotheses and concludes this case. $$\qquad (2.4)$$: $$G$$ has no simple cycle of length $$3$$ or more. We show that this case is impossible. $$G$$ does not have loops, so in this case it would contain only cycles of length $$2$$. $$G$$ would then be bipartite: its vertices could be divided into two disjoint sets $$U$$ and $$V$$ such that every edge of $$G$$ joins a vertex in $$U$$ to a vertex in $$V$$. But $$G$$ has $$5r$$ edges and one of $$\{U, V\}$$ has at most two vertices. This contradicts each vertex having degree $$2r$$ and finishes the last case.

https://math.stackexchange.com/questions/3216724/number-of-ways-an-answers-sheet-can-be-fill-with-at-least-two-consecutive-answer

# Number of ways an answers sheet can be fill with at least two consecutive answers? A student is taking a $$12$$-question multiple choice test. Each question has $$5$$ choices. How many ways can the student fill out the answer sheet so there is at least one place where two consecutive answers are the same? This is what I did: Total possible ways the student can fill the answer sheet: $$5^{12}$$ (because he has $$5$$ options in the $$12$$ questions). Total possible ways the student can fill the answer sheet with no consecutive answers: $$5 \cdot 4^{11}$$ (In the first question he can answer any of the $$5$$ options, however, in the remaining questions he can only answer $$4$$ options to avoid choosing consecutive answers.) Total possible ways the student can fill the answer sheet with at least two consecutive answers: $$5^{12}- 5 \cdot 4^{11} = 223,169,105$$. I doubt about the total possible ways the student can fill the answer sheet with no consecutive answers... is my logic right? • Write out a detailed explanation of the $5\times4^{11}$. This should either convince you that you are right, or show you where you are wrong. Do this for the first part of your answer too. – David May 7 '19 at 5:17 • Sorry, I had a typo on my first statement. – Nico May 7 '19 at 5:33 • Yes, you are correct. – N. F. Taussig May 7 '19 at 9:05 You are correct, $$5\times 4^{11}$$ is the number of ways to fill out the test with no two consecutive equal answers, so $$5^{12}-5\times 4^{11}$$ is the number of ways to fill out the test so that there are two consecutive equal answers. I am trying to do this question and am getting a different answer. Instead of making a new thread, I hope I am allowed to post my response and see where people think I'm going wrong here. If not please let me know and I will delete. Denote the answers of the multi-choice test to be A_1,A_2,..., A_12. Now, for us to have consecutive answers be the same it must be that A_1 = A_2 or A_2 = A_3 or .... or A_11 = A_12. Denote this pair of questions with the same answer to be A_s. Now, consider a new set which contains 11 elements: A_s, and our other 10 remaining answers to questions. Certainly, we can position A_s anywhere, as remember that we have set A_s to be the two consecutive answers already. Thus, we have no conditions on the remaining choices. We can re-arrange this set 11! different ways (as it is simply the re-arrangements of a set with 11 elements where order matters), and each element can have 5 different options. Thus, we have 11!x(5 x 11) = 21,954,424,000 different options. My answer seems way too big • You have not accounted for the possibility that there is more than one pair of consecutive answers that are the same. To do so, you could apply the Inclusion-Exclusion Principle, but with $11$ possible pairs of consecutive answers that could be the same, that could prove exceedingly difficult. – N. F. Taussig May 20 '19 at 0:02

https://forum.math.toronto.edu/index.php?PHPSESSID=osjucte5sjf5bgei03vl6fofd2&topic=1204.0;wap2

MAT334-2018F > MAT334--Lectures & Home Assignments About the definition of Argument (in book) (1/2) > >> Ende Jin: I found that the definition of "arg" and "Arg" in the book is different from that introduced in the lecture (exactly opposite) (on page 7). I remember in the lecture, the "arg" is the one always lies in $(-\pi, \pi]$ Which one should I use? Victor Ivrii: --- Quote ---Which one should I use? --- End quote --- This is a good and tricky question because the answer is nuanced: Solving problems, use definition as in the Textbook, unless the problem under consideration requires modification: for example, if we are restricted to the right half-plane  $\{z\colon \Re z >0\}$ then it is reasonable to consider $\arg z\in (-\pi/2,\pi/2)$, but if we are restricted to the upper half-plane  $\{z\colon \Im z >0\}$ then it is reasonable to consider $\arg z\in (0,\pi)$ and so on. Ende Jin: I am still confused. Let me rephrase the question again. In the textbook, the definition of "arg" and "Arg" are: $arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta$ which means $arg(z) \in \mathbb{R}$ while $Arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta \land \theta \in [-\pi, \pi)$ which means $Arg(z) \in [-\pi, \pi)$ While in the lecture, as you have introduced, it is the opposite and the range changes to $(-\pi, \pi]$ instead of $[-\pi, \pi)$ (unless I remember incorrectly): Arg is defined to be $Arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = (cos\theta + isin\theta)$ which means $arg(z) \in \mathbb{R}$ while arg is $arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta \land \theta \in (-\pi, \pi]$ I am confused because if I am using the definition by the book, when $z \in \{z : Re (z) > 0\}$ then $arg(z) \in (-\frac{\pi}{2} + 2\pi n,\frac{\pi}{2} + 2\pi n), n \in \mathbb{Z}$ Victor Ivrii: BTW, you need to write \sin t and \cos t and so on to have them displayed properly (upright and with a space after): $\sin t$, $\cos t$ and so on Ende Jin: Thus in a test/quiz/exam, I should follow the convention of the textbook, right?

https://www.jiskha.com/questions/848377/consider-the-differential-equation-dy-dt-y-t-2-a-show-that-the-constant-function-y1-t-0

Differential Equations Consider the differential equation: dy/dt=y/t^2 a) Show that the constant function y1(t)=0 is a solution. b)Show that there are infinitely many other functions that satisfy the differential equation, that agree with this solution when t<=0, but that are nonzero when t>0 [Hint: you need to define these functions using language like " y(t)=...when t<=0 and y(t)=...when t>0 and "] c) Why doesn't this example contradict the Uniqueness Theorem? I'm trying to do part b and after I separated and integrated I got ln|y|=(-1/t)+C I'm not sure if I can get C with the solution they gave in part a)y1(t)=0. Anyways, I get y(t)=Ce^-(1/t). I don't know where to go from there. 1. 👍 2. 👎 3. 👁 1. dy/y = dt/t^2 ln y = -1/t + c y = e^(-1/t+c) = e^c e^(-1/t) =C e^(-1/t) agree C can be anything so what if C = 0 ? then y(t) = 0 for all t if C = 0 for t</= 0 then C can be anything at all for t>0 1. 👍 2. 👎 2. If y(t) depends on what C is, then how this equation doesn't contradict the uniqueness theorem if it has many solutions? 1. 👍 2. 👎 3. Because the general solution contains an an arbitrary constant C. The value of C depends on your boundary conditions, for example if y =5 at t = 2 then 5 = C e^-(1/2) 5 = C (.606) C = 8.24 1. 👍 2. 👎 4. scroll down through this: http://hyperphysics.phy-astr.gsu.edu/hbase/diff.html 1. 👍 2. 👎 Similar Questions 1. math help!!! stuck!! pls help/explain!!! 2. Do the values in the table represent a linear function? If so, what is the function rule? (1 point) x = -2, 0, 2, 4 y = -4, 0, 4, 8 The values do not show a linear function. Yes, they show a linear 2. AP Calc Help!! Let f be the function satisfying f'(x)=x√(f(x)) for all real numbers x, where f(3)=25. 1. Find f''(3). 2. Write an expression for y=f(x) by solving the differential equation dy/dx=x√y with the initial condition f(3)=25. Please 3. Calculus Suppose that we use Euler's method to approximate the solution to the differential equation 𝑑𝑦/𝑑𝑥=𝑥^4/𝑦 𝑦(0.1)=1 Let 𝑓(𝑥,𝑦)=𝑥^4/𝑦. We let 𝑥0=0.1 and 𝑦0=1 and pick a step size ℎ=0.2. 4. Calc Consider the differential equation dy/dx=-2x/y Find the particular solution y =f(x) to the given differential equation with the initial condition f(1) = -1 1. Physic Show that the function y(x,t) = X^2 + V^2t^2 is a solution of the general wave equation. 2. Calculus for what values of r does the function y=e^rx satisfy the differential equation y''-4y'+y=0 Show steps please! Thank you! 3. Algebra Create a system of equations that includes one linear equation and one quadratic equation. Part 1. Show all work to solving your system of equations algebraically. Part 2. Graph your system of equations, and show the solution 4. Calculus Consider the differential equation dy/dx = x^4(y - 2). Find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 0. Is this y=e^(x^5/5)+4? 1. Calculus Verify that the function y(x)= x - (1/x) is a solution to the differential equation, xy'+ y= 2x I have absolutely no clue what to do with that + y Please help :c 2. Calculus BC Let y = f(x) be the solution to the differential equation dy/dx=y-x The point (5,1) is on the graph of the solution to this differential equation. What is the approximation of f(6) if Euler’s Method is used given ∆x = 0.5? 3. Calculus!! Consider the differential equation given by dy/dx = xy/2. A. Let y=f(x) be the particular solution to the given differential equation with the initial condition. Based on the slope field, how does the value of f(0.2) compare to 4. math Consider the differential equation dy/dx = -1 + (y^2/ x). Let y = g(x) be the particular solution to the differential equation dy/ dx = -1 + (y^2/ x) with initial condition g(4) = 2. Does g have a relative minimum, a relative

https://mathoverflow.net/questions/330435/automorphism-group-of-the-special-unitary-group-sun

# Automorphism group of the special unitary group $SU(N)$ Let us consider the automorphism group of the special unitary group $$G=SU(N)$$. We know there is an exact sequence: $$0 \to \text{Inn}(G) \to \text{Aut}(G) \to \text{Out}(G) \to 0.$$ For $$G=SU(2)$$, we have: • $$\text{Z}(SU(2)) =\mathbb Z_2$$, • $$\text{Inn}(SU(2)) = SO(3)$$, • $$\text{Out}(SU(2)) = 0$$, And so $$\text{Aut}(SU(2))=SO(3)$$. For $$N > 2$$, we have: • $$\text{Z}(SU(N)) =\mathbb Z_{N}$$, • $$\text{Inn}(SU(N)) = PSU(N)$$, • $$\text{Out}(SU(N)) = \mathbb Z_2$$. My question is: Does $$\text{Aut}(SU(N))=PSU(N) \times \mathbb Z_2$$? If not, does this answer depend on whether $$N$$ is odd or even? It looks to me that there is a nontrivial fibration depending on something like $$H^2(B\mathbb Z_2,PSU(N))$$ due to $$B\text{Inn}(G) \to B\text{Aut}(G) \to B\text{Out}(G) \to B^2\text{Inn}(G) \to$$ and thus $$BPSU(N) \to B\text{Aut}(G) \to B\mathbb Z_2 \to B^2PSU(N) \to$$ But I do not know how to define $$H^2(B\mathbb Z_2,PSU(N))$$, if this is a correct thing to ponder. • this may be preliminary for answering your question: mathoverflow.net/questions/40666/… - which you may already know well. – wonderich May 1 at 3:48 • In fact this completely answers the question. – abx May 1 at 4:05 • what is the answer? I am asking the total Aut of $SU(N)$? – annie heart May 1 at 4:46 • p.s. I am not asking Inn or Out of $SU(N)$. – annie heart May 1 at 4:51 • The answer is that it is a semi-direct product $\operatorname{PSU} (N)\rtimes \mathbb{Z}/2$, with $\mathbb{Z}/2$ acting on $\operatorname{PSU}(N)$ by conjugation. – abx May 1 at 5:58 Let $$G$$ be a compact simple simply connected Lie group. Then any automorphism of $$G$$ determines an automorphism of its Lie algebra $$\mathfrak{g}$$ and visa versa. So $$\mathrm{Aut}(G)$$ is naturally isomorphic to the linear group $$\mathrm{Aut}(\mathfrak{g})$$. The sequence $$1\to \mathrm{Inn}(\mathfrak{g})\to \mathrm{Aut}(\mathfrak{g})\to \mathrm{Out}(\mathfrak{g})\to 1$$ is split. Moreover, $$\mathrm{Out}(G)\cong\mathrm{Out}(\mathfrak{g})\cong \mathrm{Aut}(D_\mathfrak{g})$$ where $$D_\mathfrak{g}$$ is the Dynkin diagram of $$\mathfrak{g}$$. The upshot is: $$\mathrm{Aut}(G)\cong \mathrm{Aut}(\mathfrak{g})\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Out}(G)\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Aut}(D_\mathfrak{g}).$$ For types $$A_1, B_n, C_n, G_2, F_4, E_7, E_8$$ there are no symmetries of the Dynkin diagram. For $$A_n$$ ($$n>1$$), $$D_n$$ ($$n\not=4$$), and $$E_6$$, we have $$\mathrm{Aut}(D_\mathfrak{g})\cong \mathbb{Z}/2\mathbb{Z}$$. And in the final case of $$D_4$$, the symmetry group is the symmetric group on three letters. In particular, as stated in the comments: $$\mathrm{Aut}(\mathrm{SU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right.$$ • PSU(2) = PU(2), yes? – annie heart May 6 at 19:27 • Does it mean that the $\mathrm{Aut}(\mathrm{PU}(n))=\mathrm{Aut}(\mathrm{PSU}(n))=\mathrm{Aut}(\mathrm{SU}(n))$, since they have the same Lie algebra: $$\mathrm{Aut}(\mathrm{PSU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right. ?$$ – annie heart May 6 at 19:35 • Do we have $$\mathrm{Out}(G)\cong\mathrm{Out}(\mathfrak{g})?$$ $$\mathrm{Inn}(G)\cong\mathrm{Inn}(\mathfrak{g})?$$ $$\mathrm{Aut}(G)\cong\mathrm{Aut}(\mathfrak{g})?$$ in general? – annie heart May 6 at 19:43 • $PU(n)\cong PSU(n)$ in general. For simply-connected Lie groups, $\mathrm{Aut}(G)\cong \mathrm{Aut}(\mathfrak{g})$ and $\mathrm{Out}(G)\cong \mathrm{Out}(\mathfrak{g})$.The same argument does not work for $\mathrm{Aut}(PG)$ since $PG$ is not generally simply-connected. The inner automorphisms of the Lie algebra are the image of $G$ via the adjoint. I think you need to generally figure out the kernel. – Sean Lawton May 6 at 21:08 • I am not sure. I think it is true for simple, simply-connected compact Lie groups. "Proof:" Since $G$ is simple, the Lie algebra of $\mathrm{Inn}(\mathfrak{g})$ is $\mathfrak{g}$ itself. Thus, since $G$ is simply-connected and compact, the abstract Lie group structure on $\mathrm{Inn}(\mathfrak{g})$ is a finite central quotient of $G$. Since the center is in the kernel, it must be $PG$. And $\mathrm{Inn}(G)\cong PG$ always. $\Box$ – Sean Lawton May 6 at 21:17

https://math.stackexchange.com/questions/2723055/basis-for-eigenspace-of-identity-matrix

# Basis for eigenspace of Identity Matrix Let $A=\begin{pmatrix} 1&0\\ 0&1\\ \end{pmatrix}$. Find the bases for the eigenspaces of the matrix $A$. I know the bases for the eigenspace corresponding to each eigenvector is a vector (or system) that can scale to give any other vector contained in that said eigenspace. Thus, we see that the identity matrix has only one distinct eigenvalue $\lambda=1$. Thus the eigenvector satisfies the equation $(A-\lambda I)\vec{x}=\vec{0}$. Then we must solve $\begin{pmatrix} 0&0&0\\ 0&0&0\\ \end{pmatrix}$. I say that any vector satisfies this equation however the key says that the basis is $\lbrace (1,0),(0,1) \rbrace$. How do they come up with this solution. Thanks in advance! It is true that any vector satisfies the equation. So, our eigenspace is the whole space $\mathbb{R}^2$. Since $\{(1,0),(0,1)\}$ is a basis for $\mathbb{R}^2$, it is also a basis for this eigenspace. • Yes. This is precisely the explanation I needed. Thank you! – coreyman317 Apr 5 '18 at 8:50

https://math.stackexchange.com/questions/2467200/solve-the-recurrence-relation-tn-4tn-3-n-t3-1-n-3k-then-det

# Solve the recurrence relation $T(n)=4T(n/3) + n$, $T(3) = 1$, $n = 3^k$ then determine upper and lower bounds I would like to solve the following recurrence relation. After which I must find it's upper and lower bounds. $T(n)=4T(\frac{n}{3})+n, T(3)=1, n = 3^k$ I attempted to solve with substitution: k = 1: $T(n)=4T(\frac{n}{3})+n$ k = 2: $T(n)=4^2T(\frac{n}{3^2})+\frac{4n}{3}+n$ k = 3: $T(n)=4^3T(\frac{n}{3^3})+\frac{4^2n}{3^2}+\frac{4n}{3}+n$ General: $T(n)=4^kT(\frac{n}{3^k})+\sum_{i=1}^k(\frac{4^{i-1}}{3^{i-1}})n=4^kT(\frac{3^k}{3^k})+...=4^kT(1)+...$ Q1: Here is where my first problem arrived; I was not given a relation for T(1) so I can not finish with substitution, right? Perhaps my professor made a typo and meant to put T(1) = 1? Assuming that he didn't make a typo, I decided to try and solve the problem using Master Theorem: Master Theorem: $T(n)=aT(\frac{n}{b})+n^d$ where $a,b > 0$ and $d\ge0$ then $T(n)=\Theta(n^d)$ if $d>\log_b(a)$ $T(n)=\Theta(n^d\log(n))$ if $d=\log_b(a)$ $T(n)=\Theta(n^{\log_b(a)})$ if $d<\log_b(a)$ Using this method I get $T(n)=\Theta(n^{1.262})$ Q2: I read that $\Theta(g(n))$ is both the upper bound $O(g(n))$ and the lower bound $\Omega(g(n))$ so does that mean that the three answers to my question would be... Solution to recurrence relation: $T(n)=\Theta(n^{1.262})$ Upper bound: $O(n^{1.262})$ Lower bound: $\Omega(n^{1.262})$ • Actually, you have $T(3) = 1$ so you can just use that. – Qudit Oct 11 '17 at 7:00 • The case in when $T(3) = 1$ has the overall term of $4^{k-1}T(\frac{n}{3^{k-1}})$. When this occurs, and after the substitution $T(3) = 1$, the term becomes $4^{k-1}$. – user316270 Oct 11 '17 at 7:08 Q1: Here is where my first problem arrived; I was not given a relation for T(1) so I can not finish with substitution, right? Perhaps my professor made a typo and meant to put T(1) = 1? No, your professor was correct in providing $T(3) = 1$, given $n = 3^{k}$. As you may have noticed when using substitution method, there is generally a relationship in the power of the coefficient - in this case, $4$ - and the power of the denominator in $T()$. Realizing this, you can often quickly generalize and jump to the base condition. To explicitly illustrate: \begin{align} T(n) &= 4^{1}T(\frac{n}{3^{1}}) + n \\ &= 4^{2}T(\frac{n}{3^{2}}) + 4n + n \\ &= 4^{3}T(\frac{n}{3^{3}}) + 4^{2}n + 4n + n \\ & = \ ... \end{align} At this point, look to the value within $T()$ in order to determine the remaining number of additional substitutions need to take place. For this problem, the base case is $T(3) = 1$, and, $$3 = \frac{n}{3^{k-1}}$$ therefore, we get \begin{align} T(n) &= 4^{k-1}T(\frac{n}{3^{k-1}}) + 4^{k-2}n + ... + 4^{k - (k - 1)}n + n \\ & = 4^{k-1} + 4^{k-2}n + ... + 4^{k - (k - 1)}n + n \end{align} Now, these terms need to be summed. Re-expressing $4 = 2^{2}$ and factoring out $n$, \begin{align} T(n) & = 2^{2(k-1)} + 2^{2(k-2)}n + ... + 2^{2(k - [k - 1])}n + n \\ & = 2^{2(k-1)} + n(2^{2(k-2)} + ... + 2^{2(k - [k - 1])} + 1) \end{align} Recognizing the parenthesized terms collectively represent a geometric series, we can easily sum the terms to get \begin{align} T(n) &= 2^{2(k-1)} + n\sum_{i=1}^{k-2} \ (2^{2})^{i} \\ &= 2^{2(k-1)} + n(2^{2(k-2)+1} - 1)\end{align} Lastly, back substitute $k$ to get \begin{align} T(n) &= 4^{(log_{3}(n)-1)} + n(4^{(log_{3}(n)-2)+1} - 1) \\ &= 4^{log_{3}(n)-1}(n+1)-n \end{align} Q2: I read that Θ(g(n)) is both the upper bound O(g(n)) and the lower bound Ω(g(n)) so does that mean that the three answers to my question would be... Using case 1 of the Master Theorem as described by here, $T(n) \in \Theta(n^{1.262})$. Using theta notation alone will suffice for describing the bounds for this function, as it's implied in its definition. Consider the following graph:

https://math.stackexchange.com/questions/2216888/convergence-of-the-series-sum-n-1-infty-frac3-lnnn7-i-say-it-con

# Convergence of the series $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$; I say it converges, WebWork tells me this is incorrect. I'm almost certain that $$\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$$ converges. However, WebWork tells me that this is incorrect. I have been in that situation before but I obviously can't assume that I'm right and the computer is wrong based on that. I don't know how the system work and I'm not sure whether what I do is correct anymore. By the $p$-test, I know $$3\sum_{n=1}^{\infty}\frac{1}{n^7}$$ converges. I also know that $\ln(n)$ grow very slowly so I use the comparison $\ln(n)\le Cn$ which I think will not change that fact that the sum converges such that $$\sum_{n=1}^\infty\frac{3\ln(n)}{n^7}\le \sum_{n=1}^\infty\frac{Cn}{n^7}= C\sum_{n=1}^\infty\frac 1{n^6}\lt\infty$$ which makes me quite certain that $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$ is convergent. I also checked that \begin{align} \lim_{\epsilon \to \infty}\int_1^\epsilon\frac{3\ln(x)}{x^7}\,dx & = 3\lim_{\epsilon \to \infty}\left(-\frac{\log (x)}{6 x^6}\bigg|_{1}^{\epsilon} +\frac{1}{6}\int_1^\epsilon \frac {dx}{x^7}\right)\\ & = 3\lim_{\epsilon \to \infty}\left( -\frac{\log (x)}{6 x^6}-\frac{1}{36 x^6}\right)\bigg|_{1}^{\epsilon}\\ & = \frac 1{12}\\ \end{align} Can someone please shine some light on this for me? • Besides mixing $n$'s and $x$'s in your comparison, you are correct. The series converges. – carmichael561 Apr 4 '17 at 0:53 • WolframAlpha agrees with you that this converges. – Mark Apr 4 '17 at 0:53 • @carmichael561 Thank you I will edit that, I put that on the count of not being properly caffeinated... – user409521 Apr 4 '17 at 0:54 The series is convergent by p test and we may evaluate the “closed form”. The Riemann zeta function is defined as$$\zeta\left(s\right)=\sum_{n\geq1}\frac{1}{n^{s}},\,\textrm{Re}\left(s\right)>1.$$It is absolute convergent in the region $\textrm{Re}\left(s\right)>1$ so $$\zeta'\left(s\right)=\frac{d}{ds}\left(\sum_{n\geq1}\frac{1}{n^{s}}\right)=\sum_{n\geq1}\frac{d}{ds}\left(\frac{1}{n^{s}}\right)=-\sum_{n\geq1}\frac{\log\left(n\right)}{n^{s}}$$ hence $$S=3\sum_{n\geq1}\frac{\log\left(n\right)}{n^{7}}=\color{red}{-3\zeta'\left(7\right)}\approx0.0181.$$ • Thanks for sparking my interest for Riemann's zeta function and giving me some insight. – user409521 Apr 4 '17 at 17:21 • @InfiniteMonkey You're welcome. – Marco Cantarini Apr 4 '17 at 18:17 I would try the "Integral Test for Convergence" which says: Consider an integer $N$ and a non-negative, continuous function $f$ defined on the unbounded interval $[N, ∞)$, on which it is monotone decreasing. Then the infinite series $$\sum_N^{\infty} f(n)$$ converges to a real number if and only if the improper integral $$\int_N^{\infty} f(x) dx$$ is finite. Look at $$\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$$ and the improper integral $$\int _1^{\infty }\frac{3 \ln (n)}{n^7} = \frac{1}{12},$$ which is finite. Thus, by the Integral Test for Convergence, you can say that the infinite series $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$ converges. • The p test is more than enough. You don't need to solve that integral to know that the series converges, it is as immediate as $7 > 1$ – Ant Apr 4 '17 at 8:49 It really, really converges. The only common convergence tests I can think of that don't help are the root test and the alternating series test (the latter for the trivial reason that the summand is always positive). Ratio test: $$\lim_{n \to \infty} \frac{\log(n+1)}{n \log(n)} = 0$$ which you can do by just "limit of a product is the product of the limits" on $\frac{\log(n+1)}{\log n} \times \frac{1}{n}$. Comparison with $\frac{1}{n^2}$: $$\frac{\log n}{n^7} \leq \frac{1}{n^2}$$ if and only if $n^5 \geq \log n$, which is true for all positive $n$. Integral test someone else has covered. Cauchy condensation: it converges if and only if $$\sum_{n=0}^{\infty} \frac{n \log(2)}{2^{7n-n}}$$ does, but that is absurdly quickly convergent. • I was not aware of Cauchy condensation I'm reading about it now thanks! – user409521 Apr 4 '17 at 17:27

http://mathhelpforum.com/advanced-algebra/17574-determinant-very-hard-algebra-print.html

# Determinant with Very Hard Algebra • August 7th 2007, 02:36 AM binarybob0001 Determinant with Very Hard Algebra The problem is $ \det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ] $ Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to. $ abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b) $ There are many ways to factor out terms from this. $ abc[bc(c - b) + ac(a - c) + ab(b - a)] = $ $ abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] = $ $ abc[c^2(b - a) + b^2(a - c) + a^2(c - b)] $ There are even more ways of course. None of these ways lead me to the simple solution: $ abc(b - a)(c - a)(c - b) $ Can someone explain how to factor this? Thanks. • August 7th 2007, 03:43 AM JakeD Quote: Originally Posted by binarybob0001 The problem is $ \det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ] $ Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to. $ abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b) $ There are many ways to factor out terms from this. $ abc[bc(c - b) + ac(a - c) + ab(b - a)] = $ $ abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] = $ $ abc[c^2(b - a) + b^2(a - c) + a^2(c - b)] $ There are even more ways of course. None of these ways lead me to the simple solution: $ abc(b - a)(c - a)(c - b) $ Can someone explain how to factor this? Thanks. Multiply out \begin{aligned}(b - a)(c - a)(c - b) &= (bc -ac -ab +a^2)(c-b) \\ &= bc^2 - ac^2 -abc + a^2c -b^2c +abc + ab^2 - a^2b \\ &= bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b \end{aligned} as you wanted. • August 7th 2007, 07:28 AM binarybob0001 You can only do that if you already know the solution. What if you didn't know the solution? I would have just stopped thinking I was done. Is there a systematic way of arriving at that result? • August 7th 2007, 09:06 AM Soroban Hello, binarybob0001! Quote: The problem is: . $ \begin{bmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{bmatrix}$ Sorry I'm new. . . . . Welcome aboard! That's the best I could write the problem. . . . . Good job! It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to: . . $abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)$ There are many ways to factor out terms from this. None of these ways lead me to the simple solution: . . $abc(b - a)(c - a)(c - b)$ Can someone explain how to factor this? Here's an approach I "discovered" years ago . . . Collect terms with respect to one of the variables. Let's use $c$ . . Collect terms with $c^2$, and with $c$, and 'constants' (no $c$'s). We have: . $abc\bigg[bc^2 - ac^2 - b^2c + a^2c + ab^2 - a^2b\bigg]$ Factor "by grouping": . $abc\bigg[c^2(b-a) - c(b^2-a^2) + ab(b-a)\bigg]$ . . $= \;abc\bigg[c^2(b-a) - c(b-a)(b+a) + ab(b-a)\bigg]$ Factor out $(b-a)\!:\;\;abc(b-a)\bigg[c^2 - c(b+a) + ab\bigg]$ . . $= \;abc(b-a)\bigg[c^2 - bc - ac + ab\bigg]$ Factor "by grouping": . $abc(b-a)\bigg[c(c-b) - a(c-b)\bigg]$ Factor out $(c-b)\!:\;\;abc(b-a)(c-b)(c-a)$ . . . . ta-DAA! • August 8th 2007, 01:35 AM binarybob0001 ahah, so I just need to chose a letter, stupid me. Thank you very much. Hey, are there more problems this hard to practice on?

https://dellwindowsreinstallationguide.com/curve-fitting/

# Generating Perfect Model Data Let’s generate some model x1 and y1 data. First we will create an independent variable x1 that is a column vector ranging from 0 to 20 in steps of 2 and then we can create y1 which is a polynomial with respect to x1. We will store this in a table called data: x1=[0:2:20]'; y1=0.1*x1.^2+1.2*x1+5; data=table(x1,y1); clear x1; clear y1; For this quadratic polynomial y1=0.1*x1.^2+1.2*x1+5 of the form y1=p2(1)*x1.^2+p2(2)*x1+p2(3) the three coefficients are: p2(1)=0.1 p2(2)=1.2 p2(3)=5 p2(1) is the x.^2 coefficient p2(2) is the x coefficient and p2(3) is the constant. # Scatter Plot figure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,... 'MarkerFaceAlpha',0.5,... 'MarkerEdgeAlpha',0.5) set(gca,'FontSize',20) xlabel('x') ylabel('y') title('Polynomial') grid('minor') legend('y1',... 'Location','NorthEast') # Using polyfit to Find the Polynomial Coefficients We can use polyfit to evaluate the polynomial coefficients. Polyfit has the form [pn]=polyfit(x,y,n) Where the input arguments are the x data, y data and n, the order of polynomial. The output arguments is a row vector pn of the polynomial coefficients. In most cases we would try fitting from a first order polynomial upwards however as we know our data is a second order polynomial, as it is created from a second order polynomial we can try fitting to a second order polynomial first. [p2]=polyfit(data.x1,data.y1,2) $\displaystyle \text{p2=}\left[ {\begin{array}{*{20}{c}} {0.100000} & {1.200000} & {5.000000} \end{array}} \right]$ These are the three polynomial coefficients for the quadratic equation: $\displaystyle 0.100000{{x}^{2}}+1.200000x+5.000000$ The original polynomial was y1=0.1*x1.^2+1.2*x1+5 and the three coefficients returned match y1=p2(1)*x1.^2+p2(2)*x1+p2(3) as expected. Which as expected is a perfect match to our original equation. # Using polyval to Evaluate the Polynomial We can use polyval to evaluate the polynomial at specified x values. Let’s first create a new variable x2 and assign it as a column in a new table called datafit. x2=[1:2:19]'; datafit=table(x2); clear x2; We can use polyval to evaluate the polynomial coefficients. polyval has the form [newy]=polyval(pn,newx); We can use the function [datafit.y2]=polyval(p2,datafit.x2); To find the value of the polynomial for each x2 value. # Plotting the Fitted Data with Respect to the Raw Data We can plot this data as a scatter plot alongside the original data: figure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,... 'MarkerFaceAlpha',0.5,... 'MarkerEdgeAlpha',0.5) hold('on'); scatter(datafit.x2,datafit.y2,100,'fill',... 'MarkerEdgeColor',[255/255,0/255,0/255],... 'MarkerFaceColor',[0/255,112/255,92/255],... 'LineWidth',1,... 'MarkerFaceAlpha',0.5,... 'MarkerEdgeAlpha',0.5); hold('off'); set(gca,'FontSize',20) xlabel('x') ylabel('y') title('Polynomial') grid('minor') legend('y1','y2',... 'Location','NorthEast') It may be more common to plot the fitted curve as a line: figure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,... 'MarkerFaceAlpha',0.5,... 'MarkerEdgeAlpha',0.5) hold('on'); plot(datafit.x2,datafit.y2,... 'color','r',... 'LineWidth',3); hold('off'); set(gca,'FontSize',20) xlabel('x') ylabel('y') title('Polynomial') grid('minor') legend('y1','y2',... 'Location','NorthEast') If we want to print the equation to the chart we can use: equation=sprintf('y=(%f)x^2+(%f)x+(%f)',p2(1),p2(2),p2(3)); legend('data',equation,'location','northwest','FontSize',8); The function sprintf has the same form as fprintf but prints to a string variable opposed to outputting in the Command Window. # Underfitting – Fitting a 1st Order Polynomial to 2nd Order Polynomial Data The code above can be modified to fit to a 1st order polynomial (i.e. a straight line). Here it is obvious by eye that it is an unsuitable fit: The reader is recommended to try to use polyfit and polyval to try and fit the above curve to a 1st order polynomial (straight line). x2=[1:2:19]'; [p1]=polyfit(data.x1,data.y1,1) datafit=table(x2); clear x2; [datafit.y2]=polyval(p1,datafit.x2); The user is recommended to plot the graph above, with the updated equation in the legend. figure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,... 'MarkerFaceAlpha',0.5,... 'MarkerEdgeAlpha',0.5) hold('on'); plot(datafit.x2,datafit.y2,... 'color','r',... 'LineWidth',3); hold('off'); set(gca,'FontSize',20) xlabel('x') ylabel('y') title('Polynomial') grid('minor') equation=sprintf('y=(%f)x+(%f)',p2(1),p2(2)); legend('data',equation,'location','northwest','FontSize',8); # Overfitting – Fitting a 3rd Order Polynomial to 2nd Order Polynomial Data Here from the values of the coefficients i.e. the cubic term being extremely small 1.6509e-17 it is obvious that the polynomial is overfitted. This term can be rounded to 0 and we return with the quadratic fit. This is perfect model data so the fitting is relatively easy in real life it may be harder to distinguish data which has been over-fitted. The reader is recommended to try to use polyfit and polyval to try and fit the above curve to a 3rd order polynomial. x2=[1:2:19]'; [p3]=polyfit(data.x1,data.y1,3) datafit=table(x2); clear x2; [datafit.y2]=polyval(p3,datafit.x2); The user is recommended to plot the graph above, with the updated equation in the legend. figure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,... 'MarkerFaceAlpha',0.5,... 'MarkerEdgeAlpha',0.5) hold('on'); plot(datafit.x2,datafit.y2,... 'color','r',... 'LineWidth',3); hold('off'); set(gca,'FontSize',20) xlabel('x') ylabel('y') title('Polynomial') grid('minor') equation=sprintf('y=(%f)x^3+(%f)x^2+(%f)x+(%f)',p3(1),p3(2),p3(3),p3(4)); legend('data',equation,'location','northwest','FontSize',8); # Generating Model Data with Noise rng('default'); x1=[0:2:20]'; [m,n]=size(x1); noise=rand(m,n); y1=0.1*x1.^2+1.2*x1+5+noise; data=table(x1,y1); clear x1; clear y1; clear noise; clear m; clear n; We see a deviation from the starting polynomial with: p2(1)=0.1 p2(2)=1.2 p2(3)=5 The deviation influences the constant term the most, followed by the x term. If we times the noise by 5, we see a larger deviation: If we times the noise by 10 we see more deviation: And if we times the noise by 25 we see an even larger deviation again: If we times the noise by 250, the original function is not recognisable:

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GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video It is currently 11 Jul 2020, 13:57 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Which of the following is an integer? Author Message TAGS: ### Hide Tags Manager Joined: 15 Aug 2008 Posts: 87 Schools: NYU Stern Class of 2012 WE 1: GM, Diversified Financial Svcs. Which of the following is an integer?  [#permalink] ### Show Tags Updated on: 07 Oct 2019, 01:36 2 21 00:00 Difficulty: 5% (low) Question Stats: 82% (00:47) correct 18% (00:41) wrong based on 465 sessions ### HideShow timer Statistics Which of the following is an integer? I. $$\frac{12!}{6!}$$ II. $$\frac{12!}{8!}$$ III. $$\frac{12!}{7!5!}$$ (A) I only (B) II only (C) III only (D) I and II only (E) I, II, and III Originally posted by NickTW on 18 Aug 2009, 19:23. Last edited by Bunuel on 07 Oct 2019, 01:36, edited 2 times in total. Renamed the topic, edited the question and added the OA. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 17070 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Which of the following is an integer?  [#permalink] ### Show Tags 08 May 2015, 15:47 6 2 Hi All, Beyond the arithmetic that sdrandom1 pointed out, these calculations fall into some patterns that you are likely to see (in some form) on Test Day. 12!/6! When dealing with INDIVIDUAL factorials, if the factorial in the numerator is GREATER than the factorial in the denominator, then you will end up with an integer. Here, since 12 > 6, 12!/6! will simplify to an integer. 12!/7!5! You might not have studied the Combination Formula yet, but this calculation is what you would end up with when answering the question "how many different combinations of 7 people can you form from a group of 12 people?" In these types of "Combination" calculations, the number of groups will always be an integer, so 12!/7!5! will simplify to an integer. GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ ##### General Discussion Manager Joined: 17 Dec 2007 Posts: 82 Re: Which of the following is an integer?  [#permalink] ### Show Tags 18 Aug 2009, 19:30 2 1 NickTW wrote: Did a search, could not find anything on this one (the question stem is pretty vague). Would appreciate some enlightenment on the shortcut I'm missing! Which of the following is an integer? I. 12! / 6! II. 12! / 8! III. 12! / 7!5! A) I only B) II only C) III only D) I and II only E) I, II, and III Source: GMAT Prep Exam. Solution : E To me the question looks straight forward. we know 12! is multiple of all numbers from 1 to 12, so dividing it by 6! or 8! is pretty much sure an integer. now coming to the 3rd options 12! by 5! leaves 12*11*10*9*8*7*6/1*2*3*4*5*6*7 which results in integer, so the answe E is right Manager Joined: 30 May 2009 Posts: 130 Re: Which of the following is an integer?  [#permalink] ### Show Tags 18 Aug 2009, 19:45 1 1 There is no shortcut persay... Clear E. 12!/6! = 12*11*10*9*8*7*6!/6! = Integer 12!/8! = 12*11*10*9*8!/8! = Integer 12!/7!*5! = 12*11*10*9*8/5*4*3*2*1 = integer So all 3 are correct. GMAT Tutor Joined: 24 Jun 2008 Posts: 2308 Re: Which of the following is an integer?  [#permalink] ### Show Tags 19 Aug 2009, 08:32 4 1 Every factorial is divisible by every smaller factorial, so 12!/6! and 12!/8! are integers. 12!/(7!*5!) is the number of ways to choose 7 things from a group of 12 if order is irrelevant, so it must also be an integer. _________________ GMAT Tutor in Montreal If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Director Joined: 08 Jun 2010 Posts: 660 Re: Which of the following is an integer?  [#permalink] ### Show Tags 03 May 2015, 01:42 case 1 and case 2 are simple but case 3 is harder. for case 3, we have to write down all the factors. Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 11083 Location: United States (CA) Re: Which of the following is an integer?  [#permalink] ### Show Tags 03 Dec 2017, 17:39 6 2 NickTW wrote: Which of the following is an integer? I. 12! / 6! II. 12! / 8! III. 12! / 7!5! A) I only B) II only C) III only D) I and II only E) I, II, and III Before actually solving this problem, let's review how factorials can be expanded and expressed. As as example, we can use 5!. 5! could be expressed as: 5! 5 x 4! 5 x 4 x 3! 5 x 4 x 3 x 2! 5 x 4 x 3 x 2 x 1! Understanding how this factorial expansion works will help us work our way through each answer choice, especially answer choices 1 and 2. I. 12!/6! Since we know that factorials can be expanded, we now know that: 12! = 12 x 11 x 10 x 9 x 8 x 7 x 6! Plugging this in for answer choice 1, we have: (12 x 11 x 10 x 9 x 8 x 7 x 6!)/6! = 12 x 11 x 10 x 9 x 8 x 7, which is an integer. II. 12!/8! Once again, since we know that factorials can be expanded, we now know that: 12! = 12 x 11 x 10 x 9 x 8! Plugging this in for answer choice 2, we have: (12 x 11 x 10 x 9 x 8!)/8! = 12 x 11 x 10 x 9, which is an integer. III. 12!/(7!5!) Once again, since we know that factorials can be expanded, we now know that: 12! = 12 x 11 x 10 x 9 x 8 x 7! Plugging this in for answer choice 3 gives us: (12 x 11 x 10 x 9 x 8 x 7!)/(7!5!) (12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1) (12 x 11 x 10 x 9 x 8)/(12 x 10 x 1) 11 x 9 x 8, which is an integer. We see that the quantities in Roman numerals I, II and III are all integers. Alternate solution: For any positive integers m, n and p, 1) If m > n, then m!/n! is always an integer. 2) If m = n + p, then m!/(n!p!) is always an integer (which is in fact mCp). From the above two facts, we see that all three quotients in the Roman numerals must be integers. _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 214 REVIEWS 5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE NOW WITH GMAT VERBAL (BETA) See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews Senior Manager Joined: 05 Feb 2018 Posts: 440 Re: Which of the following is an integer?  [#permalink] ### Show Tags 20 Jan 2019, 08:00 NickTW wrote: Which of the following is an integer? I. 12! / 6! II. 12! / 8! III. 12! / 7!5! A) I only B) II only C) III only D) I and II only E) I, II, and III For III, I went through and crossed out everything that's reduced to visualize... 12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1 So you're left with 12*11*2*3 (from taking 5 and 3 from 10 and 9) = 792. Makes sense. ScottTargetTestPrep As a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the result not an integer? Am I right to think this has to do with the number of prime factors? Because 12!/8!5! and 12!/9!5! are also integers, while 12!/10!5! isn't. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and 9! would be 3^2 both of which we can still reduce... while 10! gives us a factor of 5 which we don't have. Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 11083 Location: United States (CA) Re: Which of the following is an integer?  [#permalink] ### Show Tags 22 Jan 2019, 17:19 energetics wrote: NickTW wrote: Which of the following is an integer? I. 12! / 6! II. 12! / 8! III. 12! / 7!5! A) I only B) II only C) III only D) I and II only E) I, II, and III For III, I went through and crossed out everything that's reduced to visualize... 12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1 So you're left with 12*11*2*3 (from taking 5 and 3 from 10 and 9) = 792. Makes sense. ScottTargetTestPrep As a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the result not an integer? Am I right to think this has to do with the number of prime factors? Because 12!/8!5! and 12!/9!5! are also integers, while 12!/10!5! isn't. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and 9! would be 3^2 both of which we can still reduce... while 10! gives us a factor of 5 which we don't have. So you ask a really great qestion. So, as you've shown above, when m < n + p, there is no hard and fast rule to determien when m!/(n!p!) is an integer and when it is not. _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 214 REVIEWS 5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE NOW WITH GMAT VERBAL (BETA) See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews Non-Human User Joined: 09 Sep 2013 Posts: 15412 ### Show Tags 07 Oct 2019, 01:36 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Which is integer?   [#permalink] 07 Oct 2019, 01:36

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GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 19 May 2019, 10:02 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # The figure above represents a semicircular archway over a flat street. Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 55150 The figure above represents a semicircular archway over a flat street.  [#permalink] ### Show Tags 22 Jun 2018, 00:30 20 00:00 Difficulty: 5% (low) Question Stats: 90% (01:35) correct 10% (01:47) wrong based on 655 sessions ### HideShow timer Statistics The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center? A. √2 B. 2 C. 3 D. 4√2 E. 6 NEW question from GMAT® Official Guide 2019 (PS05957) Attachment: PS05957_f009.jpg [ 34.62 KiB | Viewed 6372 times ] _________________ Senior Manager Joined: 13 Feb 2018 Posts: 270 GMAT 1: 640 Q48 V28 The figure above represents a semicircular archway over a flat street.  [#permalink] ### Show Tags 22 Jun 2018, 01:41 Let's disturb Pythagoras once again: $$h^2+2^2=6^2$$ h=$$\sqrt{32}$$ In My Opinion Ans: D e-GMAT Representative Joined: 04 Jan 2015 Posts: 2845 Re: The figure above represents a semicircular archway over a flat street.  [#permalink] ### Show Tags 22 Jun 2018, 05:24 2 Solution Given: • The figure depicts a semi-circular archway over a flat street • The semi-circle has a center at O and a radius of 6 feet To find: • The length of the height h, in feet, of the archway 2 feet from its center Approach and Working: If we add the point O and A, we get a right-angled triangle OAB, right-angled at point B and hypotenuse AO As the radius is 6 feet, we can say • AO = 6 feet Applying Pythagoras Theorem in triangle OAB, we can write: • $$AO^2 = AB^2 + BO^2$$ Or, $$6^2 = h^2 + 2^2$$ Or, $$h^2 = 36 – 4 = 32$$ Or, $$h = 4\sqrt{2}$$ Hence, the correct answer is option D. _________________ CEO Joined: 12 Sep 2015 Posts: 3716 Re: The figure above represents a semicircular archway over a flat street.  [#permalink] ### Show Tags 23 Jun 2018, 09:52 1 Top Contributor Bunuel wrote: The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center? A. √2 B. 2 C. 3 D. 4√2 E. 6 NEW question from GMAT® Official Guide 2019 (PS05957) Attachment: PS05957_f009.jpg Let's add the radius of 6 feet to the diagram to get: From here, we can see the right triangle hiding within the diagram, which means we can apply the Pythagorean Theorem. Se can write: h² + 2² = 6² Simplify: h² + 4 = 36 So, we get: h² = 32 This means h = √32 Check the answer choices. . . √32 is not among them. Looks like we need to simplify √32 We'll use the fact that √(xy) = (√x)(√y) So, √32 = √[(16)(2)] = (√16)(√2) = (4)(√2) = 4√2 Check the answer choices. . . D RELATED VIDEO (simplifying roots) _________________ Test confidently with gmatprepnow.com Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 6133 Location: United States (CA) Re: The figure above represents a semicircular archway over a flat street.  [#permalink] ### Show Tags 25 Jun 2018, 12:06 1 Bunuel wrote: The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center? A. √2 B. 2 C. 3 D. 4√2 E. 6 Since the radius is 6 feet, we can create a right triangle with hypotenuse of 6, leg of 2, and leg of h; thus, we have: 2^2 + h^2 = 6^2 4 + h^2 = 36 h^2 = 32 h = √16 x √2 = 4√2 _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Senior Manager Joined: 10 Apr 2018 Posts: 258 Location: United States (NC) Re: The figure above represents a semicircular archway over a flat street.  [#permalink] ### Show Tags 02 Aug 2018, 13:30 if the radius is 6 max height of the arch can be 6 from the center of semi-circle. So height asked in question which is two feet away from center is definitely less than 6 . Also point 2 feet away from center to end of semicircle the dist is 4 so height is little more than 4 . So the only answer choice that matches it is D _________________ Probus ~You Just Can't beat the person who never gives up~ Babe Ruth Re: The figure above represents a semicircular archway over a flat street.   [#permalink] 02 Aug 2018, 13:30 Display posts from previous: Sort by

http://gesuitialquirinale.it/rdil/polar-coordinates-pdf.html

# Polar Coordinates Pdf My questions is, how important are these topics for calc III? Do I need to re-study or is it not important?. The fact that a single point has many pairs of polar coordinates can cause complications. location than conventional Cartesian coordinates. 3) Rectangular coordinates of point P are given. x t y t9cos , 9sin 16. We now proceed to calculate the angular momentum operators in spherical coordinates. admiral calls a polar coordinate (in 𝑟,𝜃)form and then the defending admiral declares whether the shot was a hit or a miss. To specify a clockwise direction, enter a negative value for the angle. The ordered pairs, called polar coordinates, are in the form $$\left( {r,\theta } \right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$ - axis (polar axis), going counter-clockwise. Polar - Rectangular Coordinate Conversion Calculator. 5 MM Graph Paper. Practice Problem: Convert the following sets of rectangular coordinates into polar coordinates. TrigCheatSheet. Apr 11, 2014 - Explore brittanykaye911's board "polar coordinates", followed by 154 people on Pinterest. Yes office hours Wednesday 2/20 2-4pm SC 323. In this note, I would like to derive Laplace's equation in the polar coordinate system in details. examples to convert image to polar coordinates do it explicitly - want a slick matrix method I thought using the method used above. CARTESIAN & POLAR COORDINATES In fact, as a complete counterclockwise rotation is given by an angle 2π, the point represented by polar coordinates (r, θ) is also represented by (r, θ+ 2nπ) and (-r, θ+ (2n + 1)π) where n is any integer. For any point P consider the two distances:. 3 WS Polar Coordinates (Answers). The area of a region in polar coordinates defined by the equation $$r=f(θ)$$ with $$α≤θ≤β$$ is given by the integral $$A=\dfrac{1}{2}\int ^β_α[f(θ)]^2dθ$$. Polar coordinates use a distance and an angle to locate a point. You can see this by just drawing unit vectors at each point on, say, a circle: (draw). 2 : Apr 12, 2018, 11:37 AM. Polar coordinates describe the distance from P to to a special point O, called the pole or origin and the angle that the line segment PO makes with a special ray called the polar axis. Sign up to join this community. When you drag the red point, you change the polar coordinates $(r,\theta)$, and the blue point moves to the corresponding position $(x,y)$ in Cartesian coordinates. 1 Exponential Equations Blank. In case n = 3, the polar coordinates (r,θ,φ) are called spherical coordinates, and we have y = x1, x = x2, z = x3, r2 = x2 + y2 + z2, x = rsinφsinθ, y = rsinφcosθ, and x = rcosφ, so we can take r3 = r, φ2 = θ. It provides resources on how to graph a polar equation and how to find the area of the shaded. Polar coordinates The representation of a complex number as a sum of a real and imaginary number, z = x + iy, is called its Cartesian representation. y2 4y 8x 20 0 y 2 2 4 2 x 3 2 23. There are three types of polar graph that are Large Single Polar Graph which has thirty marks for r in increment of five degrees, Smaller (Double) Polar Graph which has two polar graphs on one page, each with twenty scale marks for r increment of 5 degrees and Combined Cartesian and Polar has three pages here, One is a large cartesian grid, one a large polar grid and the third one has one. Draw a horizontal line to the right to set up the polar axis. Polar Graph Paper. The principal reason for this is the artificial expansion of the natural conic shapes of the spaces into a cylindrical shape. CONIC SECTIONS IN POLAR COORDINATES If we place the focus at the origin, then a conic section has a simple polar equation. Plane Curvilinear Motion Polar Coordinates (r -θ) The particle is located by the radial distance r from a fixed point and by an angular measurement θto the radial line. In polar coordinates, every point is located around a central point, called the pole, and is named (r,nθ). The value of r can be positive, negative, or zero. requirement for the generalized coordinates is that they span the space of the motion and be linearly independent. This introduction to polar coordinates describes what is an effective way to specify position. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. dinates and the polar direction vectors is that the polar direction vectors change depending on where they are relative to the origin. The small change r in rgives us two concentric circles and the small change in gives us an angular wedge. Graph the point P, (r; ) = 3;ˇ 3. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Let suppose we have a small change in rand. Graphs in Polar Coordinates Purpose The purpose of this lab is to help you become familiar with graphs in polar coordinates. Student information Link. Therefore r˙(t) = (˙rcosθ − rθ˙sinθ)i + (˙rsinθ + rθ˙cosθ)j. (See Figure 9. d is the perpendicular distance from the line to the origin. 6) Area in Polar Coordinates (Fig. Describe planar motion and solve motion problems by defining parametric equations and vector-valued functions. We need to subtract 960 by 180k, such that the result is between 0 and 180. It is sometimes convenient to refer to a point by name, especially when this point occurs in multiple \draw commands. It can be printed and done on paper or completed at Socrative. However, it still is a useful tool to give you an introduction to the concepts pertaining to polar coordinates. r = secθcscθ ⇒ 24. Looking for polar graph paper? We've got just what you need. By default, angles increase in the counterclockwise direction and decrease in the clockwise direction. This calculator converts between polar and rectangular coordinates. To convert from Polar coordinates to Cartesian coordinates, draw a triangle from the horizontal axis to the point. All the terms above are explained graphically there. txt) or read online for free. This OER repository is a collection of free resources provided by Equella. The defending admiral records the location of. os¶z y: Islh1ï/2 3cos0 Ð=3sm0 77t 57t 10) 2,— Convert each pair of rectangular coordinates to polar coordinates where r > 0 and O < 2m 11) - 31 13). That leads to the pre-factor (2/R^2). Sketching Polar Curves Examples. Graph the point Q, (r; ) = 2; ˇ 6. c Double Integrals in Polar Coordinates (r; ) Let us suppose that the region boundary is now given in the form r = f( ) or = h(r), and/or the function being integrated is much simpler if polar coordinates are used. 2-3-17: We continued to convert between polar and rectangular equations. These will all be positive X,Y rectangular coordinates in Quadrant I of the Cartesian plane (X headed right from 0 and Y headed up from 0). ) The graph of = , where is a constant, is the line of inclination. The angular dependence of the solutions will be described by spherical harmonics. Suppose that X is a random vector with joint density function f X(x). PreCalculus. Key Concept: Constellations can be represented graphically. All the coordinates are made beforehand. i have this function f(r,theta) which i want to graph in polar/cylindrical coordinates. In polar coordinates the position of an object $$R$$ distance from the origin as represented in the diagram above is modelled $$\mathbf{r} = R \hat{r}$$ The velocity and acceleration in polar coordinates is derived by differentiating the position vector. Some useful properties about line integrals: 1. Michael VanValkenburgh To make it easier to type and easier to read, this handout will focus on the computational aspects of integration in polar coordinates. If you were to add a true position characteristic it would look like this. However, we can use other coordinates to determine the location of a point. However, we can still rotate around the system by any angle we want and so the coordinates of the origin/pole are (0,θ). Download the pdf file and print. Please read through this supplement before going to quiz section for the polar worksheet on Thursday. Let r1 denote a unit vector in the direction of the position vector r , and let θ1 denote a unit vector perpendicular to r, and in the direction of increasing θ, see Fig. Show Instructions. 2) Equal amounts of the Primaries produce white. r = 2 and θ= 30°, so P is located 2 units from the origin in the positive direction on a ray making a 30°angle with the polar axis. Unit Six Precalculus Practice Test Vectors & Polar Graphs Page 3 of 6 14. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. Cartiesian Coordinate System. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. ) Abbreviated podcast notes on lecture 5. Use a double integral in polar coordinates to calculate the area of the region which is common to both circles r= 3sin and r= p 3cos. Solution; The Cartesian coordinate of a point are $$\left( { - 8,1} \right)$$. 5 Polar Coordinates and Graphs-1. 4 Many problems are more easily stated and solved using a coordinate system other than rectangular coordinates, for example polar coordinates. A polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. x t y t9cos , 9sin 16. You can copy that worksheet to your home. 2 We can describe a point, P, in three different ways. 686 CHAPTER 9 POLAR COORDINATES AND PLANE CURVES The simplest equation in polar coordinates has the form r= k, where kis a positive constant. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. This is the xy-plane. Find the distance between the points. a) r=3secθ b) r=−3sinθ c) rcsc 1θ= 5) Convert the rectangular equation to polar form. 9) ( , ) 10) ( , ) Two points are specified using polar coordinates. Different radials and degree over them are known as the polar paper. The old vvvv nodes Polar and Cartesian in 3d are similar to the geographic coordinates with the exception that the angular direction of the longitude is inverted. And polar coordinates, it can be specified as r is equal to 5, and theta is 53. …Polar-coordinates are entered…in a magnitude direction format. Getting Started To assist you, there is a worksheet associated with this lab that contains examples and even solutions to some of the exercises. The answer is: (r,θ) Polar = (p x2 +y2, arctan y x) Polar Meanwhile, for a point given by Polar coordinates, (r,θ) Polar, we need to specify the coordinates in Cartesian form in terms of the Polar data r and θ. 3) Rectangular coordinates of point P are given. Customize Polar Axes. On the first region we would have −2 6 x 6 2 and √ 4−x2 6 y 6 √ 9−x2, on the second region −3 6 x 6 2. First, fix an origin (called the pole) and an initial ray from O. My knows are (R, r, theta, Phi) My unknowns are (Phi1, R1) If you guys are up for it, could you assist me in establishing a formula for R1 and Phi1 in terms of R, r, theta, and Phi?. A polar coordinate graph paper that’s perfect for when you need to compare two graphs that have minor differences. In addition to, It converts complex number into polar form and vice versa. 5,13) into polar coordinates. Graphing Worksheets for Practice. But what about r f(T)? At first you might think dr dT is the slope of the tangent line to the curve but consider r = constant e. The unit tangent vector to the curve is then Tˆ = ˙xˆı+ ˙y ˆ (2) where we have used a dot to denote derivatives with respect to s. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. 5 , 0 1 2 ≤ ≤θ π. Polar Coordinates Polar coordinates of a point consist of an ordered pair, r θ( , ), where r is the distance from the point to the origin, and θ is the angle measured in standard position. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, Calculating a limit in two variables by going to polar coordinates. Use a double integral in polar coordinates to calculate the area of the region which is common to both circles r= 3sin and r= p 3cos. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. The rectangular coordinates for P (5,20°) are P (4. Check out our many other free graph/grid paper styles. pdf (392 KB) Equella is a shared content repository that organizations can use to easily track and reuse content. (i) Plot each point. Defining Polar Coordinates. Large single polar graph--Thirty scale marks for r in increments of five degrees. It can be found by the "gradient in polar coordinates" googling. TrigCheatSheet. Polar coordinates with polar axes. 841d f0, 2pd u3 5 cos21s2y3d 5 0. Review: Polar coordinates Definition The polar coordinates of a point P ∈ R2 is the ordered pair (r,θ) defined by the picture. We see this general pattern in the circle of gure 2. This easy-to-use packet is full of stimulating activities that will give your students a solid introduction to polar coordinates and trigonometric form!. their direction does not change with the point r. Ciencia y Tecnología, 32(2): 1-24, 2016 - ISSN: 0378-0524 3 II. POLAR COORDINATES (OL]DEHWK :RRG DEFINITION OF POLAR COORDINATES. The angular dependence of the solutions will be described by spherical harmonics. The divergence We want to discuss a vector fleld f deflned on an open subset of Rn. However, in polar coordinates we have u(r,θ) = r sinθ r2 = sinθ r so that u r = − sinθ r2, u. If we restrict rto be nonnegative, then = describes the. Laplace’s equation in polar coordinates, cont. In case n = 3, the polar coordinates (r,θ,φ) are called spherical coordinates, and we have y = x1, x = x2, z = x3, r2 = x2 + y2 + z2, x = rsinφsinθ, y = rsinφcosθ, and x = rcosφ, so we can take r3 = r, φ2 = θ. 6) Area in Polar Coordinates (Fig. In this unit we explain how to convert from Cartesian co-ordinates to polar co-ordinates, and back again. 1 Background on Polar Coordinates. 7 - Polar Coordinates ° 3,225 3,0. The polar coordinate system is formed by fixing a point, O, which is the pole (or origin). The small change r in rgives us two concentric circles and the small change in gives us an angular wedge. To convert from Cartesian co-ordinates to polar use the transformation $y=r\sin { \theta }$ and [m. In this system coordinates for a point P are and , which are indicated in Fig. See Large Polar Graph Paper. Mechanics 1: Polar Coordinates Polar Coordinates, and a Rotating Coordinate System. 5 MM Graph Paper. Viewed 11k times 3. Trigonometry - Trigonometry - Polar coordinates: For problems involving directions from a fixed origin (or pole) O, it is often convenient to specify a point P by its polar coordinates (r, θ), in which r is the distance OP and θ is the angle that the direction of r makes with a given initial line. The ordered pairs, called polar coordinates, are in the form $$\left( {r,\theta } \right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$ – axis (polar axis), going counter-clockwise. The polar coordinate system is extended into three dimensions with two different coordinate systems, the cylindrical and spherical coordinate system. Cartesian/Polar Coordinates Junior high school The connection between Cartesian coordinates and Polar coordinates is established by basic trigonometry. The transformation from Cartesian coordinates to spherical coordinates is. For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. In this unit we explain how to convert from Cartesian co-ordinates to polar co-ordinates, and back again. This article explains how to convert between polar and cartesian coordinates and also encourages the creation of some attractive curves from some relatively easy equations. The point P has. For any point P consider the two distances:. In the polar coordinate system, the ordered pair will now be (r, ). 4 Many problems are more easily stated and solved using a coordinate system other than rectangular coordinates, for example polar coordinates. These charts print on a standard sheet of 8 1/2 x 11 paper. Home Decorating Style 2020 for Polar Coordinate System In Autocad Pdf, you can see Polar Coordinate System In Autocad Pdf and more pictures for Home Interior Designing 2020 4680 at Manuals Library. find the x and y coordinates of a point (r, θ)), we use the following formulas: x = r cos θ, y = r sin θ. Polar Coordinates (r-θ)Ans: -0. First, let's get some preliminaries out of the way. Free Polar Graph Paper Template. We make the convention (−r,θ) = (r,θ +π). Determine a set of polar coordinates for the point. The orientation of a plane curve can be represented by arrows drawn along the curve. (− −5, 5) 9. (iii) Find the Cartesian coordinates of the point. Spherical Coordinates and the Angular Momentum Operators. Students will plot points in the polar coordinate system, convert coordinates and convert equations from rectangular to polar form and vice versa. b) Set up an expression with two or more integrals to find the area common to both curves. 1 New Optional Features pgfplots has been written with backwards compatibility in mind: old TEX les should compile without modi cations and without changes in the appearance. 6 Complex Polar Coordinates (slides, 4-to-1). If we restrict rto be nonnegative, then = describes the. For coordinate conversions: Example 2: Find the rectangular coordinate for the point whose polar coordinates are (a ) 4 5, 3 (b ) 5 4, 6 Example 3: Convert the following rectangular coordinate into four different, equivalent polar coordinates. to describe using polar coordinates. 442, C 5 s21, 0. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. 197t 12 16) x = — I pairs of polar coordinates that describe the same point as the provided polar (-q 1-71/2) Convert each pair of polar coordinates to rectangular coordinates. pdf (condensed podcast notes, 4 slides to a page) Presentations (slides without audio) on lecture 5. 3 Double Integrals in Polar Coordinates In Chapter 10, we explored polar coordinates and saw that in certain situations they simplify problems considerably. R Pr( converts a rectangular form to r in polar coordinates. But many teachers might prefer that you measure angles by yourself using a protractor on blank paper. Lecture #6: The Jacobian for Polar Coordinates Recall. Polar coordinates (Introduction and conversion) Sketching polar curves. For problems 5 and 6 convert the given equation into an equation in terms of polar coordinates. The second point lies on the positive 'y' axis, so the angle in polar coordinates is. The fact that a single point has many pairs of polar coordinates can cause complications. The polar coordinates (r,θ) are defined by r2 = x2 + y2, (2) x = rcosθ and y = rsinθ, so we can take r2 = r and φ2 = θ. The conversion from polar to rectangular coordinates is the same idea as converting rectangular form to polar form in complex numbers. Because Dis a circular disk, we will set up the integral in polar coordinates. Source: Wikipedia - Polar Coordinate System. Cartesian coordinate system: start with xand yaxes. To Convert from Cartesian to Polar. Please read through this supplement before going to quiz section for the polar worksheet on Thursday. (iii) Find the Cartesian coordinates of the point. If there are three variables, the graph is 3D. Polar coordinates part 4 This is a continuation of the polar coordinates part 3. The azimuthal angle, now designated as ϕ, specifies the rotational orientation. Solution: The function that we need to use in this example is G, which converts the pair of rectangular coordinates (x,y) into the polar coordinates (r,!). 2 Calculus In The Polar Coordinate System Contemporary Calculus 4 Area in Rectangular Coordinates (Fig. 3 WS Polar Coordinates (Answers). Convert the following equation of a circle to polar coordinates: 4x2 + 3 2 x +4y2 +1y. 1 Polar Coordinates and Rectangular Coordinates In astronomical calculations, polar coordinate systems are usually used. There are three types of polar graph that are Large Single Polar Graph which has thirty marks for r in increment of five degrees, Smaller (Double) Polar Graph which has two polar graphs on one page, each with twenty scale marks for r increment of 5 degrees and Combined Cartesian and Polar has three pages here, One is a large cartesian grid, one a large polar grid and the third one has one. I still did that, but I also tried odd things. You should be familiar with the Cartesian Coordinate System, also called rectangular coor- dinates, and with the definitions of sin and cos. Polar coordinates with polar axes. in polar from f(r,θ ) = 1 r D E T 3 2 1 cos(T) r 1 r r ´ µ. The Cartesian coordinate of a point are $$\left( {2, - 6} \right)$$. Precalculus. The polar coordinates of a point are given. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates. The transformation from spherical coordinates to Cartesian coordinate is. Polar Coordinates-measures the distances (and direction) from the origin (radius)& the circle •• (r, f), (radius): •• ndusionf Rectangular Coordinates deal with horizontal & vertical distances, whereas polar coordinates deal with diagonal & circular distances. In these notes, we want to extend this notion of different coordinate systems to consider arbitrary coordinate systems. The graph of an equation in polar coordinates is the set of points which satisfy the equation. edu is a platform for academics to share research papers. The small change r in rgives us two concentric circles and the small change in gives us an angular wedge. In this system coordinates for a point P are and , which are indicated in Fig. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. If we restrict rto be nonnegative, then = describes the. 23 17_2_polar_coordinates. You can modify certain aspects of polar axes in order to make the chart more readable. Selection File type icon File name Description Size Revision Time User; Ċ: D32. Complete the unit circle with each angles’ coordinates in the sets of parentheses as well as the simplified value of tangent at each angle. a) Find the polar coordinates of the points of intersection between the two curves. This de nition is worded as such in order to take into account that each point in the plane can have in nitely many representations in polar coordinates. Spherical polar coordinates. This is the xy-plane. Polar Coordinates (r-θ)Ans: -0. Therefore r˙(t) = (˙rcosθ − rθ˙sinθ)i + (˙rsinθ + rθ˙cosθ)j. But most commercial motion control cards do not support the polar coordinate, so this paper presents a program module based on polar coordinate system, which can be integrated into computer numeric control (CNC) controller based on motion control cards. Shade the. (b) Compute the Christoffel symbols of S in polar coordinates. In spherical polar coordinates we describe a point (x;y;z) by giving the distance r from the origin, the angle anticlockwise from the xz plane, and the angle ˚from the z-axis. 5: Polar Coordinates Polar coordinate system, introduced by Isaac Newton, is often more convenient in some applications than the more traditional Cartesian, or rectangular, coordinate system. And you'll get to the exact same point. G15 and G16 G-Codes [Polar Coordinates and CNC Bolt Circles] CNCCookbook's G-Code Training What are Polar and Cartesian Coordinates? Until this point, we've strictly been using Cartesian Coordinates where X, Y, and Z represent distances from part zero (absolute coordinates) or from the current position (relative coordinates). Consider the top which is bounded above by z= p 4 x2 y2 and bounded below by z= p x2 + y2, as shown below. Choose a point in the plane that is called the pole (origin) and labeled O. 6 Graphs of Rational Functions. Polar coordinates part 4 This is a continuation of the polar coordinates part 3. In this section, we discuss how to graph pdfedit ubuntu package equations in polar coordinates on the. If we wish to relate polar coordinates back to rectangular coordinates (i. The graph above shows symmetry with respect to the y-axis. 4 2D Elastostatic Problems in Polar Coordinates Many problems are most conveniently cast in terms of polar coordinates. 8, as outlined in the. Spherical coordinates system (or Spherical polar coordinates) are very convenient in those problems of physics where there no preferred direction and the force in the problem is spherically symmetrical for example Coulomb's Law due to point. SPHERICAL POLAR COORDINATES. It is sometimes convenient to refer to a point by name, especially when this point occurs in multiple \draw commands. Double Integrals in Polar Coordinates “Integrating Functions over circular regions” Suppose we want to integrate the function f(x,y) = x2 over the fol-lowing region: 2 3 We would have to break the region up into three pieces. Mon Nov 11 - I retaught graphing roses and then we began converting from polar form to rectangular and rectangular to polar. Michael VanValkenburgh To make it easier to type and easier to read, this handout will focus on the computational aspects of integration in polar coordinates. 3 WS Polar Coordinates (Answers). Conversion: Rectangular to Polar/ Polar to Rectangular 2011 Rev by James, Apr 2011 1. [See how to convert rectangular and polar forms in the complex numbers chapter. 841 cos us d 5 2y3, 3 cos sud 5 2, 1 2 3 cos sud 5 21, s2r, u 1 pd fsud 5 2gsu 1 pd, p, sr, ud. Convert the following equation of a circle to polar coordinates: 2x2 +3x+2y2 + −5y = 7 7. We know sine starts at zero, and then grows until the function reaches a height of one at ˇ=2. SCHROEDINGER'S EQUATION IN SPHERICAL POLAR COORDINATES The magnitude of a central force on an object depends on only the distance of that object. a) Find the polar coordinates of the points of intersection between the two curves. Most of the things we've done can also be done in the polar, cylindrical, and spherical coordinate as well. Describe the graph. However, we can use other coordinates to determine the location of a point. 21 Locating a point in polar coordinates Let’s look at a specific example. ) The graph of = , where is a constant, is the line of inclination. Thus its area will be Z 2π 0 R2 2 dϑ = R2 2 x 2π 0. 54 Example 1 – Using Polar Coordinates to Describe a Region Use polar coordinates to describe each region shown in. This is an advantage of using the polar form. Polar Rectangular Regions of Integration. Convert each pair of rectangular coordinates to polar coordinates where r and. We basically use a 2D formation having two coordinates x and y, if you are wishing to create graph points on a coordinate plane then below we are providing instructions of doing that. θ π 4 is the straight line through the origin pole making an angle of π 4 University of Calgary MATH 267 - Summer 2019 Math267-Polar-Coordinates-Double-Integrals. We shall show how easy it becomes using polar coordinates instead. Homework 2: Spherical Polar Coordinates Due Monday, January 27 Problem 1: Spherical Polar Coordinates Cartesian coordinates (x,y,z) and spherical polar coordinates (r,θ,ϕ) are related by x = r sinθ cosϕ y = r sinθ sinϕ z = r cosθ. Example 3 2Given the ellipse with equation 9 x2 + 25 y = 225, find the major and minor axes, eccentricity, foci and vertices. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. Viewed 11k times 3. If we restrict rto be nonnegative, then = describes the. Symmetry with. 25 and also 12? How about 17 and 13? Good times. View Notes - Polar-Coordinate-System. 5, 30°), (-1. • θis measured from an arbitrary reference axis • e r and eθ are unit vectors along +r & +θdirns. I'm always amazed by what my students come up with on this one - I've even had a student who designed a penguin using polar equations!. Extensions Graphing the polar equations will help the students to make the connection when they are learning to change polar coordinates to rectangular coordinates and back The students can visually see the points on the polar axis and compare the point on the rectangular axis. This discussion is critical for you to understand in order to correctly determine the polar coordinates. 1 Illustrating polar coordinates. The graphing worksheets are randomly created and will never repeat so you have an endless supply of quality graphing worksheets to use in the classroom or at home. d) ˜˝3, ˝Π 6 ˚. Then a number of important problems involving polar coordinates are solved. Convert the following equation to polar coordinates: y = − 4 3 x 6. Find the distance between the points. pdf View Download. In particular, how the angle increases counter-clockwise and how the radius rincreases going away from the origin. Convert the equation of the circle r= 2sinto rectangular coordinates and nd the center and radius of the circle. The polar coordinate system (r, θ) and the Cartesian system (x, y) are related by the following expressions: With reference to the two-dimensional equ ations or stress transformation. Polar coordinates are not unique. In a rectangular coordinate system, we were plotting points based on an ordered pair of (x, y). 3 convert the equation from polar to rectangular and draw the graph (a) r = 2secθ (b) θ = π/3 (c* graph only) r = θ 4 find the slope of the tangent line at the given value of θ (a) r = 3, θ = π/6 (b) r = 1+3sinθ, θ = π/4 5* What is the distance formula for two points in polar coordinates, (r1, θ1) & (r2, θ2) ?. This is the result of the conversion to polar coordinates in form. Arc length of polar curves. r = 2 and θ= 30°, so P is located 2 units from the origin in the positive direction on a ray making a 30°angle with the polar axis. z axis up, θ is sometimes called the zenith angle and φ the azimuth angle. We begin with a brief review of polar coordinates. Polar sun path chart program This program creates sun path charts using polar coordinate for dates spaced about 30 days apart, from one solstice to the next. Here, the two-dimensional Cartesian relations of Chapter 1 are re-cast in polar coordinates. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations. The rst coordinate is the distance of the point from the origin (0;0), and the second coordinate is the angle, in standard. Creating Constellations on a Coordinate Plane Grade Level/Subject: Science or math (Could be simplified to 3rd grade level by making the coordinates all positive. If we restrict rto be nonnegative, then = describes the. We see this general pattern in the circle of gure 2. We interpret r as the distance from the sun and θ as the planet's angular bearing, or its direction from a fixed point on the sun. $\begingroup$ See answer here with diagram showing the directions and magnitudes of the changes in velocity (in polar form). This is the xy-plane. If C is a circle of radius R, then its polar equation is f(ϑ) = R where 0 6 ϑ 6 2π. 8 Polar Equations of Conics We have seen that geometrically the conic sections are related since they are all created by intersecting a plane with a right circular cone. 442, C 5 s21, 0. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. A point P is located at (r,θ) in a polar coordinate system if the distance from P. Showing top 8 worksheets in the category - Number Planes. Therefore r˙(t) = (˙rcosθ − rθ˙sinθ)i + (˙rsinθ + rθ˙cosθ)j. However, doing the math is the tricky part. We shall show how easy it becomes using polar coordinates instead. Polar - Rectangular Coordinate Conversion Calculator. We basically use a 2D formation having two coordinates x and y, if you are wishing to create graph points on a coordinate plane then below we are providing instructions of doing that. However, the Coriolis acceleration we are discussing here is a real acceleration and which is present when rand both change with time. (5, 960°) SOLUTION: Let P(r, θ) = (5, 960°). A point P in the plane, has polar coordinates (r; ), where r is the distance of the point from the origin and is the angle that the ray jOPjmakes with the positive x-axis. This is an advantage of using the polar form. Then we will use these formulas to convert Cartesian equations to polar coordinates, and vice versa. Changing the solvent has little. Frame of Reference In the polar coordinate system, the frame of reference is a point O that we call the pole and a ray that. polar coordinates pl (plural only) ( mathematics ) The coordinates of a point in a plane, measured as its Cartesian distance from the origin and the angle measured anticlockwise / counterclockwise from the x -axis to a line joining the point to the origin. Graphing in Polar Coordinates Jiwen He 1 Polar Coordinates 1. We need to subtract 960 by 180k, such that the result is between 0 and 180. Eliminate the parameter and identify the graph of the parametric curve. The equations of the 10 - and 20 - radius circles are r = 10 and r = 20, respectively. A point in polar coordinates requires an angle a, in degrees, and distance from the origin, r. It’s easy to convert rectangular coordinates to polar coordinates when the angle of the polar coordinate is 0°, 30°, 45°, 60°, or 90°. Spherical polar coordinates provide the most convenient description for problems involving exact or approximate spherical symmetry. Lesson 6: Polar, Cylindrical, and Spherical coordinates 1. How does this compare with the computation in the usual coordinate system ψ(x,y) = (x,y,0)? Answer: In order to compute the Christoffel symbols, first we need to compute the partials of E, Fand G: E ρ = 0 E θ = 0 F ρ = 0 F θ = 0 G ρ = 2ρ G θ = 0. Symmetry with. r (x ;y)=( rcos( ) sin( )) =ˇ 6 =ˇ 3 Polar coordinates are related to ordinary (cartesian) coordinates by the formulae x = r cos( ) y = r sin( ) r = p x 2+ y = arctan(y=x):. Spherical-polar coordinates. Search this site. Superposition of separated solutions: u = A0=2 + X1 n=1 rn[An cos(n ) + Bn sin(n )]: Satisfy boundary condition at r = a,. In fact, we will look at how to calculate the area given one polar function, as well as when we need to find the area between two polar curves. To find the game you're looking for, use the filter below. For example, the behavior of the drum surface when you hit it by a stick would be best described by the solution of the wave equation in the polar coordinate system. pdf), Text File (. Elasticity equations in polar coordinates (See section 3. d is the perpendicular distance from the line to the origin. In this section, we explore the question of how to quantize a system in curvilinear coordinates, using plane polar coordinates as an example. The r represents the distance you move away from the origin and θ represents an angle in standard position. Lecture 36: Polar Coordinates A polar coordinate system, gives the co-ordinates of a point with reference to a point Oand a half line or ray starting at the point O. Review: Polar coordinates Definition The polar coordinates of a point P ∈ R2 is the ordered pair (r,θ) defined by the picture. b) Show that the area of R is 1 (9 3 2) 16 − π. We will look at polar coordinates for points in the xy-plane, using the origin (0;0) and the positive x-axis for reference. And if we talking about polar paper for maths so this is a type graph paper which is used in many projects and also. A location is defined by its distance in x,y from the origin point. POLAR COORDINATE GRAPH PAPER Author: Frances Elizabeth Blount Created Date: 10/26/2007 9:37:15 AM. The easiest kind of region R to work with is a rectangle. 5: Polar Coordinates Polar coordinate system, introduced by Isaac Newton, is often more convenient in some applications than the more traditional Cartesian, or rectangular, coordinate system. Convert Rectangular to Polar Coordinates Polar Axis If a point P has rectangular coordinates (x, y) then the polar coordinates (r, e) of P are given by and tan-I y —, when x > O tan tan — or + 1800, when x < O. Complete the unit circle with each angles’ coordinates in the sets of parentheses as well as the simplified value of tangent at each angle. Polar Coordinates - Solution Question 1 Plot the points with Cartesian coordinates A 8 p 3;8 and B 5 4;5 p 3 4 and then convert them to polar coordinates. Apr 9 - Today I handed back and we went over the unit 5 and Unit 6 tests. Write an equation for this curve in rectangular coordinates. The polar coordinate system,(r, ), is convenient if we want to consider radial distance from a fixed point (origin, or pole) and bearing (direction). The orientation of a plane curve can be represented by arrows drawn along the curve. Selection File type icon File name Description Size Revision Time User; Ċ: D32. The photochemistry of 4-haloanilines and 4-halo-N,N-dimethylanilines has been studied in apolar, polar aprotic, and protic solvents. In polar coordinates the position of an object $$R$$ distance from the origin as represented in the diagram above is modelled $$\mathbf{r} = R \hat{r}$$ The velocity and acceleration in polar coordinates is derived by differentiating the position vector. The following steps can be used for graphing polar curves: 1. Using relative coordinate, points entered by typing @x,y [Enter] Polar Coordinates Polar coordinates used when you need to draw the next points at specify angle. Polar coordinate lines. Polar, Cylindrical, and Spherical Coordinates 1. 64 Spoke Degrees. The polar coordinate system is a two-dimensional coordinate system using a polar grid: The r and θ coordinates of a point P measure respectively the distance from P to the origin O and the angle between the line OP and the polar axis. The solution is such that the stress components are in the form of a Fourier series in θ {\displaystyle \theta \,}. Polar Coordinates. Find the distance between the points. The area element in polar coordinates In polar coordinates the area element is given by dA = r dr dθ. 1 Exponential Equations Blank. Graphing in Polar Coordinates Jiwen He 1 Polar Coordinates 1. Find a different pair of polar coordinates for each point such that 0 ≤ ≤ 180° or 0 ≤ ≤ π. 2 Calculus In The Polar Coordinate System Contemporary Calculus 4 Area in Rectangular Coordinates (Fig. Polar Coordinates Polar coordinates of a point consist of an ordered pair, r θ( , ), where r is the distance from the point to the origin, and θ is the angle measured in standard position. 3 WS Polar Coordinates (Answers). The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Pre-Calculus Worksheet Name: _____ Section 10. Polar-coordinate equations for lines A polar coordinate system in the plane is determined by a Pole Pand a half-line called the polar axis, extending from Pto the right in Figure 1. However, it still is a useful tool to give you an introduction to the concepts pertaining to polar coordinates. 04 Double Integrals in Polar Coordinates. Find the polar coo dinate. Show the angle θ between two lines with slopes m 1 and m 2 is given by the equation tanθ = m 2 −m 1 1−m 2m 1 I've added some more information to the diagram, based on the hint to include the angle the lines make with the x-axis. 54 Example 1 – Using Polar Coordinates to Describe a Region Use polar coordinates to describe each region shown in. Also remember that there are three types of symmetry - y-axis, x-axis, and origin. Media in category "Polar coordinate system" The following 124 files are in this category, out of 124 total. It only takes a minute to sign up. The use of polar graph paper or circular graph paper uses, in schools and colleges math teachers, are also still making assignments that require students to make a graph and draw my own by hands. Coordinate Graph Paper PDF. 7 is self explanatory. • θis measured from an arbitrary reference axis • e r and eθ are unit vectors along +r & +θdirns. 0 International License. Polar Coordinates. This coordinate system is convenient to use when the distance and direction of a particle are measured relative to a fixed point or when a particle is fixed on or moves along a rotating arm. The polar coordinate system is an adaptation of the two-dimensional coordinate system invented in 1637 by French mathematician Ren é Descartes (1596 – 1650). But many teachers might prefer that you measure angles by yourself using a protractor on blank paper. As previously noted, the Cartesian coordinate (a,b) refers to the point a cen-timeters in the x-direction and b centimeters in the y-direction. We need to subtract 960 by 180k, such that the result is between 0 and 180. In particular, how the angle increases counter-clockwise and how the radius rincreases going away from the origin. pdf), Text File (. 5) 1,150 6) 1, 240 7) Plot 3, 4 A on the polar grid and find three additional pairs of polar coordinates that name the point if 22. Double Integrals in Polar Coordinates. A polar coordinate system, gives the co-ordinates of a point with reference to a point O and a half line or ray starting at the point O. Polar coordinates and Defining the Polar Coordinate Axes Any right triangle can be defined on a circle with radius r. This introduction to polar coordinates describes what is an effective way to specify position. r is the radius, and θ is the angle formed between the polar axis (think of it as what used to be the positive x-axis) and the segment connecting the point to the pole (what used to be the origin). (ii) Find two other pairs of polar coordinates for each point, one with r ˜ 0 and one with r ˚ 0. The first equation looks easy but there is a hidden assumption that you need to be aware of. 2 Calculus In The Polar Coordinate System Contemporary Calculus 4 Area in Rectangular Coordinates (Fig. Write the inequalities for. Polar coordinates are a set of values that quantify the location of a point based on 1) the distance between the point and a fixed origin and 2) the angle between. Frame of Reference In the polar coordinate system, the frame of reference is a point O that we call the pole and a ray that emanates from it that we call the polar axis. Since the axis of the parabola is vertical, the form of the equation is Now, substituting the val-ues of the given coordinates into this equation, we obtain. We must also know how to convert from rectangular to polar coordinates and from polar coordinates to. Apr 9 - Today I handed back and we went over the unit 5 and Unit 6 tests. 0 Unported by Lantonov. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. 841d f0, 2pd u3 5 cos21s2y3d 5 0. Use absolute polar coordinates when you know the precise distance and angle coordinates of the point. To specify a clockwise direction, enter a negative value for the angle. Math 126 Worksheet 5 Polar Coordinates Graphing Polar Curves The aim of this worksheet is to help you familiarize with the polar coordinate system. 10) It is often convenient to work with variables other than the Cartesian coordinates x i ( = x, y, z). Pre-Calculus Notes Name: _____ Section 10. 1 De ning Polar Coordinates oT nd the coordinates of a point in the polar coordinate system, consider Figure 1. Graph the point P, (r; ) = 3;ˇ 3. There are some aspects of polar coordinates that are tricky. org are unblocked. 11) ( , ), ( , ) 12) ( , ), ( , ) Critical thinking question: 13) An air traffic controller's radar display uses polar coordinates. Polar Coordinates. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. The equations of the 10 - and 20 - radius circles are r = 10 and r = 20, respectively. location than conventional Cartesian coordinates. 362 Chapter 10 Conics, Parametric Equations, and Polar Coordinates 21. The conversion from polar to rectangular coordinates is the same idea as converting rectangular form to polar form in complex numbers. Review: Polar coordinates Definition The polar coordinates of a point P ∈ R2 is the ordered pair (r,θ) defined by the picture. First measure a circle feature. We graph some of the basic functions in polar coordinates using LiveMath and a graphing calculator. Polar coordinates An alternative to using rectangular coordinates (x and y) to specify points in the plane is to specify how far the point is from the origin and the direction it lies in. The ordered pair specifies a point's location based on the value of r and the angle, θ, from the polar axis. 5355 0 -10] x = 1×4 5. Polar Curves Curves in Polar Coordinate systems are called Polar Curves, which can be written as r = f(µ) or, equivalently, as F(r;µ) = 0. Ciencia y Tecnología, 32(2): 1-24, 2016 - ISSN: 0378-0524 3 II. (b) Find the velocity of the particle in polar coordinates. The graph above shows symmetry with respect to the y-axis. More Graphing Polar Equations. Enter this lesson and corresponding worksheet covering the basics of the polar coordinate system. What happens when you divide a circle by 365. d) ˜˝3, ˝Π 6 ˚. r is a directed distance from the pole to P. As the function approaches ˇ, the value reduces back to zero. 2 : Apr 12, 2018, 11:37 AM. (it can be positive, negative, or zero. (iii) Find the Cartesian coordinates of the point. Find the volume of the region bounded by the paraboloid z= 2 4x2 4y2 and the plane z= 0. a) r=3secθ b) r=−3sinθ c) rcsc 1θ= 5) Convert the rectangular equation to polar form. Beautifull!! #2 Andre, December 19, 2009 at 12:22 p. Graph the point P, (r; ) = 3;ˇ 3. SCHROEDINGER'S EQUATION IN SPHERICAL POLAR COORDINATES The magnitude of a central force on an object depends on only the distance of that object. In other words, we need to rewrite the equation so that the denominator begins with 1. These charts print on a standard sheet of 8 1/2 x 11 paper. Example: What is (12,5) in Polar Coordinates?. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. I Computing volumes using double integrals. 686 CHAPTER 9 POLAR COORDINATES AND PLANE CURVES The simplest equation in polar coordinates has the form r= k, where kis a positive constant. k = 5 Since k is odd, we need to replace r with -r to obtain the correct polar coordinates. I Changing Cartesian integrals into polar integrals. 0 Unported by Lantonov. The area inside the polar curve r = 3 + 2cos q is-4 -2 2 4-4-2 2 4 (A) 9. Cylindrical Coordinates Transforms The forward and reverse coordinate transformations are != x2+y2 "=arctan y,x ( ) z=z x =!cos" y =!sin" z=z where we formally take advantage of the two argument arctan function to eliminate quadrant confusion. Polar coordinates An alternative to using rectangular coordinates (x and y) to specify points in the plane is to specify how far the point is from the origin and the direction it lies in. Polar coordinates and applications Let’s suppose that either the integrand or the region of integration comes out simpler in polar coordinates (x= rcos and y= rsin ). (ii) Find two other pairs of polar coordinates for each point, one with r ˜ 0 and one with r ˚ 0. Precalculus: Polar Coordinates Practice Problems 3. Let (r,θ) denote the polar coordinates describing the position of a particle. Cartesian/Polar Coordinates Junior high school The connection between Cartesian coordinates and Polar coordinates is established by basic trigonometry. Polar Graph Paper. A Cartesian coordinate system is the unique coordinate system in which the set of unit vectors at different points in space are equal. Ask Question Asked 7 years, 3 months ago. Algebra of complex numbers You should use the same rules of algebra as for real numbers,. POLAR COORDINATES CONTINUED Writing Equations in Polar Form x y r r r, , cos , sin oo T T T I can convert an equation from Rectangular Form to Polar Form. Any geometric object in the plane is a set (collection) of points, so we can describe it by a set of coordinate pairs. (i) Plot each point. To evaluate ZZ R f(x,y)dxdy proceed as follows: • work out the limits of integration if they are not. Show Step-by-step Solutions. The ranges of the variables are 0 < p < °° 0 < < 27T-00 < Z < 00 A vector A in cylindrical coordinates can be written as (2. 23 17_2_polar_coordinates. 7 is self explanatory. The fact that a single point has many pairs of polar coordinates can cause complications. Example Sketch the curve described by the polar equation. This allows you to fully utilize the paper size that you have on hand. 1 Polar Coordinates and Polar Equations OBJECTIVE 1: Plotting Points Using Polar Coordinates In this section we begin our study of the polar coordinate system. You can see this by just drawing unit vectors at each point on, say, a circle: (draw). The transformation from spherical coordinates to Cartesian coordinate is. (5, 960°) SOLUTION: Let P(r, θ) = (5, 960°). A CNC program module based on polar coordinate system Article (PDF Available) in International Journal of Advanced Manufacturing Technology 68(5-8) · September 2013 with 2,790 Reads. No office hours Tuesday 2/19. y x y x P(x,y) 0 y x y x P(x,y) 0 r a) b. You should be familiar with the Cartesian Coordinate System, also called rectangular coor- dinates, and with the definitions of sin and cos. A point P is located at (r,θ) in a polar coordinate system if the distance from P. A polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This measurement will display in the Cartesian coordinates. The polar coordinates r and φ can be converted to the Cartesian coordinates x and y by using the trigonometric functions sine and cosine: = ⁡, = ⁡. 21 Locating a point in polar coordinates Let’s look at a specific example. Basic polar coordinates are those coordinates with angles not lower than -360° and not higher than +360°. The location of P in the plane can also be described using polar coordinates. Polar Coordinates This file contains one interactive page that your students could use to practice plotting polar coordinates. First, fix an origin (called the pole) and an initial ray from O. Consider the top which is bounded above by z= p 4 x2 y2 and bounded below by z= p x2 + y2, as shown below. How does this compare with the computation in the usual coordinate system ψ(x,y) = (x,y,0)? Answer: In order to compute the Christoffel symbols, first we need to compute the partials of E, Fand G: E ρ = 0 E θ = 0 F ρ = 0 F θ = 0 G ρ = 2ρ G θ = 0. Polar coordinates are in the form r, , where is the independent variable. We want to nd another way to get to the point (x;y). The Polar Coordinate System is a different way to express points in a plane. The polar form of (a,b) is illustrated in Figure 1. Number Planes. Polar coordinates and applications Let's suppose that either the integrand or the region of integration comes out simpler in polar coordinates (x= rcos and y= rsin ). Plane polar coordinates pdf Plane polar coordinates pdf Plane polar coordinates pdf DOWNLOAD! DIRECT DOWNLOAD! Plane polar coordinates pdf Polar Coordinates r, θ in the plane are described by r distance from the origin and θ 0, 2π is the counter-clockwise angle. It is sometimes convenient to refer to a point by name, especially when this point occurs in multiple \draw commands. generally start to learn to think in terms of polar coordinates. Consider the top which is bounded above by z= p 4 x2 y2 and bounded below by z= p x2 + y2, as shown below. 2 We can describe a point, P, in three different ways. Extensions Graphing the polar equations will help the students to make the connection when they are learning to change polar coordinates to rectangular coordinates and back The students can visually see the points on the polar axis and compare the point on the rectangular axis. Double Integrals in Polar Coordinates 1. Finally, the Coriolis acceleration 2r Ö. 5355 0 -10] x = 1×4 5. Stirling's Web Site. Compass Labels on Polar Axes.

https://www.physicsforums.com/threads/integral-yielding-2-answers.727979/

1. Dec 12, 2013 ### Appleton 1. The problem statement, all variables and given/known data $\int \frac{1}{\sqrt{1-x^{2}}} dx$ I seem to be getting two incompatible answers when I substitute x with $sin \theta$ and $cos \theta$. Could someone please help me with where I'm going wrong. 2. Relevant equations 3. The attempt at a solution first attempt (the correct answer I believe): $let\ x = sin \theta$ $dx = cos \theta\ d\theta$ $\int \frac{cos \theta}{\sqrt{1-sin^{2}\theta}} d\theta$ $= \int \frac{cos \theta}{\sqrt{cos^{2}\theta}} d\theta$ $= \int \frac{cos \theta}{{cos\theta}} d\theta$ $= \int 1\ d\theta$ $= \theta$ $x = sin \theta$ $\theta = arcsin x$ second attempt (the wrong answer I believe) $let\ x = cos \theta$ $dx = -sin \theta\ d\theta$ $\int \frac{-sin \theta}{\sqrt{1-cos^{2}\theta}} d\theta$ $= \int \frac{-sin \theta}{\sqrt{sin^{2}\theta}} d\theta$ $= \int \frac{-sin \theta}{{sin\theta}} d\theta$ $= \int -1\ d\theta$ $= -\theta$ $x = cos \theta$ $\theta = -arccos x$ 2. Dec 12, 2013 ### sandyp Its simple...we know that d/dx arc sinx = (1-x^2)^-1/2 d/dx arc cos x = -(1-x^2)^-1/2 (notice the minus sign) as integral is antiderivative of a function you are getting the same values yes integral arc sinx=(minus)integral arc cos x..Hope your doubt is clarified 3. Dec 12, 2013 ### ShayanJ In fact $\int \frac{1}{\sqrt{1-x^2}}dx=\sin^{-1}{x}+C_1 \ and \ \int \frac{1}{\sqrt{1-x^2}}dx=-\cos^{-1}{x}+C_2$ If you take a look at the plots of the functions $\sin^{-1}{x}$ and $\cos^{-1}{x}$,you can see that you can get one of them by multiplying the other by a minus sign and then adding a constant(or vice versa!)...and you can see that you have the negative sign and also the constant in the results of the integral! 4. Dec 12, 2013 ### Appleton Thanks for clearing that up for me. 5. Dec 12, 2013 ### Staff: Mentor No one has mentioned this. There's a trig identity that can shed some light here. $sin^{-1}(x) + cos^{-1}(x) = \pi/2$ The two functions differ by a constant. If an indefinite integral produces two different antiderivatives, those functions can differ by at most a constant.

https://mathhelpboards.com/threads/integral-substitution-method-problem.5634/

Integral - substitution method problem Yankel Active member Hello all I am working on this integral $\int \frac{x^{2}+1}{x^{4}+1}dx$ Now, I have tried this way: $u=x^{2}+1$ after I did: $\int \frac{x^{2}+1}{\left ( x^{2}+1 \right )\left ( x^{2}-1 \right )}dx$ But I got stuck, I got: $\frac{1}{2}\cdot \int \frac{1}{u\sqrt{u-1}}dx$ I thought of making another substitution, but I tried and failed. Help required Prove It Well-known member MHB Math Helper Hello all I am working on this integral $\int \frac{x^{2}+1}{x^{4}+1}dx$ Now, I have tried this way: $u=x^{2}+1$ after I did: $\int \frac{x^{2}+1}{\left ( x^{2}+1 \right )\left ( x^{2}-1 \right )}dx$ But I got stuck, I got: $\frac{1}{2}\cdot \int \frac{1}{u\sqrt{u-1}}dx$ I thought of making another substitution, but I tried and failed. Help required Well first of all, \displaystyle \displaystyle \begin{align*} x^4 + 1 \end{align*} is NOT equal to \displaystyle \displaystyle \begin{align*} \left( x^2 + 1 \right) \left( x^2 - 1 \right) \end{align*}. I would try \displaystyle \displaystyle \begin{align*} x^4 + 1 &= x^4 + 2x^2 + 1 - 2x^2 \\ &= \left( x^2 + 1 \right) ^2 - \left( \sqrt{2} \, x \right) ^2 \\ &= \left( x^2 - \sqrt{2}\, x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) \end{align*} and so \displaystyle \displaystyle \begin{align*} \int{ \frac{x^2 + 1}{x^4 + 1}\,dx} &= \int{ \frac{x^2 - \sqrt{2}\,x + 1 + \sqrt{2}\,x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + 1} \,dx} + \sqrt{2} \int{ \frac{x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + \left( \frac{\sqrt{2}}{2} \right) ^2 - \left( \frac{\sqrt{2}}{2} \right) ^2 + 1 } \, dx} + \sqrt{2} \int{ \frac{x}{x^4 + 1} \, dx} \\ &= \int{ \frac{1}{ \left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 }\, dx} \end{align*} Now in the first integral, let \displaystyle \displaystyle \begin{align*} x + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\tan{(\theta)} \implies dx = \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta \end{align*} and in the second integral let \displaystyle \displaystyle \begin{align*} x^2 = \tan{(\phi)} \implies 2x\,dx = \sec^2{(\phi)}\,d\phi \end{align*} and the integrals become \displaystyle \displaystyle \begin{align*} \int{ \frac{1}{\left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 } \, dx} &= \int{ \frac{1}{ \left[ \frac{\sqrt{2}}{2} \tan{(\theta)} \right] ^2 + \frac{1}{2} } \cdot \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta } + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{ \tan^2{(\phi)} + 1 } \, d\phi } \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{ \frac{1}{2}\tan^2{(\theta)} + \frac{1}{2} } \, d\theta} + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{\sec^2{(\phi)}}\,d\phi} \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{ \tan^2{(\theta)} + 1 } \, d\theta} + \frac{\sqrt{2}}{2} \int{ 1 \, d\phi } \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{\sec^2{(\theta)}}\,d\theta} + \frac{\sqrt{2}}{2} \phi + C_1 \\ &= \sqrt{2} \int{ 1\,d\theta } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \, \theta + C_2 + \frac{\sqrt{2}}{2}\arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \arctan{ \left( \sqrt{2}\, x + 1 \right) } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C \textrm{ where } C = C_1 + C_2 \end{align*} Yankel Active member sorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it. I have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ? Thanks ! paulmdrdo Active member I would try this method but this is open for any correction $\displaystyle \int\frac{x^2+1}{x^4+1}dx$ by letting $\displaystyle u\,=\,x^2$ $\displaystyle du\,=\,2xdx$ $\displaystyle dx\,=\,\frac{du}{2x}$ $\displaystyle x\,=\,u^{\frac{1}{2}}$ can you continue..? ZaidAlyafey Well-known member MHB Math Helper For practice try $$\displaystyle \int \frac{1}{1+x^4}\,dx$$ $$\displaystyle \int \frac{1}{\sqrt[4]{1+x^4}}\, dx$$ $$\displaystyle \int \frac{x^2}{1+x^4}\, dx$$ They are a little bit tougher . [*] By the way I have a way to solve the integral using complex numbers , the answer is nasty . MarkFL Staff member Using partial fractions, we may write: $$\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{2}\int\frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2-\sqrt{2}x+1}\,dx$$ Completing the squares... $$\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{\sqrt{2}} \right)^2+\frac{1}{2}}+\frac{1}{\left(x-\frac{1}{\sqrt{2}} \right)^2+\frac{1}{2}}\,dx$$ Now, we may simply apply the formula: $$\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a} \right)+C$$ to get (after simplification): $$\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{\sqrt{2}} \left(\tan^{-1}\left(\sqrt{2}x+1 \right)+ \tan^{-1}\left(\sqrt{2}x-1 \right) \right)+C$$ Prove It Well-known member MHB Math Helper sorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it. I have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ? Thanks ! It's a method known as trigonometric substitution. I suggest you read what Mark has written in post #8 of this thread.

https://math.stackexchange.com/questions/1085643/convex-set-with-empty-interior-is-nowhere-dense/1096673

# Convex set with empty interior is nowhere dense? Suppose $C\subseteq\mathbb R^n$ is a convex set and $C^o=\varnothing$. Is it necessarily true that $(\overline C)^o=\varnothing$? In general, is this true if $\mathbb R^n$ is replaced by a topological vector space $X$? Or can a counterexample be found? I know that if $C^o\neq\varnothing$, then $C^o=(\overline C)^o$, so my question is whether this result can be generalized to the case when $C^o=\varnothing$. Update: It's not true in general topological vector spaces. Indeed, if $X$ is a topological vector space and $Y$ is a proper dense subspace (such examples can be constructed), then $Y$ is convex and has empty interior, but $(\overline Y)^o=X^o=X$ is clearly not empty. Yet, I'm still not sure if the result holds for finite-dimensional vector spaces (in which every proper subspace is closed and thus no proper subspace is dense, so the preceding counterexample doesn't work). • Seems obvious that a convex subset of $\mathbb R^n$ with empty interior is contained in a hyperplane, which is a nowhere dense closed set. What am I missing? – bof Dec 30 '14 at 20:54 This is true in $\mathbb{R}^n$. Say $n = 3$. If $\overline{C}$ has non empty interior that $C$ is dense in some ball hence it contains the vertices of a tetrahedron (just take a tetrahedron inside the ball and move its vertices slightly to land on a point in $C$). But now by convexity, $C$ must contain the whole solid tetrahedron which contradicts the fact that $C$ has empty interior. In higher dimensions, you can start with a generalized cube and move its vertices slightly to enter $C$. • I see that the intuitive argument works, but I still have trouble operationalizing it rigorously. I will keep trying. Thank you for the answer! – triple_sec Dec 31 '14 at 16:03 • Maybe I'm missing something, but it seems to me that you don't need to resort to "move its vertices slightly". If $C$ is dense in some ball, then $C$ must contain $4$ non-coplanar points (because no subset of a plane can be dense in a ball). Use these points as the vertices of a tetrahedron. – Dave L. Renfro Jan 8 '15 at 17:07 • That works too. – user203787 Jan 8 '15 at 18:00 Here is a rigorous—and, accordingly, a bit lengthy—operationalization of the intuition offered by the answer by @OohAah. $$\textbf{Proposition}\phantom{---}$$ Let $$C\subseteq \mathbb R^n$$ ($$n\in\mathbb N$$) be a convex set. If $$C$$ has no interior, then nor does $$\overline C$$. $$\textit{Proof}\phantom{---}$$ Suppose that $$(\overline C)^o\neq\varnothing$$. I will show that $$C^o$$ is not empty, either. Let $$\mathbf x_0\in (\overline C)^o$$. Then, there exists some $$\varepsilon>0$$ such that $$\mathbf x_0\in B(2\varepsilon,\mathbf x_0)\subseteq\overline C$$, where $$B(2\varepsilon,\mathbf x_0)$$ denotes the open ball of radius $$2\varepsilon$$ around $$\mathbf x_0$$ with respect to the Euclidean norm. Let $$\{\mathbf e_i\}_{i=1}^n$$ denote the standard basis of $$\mathbb R^n$$ and define $$\mathbf x_i\equiv \mathbf x_0+\varepsilon \mathbf e_i$$ for each $$i\in\{1,\ldots,n\}$$. Clearly, $$\{\mathbf x_i\}_{i=0}^n\subseteq B(2\varepsilon,\mathbf x_0)\subseteq \overline C.$$ That is, $$\{\mathbf x_i\}_{i=0}^n$$ is included in the closure of $$C$$, so that, for each $$i\in\{0,1,\ldots,n\}$$, there exists some $$\mathbf y_i$$ such that • $$\mathbf y_i\in C$$; and • $$\mathbf y_i\in B((2\sqrt n)^{-1}\varepsilon,\mathbf x_i)$$. For each $$i\in\{1,\ldots,n\}$$ define $$\mathbf b_i\equiv \mathbf y_i-\mathbf y_0$$. I claim that the vectors $$\{\mathbf b_i\}_{i=1}^n$$ are linearly independent. Indeed, suppose that $$\lambda_1,\ldots,\lambda_n\in\mathbb R$$ satisfy $$\sum_{i=1}^n\lambda_i\mathbf b_i=0.$$ Suppose, for the sake of contradiction, that not all of $$\{\lambda_i\}_{i=1}^n$$ are zero. Then, $$\sum_{i=1}^n|\lambda_i|>0\tag{1}.$$ In addition, \begin{align*} 0=&\,\sum_{i=1}^n\lambda_i\mathbf b_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_i-\mathbf x_0+\mathbf x_0-\mathbf y_0)\\=&\,\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\varepsilon\mathbf e_i+\mathbf x_0-\mathbf y_0), \end{align*} or \begin{align*} -\varepsilon\sum_{i=1}^n\lambda_i\mathbf e_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0). \end{align*} Clearly, the Euclidean norm of the left-hand side is $$\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}$$, so the following chain of inequalities holds true: \begin{align*} &\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}=\left\|\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0)\right\|\leq\sum_{i=1}^n|\lambda_i|\left(\|\mathbf y_i-\mathbf x_i\|+\|\mathbf x_0-\mathbf y_0\|\right)\\ \underset{\text{see (1)}}{<}&\,\sum_{i=1}^n|\lambda_i|\left(\frac{\varepsilon}{2\sqrt{n}}+\frac{\varepsilon}{2\sqrt{n}}\right)=\sum_{i=1}^n|\lambda_i|\frac{\varepsilon}{\sqrt{n}}\leq\sqrt{\sum_{i=1}^n|\lambda_i|^2}\sqrt{\sum_{i=1}^n\frac{\varepsilon^2}{n}}=\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}, \end{align*} where I used the Cauchy–Schwarz inequality. This is a contradiction, which implies that $$\lambda_1=\ldots=\lambda_n=0$$. Hence, the vectors $$\{\mathbf b_i\}_{i=1}^n$$ are linearly independent, and since $$\dim\mathbb R^n=n$$, it follows also that they constitute a basis. If $$\mathbf z\in\mathbb R^n$$, there exists a unique $$(\mu_i)_{i=1}^n\in\mathbb R^n$$ such that $$\mathbf z=\sum_{i=1}^n\mu_i\mathbf b_i$$. Define $$\|\mathbf z\|_b\equiv\sum_{i=1}^n|\mu_i|$$. Then, $$\|\cdot\|_b$$ is a norm and since $$\mathbb R^n$$ is finite-dimensional, it must be equivalent to the Euclidean norm. Therefore, there exists some $$\xi_b>0$$ such that $$\|\cdot\|_b\leq \xi_b\|\cdot\|$$. Let $$D\equiv\left\{\sum_{i=0}^n\alpha_i\mathbf y_i\,\Bigg|\,\alpha_i\geq0\text{ for all i\in\{0,1,\ldots,n\} and }\sum_{i=0}^n\alpha_i=1\right\}.$$ Since $$\{\mathbf y_i\}_{i=0}^n\subseteq C$$ and $$C$$ is convex, it follows that $$D\subseteq C$$. Define $$\mathbf w\equiv\sum_{i=0}^n\frac{1}{n+1}\mathbf y_i.$$ Clearly, $$\mathbf w\in D$$ and $$\mathbf w-\mathbf y_0=\sum_{i=1}^{n}\frac{1}{n+1}(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^{n}\frac{1}{n+1}\mathbf b_i.$$ Let $$\delta\equiv\frac{1}{n(n+1)\xi_b}>0.$$ I will show that $$B(\delta,\mathbf w)\subseteq D$$. To this end, pick any $$\mathbf z\in B(\delta,\mathbf w)$$, so that $$\|\mathbf z-\mathbf w\|<\delta.$$ Since $$\{\mathbf b_i\}_{i=1}^n$$ is a basis, there exists a unique $$(\mu_i)_{i=1}^n\in\mathbb R^n$$ such that $$\mathbf z-\mathbf y_0=\sum_{i=1}^n\mu_i\mathbf b_i$$. Then, $$\mathbf z-\mathbf w=(\mathbf z-\mathbf y_0)-(\mathbf w-\mathbf y_0)=\sum_{i=1}^n\left(\mu_i-\frac{1}{n+1}\right)\mathbf b_i.$$ Consequently, for any $$i\in\{1,\ldots,n\}$$, it follows that \begin{align*} \left|\mu_i-\frac{1}{n+1}\right|\leq\sum_{j=1}^n\left|\mu_j-\frac{1}{n+1}\right|=\|\mathbf z-\mathbf w\|_b\leq \xi_b\|\mathbf z-\mathbf w\|<\xi_b\delta=\frac{1}{n(n+1)}. \end{align*} Hence, $$0\leq\frac{(n-1)}{n(n+1)}=\frac{1}{n+1}-\frac{1}{n(n+1)}<\mu_i<\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{1}{n},$$ for each $$i\in\{1,\ldots,n\}$$ so that $$\sum_{i=1}^n\mu_i<1$$. It follows that \begin{align*} \mathbf z=\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf b_i=\mathbf y_0+\sum_{i=1}^n\mu_i(\mathbf y_i-\mathbf y_0)=\left(1-\sum_{i=1}^n\mu_i\right)\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf y_i, \end{align*} so $$\mathbf z\in D$$. Conclusion: • $$\mathbf w\in B(\delta,\mathbf w)\subseteq D\subseteq C$$; and • $$B(\delta,\mathbf w)$$ is a non-empty open set; so that • $$\mathbf w\in C^o$$. In particular, $$C^o$$ is not empty. $$\blacksquare$$ Intuitively, the points $$\{\mathbf x_i\}_{i=0}^n$$ in $$\overline C$$ span an $$n$$-dimensional simplex, whose vertices are perturbed slightly so that the new vertices do not lie in the same hyperplane and are all in $$C$$. The convex hull of these new vertices gives rise to a “distorted simplex” fully contained in $$C$$. Then, the centroid of this distorted simplex can be surrounded by a small ball still included in the distorted simplex, and hence in $$C$$.

https://www.physicsforums.com/threads/question-about-diagonalizable-matrices.705803/

Suppose that ##A## is a diagonalizable ## n \times n ## matrix. Then it is similar to a diagonal matrix ##B##. My question is, is ##B## the only diagonal matrix to which ##A## is similar? I have thought about this, but am unsure if my answer is correct. My claim is that ##B## is the only diagonal matrix to which ##A## is similar, up to a rearrangement (or permutation) of the eigenvectors of ##A## in an ordered eigenbasis for ##ℝ^{n}##. Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_{A}]_{β}## but this will be merely a rearrangement of the diagonal entries of B. Is this true? If so, how might I go about proving this conjecture of mine? BiP Related Linear and Abstract Algebra News on Phys.org Yes, it's true. A hint for the proof: the diagonal entries are the eigenvalues of ##A##. CompuChip Homework Helper If ##A = X^{-1}BX = Y^{-1}B'Y## then ##B = (Y X^{-1})^{-1} B' (Y X^{-1})##. So if A is similar to two diagonal matrices, then these matrices are similar to each other. I guess a starting point would be to check which similarity transformations keep diagonal matrices diagonal, and see what you can say about X and Y that way. HallsofIvy Homework Helper Yes, it's true. A hint for the proof: the diagonal entries are the eigenvalues of ##A##. Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders. Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders. That's exactly what the OP said: Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_A]_\beta## but this will be merely a rearrangement of the diagonal entries of B. mathwonk

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# Fourier Transform Of Triangular Pulse To do so, you just have to divide the pulse by its norm, i. Signal Transmission Through LTIC Systems. ∆(t), from Problem 3. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. The Fourier transform of a gate pulse is: a. For any parameter, ≠. otherwise. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. Therefore, the departure of the roll-off from that of the sinc function can be ascribed to aliasing in the frequency domain , due to sampling in the time domain. Re: Help me get a Fourier transform of sinc It show the Fourier Transform of sinc(x) function --> a rectangular pulse. See under Discrete Fourier transform data compression use of, 494-495 decomposition. Fourier transforms of deltas and sinusoids Fourier transform of periodic signals Preface Not Just One 1 2 3 0. 1: The Fourier transform of a triangular pulse. Then the Fourier series expansion of the output function y(t) literally gives the spectrum of the output! B. periodic relationship, 222 discrete time Fourier series (DTFT),. 8 Boxcar Pulse Bc(t) Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier Transform property u(t) !F A˝sinc! ˝ 2 Thus u(t j!˝˝) !F e A˝sinc! ˝ 2 Therefore u(t a)u (t b) !F ej!a A˝sinc! ˝ 2. 2 The Fourier Transform for Periodic Signals 4. In Fourier Transform Nuclear Magnetic Resonance spectroscopy (FTNMR), excitation of the sample by an intense, short pulse of radio frequency energy produces a free induction decay signal that is the Fourier transform of the resonance spectrum. 4 Find The Fourier Transform Of The Triangular Pulse (Fig. 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. Plot sine wave. The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the spectrum-shaped pulse. Amplitude-modulated optical pulse trains undergo the discrete Fourier transform (DFT) realized by the temporal Talbot effect in a dispersive fiber. Fourier series: Given a 2ˇ-periodic complex-valued function f (think of it as a function on an interval T of length 2ˇ), its Fourier (series) transform is the sequence of its Fourier coecients: (f) = fb= fck: k 2 Zg, which can be thought of as a complex-valued function of the discrete frequency variable k. The series produced is then called a half range Fourier series. A square wave is approximated by the sum of harmonics. The S-transform can be effectively utilized for the analysis of nonstationary data. Output kernel Figure 5. The Gaussian shape is often called a bell shape. I'm at a computer without MATLAB at the moment. d R pp T P, where R pp is the auto-correlation of p(t) d. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Still, many problems that could have been tackled by using Fourier transforms may have gone unsolved because they require integration that is difficult and tedious. T T 0 elsewhen. There are similar convolution theorems for inverse Fourier transforms. 2 p693 PYKC 10-Feb-08 E2. The Fourier Transform 1. That is: since the is an odd function, its contribution is zero. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. DSP First and it's accompanying digital assets are the result of more than 20 years of work that originated from, and was guided by, the premise that signal processing is the best starting point for the study of electrical and computer engineering. time signal. School of Health Information Sciences. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). Signal Distortion over a Communication Channel. The Fourier series of this general pulse train is:. Equation of the Day #11: The Fourier Transform. Except now we're going to build a composite wave form that is a triangle wave. (We shall use this in the assignment). In order for F(t) to be real, F (- t) = F* ( t) must hold, Example 9. Plotting a triangular signal and finding its Fourier transformation in MATLAB. A “Brief” Introduction to the Fourier Transform This document is an introduction to the Fourier transform. What is the FT of a triangle function? - To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. time signal. This is a good point to illustrate a property of transform pairs. This pulse can be used to represent an electrostatic discharge, an electromagnetic pulse, or a lightning event. The LCFBG in the. Applying the time-convolution property to y(t)=x(t) * h(t), we get: 2(ω) e-jωτ. For1secondofdatasampledat40,000. Which are the only waves that correspond/ support the measurement of phase angle in the line spectra? a. t/, use the derivative property to find the Fourier transform of x. Let's look at the triangular pulse and its Fourier transform,. Fourier transform of a triangular pulse is sinc 2, i. Note: This is why the Fourier transform of a rectangular pulse is shaped like a sinc function, and the Fourier transform of a sinc function is shaped like a rectangular pulse 3. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. t2 exp( 2 ) 2. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Realisability of One Port Network – Consider the network function H(s) which is the ratio of Laplace transform of the output R(s) to the Laplace transform of the excitation E(s). The line spectrum, obtained from the Fourier series coefficients, indicates how the power of the signal is distributed to harmonic frequency components in the series. "Fourier Series--Triangle Wave. The Gaussian shape is often called a bell shape. Click the button "Real Transform" in the Fourier graph. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. We then generalise that discussion to consider the Fourier transform. (a) below is X(Ω) = sin Ω 2 Ω 2!2. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2ˇf 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. A group of algorithms is presented generalizing the fast Fourier transform to the case of noninteger frequencies and nonequispaced nodes on the interval $[ - \pi ,\pi ]$. School of Health Information Sciences. Consequently, the square wave has a wider bandwidth. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain −𝝅 ↔ −𝝅 Option (c) 10. Sometimes there is a big spike at zero so try taking the log of it before plotting. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Then the Fourier series expansion of the output function y(t) literally gives the spectrum of the output! B. Signals & Systems - Triangular Signal Watch more videos at https://www. 2) Here 0 is the fundamental frequency of the signal and n the index of the harmonic such. 2 The Fourier Transform Having now presented the idea of the frequency response H(s)js=j! = H(j!) of linear systems, it is appropriate to recognise that in the special case of this Laplace transform value H(s)being evaluated at the purely imaginary value s = j!, it is known as the ‘Fourier transform’. x(t)= γ (t)-γ (t-T 0) a) by using a table of Fourier Transforms and Properties (this is just a rectangular pulse function of width T 0 that is not centered on the origin). , sin(x)/x] in the frequency domain. Consequently, the square wave has a wider bandwidth. Improving images by “ deconvolution ” of. In this chapter much of the emphasis is on Fourier Series because an understanding of the Fourier Series decomposition of a signal is important if you wish to go on and study other spectral techniques. FFT onlyneeds Nlog 2 (N). Conceptually, this occurs because the triangle wave looks much more like the 1st harmonic, so the contributions of the higher harmonics are less. The FT of a rectangle function is a sinc. tutorialspoint. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter. The Fourier Transform of the Box Function. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. Triangle Pulse 0 0. periodic relationship, 222 discrete time Fourier series (DTFT),. Exams are approaching, and I'm working through some old assignments. The transform of a triangular pulse is a sinc2 function. Try taking the real part of it with real(). 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. If a function is defined over half the range, say 0 to L, instead of the full range from -L to L, it may be expanded in a series of sine terms only or of cosine terms only. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. In this particular SPICE simulation, I’ve summed the 1st, 3rd, 5th, 7th, and 9th harmonic voltage sources in series for a total of five AC voltage sources. Relation of Z transform to Laplace transform. 2005-03-15. We say that f(t) lives in the time domain, and F(ω) lives in the frequency domain. We begin by discussing Fourier series. On the other hand, as Fourier transform can be considered as a special case of Laplace transform when the real part of the complex argument is zero:. Baskakov, O I; Civis, S; Kawaguchi, K. After simplification the sinc squared function is obtained as the Fourier transform of a triangular pulse with unit area. 30) yields 2 2 ( ) ˆ p p T T. FFTs of Functions. '' Figure 8 shows an example for. Discrete Fourier Transform of Windowing Functions. )2 Solutions to Optional Problems S9. More Advanced Topics Up: Fourier Series-What, How, and Why Previous: The Fast Fourier Transform Using the Fourier Transform. Fourier Transform of Useful Functions. Since f(t) is even then g(w) is real. For any parameter, ≠. It is reversible, being able to transform from either domain to the other. You can buy my book 'ECE for GATE' here https://goo. We then generalise that discussion to consider the Fourier transform. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. The Fourier transform of the triangular pulse shown in Fig. The curve should be symmetrical with respect to the origin in 1024 points. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Recall that normalized Fourier transform of triangular pulse is $sinc^{2}(f)$$. The transform of a triangular pulse is a sinc 2 function. Compare the result with pan (b). Find the Fourier Transform of a rectangular pulse that is zero everywhere except between t=0 and t=T 0 where it has a height of one:. Fast Fourier Transform(FFT) • The Fast Fourier Transform does not refer to a new or different type of Fourier transform. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. Fourier Transform of the Gaussian Konstantinos G. The reason why Fourier analysis is so important in physics is that many (although certainly. Homework Statement What is the Fourier transform of the function graphed below? According to some textbooks the Fourier transform for this function Fourier Transform (Triangular Pulse) | Physics Forums. Cvetkovic, IntechOpen, DOI: 10. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. In this tutorial numerical methods are used for finding the Fourier transform of continuous time signals with MATLAB are presented. The fast Fourier transform (FFT) is a fast algorithm for calculating the Discrete Fourier Transform (DFT). A major challenge in the data analysis process of NOM using the FT-ICR MS technique is the need to sort the entire data set and to present it in an accessible mode. Often we are confronted with the need to generate simple, standard signals ( sine, cosine , Gaussian pulse , squarewave , isolated rectangular pulse , exponential decay, chirp signal ) for. 1 The Fourier transform and series of basic signals Triangular pulse 1. The graph at the bottom of the screen is the Fourier transform of this pulse. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. (a) below is X(Ω) = sin Ω 2 Ω 2!2. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). The series does not seem very useful, but we are saved by the fact that it converges rather rapidly. Fourier Transform Fourier Transform maps a time series (eg audio samples) into the series of frequencies (their amplitudes and phases) that composed the time series. This is the principle on which a pulse Fourier transform spectrometer operates. Discrete-Time Fourier Transform (DTFT) Chapter Intended Learning Outcomes: (i) Understanding the characteristics and properties of DTFT (ii) Ability to perform discrete-time signal conversion between the time and frequency domains using DTFT and inverse DTFT. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Sometimes there is a big spike at zero so try taking the log of it before plotting. In the synthesis, we are going to obtain a network from the given network function which may be admittance function or impedance function. Fourier Transforms For additional information, see the classic book The Fourier Transform and its Applications by Ronald N. Sometimes fft gives a complex result. rectangular pulse triangular pulse periodic time function unit impulse train (model of regular sampling). The example in this figure pertains to an output sampling rate which is times that of the input signal. Recall that normalized Fourier transform of triangular pulse is [math]sinc^{2}(f)$[math]. The Fourier transform is an integral transform widely used in physics and engineering. In this article, we will discuss the fact that choice of different window functions involves a trade-off between the main lobe width and the peak sidelobe (PSL). Improving images by “ deconvolution ” of. How to get a triangular pulse? Discrete Fourier Transform using scipy. 4 Reconstruction of a triangular pulse. What do we hope to achieve with the Fourier Transform? We desire a measure of the frequencies present in a wave. Can you explain this answer? is done on EduRev Study Group by Electrical Engineering (EE) Students. In some cases, as in this one, the property simplifies things. Triangle Pulse Sinc Pulse. What is the FT of a triangle function? – To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. DAS RISIN and MAIZAN MUHAMAD. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. A Tables of Fourier Series and Transform Properties 321 Table B. periodic relationship, 222 discrete time Fourier series (DTFT),. The Fourier transform of this convolution is the product w 1 w 2 of the two transforms, each one a series of parallel walls, and differs from zero only when both factors are different from zero. We demonstrate both amplitude and phase tailoring by generating a picosecond squarelike pulse as well as trains of femtosecond pulses with a terahertz-range repetition rate from either a. The LCFBG in the. Digital signal processing (DSP) vs. For the bottom panel, we expanded the period to T=5, keeping the pulse's duration fixed at 0. Relation of Z transform to Laplace transform. Ideal and Practical Filters. Sibbett A Wollaston prism is used in the design of a polarizing Fourier. The impulse response for the triangle in Fig. The graph at the bottom of the screen is the Fourier transform of this pulse. † The Fourier series is then f(t) = A 2 ¡ 4A …2 X1 n=1 1 (2n¡1)2 cos 2(2n¡1)…t T: Note that the upper limit of the series is 1. Signals & Systems - Triangular Signal Watch more videos at https://www. Exams are approaching, and I'm working through some old assignments. Signal Distortion over a Communication Channel. Fourier transform theorem table 4 1 jpg bax blog fourier transform table rh baxtyfraze blo com Pdf Fourier Transforms And Their Application To Pulse Amplitude. I One-Dimensional Fourier Transform. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain −𝝅 ↔ −𝝅 Option (c) 10. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). Slow Fourier Transforms Consider a general 1D Fourier transform relating two vectors of length : contains the values in real-space contains the frequency components The Slow Fourier Transform : Suppose that we precompute and store the coefficients c(j,k) = exp(i 2pi jk/n). I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. I'm at a computer without MATLAB at the moment. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. Signal Energy and Energy Spectral Density. Minimalist Tall Rectangular Planter On 4 Or 6 Zebra Onion Grass. F(ω) is just another way of looking at a function or wave. A LABORATORY DEMONSTRATION OF HIGH-RESOLUTION HARD X-RAY AND GAMMA-RAY IMAGING USING FOURIER-TRANSFORM TECHNIQUES David Palmer and Thomas A. A two parts tutorial on Fourier series. For voltage signals, the power per unit frequency is proportional to |H(f)|2 and is called the power spectrum or spectral power density of h(t). I One-Dimensional Fourier Transform. Sampling theorem. 1 Fourier Series Representation of Periodic Signals 3. 900 950 1000 1050 1100 1150 0. It is reversible, being able to transform from either domain to the other. Then using the relation: one can show that. PDF | Fourier transform ultrashort optical pulse shaping using a single linearly chirped fiber Bragg grating (LCFBG) is proposed and experimentally demonstrated in this letter. Solution: g(t) is a triangular pulse of height A, width W, and is centered at t 0. Therefore the FT of a triangle function is the product of two identical sincs, or a sinc^2. 4*Pi); # The base function is f0 = f restricted to [0,T] f0:=x->x; # Then f(x)=f0(trw(x,T))=trw(x,T) # INTEGRATION STEPS FOR FOURIER COEFFICIENTS # Possible also by hand using an integral table. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Note that as long as the definition of the pulse function is only motivated by the time-domain experience of it, there is no reason to believe that the oscillatory interpretation (i. Cosine waves c. Plotting a triangular signal and finding its Fourier transformation in MATLAB. The Fourier Transform In this appendix, the most important facts pertaining to the Fourier trans­ form are surveyed without proof. NEW! Updated labs, visual demos, an update to the existing chapters, and hundreds of new homework problems and solutions. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. Properties of the Fourier Transform • Linearity: • Let and • then • Time Scaling: • Let • then Compression in the time domain results in expansion in the frequency domain Internet channel A can transmit 100k pulse/sec and channel B can transmit 200k pulse/sec. Signals & Systems - Triangular Signal Watch more videos at https://www. Variable position. Since the pulse is getting narrower and narrower in the limit, the multiplication with this other function has less and less effect, until it has no effect once the. This is the example given above. The Fourier transform (English pronunciation: / ˈ f ʊr i eɪ /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Signals & Systems - Reference Tables 1 Table of Fourier Transform Pairs Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform. Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. t2 exp( 2 ) 2. As is an even function, its Fourier transform is Alternatively, as the triangle function is the convolution of two square functions ( ), its Fourier transform can be more conveniently obtained according to the convolution theorem as:. Periodicity of the Fourier transform; Fourier transform as additive synthesis. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the. Note that as long as the definition of the pulse function is only motivated by its behavior in the time-domain experience, there is no reason to believe that the oscillatory interpretation (i. Another representation of signals that has been found very useful is frequency domain representation. We can sample a function and then take the FFT to see the function in the frequency domain. The triangle peak is at the integral of the signal or sum of the sequence squared. ), the frequency response of the interpolation is given by the Fourier transform, which yields a sinc function. Single-pulse, Fourier-transform spectrometer having no moving parts M. Fourier transform: F(s) = - f(x)e-'2n"dr Here, two sidebands have been introduced (in addition to the carrier frequency), and the signal of frequency s has been shifted into the region of the carrier frequency, so. Today I want to start getting "discrete" by introducing the discrete-time Fourier transform (DTFT). Now, if we're given the wave function when t=0, φ(x,0) and the velocity of each sine wave as a function of its wave number, v(k), then we can compute φ(x,t) for any t by taking the inverse Fourier transform of φ(x,0) conducting a phase shift, and then taking the Fourier transform. profile closer to Gaussian. DFT needs N2 multiplications. Of course, we must sample often enough to avoid losing content. P d P2 f , where is Fourier Transform of p(t) c. 5 The multiplication Property 4. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first row is the graph of the unit pulse function and its Fourier transform , a function of frequency. For example, a rectangular pulse in the time domain coincides with a sinc function [i. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2ˇf 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. The actual Fourier transform are only the impulses. As in the case of ideal sampling, the spectrum contains uniformly spaced (scaled) copies of 𝑋(𝑓), with a spacing of 1⁄𝑇 Hz. † The Fourier series is then f(t) = A 2 ¡ 4A …2 X1 n=1 1 (2n¡1)2 cos 2(2n¡1)…t T: Note that the upper limit of the series is 1. 8 Boxcar Pulse Bc(t) Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier Transform property u(t) !F A˝sinc! ˝ 2 Thus u(t j!˝˝) !F e A˝sinc! ˝ 2 Therefore u(t a)u (t b) !F ej!a A˝sinc! ˝ 2. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Cosine waves c. Moreover, the amplitude of cosine waves of wavenumber in this superposition is the cosine Fourier transform of the pulse shape, evaluated at wavenumber. Mapping s plane semiellipse to Z plane. The corresponding intensity is proportional to this transform squared, i. To do so, you just have to divide the pulse by its norm, i. 3 Properties of the Continuous-Time Fourier Transform 4. Prince California Institute of Technology, Pasadena, CA 91125 Abstract A laboratory imaging system has been developed to study the use of Fourier-transform techniques in high-resolution hard x-ray andγ. The dotted-line is a sinc function that doesn't apply to this question, but gives the notion that this transform has something to do with the transform of a square pulse (i. Ideal and Practical Filters. F(ω) is called the Fourier Transform of f(t). Plot sine wave. 4 The Convolution Property 4. Today I want to follow up by discussing one of the ways in which reality confounds our expectations and causes confusion. Fn = 6 shows that as T/t increases the lines get closer together and the spectrum begins to look like that of the Fourier Transform. They are widely used in signal analysis and are well-equipped to solve certain partial. Signal Distortion over a Communication Channel. profile closer to Gaussian. periodic relationship, 222 discrete time Fourier series (DTFT),. Fourier series and square wave approximation Fourier series is one of the most intriguing series I have met so far in mathematics. The actual Fourier transform are only the impulses. The primary reason for it’s success is that it plots any course of constant bearing (angle w. I would guess it would be a triangle because I think the transform of a triangular pulse centered at zero is something like T sinc 2 (pi * f * T) I would restate your original question as follows: What is the transform of the following periodic rectangular function: per-rect(t) which has period T=T1+T2; is 1 during each T1 duration of its. 5 Signals & Linear Systems Lecture 10 Slide 3 Connection between Fourier Transform and Laplace. 1 The Fourier transform and series of basic signals Triangular pulse 1. 6: Roll-off of the rectangular-window Fourier transform. We describe a pulse-shaping technique that uses second-harmonic generation with Fourier synthetic quasi-phase-matching gratings. time signal. In the case of natural sampling, however, the spectral copies have different scaling factors: 𝑃(𝑘⁄𝑇)/𝑇. For above triangular wave: The square wave has much sharper transition than the triangular wave. These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. PSfrag replacements PROBLEM: Let x. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Inverse Z transform using inversion integral. Click the button "Real Transform" in the Fourier graph. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. Fourier Transforming the Triangular Pulse. Fourier Slice Theorem or central Slice Theorem. Translation (that is, delay) in the time domain goes over to complex phase shifts in the frequency domain. , use ^ M(n)/k^ M kinstead of ^ M(n). Then i deconvoluted the convoluted signal using triangular pulse with the deconvolution tool given in labview. 1-1) Since u(t) = 0 for t < 0, eq. Signal Transmission Through LTIC Systems. 2to create successive recon-structions of the pulse. Fourier transforms are used widely, and are of particular value in the analysis of single functions and combinations of functions found in radar and signal processing. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. Z transforms and difference equations. Time-harmonic impulse response calculations—The time-harmonic pressure generated by these triangular source geometries is proportional to the Fourier transform of the impulse response. Derpanis October 20, 2005 In this note we consider the Fourier transform1 of the Gaussian. A "Brief" Introduction to the Fourier Transform This document is an introduction to the Fourier transform. tutorialspoint. It is reversible, being able to transform from either domain to the other. Inverse Z transform using inversion integral. See Discrete Fourier Transform discrete vs. ∆(t), from Problem 3. The first example has a duty cycle of 0. The Fourier Transform: Examples, Properties, Common Pairs Square Pulse Spatial Domain Frequency Domain f(t) F (u ) 1 if a=2 t a=2 0 otherwise sinc (a u ) = sin (a u ) a u The Fourier Transform: Examples, Properties, Common Pairs Square Pulse The Fourier Transform: Examples, Properties, Common Pairs Triangle Spatial Domain Frequency Domain f(t. The Fourier transform (English pronunciation: / ˈ f ʊr i eɪ /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Sometimes fft gives a complex result. Padgett, A. But what we're going to do in this case is we're going to add them. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. Applying the time-convolution property to y(t)=x(t) * h(t), we get: That is: the Fourier Transform of the system impulse response is the system Frequency Response L7. property, we can compute the Fourier transform of the phasor: 0 ( )0 F e f fj tω = −δ This allows us to compute the Fourier transform of periodic signals. Triangular waves d. This is the example given above. (Browser settings for best viewing results) (How to cite this work). We then generalise that discussion to consider the Fourier transform. Fourier Transforms and Theorems. A) shows the original pulse and the real part of its Fourier transform. This transform pair isn't as important as the reason it is true. " From MathWorld--A Wolfram Web Resource. t/ D X1 nD1. In mathematics, the continuous Fourier transform is one of the specific forms of Fourier analysis. 003 EECS MIT discrete operator transforms. 6 depicts a resistor and capacitor in series. Using the above, we obtain: [ ] 2 0 ( ) ( )0 j nf t n. the Fourier transform function) should be intuitive, or directly understood by humans. One imagines a delta function to be a square pulse of unit area in the limit as the base of the pulse becomes narrower and narrower and higher and higher. 082 Spring 2007 Fourier Series and Fourier Transform, Slide 22 Summary • The Fourier Series can be formulated in terms of complex exponentials – Allows convenient mathematical form – Introduces concept of positive and negative frequencies • The Fourier Series coefficients can be expressed in terms of magnitude and phase. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Exams are approaching, and I'm working through some old assignments. The magnitude of the Fourier transform of a rectangular pulse equals the absolute value of a sinc. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. Existence of the Fourier Transform. 104 Chapter 5. jaj<1 (n+ 1)anu[n] 1 (1 ae j. In the second row is shown , a delayed unit pulse, beside the real and imaginary parts of the Fourier transform. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). The fundamental frequency is 50 Hz and each harmonic is, of course, an integer multiple of that frequency. Four chapters on analog signal processing systems, plus many updates and enhancements. This is a good point to illustrate a property of transform pairs. In other words, the input signal is upsampled by a factor of using linear interpolation.

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Question # (Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast $$8$$ units of vitamin A and $$10$$ units of vitamin C. Food I contains $$2$$ units/kg of vitamin A and $$1$$ unit/kg of vitamin C. Food II contains $$1$$ unit/kg of vitamin A and $$2$$ units/kg of vitamin C. It costs $$Rs 50$$ per kg to purchase Food I and $$Rs. 70$$ per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a mixture. Solution ## Let the mixture contain $$x$$ kg of Food I and $$y$$ kg of Food II. Clearly, $$x \geq 0,y \geq 0$$. We make the following table from the given data:ResourcesFoodI$$(x)$$FoodII$$(y)$$RequirementVitamins A (units/ kg)$$2$$$$1$$$$8$$Vitaminc C (units/ kg)$$1$$$$2$$$$10$$Cost (Rs/ kg)$$50$$$$70$$Since the mixture must contain at least $$8$$ units of vitamin A and $$10$$ units of vitamin C, we have the constraints:$$2x + y \geq 8$$$$x + 2y \geq 10$$Total cost $$Z$$ of purchasing $$x$$ kg of food I and $$y$$ kg of Food II is$$Z = 50x 70y$$Hence, the mathematical formulation of the problem is:Minimise $$Z = 50x + 70y ... (1)$$subject to the constraints :$$2x + y \geq 8 ... (2)$$$$x + 2y \geq 10 ... (3)$$$$x, y \geq 0 ... (4)$$Let us the graph the inequalities $$(2)$$ to $$(4)$$. The feasible region determined by the system is shown in the Fig. Here again, observe that the feasible region is unbounded.Let us evaluate $$Z$$ at the corner points $$A(0,8), B(2,4)$$ and $$C(10,0)$$.Corner Point$$Z = 50x + 70y$$$$(0, 8)$$$$560$$$$(2, 4)$$$$380\leftarrow Minimum$$$$(10, 0)$$$$500$$In the table, we find that smallest value of $$Z$$ is $$380$$ at the point $$(2,4)$$. Can we say that the minimum value of $$Z$$ is $$380$$? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality$$50x + 70y < 380$$ i.e., $$5x + 7y < 38$$to check whether the resulting open half plane has any point common with the feasible region. From the Fig, we see that it has no points in common.Thus, the minimum value of $$Z$$ is $$380$$ attained at the point $$(2, 4)$$. Hence, the optimal mixing strategy for the dietician would be to mix $$2$$ kg of Food I and $$4\ kg$$ of Food II,and with this strategy, the minimum cost of the mixture will be $$Rs. 380$$.Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More

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# Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$? Let $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combination of $a,b$. My question is : can we explicitly determine how many positive integers are there which can not be written as a non-negative integer combination of $a,b$ ? • @B.Goddard $a, b$ are assumed to be relatively prime. Also, the linear combinations here are assumed to have non-negative coefficients. That changes the problem quite a bit from the one you describe. – Arthur Jun 20 '18 at 13:11 • The most explicit I can get is a summation. If $a\lt b$ then there are: $$\sum_{k=1}^{a-1}\left\lfloor\frac{bk}{a}\right\rfloor$$ such numbers. Although you should definitely double-check this result. – N. Shales Jun 20 '18 at 16:19 • Having thought a bit more I further conjecture that the result simplifies to $$\frac{1}{2}(a-1)(b-1)\, .$$ – N. Shales Jun 20 '18 at 17:16 • @N.Shales your conjecture is correct: \begin{align*} R &= \sum_{k=1}^{a-1} \left\lfloor \frac{bk}{a} \right \rfloor\\ &= \sum_{k=1}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor \\ &= \sum_{k=1}^{a-1} b+\left\lfloor \frac{-kb}{a} \right\rfloor \\ &= \sum_{k=1}^{a-1} b-1-\left\lfloor \frac{kb}{a} \right\rfloor \\ &= (a-1)(b-1) - \sum_{k=1}^{a-1} \left\lfloor \frac{kb}{a} \right\rfloor \\ &= (a-1)(b-1) - R \end{align*} So that $2R = (a-1)(b-1)$. The key is the equality $\lfloor (-kb)/a\rfloor = -\lfloor kb/a \rfloor - 1$ which holds when $a$ does not divide $kb$. – Yong Hao Ng Jun 28 '18 at 8:20 • @N.Shales I saw it a few days ago but it took me a while before I found a not very involved solution to the floor-sum. – Yong Hao Ng Jun 29 '18 at 13:11 Let $N = ab$, we count the number of integers $0\leq t \leq N$ that can be written as $$t = ma+nb$$ such that $m,n\geq 0$. Let there be $R$ such numbers. Since the largest non-representable (positive)-integer is $ab-a-b$, all non-representable numbers are in $[0,N]$ and hence we know there are $N+1 - R$ of them. If $n\geq a$, then we replace it with $$ma+nb = (m+b)a + (n-a)b = m'a+n'b$$ So with repeated application we may assume that $0\leq n < a$. We require $$0 \leq ma+nb \leq N \implies 0 \leq ma \leq N-nb = (a-n)b$$ Hence we get the bounds of $m$ as $$0 \leq m \leq \left \lfloor \frac{(a-n)b}{a}\right \rfloor$$ This means the number of integers we can represent for a given $n$ is $\left \lfloor \frac{(a-n)b}{a} \right \rfloor + 1$. Now summing this over all possibilities $n=0,1,\dots a-1$, the number of representable integers $R$ is \begin{align*} R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1 &= \left\lfloor \frac{(a)b}{a} \right\rfloor + \left\lfloor \frac{(a-1)b}{a} \right\rfloor + \cdots + \left\lfloor \frac{(2)b}{a} \right\rfloor + \left\lfloor \frac{(1)b}{a} \right\rfloor + a\\ R &= a + b + \sum_{k=1}^{a-1} \left\lfloor \frac{kb}{a} \right\rfloor \\ \end{align*} Next we observe that if $a$ does not divide $kb$, then for some integer $t$ $$t < \frac{kb}{a} < t+1 \Longleftrightarrow -t-1 < \frac{-kb}{a} < -t$$ In particular this applies for $1\leq k\leq a-1$, since $\gcd(a,b)=1$ and $a>1$. We use the inequalities to obtain $$\left \lfloor \frac{-kb}{a}\right \rfloor = -t-1 = - \left \lfloor \frac{kb}{a}\right \rfloor -1,$$ This lets us get another equation of $R$ as \begin{align*} R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1 &= \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor + b + 1 \\ &= a(b+1) + \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor\\ &= a(b+1) + \sum_{k=1}^{a-1}\left\lfloor \frac{-kb}{a} \right\rfloor \\ &= a(b+1) + \sum_{k=1}^{a-1}-\left\lfloor \frac{kb}{a} \right\rfloor -1\\ R &= ab + 1 - \sum_{k=1}^{a-1}\left\lfloor \frac{kb}{a} \right\rfloor \end{align*} Therefore adding both equations of $R$, we have $$2R = ab + a + b + 1 = (a+1)(b+1)$$ giving us $R=(ab+a+b+1)/2$. Notice that $R$ is not an integer if and only if $a,b$ are both even, which is not possible since $\gcd(a,b)=1$. Finally, the number of integers considered is $ab+1$, so the number of integers not representable is $$ab+1 - \frac{ab+a+b+1}{2} = \frac{ab -a - b+1}{2} = \frac{(a-1)(b-1)}{2}$$ Proposition. Let $a,b$ be relatively prime integers greater than $1,$ and let $m=ab-a-b.$ If $x\in\{0,1,2,\dots,m\},$ then exactly one of the two numbers $x$ and $m-x$ can be expressed as a linear combination of $a$ and $b$ with nonnegative integer coefficients. It follows that exactly half the elements of $\{0,1,2,\dots,m\}$ can be expressed as nonnegative integral linear combinations of $a$ and $b;$ that is, there are exactly $\frac{m+1}2=\frac{(a-1)(b-1)}2$ numbers which cannot be so expressed.

https://math.stackexchange.com/questions/1642495/how-many-answers-to-3x-2y-5

# How many answers to $|3^x-2^y|=5$? How many answers are there to the equation $|3^x-2^y|=5$ such that $x$ and $y$ are positive integers? Are there infinite? I've found $(2,2)$, $(3,5)$, and $(1,3)$. It seems to explode with larger values, but it's not a steady increase and there seems to be many dips. Do we KNOW that there are no large values for $x$ and $y$ where a power of 3 comes close to a power of 2? • Its a putnam problem . – DeepSea Feb 5 '16 at 23:12 • Kf-Sansoo, had to google "putnam problem." Do you mean literally one, or one that fits the general qualifications? Do I need to add a tag? If it is one that's been mentioned before, could you give a link? – Elem-Teach-w-Bach-n-Math-Ed Feb 5 '16 at 23:18 • – Lucian Feb 6 '16 at 0:01 • @Kf-Sansoo: so what? Does that mean the question shouldn't be discussed on MSE? Does it mean that there is helpful information about the question to be found elsewhere? Or what? – Rob Arthan Feb 6 '16 at 0:01 • @Lucian: So, as it's conjectured that $Ax^n-By^m=C$ has a finite # of solutions for $(x,y,n,m)$ then $3^n-2^m=5$ can be conjectured to also have a finite # or solutions for $(n,m)$? In other words, it's unproven in a more general sense, so with the greater constraints it should be even less likely? If there are finite answers, my question could then be rephrased, "Is there a 4th answer?" – Elem-Teach-w-Bach-n-Math-Ed Feb 6 '16 at 0:18 There are only a finite number of solutions. It was proved by Pillai that $a^x - b^y = k$ where $a,b,k$ are fixed positive integers, $a > 1, b > 1, k \neq 0,$ with positive integer variables $x,y,$ has finitely many solutions. This is from page 51 in Shorey and Tijdeman, Exponential Diophantine Equations. The two papers by Pillai are 1931 and 1936. Both are in the Journal of the Indian Mathematical Society. A detail: if $k$ is larger than some bound that depends on $a,b,$ there is only one solution. Since we have more than one solution for $k = -5,$ it appears Pillai's bound is not tight enough to finish this problem. We just know one solution for $k=5.$ • I'll probably end up going with this as best answer. I've got an addendum though for your perusal, if I may. Since $|3^x-2^y|=5$ is proven to have finite solutions, I'd like to carry this knowledge to similar problems: i.e. $|(3^x)(5^y)-2^z|=7$, $|(2^x)(3^y)-5^z|=7$, $|(2^x)(3^y)-5^z|=7$. Namely working my way up to prime solutions such as $|(2^r)(7^s)(11^t)(13^u)(23^v)-(3^w)(5^x)(17^y)(19^z)|=29$ (or similarly creating primes less than the square of the largest prime used; in this case 23). Any thoughts? Probably throw this out there as a new ?. – Elem-Teach-w-Bach-n-Math-Ed Feb 8 '16 at 22:39 Here's an elementary self-contained argument that there is no solution with $y>5$. A power of $3$ is congruent to either $1$ or $3 \bmod 8$, so once $y \geq 3$ we must have $3^x - 2^y = -5$. Once $y \geq 6$, we then have $3^x \equiv -5 \bmod 2^6$, and thus $x \equiv 11 \bmod 16$. But then $3^x + 5 \equiv 12 \bmod 17$, and no power of $2$ is congruent to $12 \bmod 17$ (the powers of $2 \bmod 17$ are $2,4,8,-1,-2,-4,-8,1,2,4,8,-1$ etc.), QED. There is a large literature on such Diophantine questions. One key phrase is "$S$-unit equations". In general it has been known for some time that there are finitely many solutions, and indeed for equations of the form $$\prod_i p_i^{n_i} - \prod_j q_j^{m_j} = r$$ this already follows from Thue's theorem (1909); and by now we even have effective algorithms known to find all solutions. There's still no elementary technique known in general, but in your case (where only the primes 2,3,5) appear an elementary solution is contained in a 1976 paper This is not a complete answer, but I'd just like to note a connection with the concept of irrationality measure. First recall that $\alpha \in \mathbf{R}$ has irrationality measure $\mu(\alpha)$ if, for all $r>\mu(\alpha)$, there are only finitely many pairs of integers $p,q$ such that $$\left| \alpha - \frac{p}{q} \right| < \frac{1}{q^r}.$$ Note that $\mu(\alpha)=\infty$ is allowed (but we have $\mu(\alpha)=2$ for all but a measure-zero set of reals $\alpha$). With this definition out of the way, note that $$\left| 3^x - 2^y \right| = 5$$ implies $$3^x = 2^y \pm 5.$$ Take logs: $$\log(3^x) = \log(2^y) + \log(1 \pm 5 \cdot 2^{-y}).$$ So by the Taylor expansion of $\log(1+t)$, $$\left| \log(3^x) - \log(2^y) \right| < 5 \cdot 2^{-y}+\frac{5^2 \cdot 2^{-2y}}{2}+\cdots < 6 \cdot 2^{-y}$$ for $y$ sufficiently large. Hence, using $\log(3^x) = x \log 3$ and $\log(2^y) = y \log 2$ $$\left| \frac{\log 3}{\log 2} - \frac{y}{x} \right| < \frac{6}{x \log 2} \cdot 2^{-y}.$$ Since the RHS decreases much faster than anything polynomial in $x$ (at least for $y/x$ within some fixed small interval around $\log(3)/\log(2)$), the existence of infinitely many solutions to your equation would imply that $\mu(\log(3)/\log(2))$ were infinite. It is probably known that this latter statement is false, probably by methods similar to the proof in the Shorey/Tijdeman book mentioned by Will Jagy.

http://nl.mathworks.com/help/matlab/ref/istriu.html?s_tid=gn_loc_drop&nocookie=true

# istriu Determine if matrix is upper triangular ## Syntax • `tf = istriu(A)` example ## Description example ````tf = istriu(A)` returns logical `1` (`true`) if `A` is an upper triangular matrix; otherwise, it returns logical `0` (`false`).``` ## Examples collapse all ### Test Upper Triangular Matrix Create a 5-by-5 matrix. `A = triu(magic(5))` ```A = 17 24 1 8 15 0 5 7 14 16 0 0 13 20 22 0 0 0 21 3 0 0 0 0 9``` Test `A` to see if it is upper triangular. `istriu(A)` ```ans = 1 ``` The result is logical `1` (`true`) because all elements below the main diagonal are zero. ### Test Matrix of Zeros Create a 5-by-5 matrix of zeros. `Z = zeros(5);` Test `Z` to see if it is upper triangular. `istriu(Z)` ```ans = 1 ``` The result is logical `1` (`true`) because an upper triangular matrix can have any number of zeros on the main diagonal. ## Input Arguments collapse all ### `A` — Input arraynumeric array Input array, specified as a numeric array. `istriu` returns logical `0` (`false`) if `A` has more than two dimensions. Data Types: `single` | `double` Complex Number Support: Yes collapse all ### Upper Triangular Matrix A matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero. For example, the matrix $A=\left(\begin{array}{cccc}1& -1& -1& -1\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -2& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -3\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right)$ is upper triangular. A diagonal matrix is both upper and lower triangular. ### Tips • Use the `triu` function to produce upper triangular matrices for which `istriu` returns logical `1` (`true`). • The functions `isdiag`, `istriu`, and `istril` are special cases of the function `isbanded`, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, ```istriu(A) == isbanded(A,0,size(A,2))```.

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Mathematics # If $\displaystyle\int x^5e^{-x^2}dx=g(x)\cdot e^{-x^2}+C$ then the value of $g(-1)$ is? $-\dfrac{5}{2}$ ##### SOLUTION Put $x^2=t$ $2xdx=dt$ $\displaystyle\int t^2e^{-t}\dfrac{dt}{2}$ $=\dfrac{1}{2}[-t^2\cdot e^{-t}+2\displaystyle\int te^{-1}dt]+c$ $=\dfrac{1}{2}[-t^2\cdot e^{-t}-2te^{-t}+\displaystyle\int 2e^{-t}dt]+c$ $=\dfrac{1}{2}(-t^2e^{-t}-2(te^{-1}+e^{-t}))+c$ $=\dfrac{-(x^4+2x^2+2)e^{-x^2}}{2}+c$ $g(x)=\dfrac{-(x^4+2x^2+2)}{2}$ $g(-1)=-\dfrac{5}{2}$. Its FREE, you're just one step away Single Correct Medium Published on 17th 09, 2020 Questions 203525 Subjects 9 Chapters 126 Enrolled Students 86 #### Realted Questions Q1 Subjective Hard Evaluate the given integral. $\displaystyle\int { { e }^{ x } } \left( \cfrac { \sin { x } \cos { x } -1 }{ \sin ^{ 2 }{ x } } \right) dx$ 1 Verified Answer | Published on 17th 09, 2020 Q2 Subjective Medium Evaluate the given integral. $\int { x\sec ^{ 2 }{ x } } dx\quad \quad$ 1 Verified Answer | Published on 17th 09, 2020 Q3 Single Correct Hard The value of the definite integral $\int_\limits{ 1}^e( (x+1)e^x\ln x) dx$ is- • A. $e$ • B. $e^e(e-1)$ • C. $e^{x+1}$ • D. $e^{e+1}+e$ 1 Verified Answer | Published on 17th 09, 2020 Q4 Single Correct Hard The integral $\displaystyle \int (1+x-\displaystyle \frac{1}{x})e^{x+\frac{1}{x}} dx$ is equal to • A. $(x-1)e^{x+ \frac{1}{x}} +c$ • B. $(x+1)e^{x+\frac{1}{x}} +c$ • C. $-xe^{x+\frac{1}{x}} +c$ • D. $xe^{x+\frac{1}{x}} +c$ The average value of a function f(x) over the interval, [a,b] is the number $\displaystyle \mu =\frac{1}{b-a}\int_{a}^{b}f\left ( x \right )dx$ The square root $\displaystyle \left \{ \frac{1}{b-a}\int_{a}^{b}\left [ f\left ( x \right ) \right ]^{2}dx \right \}^{1/2}$ is called the root mean square of f on [a, b]. The average value of $\displaystyle \mu$ is attained id f is continuous on [a, b].

https://www.physicsforums.com/threads/simple-integral-but-a-question-of-methodology.164031/

# Simple integral - but a question of methodology ianb simple integral -- but a question of methodology $$\int \frac{(x-1)}{2} dx$$ First, try substitution. Let $$u = x - 1$$, and $$du = dx$$ $$\int \frac{u}{2} du$$ $$\frac{1}{2} \int u \ du$$ $$\frac{1}{2} \times \frac{u^{2}}{2}$$ $$\frac{1}{2} \times \frac{(x-1)^2}{2}$$ $$\frac{(x-1)^{2}}{4} + C$$ Or we could do this: $$\int \frac{(x-1)}{2} dx$$ $$\int \frac{1}{2}(x-1)dx$$ $$\frac{1}{2} \int(x-1)dx$$ $$\frac{1}{2} (\frac{x^2}{2} - x) + C$$ $$\frac{1}{4} (x^2 - 2x) + C$$ Why should one of them be right? Last edited: ZioX They're both right. Antiderivatives differ by a constant. (Differentiating eliminates the constant term). Show that those two answers differ by a constant. Homework Helper ... $$\frac{1}{4} (x^2 - \frac{x}{2}) + C$$ Why should one of them be right? The last line is wrong. You've pulled out the factor 1 / 4 incorrectly. $$\frac{1}{4} \left( x ^ 2 - 2x \right)$$ Both are correct, the two results differ from each other by a constant 1/4. You should note that if F(x) is one anti-derivative of f(x), then F(x) + C is also f(x)'s anti-derivative, where C is any constant. Proof: Differentiate the LHS with respect to x, we have: (F(x) + C)' = F'(x) + C' = F'(x) + 0 (the derivative of a constant is just 0) = F'(x) = f(x) Can you get this? :) ianb Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25. Thoughts? ianb Ah, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it. Homework Helper Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25. Thoughts? Nope, they are identically the same, let F(x) + C1, and F(x) + C2 be two anti-derivatives of f(x), we have: $$\int_a^b f(x) dx = (F(b) + C_1) - (F(a) + C_1) = F(b) - F(a)$$ $$\int_a^b f(x) dx = (F(b) + C_2) - (F(a) + C_2) = F(b) - F(a)$$, they are still the same. The constant will automatically vanish when you subtract the two terms. :) ianb Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1.7. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors? Last edited: ianb I'm sorry, I should have been clearer. Message edited accordingly. Homework Helper Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors? Uhm, well, no, they are not different. Well, here we go: $$\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \left( \frac{(x - 1) ^ 2}{4} \right) \right|_0 ^ 1 = \frac{(1 - 1) ^ 2}{4} - \frac{(0 - 1) ^ 2}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$$ $$\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \frac{1}{4} \left( x ^ 2 - 2x \right) \right|_0 ^ 1 = \frac{1}{4} \left( (1 ^ 2 - 2 \times 1) - (0 ^ 2 - 0) \right) = \frac{1}{4} (-1 - 0) = -\frac{1}{4}$$ They are the same. :) ianb Wow. I'm embarrassed to say the least. You, sir, rock my world.

https://math.stackexchange.com/questions/3151946/fewest-steps-to-reach-200-from-1-using-only-1-and-%C3%972/3152029

# Fewest steps to reach $200$ from $1$ using only $+1$ and $×2$ This is a problem from the AMC 8 (math contest): A certain calculator has only two keys $$[+1]$$ and $$[\times 2]$$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “$$9$$” and you pressed $$[+1]$$, it would display “$$10$$”. If you then pressed $$[\times 2]$$, it would display “$$20$$”. Starting with the display “$$1$$”, what is the fewest number of keystrokes you would need to reach “$$200$$”? Intuitively I worked back from $$200$$, dividing by $$2$$ until I reached an odd number, subtracting $$1$$ when I did, etc..to reach the correct answer of $$9$$ steps. However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically. The best I can come up with is that beyond the first step from $$1$$ to $$2$$, multiplication by $$2$$ is always going to yield a larger step than addition by $$1$$ and therefore I should take as many $$[\times 2]$$ steps as I can. This doesn't feel rigorous enough, though. EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution. • Do you want the fewest steps to get to exactly $200$ or at least $200$? – John Douma Mar 17 at 19:37 • @JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice. – TonyK Mar 17 at 19:43 • @TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise. – John Douma Mar 17 at 19:44 • @JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$. – jeremy radcliff Mar 17 at 19:48 Look at what the operations $$+$$ and $$\times$$ do to the binary expansion of a number: • $$\times$$ appends a $$0$$, and increases the length by one, leaving the total number of $$1$$'s unchanged; • if the final digit is $$0$$, then $$+$$ increases the number of $$1$$'s by one, but doesn't change the length; • if the final digit is $$1$$, then $$+$$ doesn't increase the total number of $$1$$'s (it may in fact decrease it), and doesn't increase the total length by more than $$1$$. Therefore, with a single key press: • you can increase the length by one, but this won't increase the number of $$1$$'s; • you can increase the number of $$1$$'s by one, but this won't increase the length. The binary expansion of $$200$$ is $$200_{10}=11001000_2$$. This has three $$1$$'s, and a length of eight. Starting from $$1$$, we must increase the length by seven, and increase the number of $$1$$'s by two. So this requires at least nine steps. • Beautiful, this allows you to determine the optimal solution and path for an arbitrary number from it's binary representation. – Jared Goguen Mar 18 at 1:34 • There actually exist two ways of getting $200$ using nine steps. $$1+1+1\times2\times2\times2+1\times2\times2\times2=200\\ 1\times2+1\times2\times2\times2+1\times2\times2\times2=200$$ – Kay K. Mar 18 at 2:13 • @KayK.: Yes, but only because there are two ways of getting to 2. The rest is uniquely determined. – TonyK Apr 4 at 16:05 You can proceed by induction on $$n$$, showing that the quickest way to reach any even number $$2n$$ involves doubling on the last step, which is clearly true for the base case $$n=1$$ (where doubling and adding $$1$$ have a tomato-tomahto relationship). Now if the last step to reach $$2n+2$$ isn't doubling, it can only be adding $$1$$ from $$2n+1$$. But $$2n+1$$ can only be reached by adding $$1$$ from $$2n$$, at which point the inductive hypothesis says the next previous number was $$n$$. But you can get from $$n$$ to $$2n+2$$ more quickly in two steps: add $$1$$, then double. So the last step in the quickest route to $$2n+2$$ is doubling from $$n+1$$. • This is nice as it formalizes the OPs intuition, whereas the binary representation answer (which is super slick) might feel a little out of the blue. – jacob1729 Mar 17 at 22:55 Setting the display to binary base, $$[\times2]$$ inserts a $$0$$ to the right and $$[+1]$$ increments; if the rightmost digit is a zero, it just turns it to a $$1$$. Using these rules you build a number of $$o$$ ones and $$z$$ zeroes in $$o-1+z$$ keystrokes, starting from $$1$$. This seems close to optimal. • you just exactly reproduces the binary 200, why should we think it is not optimal? – dEmigOd Mar 17 at 19:54 • @dEmigOd: I didn't prove that inserting the bits one by one with $\times2$ or $\times2+1$ is optimal. – Yves Daoust Mar 17 at 19:57 I'll make a try Since $$200=2^7+2^6+2^3$$ you will need at least $$8$$ steps to reach $$200$$ (since we start from $$1$$ and we get a number of the form $$2^a+...+2^l$$) so it remains to show that $$8$$ steps are not enough. Maybe you could try to show that if there was a solution with $$8$$ steps then it would contain only one $$+1$$ which contradicts the fact that in $$200=2^7+2^6+2^3$$ we have two $$+$$ Working backwards, use the tree diagram and stop lagging branches (e.g. if $$\color{red}{99}$$ in step $$3$$ leads to $$1$$, then it takes one more operation than $$\color{red}{99}$$ in step $$2$$): $$\hspace{3cm}$$

http://hurstbournefamilycare.com/letter-from-ewv/aadcea-area-of-a-sector-in-radians

There are three formulas for calculating the area of a sector. The formula for the area of a sector is: A = r² * θ / 2. Find the central angle of a sector whose radius is 56 cm and area, is 144 cm2. Geometric skills. How to use the calculator Enter the radius and central angle in DEGREES, RADIANS or both as positive real numbers and press "calculate". Area of a sector when the central angle is given in degrees If the angle of the sector is given in degrees, then the formula for the area of a sector is given by, Area of a sector = (θ/360) πr2 A = (θ/360) πr2 Find the radius of a semi – circle with the area of 24 inches squared. Math A level Syllabus, 2016. The area of a sector of a circle with a radius of 5 cm, with an angle of 2 radians, is 25 cm 2. Then, the area of a sector of circle formula is calculated using the unitary method. 40pie units squared. Show that 2θ-3sinθ=0. The area of the minor segment as shown in Fig 5 . Area of a cyclic quadrilateral. Area of sector of circle = (1/2)r²Ө , Ө must be in radians. Example 1 Find the arc length and area of a sector of a circle of radius … Step 1: Find the area of the circle. Katelyn is making a semi-circular design to put on one of her quilts. Thus, if you are not sure content located Radians, Arc Length and Sector Area Radians Radians are units for measuring angles. They can be used instead of degrees. Show that 2θ-3sinθ=0. If the radius of the circle is , what is the area of the semi-circular design? With the help of the community we can continue to Each of these formula is applied depending on the type of information given about the sector. Yes, both the semicircle and the quadrant are special types of the sector of a circle with angles $$180^{\circ}$$ and $$90^{\circ}$$ respectively. Find the area of a sector if the radius of the circle is 4, and the angle of the sector is  radians. In simple words, the area of a sector is a fraction of the area of the circle. We can find the area of a sector of a circle in a similar manner. Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Jul 2009 448 89. Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; Displaying top 8 worksheets found for - Area Of A Sector In Radians. Both can be calculated using the angle at the centre and the diameter or radius. So in the below … The area of the circle is equal to the radius square times . as shown in Fig 4, we consider the sector as a fraction of the circle hence: Area of a segment. Introducing Radians; 9. If the diameter of a circle is 6, find the area of a sector with a sector angle of . Varsity Tutors LLC , where  is the angle measure of the sector in radians, and  is the radius of the circle. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing The full angle is 2π in radians, or 360° in degrees, the latter of which is the more common angle unit. The measure of central angle XYZ is 1.25 radians. The formulas to find the area of a sector in Degrees (D°) or Radians (R°) are shown below: Area (Degrees) = πr2 x θ/360 Area (Radians) = ½r2θ r, D° r, R° r, s r, A D°, s Use 3.14 for . Track your scores, create tests, and take your learning to the next level! Area of a sector given the arc length. The non-shaded area of the circle shown below is called a SECTOR. Area of an ellipse. ChillingEffects.org. The angle AOB is in radians. Area of sector. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! What is the area of Fiona's circle? An arc of a circle when bounded by two radii with the centre of the circle then we get a slice of a circle. Find the radius of a sector whose area is 47 meters squared and central angle is 0.63 radians. Use … Send your complaint to our designated agent at: Charles Cohn Circle sector area calculator - step by step calculation, formulas & solved example problem to find the area of circle sector given input values of corcle radius & the sector angle in degrees in different measurement units between inches (in), feet (ft), meters (m), centimeters (cm) & millimeters (mm). Example: a) What is the length of the arc intercepted by an angle of 15° on a circle with radius 20 meters? Use prior knowledge on length of circumference and area of circle to deduce formulae to calculate arc length and sector area. Given, the length of the arc, the area of a sector is given by. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! The number inside the sector is the area. degree radian; area S . If r is in "m", the area will be in "m" 2. Now, this looks messy, but we can simplify it to get: Next, use your calculator to find a decimal answer, and then round to get our final answer. Find the area of a sector whose arc is 8 inches and radius, is 5 inches. a And then we just can solve for area of a sector by multiplying both sides by 81 pi. Find the area of the sector with radius 7\ "cm" and central angle 2.5 radians. There are two types of sectors, minor and major sector. I remember this formula as it is quite easy to remember. SECTORS . The portion of the circle's circumference bounded by the radii, the arc, is part of the sector. Perimeter of sector = r + 2r = r( + 2) Where is in radians If angle is in degrees, = Angle × π/(180°) Let us take some examples: Find perimeter of sector whose radius is 2 cm and angle is of 90° First, We need to convert angle in radians = Angle in degree × π/(180°) = 90° × π/(180° ) = π/4 Convert degrees into radians and viceversa. What is the area of the shaded sector? Fiona draws a circle with a diameter of 14 meters. A Terminal side Vertex B Initial Side C B, ABC, CBA, and are all notations for this angle. Questionnaire. which specific portion of the question – an image, a link, the text, etc – your complaint refers to; (see diagrams below) There is a lengthy reason, but the result is a slight modification of the Sector formula: I have managed to get: 3=½r²θ and 2=½r²sinθ Therefore: ½r²θ-3=0 and ½r²sinθ-2=0 But I'm unsure where to go from there. Area of a circle. Find the area of a sector if the radius is 1 and the angle of sector is  radians. Find the area of a sector with the radius of 1 and angle of . In order to calculate the area of a sector, you need to know the following two parameters: With the above two parameters, finding the area of a circle is as easy as ABCD. Infringement Notice, it will make a good faith attempt to contact the party that made such content available by 3. The arc length formula is used to find the length of an arc of a circle; ℓ = rθ ℓ = r θ, where θ θ is in radian. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. is obtained by the arc AB of the centre O is given by: where is in radians. Finding a Missing Angle With the Sine Rule; 13. Section 4.2 – Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the Express the answer in terms of . Side of polygon given area. r O 1 radian is the size of the angle formed at the centre of a circle by 2 radii which join the ends of an arc equal in length to the radius. Some of the worksheets for this concept are Arc length and sector area, Area of a sector 1, L 2r, Find the area of the shaded sector in the following, Radians arc length and area of a sector, Radians, Mcr3ui radian work, Area and arc length of a sector. Exercise worksheet on 'Find the area of a sector of a circle when the angle is given in radians.' Figure 6. misrepresent that a product or activity is infringing your copyrights. 222 r Ar r Note, to use this formula, the measure of the central angle must be given in radians. Calculate the area of the sector shown below. Calculate the area of a sector with a radius of 10 yards and an angle of 90 degrees. A-Level Maths Edexcel C2 June 2008 Q7b ExamSolutions Northeastern University, Bachelor of Science, Industrial Engineering. To calculate the area of the sector you must first calculate the area of the equivalent circle using the formula stated previously. The non-shaded area of the circle shown below is called a SECTOR. If the angle of the sector is given in degrees, then the formula for the area of a sector is given by. Find the area of a sector with a radius and angle of . © 2007-2020 All Rights Reserved, Computer Science Tutors in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in San Francisco-Bay Area. A sector is created by the central angle formed with two radii, and it includes the area inside the circle from that center point to the circle itself. In this example the sector subtends a right-angle (900) at the centre of the circle. Area of a sector of a circle. A-level : area and arc length of a sector tutorial In this tutorial you are shown how to find the area of a sector and arc length when the angle is in degrees or radians. Problem Solving With the Cosine Rule; 15. Find the angle of a sector whose arc length is 22 cm and area, is 44 cm2. When the angle at the centre is 360°, area of the sector, i.e., the complete circle = πr² When the angle at the center is 1°, area of the sector = Thus, when the angle is θ, area of sector, OPAQ = Write the formula for the area of a sector in radians. Using this formula, and approximating , the area of the circle is . Sep 2, 2009 #2 For #1. Recognize parts of a circle and use appropriate terminology. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such The circular face of a watch has an area that measures between 800 and 900 square millimeters. We know that the area of the whole circle is equal to πr². The ratio of the area of the sector to the area of the full circle will be the same as the ratio of the angle θ to the angle in a full circle. When angle of the sector is 360°, area of the sector i.e. Area of a regular polygon. Find the area of the sector. How to find the area of a sector whose central angle is in radian: formula, 1 example, and its solution. Section 4.2 – Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the

http://mathhelpforum.com/calculus/125089-piecewise-function-problem.html

1. ## Piecewise Function problem Let f(x) = (0) if x< 0 (x) if 0 <= x <= 1 (2 - x) if 1 < x <= 2 (0) if x > 2 and g(x) = [integrate] f(t)dt on [0, x] i) Find an expression for g(x) similar to the one for f(x). ii) Where is f differentiable? Where is g differentiable? I'm pretty much stuck, and don't know how to start this question. I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x). But from that, how do I figure out an expression for g(x) from the piecewise function? Any and all help is appreciated here. 2. Originally Posted by Tulki Let f(x) = (0) if x< 0 (x) if 0 <= x <= 1 (2 - x) if 1 < x <= 2 (0) if x > 2 and g(x) = [integrate] f(t)dt on [0, x] i) Find an expression for g(x) similar to the one for f(x). ii) Where is f differentiable? Where is g differentiable? I'm pretty much stuck, and don't know how to start this question. I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x). But from that, how do I figure out an expression for g(x) from the piecewise function? Any and all help is appreciated here. i) for $\displaystyle x < 0$, $\displaystyle f(x) = 0$ and $\displaystyle g(x) = \int f(t) ~dt = 0$ for $\displaystyle 0 \leq x \leq 1$, $\displaystyle f(x) = x$ and $\displaystyle g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{x} t ~dt = \frac{1}{2}x^2$ for $\displaystyle 1 < x \leq 2$, $\displaystyle f(x) = 2 - x$ and $\displaystyle g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{1} f(t) ~dt + \int_{1}^{x} f(t) ~dt=$$\displaystyle \int_{0}^{1} t ~dt + \int_{1}^{x} (2-t) ~dt= \frac{1}{2} + (2x - 2) - (\frac{1}{2}x^2 - \frac{1}{2})$ for $\displaystyle x > 2$ 3. Hello, Tulki! Here's part (a). Did you make a sketch? $\displaystyle \text{Let }\:f(x) \;=\;\left\{\begin{array}{cccc}0 && x < 0 \\ \\[-4mm] x && 0 \leq x \leq 1 \\ \\[-4mm] 2 - x && 1 < x \leq 2 \\ \\[-4mm] 0 && x > 2 \end{array}\right\}$ and: .$\displaystyle g(x) \:=\: \int_0^x f(t)\,dt$ a) Find an expression for $\displaystyle g(x)$ similar to the one for $\displaystyle f(x).$ $\displaystyle g(x)$ gives the area under the graph from $\displaystyle 0\text{ to }x$ The graph looks like this: Code: | | 1+ * | *:::* | *:::::::* | *:::::::::| * - * * * - - - * - + - * * * - - 0 1 x 2 Then: .$\displaystyle \displaystyle g(x) \;=\;\left\{ \begin{array}{cccc} 0 && x < 0 \\ \\[-3mm] \int^x_0t\,dt && 0 \leq x \leq 1 \\ \\[-2mm] \frac{1}{2} + \int^x_1 (2-t)\,dt && 1 < x \leq 2 \\ \\[-3mm] 1 && x > 2 \end{array} \right\}$ 4. Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved. For x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area). For x between 0 and 1, y= x so the "area under the curve" is the area of a right triangle with base x and height x: its area is $\displaystyle \frac{1}{2}x^2$. For x between 1 and 2, y= 2- x and we can break the "area under the currve" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is $\displaystyle \frac{1}{2}(1)(1)= \frac{1}{2}$. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is $\displaystyle \frac{2-x+ 1}{2}x= \frac{(3-x)x)}{2}= \frac{3x- x^2}{2}$. The total area is $\displaystyle \frac{1}{2}+ \frac{3x- x^2}{2}= \frac{1+ 3x- x^2}{2}$. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1. For x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1. 5. Thank you very much, all of you! I suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect.

https://math.stackexchange.com/questions/4507765/dense-subset-of-sn-1-with-antipodal-complement

# Dense subset of $S^{n-1}$ with antipodal complement Let $$n \in \mathbb{N}$$, and let $$S^{n} \subseteq \mathbb{R}^{n+1}$$ be the unit $$n$$-sphere. Does there exist a dense subset $$D$$ of $$S^n$$ such that $$\textbf{x} \in D \implies -\textbf{x} \notin D$$? This is true for $$n=1$$. Note that $$S^1 \cong A := [0,1]/\mathbb{Z}$$, where the equivalent question is: does there exists a dense subset $$D'$$ of $$A$$ such that $$x \in D \implies x \pm \frac{1}{2} \notin D'$$? Consider the function $$f : [0,1) \to \{0,1\}$$, $$f(x)= \begin{cases} 0 & x< \frac{1}{2},\text{ any decimal expansion of } x \text{ has finitely many 3s} \\ 1 & x < \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s} \\ 1 & x = \frac{1}{2} \\ 1 & x > \frac{1}{2}, \text{ any decimal expansion of } x \text{ has finitely many 3s}\\ 0 & x > \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s} \end{cases}$$ Then $$D' = f^{-1}(\{0\})$$ is a suitable dense set. However, I am not sure how to approach the cases $$n >1$$, since $$S^n$$ is no longer homeomorphic to a very simple object and my topological toolkit is quite unsophisticated. If the question can be answered with elementary methods, I welcome any hints. I am not even sure whether or not I expect such a set $$D$$ to exist for $$n> 1$$. • It is also equivalent to ask the analogue of the question in $\Bbb R^n$, if you’ve figured out what the antipodes look like after stereographic projection Aug 7 at 14:42 • I don't have time for a full answer, but you could consider the stereographic projection $\mathbb{S}^n\setminus\{(0,\ldots,0,1)\}\rightarrow\mathbb{R}^n$, then look at where the intersection $H_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$ with $\mathbb{S}^n$ lands, and then consider the primages of their intersection with $\mathbb{Q}^n$ and $\mathbb{R}^n\setminus\mathbb{Q}^n$ respectivly. Aug 7 at 14:58 • @SamuelAdrianAntz I almost understand your comment, but it's not quite clear. I am a little suspicious of your last step because note that $D$ quite naturally bijects with its complement, so both are uncountable. I understand the stereographic projection to be a bijection, so the preimage of $\mathbb{Q}^n$ would be countable? Or have I misunderstood? Aug 7 at 15:19 • @FShrike That's certainly a possible approach. But even if I could figure out where they antipodes go, I am not certain that this simplifies matters much (or enough for me). After all, $\mathbb{R}$ is still much simpler than $\mathbb{R}^n$, $n>1$. Aug 7 at 15:28 This answer is an expansion of my comment, but I simplified it a bit without using the stereographic projection. First of all, the obvious way of taking the intersection of $$\mathbb{Q}^{n+1}$$ with $$\mathbb{S}^n\subset\mathbb{R}^{n+1}$$ won't work, as it can even be empty as explained in this answer here. Consider the half-spaces $$\mathbb{H}_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$$ and the (surjective) projections: $$\operatorname{pr}_\pm\colon \mathbb{H}_\pm^{n+1}\cap\mathbb{S}^n \twoheadrightarrow\mathbb{D}^n, (x_0,x_1,\ldots,x_n)\mapsto(x_1,\ldots,x_n)$$ flattening down both half-spheres. The subsets $$A=\operatorname{pr}_-^{-1}(\mathbb{D}^n\cap\mathbb{Q}^n)$$ and $$B=\operatorname{pr}_+^{-1}(\mathbb{D}^n\cap(\mathbb{R}^n\setminus\mathbb{Q}^n))$$ are both dense (in the closure of their respective half-sphere) as preimages of dense sets under continuous maps. $$A\cup B$$ is therefore dense in $$\mathbb{S}^n$$ as the closure operator commutes with finite unions (See here). For $$x\in A$$, we have $$x_0<0$$, therefore $$-x\notin A$$, as well as $$x_1,\ldots,x_n\in\mathbb{Q}$$, therefore $$-x\notin B$$. For $$y\in B$$, we have $$y_0>0$$, therefore $$-y\notin B$$, as well as $$y_1,\ldots,y_n\notin\mathbb{Q}$$, therefore $$-y\notin A$$. • So one can use effectively the same idea as for $n=1$. Fair enough. I convinced myself it would be much more difficult and didn't try it. Also, your profile description is fairly entertaining. Aug 7 at 21:26 • As a minor correction, $A$ and $B$ are images, not preimages. Aug 7 at 21:33 • Thanks a lot! :-) The mistake with $A$ and $B$ is also corrected now. Aug 7 at 21:40 • No worries, and thanks for the answer. I'm sorry, but I think I have just observed another typo. I think the projection maps surject onto $D_1(0)$ in $\mathbb{R}^{n-1}$, the interior of the $(n-1)$-sphere. Aug 7 at 21:45 • Oh, yes! Honestly, I'm really not that good in writing down my thoughts in the correct notation. Aug 7 at 21:47

http://mathhelpforum.com/calculus/159154-difficult-trigonometric-integral.html

# Math Help - Difficult Trigonometric Integral 1. ## Difficult Trigonometric Integral I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps. $\int {sin^2 x \over \sqrt {1+cosx}}$ Thanks for the help! 2. Originally Posted by Tylerzzz I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps. $\int {sin^2 x \over \sqrt {1+cosx}}$ Substitute $u = \cos x$. Then $du = -\sin x\,dx$ and the integral becomes $-\int \frac{\sin x}{\sqrt {1+cosx}}(-\sin x)dx = -\int \frac{\sqrt{1-u^2}}{\sqrt{1+u}}du = -\int\sqrt{1-u}\,du = \frac23(1-u)^{3/2} = \frac23(1-\cos x)^{3/2}$ (plus a constant). 3. Wow! I never really thought about solving it that way. Thanks a lot! 4. EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed] 5. Originally Posted by TheCoffeeMachine EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed] You had me worried for a moment. 6. Hello, Tylerzzz! Another approach . . . $\displaystyle \int \frac{\sin^2\!x}{\sqrt{1+\cos x}}\,dx$ We have: . $\dfrac{\sin^2\!x}{\sqrt{1+\cos x}}$ Multiply by $\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}\!:\;\;\dfrac{\sin^2\!x}{\sqrt{1+\cos x}} \cdot\dfrac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{1-\cos^2\!x}}$ . . . $\;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{\sin^2\!x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sin x} \;=\;\sin x\sqrt{1-\cos x}$ The integral becomes: . $\displaystyle \int(1-\cos x)^{\frac{1}{2}}(\sin x\,dx)$ Let: . $u \:=\:1-\cos x \quad\Rightarrow\quad du \:=\:\sin x\,dx$ Substitute: . $\displaystyle \int u^{\frac{1}{2}}\,du \;=\;\tfrac{2}{3}u^{\frac{3}{2}} + C$ Back-substitute: . $\frac{2}{3}(1-\cos x)^{\frac{3}{2}} + C$ 7. Originally Posted by Opalg You had me worried for a moment. Haha! In my defence, I blame the UCAS for the plight of my eyesight today.

http://mathhelpforum.com/pre-calculus/40393-need-help-half-life-problems-using-logs.html

# Thread: Need Help for Half Life Problems using Logs 1. ## Need Help for Half Life Problems using Logs Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help. The equation that I got was: N = No (1/2) ^ t N is final amount, No is the initial ammount and t is time. I was given the data: Time (min) 10 20 30 40 Amount (g) .83 .68 .56 .46 and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life. How do I do this? I have no clue... 2. Do you have Excel?. You can do all sorts of regressions with it. Also, many calculators do regressions. Do you have one?. If so, what is it. Maybe I can step you through it. Anyway, using Excel I got an equation of $y=1.0091e^{-.0196x}$ This has an R^2 of 1, so it is as good as it gets. Enter in x=10, do you get .83?. 3. hi thanks for replying, I have a TI-83+. 4. I would have to re-familiarize myself. I suppose the 83 does exponential regressions. I am not sure. 5. it does, i did it before, I think. I just forgot how to do it. I think it's storing the data into 2 list and then compare them or something... but then, how do I set the equations up? I mean that N = No (1/2) ^ t thing , hate math final, lol. Maybe you can show me how you did it in excel? 6. You can find the half life by using the equation $k=\frac{-1}{T}ln(2)$ We know k=-.0196 So, $-.0196=\frac{-1}{T}ln(2)$ $t=35.36$ We could also do it using the equation. Half of 1.0091 is .50455. Because 1.0091 is the amount at t=0 So, $.50455=1.0091e^{-.0196t}$ $t=35.36$ Same as before. To find a regression on Excel, I make the graph, then add a trendline. Click on Insert, then Chart, then Scatter Highlight your data which is entered in columns A and B and proceed. It is pretty much self-explanatory. Then onece you have the graph, click on Chart, Add Trendline, Exponential, Click on Options and check the box that says to display the equation on the chart. There you have it. 7. Originally Posted by Billwaa Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help. The equation that I got was: N = No (1/2) ^ t N is final amount, No is the initial ammount and t is time. I was given the data: Time (min) 10 20 30 40 Amount (g) .83 .68 .56 .46 and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life. How do I do this? I have no clue... $N = N_0 \left ( \frac{1}{2} \right ) ^{kt}$ $ln(N) = ln(N_0) - [k~ln(2)]t$ Put t on the horizontal axis, ln(N) on the vertical axis, and do a linear regression. The slope will be $k~ln(2)$ which you can then solve for k, and the intercept will be $ln(N_0)$. -Dan 8. thanks guys, this clear it up a little bit. ^_^ 9. Originally Posted by Billwaa hi thanks for replying, I have a TI-83+. To do this on the 83+, input the data into two lists: L1: {10,20,30,40} L2: {.86,.68,.56,.46} Now, hit [STAT] and scroll over to [CALC] Select option [0]:ExpReg. This is an exponential regression. The solution will have the form of $y=a\times b^x$. At the home screen, you should now see: ExpReg Now hit [2nd][1][,][2nd][2] You should now have the following on the screen: $\text{ExpReg } L_1\text{,}L_2$ Hit [ENTER] You'll get the regression equation: This is what the screen should say: ExpReg \begin{aligned} y&=a*b^x \\ a&=1.00914514 \\ b&=.9805442185 \end{aligned} Note that TheEmtpySet's answer was $y=1.0091e^{-.0196x}$ We can rewrite the answer that the calculator gave us to get his solution. Note that $b^x= e^{x\ln(b)}$. Thus, $.9805442185^x=e^{x\ln(.9805442185)}=e^{-.0196475365x}$ Rounding the regression equation, we get: $\color{red}\boxed{y=1.0091e^{-.0196x}}$ Hope this makes sense!

https://math.stackexchange.com/questions/2920725/is-the-limit-of-this-infinite-step-construction-an-equilateral-triangle

# Is the limit of this infinite step construction an equilateral triangle? Just for fun (inspired by sub-problem described and answered here): Let's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute: Now we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute: Now we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute: Repeate the same process infinite number of times. Can we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle. • Regarding the title question, here's a more complicated procedure: 1. Prove whether P=NP or not. 2. Draw an equilateral triangle with compass and straightedge (or any other method). – user253751 Sep 18 '18 at 1:32 • @immibis Very funny :) – Oldboy Sep 18 '18 at 5:08 • @user202729 Title changed as suggested. Thanks! – Oldboy Sep 18 '18 at 5:09 • @immibis On second thought, your procedure is not more complicated. Assuming P=NP is true or not the proof will always take finite time. And my “construction” takes infinite number of steps, each step taking constant time. Just kidding :) – Oldboy Sep 18 '18 at 21:39 • However the steps are all the same, which make it pretty simple :) – user253751 Sep 18 '18 at 22:57 Think about what happens to the maximum difference between angles over time. For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $\vert y-x\vert$. Then when we move one of the $y$-angled points, our new triangle will have angles $$y, {x+y\over 2}, {x+y\over 2}$$ since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is $$\left\vert {y\over 2}-{x\over 2}\right\vert={1\over 2}\vert y-x\vert.$$ So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $\vert y-x\vert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal. $^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, ${1\over 2}$): if $r\in(-1,1)$ then for any $a$ we have $$\lim_{n\rightarrow\infty}ar^n=0.$$ Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero! • I have also noticed that the difference between angles decreases. But having a decreasing sequence does not mean that the sequence converges to zero. You have proved the key point! – Oldboy Sep 18 '18 at 5:18 • Angles $x$ and $y$? Aaah mine eyes! – user332714 Sep 18 '18 at 7:08 By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $\theta_i$, so that the base angles are $\frac12(\pi - \theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $\theta_{i} = \frac12(\pi-\theta_{i-1})$. Thus, \begin{align}\theta_n &= -\frac12\theta_{n-1} + \frac12\pi \\[6pt] &=\frac12\left(-\frac12(\pi-\theta_{n-2})+\pi\right) = \frac14\theta_{n-2}+\frac12\pi-\frac14\pi \\[6pt] &= \cdots \\[6pt] &= \left(-\frac12\right)^{n}\theta_0 \;-\; \sum_{i=1}^n\left(-\frac12\right)^{n}\pi \\[6pt] \lim_{n\to\infty}\theta_n &= 0\cdot\theta_0 \;-\; \frac{(-1/2)}{1-(-1/2)}\pi \\ &=\frac{\pi}{3} \end{align} Thus, in the limit, the triangle becomes equilateral. $\square$ Assume WLOG that the initial triangle is isoceles. Let $\alpha$ be the apical angle, and let $\beta$ be a remaining angle. Then the transformation in question sends $$\begin{bmatrix}\alpha \\ \beta \end{bmatrix}\mapsto \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} \begin{bmatrix}\alpha \\ \beta \end{bmatrix}\text{.}$$ Let $\mathsf{X}$ be the $2\times 2$ transformation matrix on the rhs. $\mathsf{X}$ has characteristic polynomial $x^2-\tfrac{1}{2}x-\tfrac{1}{2}=0.$ By the Cayley–Hamilton theorem, $$\mathsf{X}^2=\tfrac{1}{2}\mathsf{X}+\tfrac{1}{2}\text{.}$$ Therefore we have a Sylvester formula $$f(\mathsf{X})=f(1)\left(\frac{1+2\mathsf{X}}{3}\right)+f(-\tfrac{1}{2})\left(\frac{2-2\mathsf{X}}{3}\right)$$ for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus, $$\mathsf{X}^n=\frac{1+2\mathsf{X}}{3}+(-\tfrac{1}{2})^n\left(\frac{2-2\mathsf{X}}{3}\right)\text{.}$$ The second term converges to zero, so $$\begin{split} \lim_{n\to\infty}\mathsf{X}^n&=\frac{1+2\mathsf{X}}{3}\\ &=\frac{1}{3}\begin{bmatrix}1 & 2 \\ 1 & 2\end{bmatrix}\\ &=\frac{1}{3}\begin{bmatrix} 1\\ 1\end{bmatrix}\begin{bmatrix}1&2\end{bmatrix}\text{,} \end{split}$$ $$\lim_{n\to\infty} \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}^n \begin{bmatrix}\alpha \\ \beta \end{bmatrix}=\begin{bmatrix}\tfrac{\alpha+2\beta}{3}\\ \tfrac{\alpha+2\beta}{3}\end{bmatrix}\text{.}$$ i.e., the apical and side angles approach equality as the operation is repeated.

https://www.physicsforums.com/threads/definite-integral-of-step-function.819626/

# Homework Help: Definite integral of step function Tags: 1. Jun 18, 2015 ### Byeonggon Lee I need to prove whether this expression is true or false: $\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx = \frac{n(n-1)}{2}$ I'm so confused because as I know, definite integral is possible only when the target function is continuous in closed interval. In this case, function $[x]$ should be continuous in interval $[k-1,k]$ but actually [x] is not continuous in $[k,k-1]$ $[x]=k-1$ when $k-1\leq x<k$, $[x]=k$ when $x=k$ So I thought that this expression is false, but in my book's answer, it is true: """ Because$[x]=k-1$ when $k-1\leq x<k$, $[x]=k$ when $x=k$ $\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx$ $=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx+...+\int_{n-1}^{n}[x]dx$ $=\int_{0}^{1}0dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx+...+\int_{n-1}^{n}(n-1)dx$ $=0+\bigg[x\bigg]_{1}^{2}+\bigg[2x\bigg]_{2}^{3}+...+\bigg[(n-1)x\bigg]_{n-1}^{n}$ $=0+1+2+...+(n-1)$ $=\frac{n(n-1)}{2}$ """ The book also says that I need to think interval as $k\leq x<k+1$ and I don't understand how is it possible to omit equal sign at $x<k+1$ 2. Jun 18, 2015 ### DEvens An integral is an area. The troublesome points at the end of the interval mean you are uncertain about the area at a point. What is the difference in the area under a curve if you move one single point up or down by 1? Another way to approach it is this. Think of the integral as a limit of a sum of areas. It happens that in this case you can very easily construct the limit because the function you are looking at is piece-wise constant. 3. Jun 18, 2015 ### Ray Vickson A definite integral certainly can exist for a discontinuous function. Jump discontinuities in f(x) at an interval's endpoint a or b does not affect the area under the curve y = f(x) from x = a to x = b. In other words, $$\int_{[a,b]} f(x) \, dx = \int_{(a,b]} f(x) \, dx = \int_{[a,b)} f(x) \, dx = \int_{(a,b)} f(x) \, dx$$ We can denote all four of these by the common symbol $\int_a^b f(x) \, dx$. 4. Jun 18, 2015 ### RUber You can read a little more on this idea at https://en.wikipedia.org/?title=Riemann_integral. For example, to your point on continuity: "A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure)." Set of measure zero means a finite (or countably infinite) set of distinct points. In the case of the step function is discontinuous at the integers which for a maximum n is finite, and without a maximum is countably infinite. Both sets are measure zero, so you are okay to integrate. 5. Jun 18, 2015 ### WWGD Just to nitpick, or to add a bit: while in this context of step functions discontinuities are finite, you may have in other cases uncountably-infinite sets of measure zero, e.g., the Cantor set. 6. Jun 18, 2015 ### SammyS Staff Emeritus I believe your function is also called the floor function, $\displaystyle \ \lfloor x\rfloor \,,\$ and the greatest integer function, $\displaystyle \ [\![ x]\!]\ .\$ On the interval $\displaystyle \ [k-1\,,\ k)\,,\$ we have $\displaystyle \ [x]=k-1\ .$ Take the limit: $\displaystyle \ \lim_{a\to k^-} \int_{k-1}^a [x]\,dx \ . \$

https://math.stackexchange.com/questions/367473/f-mathbbr-to-mathbbr-satisfies-x-2fx-x1fx-1-3-evaluate/367476

# $f: \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$ The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$. • What math contest is this from? Apr 20, 2013 at 17:38 • @Potato: KöMaL, Problem B. 4538. Apr 28, 2013 at 14:44 • Another example of a question that should have been closed before anybody answered it! Apr 30, 2013 at 11:03 Hint 1: $3=(x+1)-(x-2)$ $(x-2)f(x)-(x+1)f(x-1)=3\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$ Hint 2: Is there a function closely related to $f$ that would verify a simpler equation? $g(x)=f(x)+1$ $(x-2)g(x) - (x+1)g(x-1)=0 \Leftrightarrow (x-2)g(x)=(x+1)g(x-1) \Leftrightarrow g(x)=\cfrac{x+1}{x-2}g(x-1)$ $g(x)=\cfrac{x+1}{x-2}g(x-1)=\cfrac{x+1}{x-2}\cfrac{x+1-1}{x-2-1}g(x-1-1) = \left(\prod\limits_{k=0}^{n-1} \cfrac{x+1-k}{x-2-k}\right) g(x-n)$ Now give the good value to $n$, simplify the product and you'll have the expression of $g$ you need. • Yeah right. Some terms cancel each other >_< Apr 20, 2013 at 20:42 • What is "the good value"? Apr 28, 2013 at 14:31 • The value sot hat you can evaluate $g(x-n)$ Apr 28, 2013 at 14:39 • Big fan of answers with this format +1 May 1, 2013 at 3:26 • Same here, format+1. May 5, 2013 at 1:11 Let $x-1=y$, then, $$(y-1)f(y+1)-(y+2)f(y)=3$$ $$\implies f(y+1)=\frac{3+(y+2)f(y)}{y-1} \;\;\;\;\;(1)$$ Lemma: $\forall \; n \geq 2 \in \mathbb{N}$, $f(n)=n(n-1)(n+1)-1$. Base Case: If $n=2$ then $f(n)=1 \cdot 2 \cdot 3 -1=5$ which is true by information provided in the question. Inductive Step: Assume for $n=k$ that $f(k)=k(k-1)(k+1)-1$. Then by equation $(1)$, $$f(k+1)=\frac{3+(k+2)(k(k-1)(k+1)-1)}{k-1}$$ $$=\frac{k^4+2k^3-k^2-3k+1}{k-1}$$ $$=\frac{(k-1)(k^3+3k^2+2k-1)}{k-1}$$ $$=k^3+3k^2+2k-1$$ $$=k(k^2+3k+2)-1$$ $$=k(k+1)(k+2)-1$$ Thus completing the induction. Hence our Lemma is true and $f(n)=n(n-1)(n+1)-1 \; \forall \; n \geq 2 \in \mathbb{N}$ In particular, if $n=2013$, $f(2013)=2013 \cdot 2012 \cdot 2014-1=8157014183$ (I confess I used a calculator). P.S I found the lemma by trying small cases. The conditions allow you to calculate $f(x+1)$ if you know $f(x)$. Try calculating $f(3), f(4), f(5), f(6)$, and looking for a pattern. Running the following Mathematica program: f = DifferenceRoot[Function[{f, x}, {(x - 2) f[x] - (x + 1) f[x - 1] == 3, f[2] == 5}]]; f[2013] We get the answer = 8157014183.

https://math.stackexchange.com/questions/2208230/which-of-the-following-complex-numbers-is-equivalent-to-frac3-5i82i

Which of the following complex numbers is equivalent to $\frac{3-5i}{8+2i}$? I encountered this question in one of my SAT practice tests. I know the answer is option C, however the only way I got the answer was by trial and error of trying multiple ways to simplify the equation and I ended up rationalising it to get answer choice C. Is there any other, perhaps easier or more direct method, that I can use to solve these types of questions? • Mutiplying top and bottom by the conjugate of the denominator is the standard approach. No need for trial and error -- the standard method will always work. – quasi Mar 29 '17 at 6:36 • I ended up rationalising it Don't know that there is anything more direct than that. You don't even need to carry out all calculations, just figure out the sign of the imaginary part to decide between C and D. (That's assuming you discarded choices A, B upfront, as expected) – dxiv Mar 29 '17 at 6:39 • As a multiple choice question, there will be a minus sign, but A is likely to be wrong – Henry Mar 29 '17 at 6:45 • As a multiple choice question ,you can compare $|z|=|\dfrac{3-5i}{8+2i}|=\dfrac{|3-5i|}{|8+2i|}=\dfrac{\sqrt{2(17)}}{\sqrt{4(17)}}=\dfrac{\sqrt2}{2}$ – Khosrotash Mar 29 '17 at 6:56 • @Khosrotash: interesting proposal, but unfortunately it leads to harder computation than the standard way (because of $(7^2+23^2)/34^2$) :-( – Yves Daoust Mar 29 '17 at 7:10 Two ways to approach this problem. First: As quasi suggests in the comments, multiply by the conjugate of the denominator. \begin{align} \frac{3-5i}{8+2i} & = \frac{3-5i}{8+2i} \times \frac{8-2i}{8-2i} \\ & = \frac{24-40i-6i-10}{8^2+2^2} \\ & = \frac{14-46i}{68} = \frac{7-23i}{34} \end{align} The second approach, given that it's a multiple choice problem, is to multiply each of the answers by $8+2i$ and see if you obtain $3-5i$. I think the first approach is simpler, but they'll both work. Multiplying the top and bottom by the complex conjugate is how you handle this. That gives: $$\frac{3-5i}{8+2i}\cdot\frac{8-2i}{8-2i}=\frac{(3-5i)(8-2i)}{8^2+2^2}=\frac{7-23i}{34}$$ Notice that at the second step we are guaranteed to have a real denominator because $(a+bi)(a-bi)=a^2+b^2$ is always real. I think the method you are using easier one. $$\frac{3-5i}{8+2i}$$ Multiply and divide the fraction by conjugate of denominator, $$\frac{3-5i}{8+2i} \cdot \frac{8-2i}{8-2i}$$ $$\frac{24-6i-40i+10i^2}{64-4i^2}$$ $$\frac{24-6i-40i-10}{64+4}$$ $$\frac{14-46i}{68}$$ $$\frac{7-23i}{34}$$ Alternatives: Let the answer be $a+ib$. Rewrite the initial equation as $$(a+ib)(8+i2)=3-i5.$$ Expanding, you get the $2\times2$ system $$\begin{cases} 8a-2b=3,\\ 2a+8b=-5.\end{cases}$$ Then by Cramer or simply adding four times the first equation and the second $$34a=7$$ and subtracting the first from four times the second, $$34b=-23.$$ Get rid of the denominators and try the products $$(3\pm i20)(8+i2)=24\mp40+i(6\pm160)\to64-i154,$$ $$(7\pm i23)(8+i2)=56\mp46+i(14\pm144)\to102-i170.$$ (We select the signs that match those of $3-i5$.) Then we obtain the identity $$(7-i23)(8+i2)=34(3-i5).$$ The real way: Use the division formula $$\frac{a+ib}{c+id}=\frac{ac+bd}{c^2+d^2}+i\frac{bc-ad}{c^2+d^2},$$ giving $$\frac{14}{68}-i\frac{46}{68}.$$ Then compare to the proposed answers. As C seems to match but D is similar, double check the signs or try the product, for safety.

https://math.stackexchange.com/questions/106313/regular-average-calculated-accumulatively

# Regular average calculated accumulatively is it possible to calculate the regular average of a sequence of numbers when i dont know everything of the sequence, but just everytime i get a new number i know the total count of numbers and the average for the numbers - 1. for example: 2 3 10 the average is of course: 5 but in the last step to calculate i only have access to the previous average of 2 and 3: 2.5 the next number: 10 and the count of numbers: 3 if this is possible, how? Yes, and you can derive it from the expression for the average. Let the average of the first $n$ numbers be $\mu_n$. The formula for it is $$\mu_n = \frac{1}{n} \sum_{i=1}^n x_i$$ Then you can derive $$n \mu_n = \sum_{i=1}^nx_i = x_n + \sum_{i=1}^{n-1} x_i = x_n + (n-1)\mu_{n-1}$$ and hence, dividing by $n$, $$\mu_n = \frac{(n-1) \mu_{n-1} + x_n}{n}$$ i.e. to calculate the new average after then $n$th number, you multiply the old average by $n-1$, add the new number, and divide the total by $n$. In your example, you have the old average of 2.5 and the third number is 10. So you multiply 2.5 by 2 (to get 5), add 10 (to get 15) and divide by 3 (to get 5, which is the correct average). Note that this is functionally equivalent to keeping a running sum of all the numbers you've seen so far, and dividing by $n$ to get the average whenever you want it (although, from an implementation point of view, it may be better to compute the average as you go using the formula I gave above. For example, if the running sum ever gets larger than $10^{308}$ish then it may be too large to represent as a standard floating point number, even though the average can be represented). • As you're computing the previous sum $(n-1)\mu_{n-1}$ as part of the formula, I think this wouldn't help is the sum gets too large. Jun 23, 2016 at 1:12 • @danijar Completely true - a better approach is to keep running sums of the $x_n$ using a method that is robust to rounding error (e.g. Kahan summation) and a separate running count of $n$, and divide whenever you need the mean. That way, your error is bounded by the accuracy of your floating point type. Jun 23, 2016 at 7:44 • If you are likely to end up with numbers larger than $10^{308}$ then you either need to scale down your inputs, or use a more capacious floating point type. Jun 23, 2016 at 7:46 A very simple thought process results in the same formula for running average. If you have $N$ previous measures (of course the measures could all be different) the average you calculate is exactly the same as if all measures were the same as the computed average value. Then, computing the running average of the $N+1$ is equal to $N$ times the previously computed average plus the $N+1$ measure all divided by $N+1$. I know that this is the same as the formula posted in the other answer but no derivation with sums is needed or more obscure mathematical thinking is needed (OK, maybe not really obscure). What you are asking for is commonly called sequential estimation. A general approach is described in [Robbins, H. and S. Monro (1951). A stochastic approximation method. Annals of Mathematical Statistics 22, 400–407.] To add to the derivation of Chris Taylor, I personally like this rewriting as it goes quite intuitively (easy to remember). $$\mu_n = \mu_{n-1} + \frac{1}{n}(x_n - \mu_{n-1})$$ # algorithmically: sequential average computation avg += (x_n - avg)/n

http://www.billthelizard.com/2009/07/broken-stick-experiment.html

## Monday, July 27, 2009 ### The Broken Stick Experiment I found several interesting video lectures at the BLOSSOMS Video Library. One in particular that I enjoyed is called The Broken Stick Experiment [Flash]. In it, Professor Richard Larson asks the seemingly simple question, If you break a stick randomly in two places, what is the probability that you can form a triangle from the three pieces? If one of the pieces is too long, then the other two can't meet to form the triangle. The lecture is only 33 minutes, and Professor Larson builds up a very pretty geometric proof, so I won't spoil it by posting the answer here. I will provide code for a simulation, though (yes, this is still a programming blog). See if you can guess the answer before running the simulation or watching the video. import java.util.Arrays;import java.util.Random;public class BrokenStick { public static void main(String[] args) { Random random = new Random(); int trials = 10000; int triangles = 0; double[] breaks = new double[2]; double[] sides = new double[3]; for( int i=0; i < trials; ++i ) { breaks[0] = random.nextDouble(); breaks[1] = random.nextDouble(); Arrays.sort(breaks); sides[0] = breaks[0]; sides[1] = breaks[1] - breaks[0]; sides[2] = 1.0 - breaks[1]; Arrays.sort(sides); if( sides[2] < (sides[0] + sides[1]) ) { triangles++; } } System.out.println("Triangles: " + triangles); System.out.println("Trials: " + trials); double p = (double)triangles / trials; System.out.println("Proportion: " + p); }} The simulation breaks 10,000 sticks each into three random pieces using Java's Random class to come up with random lengths. If the longest piece of broken stick is longer than the two remaining pieces, a triangle can't be formed. Ten-thousand trials should be enough to accurately estimate the proportion of triangles made from the broken sticks to two decimal places. How close was your guess? Related: The Broken Stick Revisited rbonvall said... Thanks, I had a fun time solving it :) At least to me, this problem lends itself nicely to a geometric solution. In the a-b-c space, the possible breaks must satisfy a+b+c=1, which is a plane. Then, it reduces to find the fraction of the "positive part" of the plane that satisfies the constraints. And I always have fun drawing intersecting planes :P Bill the Lizard said... rbonvall, "And I always have fun drawing intersecting planes :P" Your comment reminds me of the first lecture (or maybe the second, I can't quite remember) in MIT's Linear Algebra video series. Gilbert Strang shows several methods for solving systems of equations, and makes a complete mess of drawing intersecting planes. He uses that as an illustration of why solving systems of equations using matrices is far superior. :) John | Retro Programming said... Thanks, it's an interesting problem. I think I've got the solution, I just need to run a program to double check my result :-) Alexey Busygin said... I think it should be 33,3%. Bill the Lizard said... Alexey, It's not 33%, but your initial guess is closer than mine was. Trevel said... I estimate at just under 25% triangle. Trying to figure if there's a flaw in my logic, but I can't pick it out... tg said... I agree with Trevel. Independent breaks which both must occur on the same half of the stick. <25% tg said... Got the right answer, but not sure if the logic is right: If two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two. These are independent events. The probability of one break on 50% of the stick is 1/2, the probability of two (independent) breaks on the same half is 1/2*1/2 = 0.25. Bill the Lizard said... tg, Wouldn't that indicate that there's a 25% chance of not forming a triangle? tg said... back to the drawing board.... Bill the Lizard said... tg, Don't feel bad, you almost had me convinced. It took me ten minutes to spot the flaw, and I've watched the video twice! :) tg said... How about one more try using the same concept (except using it correctly to represent all three lengths, as I didn't do the first go round): Probability of NOT forming P(NF) = P(a > b+c)*P(b>a+c)*P(c>b+a) As mentioned earlier P(a>b+c) = probability of two breaks on 1/2 of the stick = 0.25. P(NF) = 0.25*0.25*0.25 = 0.75 Thus P( NOT NF) = 0.25 Bill the Lizard said... tg, 0.25 * 0.25 * 0.25 = 0.015625 Back to your original statement, though, you said "If two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two." This is correct, but it's only one way that you could get one piece that's longer than the other two. You could also break the stick on opposite ends of the stick and have the middle piece be too long. tg said... B the L, you are a patient individual. One final attempt to redeem myself. The "*" was supposed to be "+". The risks of writing at work. We are looking for the Probability of (a>b+c) OR (b>a+c) OR (c>a+b). a + b + c = L P(a>b+c) = P(a>L/2) For a>L/2, two breaks must occur on the right half: P(of a break occurring on the right half) = 0.5 P(of two independent breaks occurring on the right)= P(a>L/2) = 0.5*0.5 = 0.25 For b>L/2, two independent breaks must occur on either the left quarter, the right quarter, or both breaks on the left quarter, or both on the right quarter. The probability of ONE break landing on that 50% of the stick = 0.5. The probability of another break independently landing on the same 50%of the stick = 0.5*0.5 = 0.25. c is the same as a, so P(c>b+a) = P(a>b+c) = 0.25 For any one length > L/2 = (a>b+c) OR (b>a+c) OR (c>a+b) = 0.25+0.25+0.25. If this is wrong, spare me the indignity and just delete my comments. Bill the Lizard said... tg, Now you have it! I like the approach you used because you're showing the exact opposite of what the problem asks for, the probability of NOT forming a triangle. This kind of thinking is a very powerful tool in mathematics, and is often overlooked. Sam152 said... Very interesting post, my nifty c++ program calculated about 18.9% successes after 10,000 tries. How close is this? Bill the Lizard said... Sam152, That's a little bit low; lower than rounding error would explain. Do you want to share your code? You could either post it here, or as a Stack Overflow question and I'd be happy to take a look. fiberfiend6891 said... Just a heads up - the link directly to the video has http doubled and doesn't work when clicked directly (I had to copy/paste and get rid of the double): http://http//blossoms.mit.edu/video/larson-watch.html This is my first time seeing your blog, and I absolutely LOVE it! Bill the Lizard said... fiberfiend6891, Tom Raywood said... Bill the Lizard, Hi. Great question. There is one thing I'm not clear on, though, from the description. Do obtuse triangles count here, or is the solution limited to acute? Bill the Lizard said... Tom, You could end up with an acute, obtuse, or (rarely) equilateral triangle. You've got me curious now, so I'll have to modify the code later and run another simulation to see if acute or obtuse triangles are more common. Good question! Tom Raywood said... Bill the Lizard, Good. If you limit outcomes to acute triangles, (including equilaterals), what probability do you get? Oh, and another question [intended to confirm an assumption on my part]: Is this strictly an ordered set or can any of the three pieces form the base? Bill the Lizard said... Tom, Any of the three sides could form the base (I'm assuming by base you mean the longest side). In the code I gave you can see how I work around this by calculating the length of the three sides, then immediately sorting them (on line 24) so that I always know where the longest side ends up. I'm working on an entire post to answer your other question. :) Tom Raywood said... Bill the Lizard, Gee, no I hadn't imagined that the base of the triangle had to be the largest of the three pieces. All of these unstated parameters make it very difficult to approach the problem in a systematic way. So far, yes, the triangle can be of any sort, the pieces can be moved around at will and, finally, (if I understand your last answer), the base of the triangle is expected to be whichever piece [of the stick] that's the longest. Is this correct? Is there anything else in the way of allowances and/or limitations? Bill the Lizard said... Tom, Disregard my last answer, as I had misunderstood your question. There are no constraints on what kind of triangle you can get (acute, obtuse, right, or no triangle at all) or on the order of the pieces. Any of the pieces can be the longest when you randomly select the two break-points. The only reason I sort the pieces in my code is because I need to know which piece is longest, so that I can tell if the pieces form a triangle or not. This has no impact on the final outcome. Tom Raywood said... Bill the Lizard, Whoa, hey, no, it can't be as simple as that can it? All it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick. And that's a full 50% of the time. What gives? I thought certainly the question would pose a greater challenge than that. Bill the Lizard said... Tom, Close. "All it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick." This much is true, but that doesn't happen a full 50% of the time (which was also my initial guess before I watched the full lecture). Tom Raywood said... Bill the Lizard, So true. (I'm feeling better already.) After 100 simulations with n=1,000,000 I'm seeing slightly less than 50%. More specifically I'm seeing 49.994067% and am definitely curious as to what explains this slight deviation. I think I'd rather contemplate it for a while though, that is, to see if I can figure it out myself. Tom Raywood said... Bill the Lizard, To follow up here, beginnings of a guess are as follows: For some small range [in the form of a difference], as two legs of the triangle both approach 1/2, the small difference between them cannot be bridged by the infinitesimally small remaining leg. Be nice to make a formal statement. Bill the Lizard said... Tom, Check your simulation code against mine. Your answer is off by quite a bit more than just rounding error. You can post your code here, or on Stack Overflow if you'd like me to take a look at it (provided it's in a language I know reasonably well, Java, C, C++, Python, PHP, or any dialect of Basic). Tom Raywood said... Bill the Lizard, Definitely courting tedium at this point. n R1 R2 R3 T P1 P2 P3 C1 These represent column headings in Excel. R1, R2 and R3 denote random numbers using the RANDBETWEEN function; I chose for each number the range from 1 to 10,001. T denotes the sum of R1, R2 and R3 for each row. P1, P2 and P3 denote which percentage of T is entailed of R1, R2 and R3 respectively. C1 (for condition 1) represents a check to see whether P1, P2 and P3 all three entail percentages less than 0.5, posting a 1 if they do and a 0 if they don't. n denotes the number of iterations which, as I said, I took the time to set at 1 million. Totaling the C1 column provides a sum which, divided into 1 million, quite closely approximates the probability that no 'leg of the triangle' will be greater than or equal to one half. Additionally I generated a macro which recalculates the worksheet 100 times, posts each result to a table and then finds the average of that set. You can imagine my surprise to hear that the actual probability varies significantly from what this arrangement points up. I'll take a look at that lecture this weekend. Bill the Lizard said... Tom, From your explanation it looks like you're generating three random lengths to represent the three pieces of stick. The problem with this approach is that there's nothing to make sure the three pieces all add up to the original length of the stick. Try it by generating only 2 random numbers (x and y) that represent the break points. If 10,000 is the length of the original stick, and you pick 2 random numbers between 0 and 10,000 then the lengths of the three pieces will be x, y-x, and 10,000-y when x < y, or y, x-y, and 10,000-x when x > y. (It's easier if you just select two random numbers then sort them so x < y, then just always use x, y-x, and 10,000-y for the three lengths, which is essentially what I did in my code.) I'll try to get my next post up this weekend. I have the answer to your question about acute and obtuse triangles, but proving why I'm getting the number I'm getting is proving to be a little bit complicated. Tom Raywood said... Bill the Lizard, I'll work this up and see what gives. It's definitely counterintuitive though that always having the same stick length would even matter, that is, any more than it would matter that triangle of type Q presents as q1, q2, ...qn for n of any size. Be nice to hear the why behind your assertion. Maybe the lecture will help in this regard. Bill the Lizard said... Tom, The particular value of the stick length doesn't matter as much as making sure the three pieces add up to whatever length you decide on. I used a stick length of 1.0 in my similation, but it doesn't hurt anything if you choose a length of 10,000. What really makes a difference is how you choose your random lengths. If you choose three random lengths then add them together, you don't really know the length of the stick you started with. If you start with a specific length, then make two random breaks within the boundaries of the stick, you're guaranteed to end up with three random lengths that add up to your original stick length. It's possible that I just didn't correctly understand the explanation of your spreadsheet. Without seeing the formulas you used I can't be sure that this part of your simulation isn't correct. The errors you're getting could be from another part of the calculation, but this is the most logical place to start looking. Tom Raywood said... Bill the Lizard, I suspect most people would anticipate that having a consistent or particular stick length shouldn't matter here. What's at issue is where the two breaks appear, that is, relative to overall length. A million sticks all of different lengths and all randomly broken in two places should demonstrate the same probabilistic characteristics of a million sticks all of the same length because, again, what's at issue is relative in any event. This principle can be demonstrated as follows. In my simulation each stick length is represented by T, that is, the sum of R1, R2 and R3. If I wanted/needed to ensure that each and every occurrence of T always comes out to equal a certain value, a few more columns could be inserted which accomplish this simply by adjusting the sizes of R1, R2 and R3 by some definitive amount. For example, if for one row T presented as equal to 16,000 [compared to, say, the value of 21,000 I'd predetermined as a uniform stick length], obviously all I'd have to do is increase each of the original random numbers by 21/16. So even though now all million stick lengths are the same, the actual percentage of that length occupied by each R1, R2 and R3 hasn't changed at all. The outcome would still match the 50% probability that my simulation currently produces. In short, Woe to The Foe, heh heh heh. Here are the formulas I employ for... ...{R1,R2,R3} ................ =RANDBETWEEN(1,10001) ..........{T} ................ =SUM(E2:G2), because R1, R2 and R3 occupy columns E, F and G. ...{P1,P2,P3} ................ =E2/H2; =F2/H2; and =G2/H2, because T occupies column H. .........{C1} ................ =IF((J2<0.5)*(K2<0.5)*(L2<0.5),1,0), because P1, P2 and P3 occupy columns J, K and L. And fun was had by all. Spartacus said... 1. Probability that both cuts are on opposite sides of the midpoint: 1/2 2. Probability that two cuts on opposite sides of the midpoint are less than half the stick length apart: 1/2 Therefore the probability of forming a triangle is 1/2*1/2=1/4 Here is a more difficult problem with an even more surprising result: You break the stick into two pieces, then randomly choose one of the two pieces and break it into two pieces. What is the probability that the resulting three pieces can form a triangle? You will be amazed by the solution! (Hint: the answer is not just a simple fraction like the original problem; however there is a very nice closed form) Bill the Lizard said... Spartacus, I've read about half-a-dozen variations on this problem (mostly linked from the notes on the BLOSSOMS page) but I haven't seen that one. This might take a while. :) Anonymous said... Sam152's solution is correct; it is the answer the Spartacus's variant. Bill's original problem statement was under-specified, admitting two interpretations that yield different answers. (Do you choose two cutpoints independently and uniformly on the original stick; or do you make one uniform random cut, and then make another uniform random cut on one of the pieces?) See http://www.cut-the-knot.org/Curriculum/Probability/TriProbability.shtml Anonymous said... This issue might be very old and might already been resolved. but just for clarification your value for second side is wrong and should be taken as follows: breaks[1] = random.nextDouble()*(1-breaks[0]); With this correction, the probability comes to 0.193 which is the correct solution. Bill the Lizard said... Anonymous, No, that's incorrect. The second break can take place at any point along the length of the stick. Your change restricts it to only the length of the first piece broken off. Anonymous said... Ok yes you are correct, this imposes a restriction and thus reduces the probability. the correct answer is 0.25 only and not 0.193. thanks. asha rani said... Is it ok for each triangle n What is n??

https://mathematica.stackexchange.com/questions/22094/modular-arithmetic-efficiently-calculating-the-remainders-of-factorials

# Modular arithmetic - efficiently calculating the remainders of factorials When working on this question regarding the divisibility of the sum of factorials, I decided to write some code to test "small values" of the problem using the following code. f[p_] := Total[Mod[#!, p] & /@ Range[p - 1]]; Table[Mod[f@Prime@i, Prime@i], {i, 1, 500}] Basically, what the code does is sum up all the factorials $$1!+2!+3!+\dots+(p-1)!$$ and find the remainder modulo $p$, for prime $p$. Unfortunately, my code as written takes a very long time to run. Checking the first 500 primes takes 88.280966 seconds on my computer, but checking the first 2000 primes took me about 4 hours. Is there any way to improve the code, or is it already the best we can do? As for optimizations not involving the code, I used Wilson's Theorem, which states that for all primes $p$, $$(p-1)!\equiv-1 \bmod p$$ Using the above theorem, we can modify the code as follows. h[p_] := Total@Flatten[{Mod[#!, p], PowerMod[(# - 1)!*(-1)^(#), -1, p]} & /@ Range[(p - 1)/2]]; Table[Mod[h@Prime@i, Prime@i], {i, 1, 500}] This is considerably faster than the previous code, since checking the first 500 primes takes only 25.896166 seconds. However, checking the first 2000 primes still takes an inordinately long time. This is bit faster: toPrime = 500; sums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]]; primes = Prime[Range[toPrime]]; Mod[sums[[primes - 1]], primes] Precompute factorial sums and primes. Mod is fast on lists. • you are way too humble! calculating your sum for even the first 2000 primes takes less than a second. However, is there a way to get around storing large numbers in sums in memory? It keeps crashing my computer when I try toPrime=5000. – Vincent Tjeng Mar 26 '13 at 6:03 • +1 (that's freaking fast!) Can you explain why Accumulate@FoldList[#1 #2 &, 1, Range[n] is so much quicker to Accumulate@Array[#! &, n] + 1? I really don't get it. – gpap Mar 26 '13 at 11:23 • @gpap Calculating factorial so many times costs a lot. Since we know we want all the factorials up to Prime[toPrime]-1, we ultimately gain a lot keeping the intermediate results with FoldList. – Michael E2 Mar 26 '13 at 12:04 • @MichaelE2 Yes, worked it out myself in the meantime - you multiply the previous result and don't calculate a factorial at every step. Thanks – gpap Mar 26 '13 at 12:11 • Is there any reason why you used #1 #2 & instead of Times? – J. M. is away Jun 15 '15 at 12:53 Let $x \equiv r_1 \bmod p$ and $y \equiv r_2 \bmod p$. Then, $x y \equiv r_1 r_2 \bmod p$. So, we can compute the sum of the factorials mod p using: f[p_] := Mod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p] Let's compare this to the naive implementation: t[p_] := Mod[Sum[k!, {k, p-1}], p] For example: f[Prime[500]] //AbsoluteTiming t[Prime[500]] //AbsoluteTiming {0.00058, 1813} {0.085628, 1813} The nice thing about using Mod[Times[##], p] as the FoldList function is that the output should be a machine number unless you are working with very large primes. That means that f can be compiled: fc = Compile[{{p, _Integer}}, Mod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p], RuntimeAttributes->{Listable} ]; Let's compare for a large prime: f[Prime[10^6]] //AbsoluteTiming fc[Prime[10^6]] //AbsoluteTiming {1.83041, 9308538} {1.05449, 9308538} Faster, but the real difference is that fc is Listable. Hence, comparing timings on a list shows a much larger difference. For example: r1 = f /@ Prime[Range[5000, 5500]]; //AbsoluteTiming r2 = fc[Prime[Range[5000, 5500]]]; //AbsoluteTiming r1 === r2 {2.06691, Null} {0.395402, Null} True Finally, since the list of all factorials is not stored anywhere, the memory used is much more manageable. Here is the memory and timing for the first 5000 primes: r1 = fc[Prime[Range[5000]]]; //MaxMemoryUsed //AbsoluteTiming {3.03463, 462848} This is quite a bit smaller than @MichaelE2's answer: MaxMemoryUsed[ toPrime = 5000; sums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]]; primes = Prime[Range[toPrime]]; r2 = Mod[sums[[primes - 1]], primes] ] //AbsoluteTiming r1 === r2 {2.40441, 4064607008} True The compiled answer uses .462KB while the approach where all the factorials are precomputed takes 4GB, and the timing is not too different.

https://math.stackexchange.com/questions/2496932/while-plotting-polar-coordinate-system-graphs-should-we-include-the-negative-of

# While plotting polar coordinate system graphs, should we include the negative of angle $\theta$? While plotting polar coordinate system graphs, should we include the negative of angle $\theta$? For example, while plotting $r=\theta$ for negative values the spiral comes out to be reverse of the graph plotted using positive values. Upon searching in google I find only one spiral. So are both spirals part of the curve? • There's no reason to omit negative values of $\theta$, other than that the pictures are sometimes less pretty. – Aaron Montgomery Oct 30 '17 at 17:46 • Negative values of $\theta$ are perfectly fine. Often times when doing integrals in polar coordinates, it is convenient to use a parameterization with negative values rather than positive ones. – superckl Oct 30 '17 at 17:49 It's not really about $\theta$, which of course can be both positive or negative. Rather it's about the values of $r$ — should we allow negative values of $r$ or not? And this is not really a mathematical question, but rather a matter of convention. Most textbooks I've seen allow both positive and negative values of $r$. However, some require that $r$ must be non-negative ($r\ge0$).

https://math.stackexchange.com/questions/4481618/how-many-points-uniquely-define-a-square

# How many points uniquely define a square? I was wondering how many points uniquely define a square. Now it is clear to me that if we have $$n$$ random points on a square then this does not necessarily uniquely define the square (for example they may all lie on the same edge). My question was more the following: supposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick to uniquely define the square, and which points should I pick? Edit: the points need not lie on the vertices. • Isn’t it 3? Pick three corner points. Jun 27 at 18:17 • @EDX Which four points are enough? If you pick the four corners, then I can for example draw a different square for which those four points lie on the midpoints of the sides, so that's not enough. Jun 27 at 18:23 • @EDX I think it depends on how we understand the question. If we take a square and pick the four corner points, there will actually be infinitely many squares containing those points, although they won't be in the corners. Jun 27 at 18:24 • @subrosar yes the points need not lie on the corners – JLB Jun 27 at 18:24 • I can not think of a way to have four points uniquely define a square, but five points should suffice. so long as exactly one set of the three points are colinear. From the three that are colinear we can tell the orientation of the square as any other edges must come at 90 degree angles to it. From there you should be able to draw lines which the edges will fall upon. You should be able to then find the distance between opposite lines which will be the edge length and you should be able to complete the square from there. Jun 27 at 18:37 Four points can be enough to unambiguously define a square. Take two of the corners, an additional point from the edge between those corners, and an additional point from the middle of the edge opposite the edge the first three points lied on. A second observer coming to look at these points having been told they all lie on the same square will learn from the three colinear points the orientation of the square and that these lie on the same edge and additional edges will either be perpendicular or parallel to this. They will then note that since if drawing a perpendicular from the fourth point to this edge it lands in the middle of the edge formed by our colinear points that this must be on the opposite edge. Next, since they know this opposite edge must be parallel to the original edge we now know the lengths of the edges must be equal to this minimum distance between our fourth point and the edge formed by our three colinear points. Finally, having noted that this length is equal to the length between our colinear points that these two must have been corners of our square. • That's easier than my construction; well done! Jun 27 at 18:58 Four points are enough, if chosen correctly. In particular, I claim that the only square which passes through the four points $$A(0,0)$$, $$B(0,0.3)$$, $$C(0.4,0)$$, and $$D(1,1)$$ is the unit square with corners $$(0,0), (0,1), (1,0), (1,1)$$. The solution lets you find many more similar examples. To prove this claim, here are some inequalities about the distances between points on a square with side length $$s$$: • Two points on the same side are at a distance between $$0$$ and $$s$$; • Two points on adjacent sides are at a distance between $$0$$ and $$\sqrt2 s$$; • Two points on opposite sides are at a distance between $$s$$ and $$\sqrt 2 s$$. We can check that if $$A,B,C,D$$ lie on a square, then they can't all lie on two adjacent sides of the square: there's no two lines that contain all four points and intersect at right angles. (It's enough to check that $$AB$$ is not perpendicular to $$CD$$, $$AC$$ is not perpendicular to $$BD$$, and $$AD$$ is not perpendicular to $$BC$$.) So there must be some pair of points on opposite sides. The six distances between the points are $$AB=0.3$$, $$AC=0.4$$, $$BC=0.5$$, $$CD \approx 1.16$$, $$BD \approx 1.22$$, and $$AD \approx 1.41$$. So we see that if any two of $$A,B,C$$ were on opposite sides, the side length $$s$$ would be at most $$0.5$$, but $$D$$ is more than $$0.5 \sqrt2 \approx 0.71$$ away from all the rest. Therefore $$A,B,C$$ are all on two adjacent sides of the squares. In particular, one side of the square must contain two of $$A,B,C$$. Can a side of the square contain $$B$$ and $$C$$? No: line $$BC$$ separates $$A$$ from $$D$$, which a side of the square can't do. If we assume $$A,B$$ are on one side of the square and $$C$$ is on an adjacent side, or that $$A,C$$ are on one side of the square and $$B$$ is on an adjacent side, we get the same conclusion: the square has a corner at $$A$$ with one side containing segment $$AB$$ and one side containing segment $$AC$$. Since $$D$$ does not lie on either line, it must be on one of the other sides of the square: a side parallel to $$AB$$, and a side parallel to $$AC$$. We get the same square and try both. • If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AB$$, those lines must contain three sides of the square. The only such square is the unit square. • If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AC$$, those lines must contain three sides of the square. The only such square is the unit square. Three points are never enough. Take three points $$A,B,C$$. If they form a right triangle, we can draw ever-bigger squares with that right angle as a corner, containing all three. If they form an obtuse triangle, that's even better; suppose without loss of generality that the altitude from $$C$$ lands on line $$AB$$ at a point $$H$$ outside segment $$AB$$. Then we can draw ever-bigger squares with one corner at $$H$$ containing all three points. For an acute triangle, begin by rotating the points arbitrarily. Let $$x_{\min}, x_{\max}, y_{\min}, y_{\max}$$ be the lowest and highest $$x$$- and $$y$$-coordinates among the three points; exclude the finitely many orientations where there are ties. Draw the four lines $$x = x_{\min}, x=x_{\max}, y = y_{\min}, y = y_{\max}$$; each of them passes through one point and together they contain all three (otherwise we'd get an obtuse angle). One of the points lies on two of the lines. Without loss of generality, $$A$$ lies on $$x = x_{\max}$$ and $$y = y_{\max}$$ and also $$x_{\max} - x_{\min} \ge y_{\max} - y_{\min}$$. Then erase the line through $$y = y_{\max}$$ and instead draw the line through $$y = y_{\min} + x_{\max} - x_{\min}$$. These four lines define a square containing all three points. Since we get this for all but finitely many ways to rotate $$A,B,C$$ (equivalently, for all but finitely many choices of a horizontal and vertical direction) we get infinitely many squares. • Ahhhhh now I see why three doesn’t suffice. Jun 27 at 20:19 If you specify what points you chose, then two are sufficient: choose opposite vertices. A diagonal can then be drawn between them, and a perpendicular bisector of the same length can be drawn, with the ends being the other two vertices. Then the sides can be drawn between them. • Sometimes the simple answer is the correct answer… “supposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick” – Matt Jun 28 at 7:27 • @Matt There is still some ambiguity. If Alice wants to communicate to Bob what square she has, two points being sufficient requires not only that Alice can choose which points to tell Bob, but that Bob knows that Alice chose those points. If all that Bob knows is that these two points are on the square, but doesn't know that they are opposite vertices, then Bob will not know which square they're from. Jun 29 at 0:55 • I understand your point, but also Bob knows it is a square and not some other completely irregular shape (otherwise the problem would be impossible). So it is reasonable Bob could know the two points are opposite vertices – Matt Jun 29 at 6:42 • Acccumulation : you require some communication between two persons must succeed and Matt : you assume such persons are perfectly reasonable in optimizing. But OP only is about one person figuring it out on its own. If that person knows about his own invented conventions (one could claim that to be cheating), two points would be enough. Jul 3 at 22:30 Two points are enough if they are ordered - a corner A, and the corner B clockwise from A. This is the minimum, as a square has 4 degrees of feeedom - center, size, and orientation. A slighly different question is: whats the minimum number of points such that one and only one square will lie on them. 5 is enough - four corners and somewhere on an edge. Four might be enough.

http://openstudy.com/updates/5606ed62e4b033021d8155a6

## anonymous one year ago Does the series $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}$converge? 1. anonymous Using the fact that $$\text{lcm}\{n,n+1\}=\dfrac{n(n+1)}{\text{gcd}\{n,n+1\}}$$, I'm almost tempted to say that $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}\sim\sum_{n\ge1}\frac{1}{n^2}$ but I don't think I can jump to that conclusion just yet. 2. mathmate We agree that lcm(n,n+1) is actually n(n+1). (The gcd is 1, by Euclid's algorithm) Since S=$$\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}=\sum_{n\ge1}\frac{1}{n(n+1)}<\sum_{n\ge1}\frac{1}{n^2}$$ The strict inequality holds for each and every term, and since $$\sum_{n\ge1}\frac{1}{n^2}$$ is a geometric series that is known to converge (=pi^2/6 prove it), so by comparison, S<pi^2/6 so S converges. 3. mathmate * series, not geometric 4. anonymous Right, since $$n$$ and $$n+1$$ are consecutive, that makes their lcm very easy to determine. And we can find the value of the series while we're at it quite easily: $\sum_{n\ge1}\frac{1}{n(n+1)}=\sum_{n\ge1}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1$Neat! 5. ganeshie8 Nice! any two consecutive integers are coprime, gcd(n, n+1) = gcd(n, 1) = 1 so lcm(n, n+1) = n(n+1) 6. anonymous consider that the least common multiple of $$n,n+1$$ is obviously $$n(n+1)$$ since $$n+1-n=1$$ and thus they must be coprime. this reduces to $$\sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\frac{n+1-n}{n(n+1)}=\sum_{n=1}^\infty\left(\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\right)\\\quad =\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$$ which is obviously telescoping and since $$1/(n+1)\to0$$ it converges 7. imqwerty well i read that a series that converges has got some bounds and that it approaches a specific value. so in this case its going like 1/2 , 1/6, 1/12 , 1/20..... so u mark a line of length 1 on a number line and u plot the point 1/2 u knw 1/2 is still left between 1/2 and 1 and then u add 1/6 still 1/3 is left so u have bounds and u knw that it can never go > 1 so yea it converges... https://en.wikipedia.org/wiki/Convergent_series 8. anonymous Could we make a more general claim here? Numerically, it would seem that $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+k\}}$also converges to $$1$$ for $$k\in\mathbb{N}$$, though much more slowly. 9. anonymous for prime $$k$$ the problem is barely any harder -- you merely have to be careful about the multiples of $$k$$: $$\sum_{n=1}^\infty\frac1{[n,n+k]}=\sum_{m=1}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\\quad=\sum_{m=1}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)$$ 10. ParthKohli $\gcd (n,n+k) \le k$$\Rightarrow {\rm lcm }(n,n+k) = \frac{n(n+k)}{\gcd(n,n+k)} \ge \frac{n(n+k)}{k}$$\Rightarrow \frac{1}{{\rm lcm} (n,n+k) }\le \frac{k}{n(n+k)}$We can sum the right-side telescopically so I guess it is true that the sum is convergent? 11. anonymous yeah, showing its convergence along those lines is exactly what I figured 12. anonymous the sum of the terms for $$n$$ coprime to $$k$$ is accomplished with relatively little effort $$\sum_{n=mk+1}^{(m+1)k-1}\frac1{n(n+k)}=\frac1k\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)$$ so we get: $$\frac1k\sum_{m=0}^\infty\left(\frac1{mk+1}-\frac1{(m+2)k-1}+\frac1{(m+1)(m+2)}\right)$$now split it up and telescope, I suppose? $$\sum\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)=1+\frac1{k+1}\\\sum\left(\frac1{m+1}-\frac1{m+2}\right)=1$$ so we get $$\frac1k\left(2+\frac1{k+1}\right)=\frac2k+\frac1{k(k+1)}=\frac3k-\frac1{k+1}$$ for prime $$k$$ 13. anonymous oops, ignore that -- I read $$(m+2)k+1$$ rather than $$(m+2)k-1$$, so that series does not quite telescope so cleanly 14. anonymous our problem term for prime $$k$$ is this: $$\frac1{mk+1}-\frac1{(m+2)k-1}$$ 15. anonymous Would that I could hand out more medals... 16. thomas5267 How do you show $$\gcd (n,n+k) \leq k$$ without using Bézout's Identity? 17. anonymous i mean, if (positive) $$d$$ divides (positive) $$n$$ and $$n+k$$ then it also divides their difference, so $$d$$ must divide $$k$$. it follows that the greatest common divisor is at most $$k$$ 18. anonymous you don't necessarily need the full strength of Bezout's identity, just the fact that divisibility is preserved by sums which is very elementary 19. ganeshie8 $$\gcd(n,n+k)=\gcd(n.k)\le k$$ 20. thomas5267 \begin{align*} \sum_{n=1}^\infty\frac1{[n,n+k]}&=\sum_{m=0}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\ &=\sum_{m=0}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)\\ \end{align*} \begin{align*} &\phantom{{}={}}\sum_{m=0}^\infty\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k-1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}+\frac{1}{(m+1)k+1}-\frac{1}{(m+1)k-1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}-\frac{2}{(m+1)^2k^2-1}\right)\\ &=1-\sum_{m=0}^\infty\frac{2}{(m+1)^2k^2-1} \end{align*} I think the sum converges but I couldn't prove it. 21. anonymous Comparing to the series $$\sum\frac{1}{n^2}$$ would suffice for convergence. It has a somewhat daunting closed form: $\sum_{m=0}^\infty \frac{2}{(m+1)^2k^2-1}=\frac{k-\pi\cot\dfrac{\pi}{k}}{2k}$ 22. anonymous (closed form courtesy of Mathematica) 23. thomas5267 I thought $$\dfrac{2}{(m+1)k^2-1}\geq\dfrac{1}{n^2}$$ for some reason! $\min(k)=2\\ \frac{2}{4n^2-1}-\frac{1}{n^2}=\frac{2n^2-4n^2+1}{n^2(4n^2-1)}=\frac{-2n^2+1}{n^2(4n^2-1)}\leq0$ 24. anonymous $\sum_{m\ge0}\frac{1}{(m+1)^2k^2-1}=\sum_{m\ge1}\frac{1}{m^2k^2-1}\le\sum_{m\ge1}\frac{1}{m^2}$ since $1\le k^2-1~~\implies~~ m^2\le m^2k^2-1~~\implies~~\frac{1}{m^2k^2-1}\le\frac{1}{m^2}$(The first inequality is true since $$k\ge2$$.)

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# Ten people are seated at a rectangular table - Permutations homework I got the following question for homework. Ten people are to be seated at a rectangular table for dinner. Tanya will sit at the head of the table. Henry must not sit beside either Wilson or Nancy. In how many ways can the people be seated for dinner? The approach I have used is to take $9!$ and subtract it from the seats where Henry, Wilson or Nancy can sit. I end up getting the answer $211680$ but for some reason, the back of the book says $201 600$. What am I doing incorrectly? - This looks like an error in the book: your answer appears to be correct. We can count the allowed seatings directly. Suppose that Henry sits next to Tanya. Then there is one seat forbidden to Wilson and Nancy, so there are $\binom72$ ways to choose seats for Wilson and Nancy. There are $2$ ways to seat Wilson and Nancy in those $2$ seats and $6!$ ways to seat the unnamed people. Finally, Henry can be on either side of Tanya, so there are altogether $2\cdot2\cdot6!\cdot\binom72$ acceptable seatings with Henry next to Tanya. Now suppose that Henry is not seated next to Tanya. Then there are $2$ seats forbidden to Wilson and Nancy, so there are $\binom62$ ways to choose their seats. As before there are $2$ ways to seat them in those two seats and $6!$ ways to seat the unnamed people. Finally, there are $7$ possible choices for Henry’s seat, so there are altogether $7\cdot2\cdot6!\cdot\binom62$ acceptable seatings with Henry not next to Tanya. Altogether, then, there are $$2\cdot6!\left(2\binom72+7\binom62\right)=1440(42+105)=211,680$$ acceptable seatings. - Using inclusion-exclusion, it's possible to see where the error in the back of the book may have come from. With Tanya assigned to the head of table, there are $9!$ ways to seat the other people with no restrictions. Among these, there $2\cdot8!$ ways in which Henry sits beside Wilson (the factor of $2$ corresponds to which one sits to the left of the other), and another $2\cdot8!$ ways in which Henry sits beside Nancy. Subtracting these, we get the book's answer $$9!-2\cdot8!-2\cdot8!=201600$$ But this double-subtracts the $2\cdot7!$ ways in which Henry sits between Wilson and Nancy. So the correct answer is $$9!-2\cdot8!-2\cdot8!+2\cdot7!=211680$$ -

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02 dez # system of linear equations problems Many problems lend themselves to being solved with systems of linear equations. Case 2: Parallel Lines The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. Materials include course notes, lecture video clips, JavaScript Mathlets, a quiz with solutions, practice problems with solutions, a problem solving video, and problem sets with solutions. Once you do that, these linear systems are solvable just like other linear systems. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. Solve age word problems with a system of equations. The problem also says that Mia will be twice as old as Shaheena. This example shows one iteration of the gradient descent. When you solve systems with two variables and therefore two equations, the equations can be linear or nonlinear. There can be any combination: 1. But let’s say we have the following situation. Systems of linear equations can be used to model real-world problems. Solving using Matrices by Row Operations. Quiz 3. Just select one of the options below to start upgrading. Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. At the first store, he bought some t-shirts and spent half of his money. A system of linear equations is called homogeneous if the constants$b_1, b_2, \dots, b_m$are all zero. “Systems of equations” just means that we are dealing with more than one equation and variable. Wouldn’t it be cl… Consider the nonlinear system of equations Cramer's Rule. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you … Word Problems with Systems of Linear Equations: This worksheet is designed for Algebra students to practice those dreaded word problems that are solved with linear systems of equations. Is the point$(1 ,3)$a solution to the following system of equations… Solution of a non-linear system. No Problem 2. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. Solving using an Augmented Matrix. Solve simple cases by inspection. Problem 1. Find them out by checking. Systems of Equations - Problems & Answers. Systems of linear equations are a common and applicable subset of systems of equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: System of equations word problem: walk & ride, Practice: Systems of equations word problems, System of equations word problem: no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite solutions), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. By … System of Linear Equations Word Problems Calvin went to Chicago's Magnificent Mile to do some Christmas shopping. A system of linear equations is a set of two or more linear equations with the same variables. Is the point$(0 ,\frac{5}{2})$a solution to the following system of equations? SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. To solve the system of equations, you need to find the exact values of x and y that will solve both equations. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Below is an example that shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. You really, really want to take home 6items of clothing because you “need” that many new things. Systems of Linear Equations. Solve the following system of equations by elimination. For problems 1 – 3 use the Method of Substitution to find the solution to the given system or to determine if the system … 9,000 equations in 567 variables, 4. etc. Gradient descent can also be used to solve a system of nonlinear equations. Looking for fun activities to teach kids about Solving Word Problems Involving Linear Equations and Linear Inequalities? Systems of Linear Equations and Problem Solving. 6 equations in 4 variables, 3. (5.2.3) – Solve mixture problems with a system of linear equations. In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. ... Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up! This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. HIDE SOLUTIONS. 1. Systems of 2 linear equations - problems with solutions Test. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. Section 8.1, Example 4(a) Solve graphically: Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. In your studies, however, you will generally be faced with much simpler problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. For example, the sets in the image below are systems of linear equations. This premium worksheet bundle contains a printable fact file and 10 fun and engaging worksheets to challenge your students and help them learn about Solving Word Problems Involving Linear Equations and Linear Inequalities. Substitution Method. Problem 1 Two of the following systems of equations have solution (1;3). B. Determining the value of k for which the system has no solutions. To use Khan Academy you need to upgrade to another web browser. So a System of Equations could have many equations and many variables. There are 7 questions. Donate or volunteer today! A large pizza at Palanzio’s Pizzeria costs$6.80 plus $0.90 for each topping. Our mission is to provide a free, world-class education to anyone, anywhere. System of NonLinear Equations problem example. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. This section provides materials for a session on solving a system of linear differential equations using elimination. Khan Academy is a 501(c)(3) nonprofit organization. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. The statement above gives the following equation m + 10 = 2 × (s + 10) m + 10 = 2 × s + 2× 10 m + 10 = 2s + 20 You now have a system of linear equation to solve m + s = 40 equation 1 m + 10 = 2s + 20 equation 2 Use equation 1 to solve for m m + s = 40 m + s - s = 40 - s m = 40 - s If you're seeing this message, it means we're having trouble loading external resources on our website. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. One application of systems of equations are mixture problems. This section covers: Systems of Non-Linear Equations; Non-Linear Equations Application Problems; Systems of Non-Linear Equations (Note that solving trig non-linear equations can be found here).. We learned how to solve linear equations here in the Systems of Linear Equations and Word Problems Section.Sometimes we need solve systems of non-linear equations, such as those we see in conics. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. System of linear equations System of linear equations can arise naturally from many real life examples. Add the second equation to the first equation and solve for x. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. set of two (or more) equations which have two (or more) variables Solving Systems of Linear Equations. It has 6 unique word problems to solve including one mixture problem … If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. Simultaneous equations (Systems of linear equations): Problems with Solutions. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\left( {2 - x} \right) = 3x + 2$$, $$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$$, $$\displaystyle \frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}$$, $$\displaystyle \frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}$$, $$\displaystyle \frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}$$, $$\displaystyle \frac{{5x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5}{3}$$. So far, we’ve basically just played around with the equation for a line, which is . 1. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Solve each of the following equations and check your answer. Section 7-1 : Linear Systems with Two Variables. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve age word problems with a system of equations. A solution of the system (*) is a sequence of numbers$s_1, s_2, \dots, s_n$such that the substitution$x_1=s_1, x_2=s_2, \dots, x_n=s_n$satisfies all the$m$equations in the system (*). Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. You discover a store that has all jeans for$25 and all dresses for 50. Improve your math knowledge with free questions in "Solve systems of linear equations" and thousands of other math skills. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! A system of linear equations is a system made up of two linear equations. Solution: Rewrite in order to align the x and y terms. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . 2 equations in 3 variables, 2. Generally speaking, those problems come up when there are two unknowns or variables to … Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. If you're seeing this message, it means we're having trouble loading external resources on our website. The same rules apply. a)\begin{array}{|l} x + y = 5 \\ 2x - y = 7; \end{array}\$ In "real life", these problems can be incredibly complex. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. Answer: x = .5; y = 1.67. Solving using Matrices by Elimination. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables.

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Learn faster with a math tutor. 9th - 10th grade. The two pairs of congruent sides may be, but do not have to be, congruent to each other. Get better grades with tutoring from top-rated professional tutors. This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2-D geometry. 3) Are opposite angles of a parallelogram congruent? Solve for x. (10 Properties of parallelogram: Opposite sides of parallelogram are equal . The diagonals of a parallelogram bisect each other. Sides of A Parallelogram The opposite sides of a parallelogram are congruent. (By definition). The angles of a parallelogram are congruent. Ask yourself which approach looks easier or quicker. Line segments XY and ZW are also congruent. One way all sides of the two parallelograms could be congruent would be if $ABCD$ and $EFGH$ are squares with the same side length: in this case they would be congruent. Properties of parallelograms are as follows: i. Parallelogram Properties DRAFT. Tags: Question 19 . CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. Is the quadrilateral a parallelogram? . Give a reason. Expand Image In fact, one method of proving a quadrilateral a rhombus is by first proving it a parallelogram, and then proving two consecutive sides congruent, diagonals bisecting verticies, etc. A parallelogram is a quadrilateral that has opposite sides that are parallel. If one side is longer than its opposite side, you do not have parallel sides; no parallelogram! Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Opposite sides are parallel and congruent. It is a quadrilateral with two pairs of parallel, congruent sides. Opposite angles are congruent. High School: Geometry » Congruence » Prove geometric theorems » 11 Print this page. Parallelogram Properties DRAFT. A parallelogram is defined to be a quadrilateral with 2 pairs of opposite sides parallel. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. Consecutive angles are supplementary (A + D = 180°). The two diagonals of a parallelogram bisect each other. A parallelogram is a quadrilateral with two pairs of parallel sides. 5) Does a diagonal of a parallelogram bisect a pair of opposite angles? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. Yes, if both pairs of opposite angles are congruent, then you have a parallelogram. The diagonals of an isosceles trapezoid are congrent. 0 Why are these two lines not congruent (and other ways to figure out if other shapes are not congruent) Solve for x. HELP ASAP 30 points Part 1 out of 2 To repair a large truck or bus, a mechanic might use a parallelogram lift. 2 years ago. One has to be on the lookout for double negatives. 62% average accuracy. Opposite sides are congruent. You can draw parallelograms. The opposite angles are congruent. 60 seconds . A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . If yes, state how you know. There are two ways to go about this. If one angle is right, then all angles are right. The diagonals of a rectangle are the bisectors of the angles. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. An equilateral quadrilateral is a square. That is true. Here are some important things that … Try to move the vertices A, B, and D and observe how the figure changes. The opposite sides of a parallelogram are congruent so we will need two pairs of congruent segments: Now if we imagine leaving $\overline{AB}$ fixed and ''pushing down'' on side $\overline{CD}$ so that these two sides become closer while side $\overline{AD}$ and $\overline{BC}$ rotate clockwise we get a new parallelogram: There is one right angle in a parallelogram and it is not a rectangle. If a quadrilateral has three angles of equal measure, then the fourth angle must be a right angle. The converse is also true that if opposite sides of a quadrangle are equal then its a parallelogram. 62% average accuracy. The figure is a parallelogram. The diagonals of a trapezoid are perpendicular. For our parallelogram, we will label it WXYZ, but you can use any four letters as long as they are not the same as each other. A rhombus is a parallelogram but with all four sides equal in length; A square is a parallelogram but with all sides equal in length and all interior angles 90° A quadrilateral is a parallelogram if: Both pairs of opposite sides are parallel. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Tags: Question 19 . The diagonals of a quadrilateral_____bisect each other, If the measures of 2 angles of a quadrilateral are equal, then the quadrilateral is_____a parallelogram, If one pair of opposite sides of a quadrilateral is congruent and parallel, then the quadrilateral is______a parallelogram, If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram, To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel, The diagonals of a rectangle are_____congruent, The diagonals of a parallelogram_______bisect the angles, The diagonals of a parallelogram______bisect the angles of the parallelogram, A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram, The diagonals of a rhombus are_______congruent, A rectangle______has consecutive sides congruent, A rectangle_______has perpendicular diagonals, The diagonals of a rhombus_____bisect each other, The diagonals of a parallelogram are_______perpendicular bisectors of each other, Consecutive angles of a quadrilateral are_______congruent, The diagonals of a rhombus are______perpendicular bisectors of each other, Consecutive angles of a square are______complementary, Diagonals of a non-equilateral rectangle are______never angle bisectors, A quadrilateral with one pair of congruent sides and one pair of parallel sides is_____a parallelogram. B) The diagonals of the parallelogram are congruent. The opposite angles of a parallelogram are supplementary. Studying the video and these instructions, you will learn what a parallelogram is, how it fits into the family of polygons, how to identify its angles and sides, how to prove you have a parallelogram, and what are its identifying properties. Reason for statement 8: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Triangles can be used to prove this rule about the opposite sides. Use a straightedge (ruler) to draw a horizontal line segment, then draw another identical (congruent) line segment some distance above and to one side of the first one, so they do not line up vertically. Opposite angles are congruent. Parallelogram definition A quadrilateral with both pairs of opposite sides parallel. 1981 times. Opposite sides are congruent. If both pairs are congruent, you have either a rhombus or a square. SURVEY . 2. A parallelogram is a quadrilateral that has opposite sides that are parallel. Check for any one of these identifying properties: You can use proof theorems about a plane, closed quadrilateral to discover if it is a parallelogram: You have learned that a parallelogram is a closed, plane figure with four sides. You can have almost all of these qualities and still not have a parallelogram. Opposite angles are congruent. The opposite sides of a parallelogram are congruent. Local and online. Go with B. Give a reason. The diagonals of a rectangle bisect eachother. Theorem: If ABCD is a parallelogram then prove that its opposite sides are equal. We already mentioned that their diagonals bisect each other. The diagonals of a kite are the perpendicular bisectors of each other. More generally, a quadrilateral with 4 congruent sides is a rhombus. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram Always To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. , all 4 sides are congruent (definition of a rhombus). Get better grades with tutoring from top-rated private tutors. Check Next Opposite angles are congruent. Want to see the math tutors near you? Property that is characteristic of a parallelogram is that opposite sides are congruent. The diagonals of a quadrilateral are perpendicular and the quadrilateral is not a rhombus. Opposite angels are congruent (D = B). Prove that opposite sides of a parallelogram are congruent. Other shapes, however, are types of parallelograms. Which statement can be used to prove that a given parallelogram is a rectangle? 4) Do the diagonals of a parallelogram bisect each other? Observe that at any time, the opposite sides are parallel and equal. Now consider just the interior angles of parallelograms, ∠W, ∠X, ∠Y, and ∠Z. Theorem 1: Opposite Sides of a Parallelogram Are Equal In a parallelogram, the opposite sides are equal. In this lesson we will prove the basic property of a parallelogram that the opposite sides in a parallelogram are equal. Q. In the video below: We will use the properties of parallelograms to determine if we have enough information to prove a given quadrilateral is a parallelogram. The four line segments making up the parallelogram are WX, XY, YZ, and ZW. Notice that line segments WX and YZ are congruent. Use this to prove that the quadrilateral must be a parallelogram. Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! The first rhombus above is a square while the second one has angles of 60 and 120 degrees. If only one set of opposite sides are congruent, you do not have a parallelogram, you have a trapezoid. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. These geometric figures are part of the family of parallelograms: For such simple shapes, parallelograms have some interesting properties. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. Give the given and prove and prove this. A parallelogram does not have other names. The interior angles are ∠W, ∠X, ∠Y, and ∠Z. Make sure that second line segment is parallel to (or equidistant from) the first line segment. Yes. The consecutive angles of a parallelogram are supplementary. The opposite sides are equal and parallel; the opposite angles are also equal. 1-to-1 tailored lessons, flexible scheduling. 1981 times. Connecting opposite (non-adjacent) vertices gives you diagonals WY and XZ. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . 2 years ago. Start at any vertex (corner). The area of a parallelogram is twice the area of a triangle created by one of its diagonals. We will show that in that case, they are also equal to each other. That the other pair of opposite sides are congruent. Opposite sides are congruent (AB = DC). Is the quadrilateral a parallelogram? Take a rectangle and push either its left or ride side so it leans over; you have a parallelogram. In today's lesson, we will show that the opposite sides of a parallelogram are equal to each other. Let’s play with the simulation given below to better understand a parallelogram and its properties. This means a parallelogram is a plane figure, a closed shape, and a quadrilateral. The parallelogram has the following properties: Opposite sides are parallel by definition. An equilateral parallelogram is equiangular. Proving That a Quadrilateral is a Parallelogram, Opposite sides are parallel -- Look at the parallelogram in our drawing. The bad in Answer A is due to your teachers written grammar. Is it possible to prove a quadrilateral a parallelogram with two consecutive and two opposite congruent sides? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. That’s a wrap! A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. In the figure given below, ABCD is a parallelogram. The figure shows a side view of the li … ft. FGKL, GHJK, and FHJL are parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The Angle-Side-Angle Triangle Congruence Theorem can be used to prove that, in a parallelogram, opposite sides are congruent. There are more than two right angles in a trapezoid. Opposite sides are congruent -- The base side (Y Z Y Z) and the top side (W X W X) of our parallelogram are equal in length (congruent); the left side (XY X Y) and right side (ZW Z W) are also congruent To be a parallelogram, the base and top sides must be parallel and congruent, and … The opposite sides of parallelogram are also equal in length. CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv… Parallelogram law. Parallelogram Theorem #2: The opposite sides of a parallelogram are congruent. SURVEY . View Untitled document (3).pdf from MATH 100 at Basha High School. If the four sides do not connect at their endpoints, you do not have a closed shape; no parallelogram! Mathematics. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. The bottom (base) side, Opposite sides are congruent -- The base side (. C) The diagonals of the parallelogram bisect the angles. 60 seconds . Get help fast. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. G 2 T: + 3 5 6 8 17 K 3 Which angles are congruent to 21? <2 2 are congruent to 21. As with any quadrilateral, the interior angles add to 360°, but you can also know more about a parallelogram's angles: Using the properties of diagonals, sides, and angles, you can always identify parallelograms. Opposite sides are congruent. One interesting property of a parallelogram is that its two diagonals bisect each other (cut each other in half). Or: Both pairs of opposite sides are congruent. To explore these rules governing the sides of a parallelogram use Math Warehouse's interactive parallelogram. Triangles can be used to prove this rule about the opposite sides. D) The opposite angles of the parallelogram are congruent. Opposite sides of a … ∴ ∴ AB = CD A B = C D and AD= BC A D = B C Properties of a parallelogram. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. Opposite angles are equal (congruent) to each other; Any two adjacent angles of a parallelogram add up to, This means any two adjacent angles are supplementary (adding to, A closed shape (it has an interior and exterior), A quadrilateral (four-sided plane figure with straight sides), Two pairs of congruent (equal), opposite angles, Two pairs of equal and parallel opposite sides, If the quadrilateral has bisecting diagonals, it is a parallelogram, If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram, If the quadrilateral has consecutive supplementary angles, it is a parallelogram, If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. To be on the lookout for double negatives proof 2 here ’ use. 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That has opposite sides are not congruent with the simulation given below to better understand parallelogram! Δabd and ΔCDB are congruent has the following properties: two pairs of parallel, then quadrilateral. Where the two diagonals bisect each other ( cut each other that case, are... Are equal and parallel ; the opposite sides are equal then its diagonals divide the quadrilateral is quadrilateral! The bottom ( base ) side, you do not have a parallelogram also. Next is it possible to prove this rule about the opposite sides not. Side, you do not have to be a quadrilateral in which both pairs of parallel, then fourth. It leans over ; you have a trapezoid and name the appropriate geometric are... Two ways to go about this and so must the left and side... Quadrangle are equal side so it leans over ; you have a trapezoid top must! That in that case, they are also equal to each other and each one separates the are... 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https://math.hecker.org/2011/01/20/linear-algebra-and-its-applications-exercise-1-5-19/?replytocom=8181

## Linear Algebra and Its Applications, Exercise 1.5.19 Exercise 1.5.19. For the following two matrices, specify the values of $a$, $b$, and $c$ for which elimination requires row exchanges, and the values for which the matrices in question are singular. $A = \begin{bmatrix} 1&2&0 \\ a&8&3 \\ 0&b&5 \end{bmatrix} \quad A = \begin{bmatrix} c&2 \\ 6&4 \end{bmatrix}$ Answer: For the first matrix a row exchange would be necessary if the first step of elimination caused the second entry of the second row, i.e., in the (2,2) position, to become 0. This would occur if we multiplied the first row by 4 and subtracted it from the second row, as we would then have $8 - 4 \cdot 2 = 0$ in the (2,2) position. Since the entry in the (1,1) position is 1, multiplying the first row by 4 would be necessary if $a$ were 4: $\begin{bmatrix} 1&2&0 \\ 4&8&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&0&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&b&5 \\ 0&0&3 \end{bmatrix}$ If we had both $a = 4$ and $b = 0$ then the matrix would be singular: $\begin{bmatrix} 1&2&0 \\ 4&8&3 \\ 0&0&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&0&3 \\ 0&0&5 \end{bmatrix}$ However, these are not the only values of $a$ and $b$ for which the matrix is singular. Suppose we proceed with elimination. The first step would be to multiply the first row by $l_1 = a$ and subtract the result from the second row: $\begin{bmatrix} 1&2&0 \\ a&8&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&b&5 \end{bmatrix}$ Assuming that $a \ne 4$ then the next step of elimination would multiply the second row of the matrix by $l_2 = b/(8-2a)$ and subtract the result from the third row: $\begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&0&5-3b/(8-2a) \end{bmatrix}$ If the (3,3) entry of the resulting matrix above is zero then the original matrix is singular. In this case we have $3b/(8-2a) = 5$. Multiplying both sides by $(8-2a)$ we have $3b = 5 \cdot (8-2a) = 40 - 10a$ or $10a + 3b = 40$. So the first matrix is singular for all values of $a$ and $b$ for which $10a+3b=40$. This includes the case $a = 4, b=0$ mentioned above, the case $a=0, b = 40/3$, the case $a=5, b=-10/3$, and an infinity of others on the line $b=-\frac{10}{3}a + \frac{40}{3}$. For the second matrix a row exchange would be necessary if $c = 0$: $\begin{bmatrix} 0&2 \\ 6&4 \end{bmatrix} \rightarrow \begin{bmatrix} 6&4 \\ 0&2 \end{bmatrix}$ On the other hand, if c = 3 then the first elimination produces zeroes in both entries of the second row, and the matrix is singular: $\begin{bmatrix} 3&2 \\ 6&4 \end{bmatrix} \rightarrow \begin{bmatrix} 3&2 \\ 0&0 \end{bmatrix}$ UPDATE: Prompted by Theodore’s comment, added a section deriving the complete set of $a$ and $b$ for which the first matrix is singular. NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang. If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books. This entry was posted in linear algebra. Bookmark the permalink. ### 2 Responses to Linear Algebra and Its Applications, Exercise 1.5.19 1. Theodore says: I also found a = 5 and b = -10/3 to also cause a singular matrix. My question is how do we find all values that can cause singularity • hecker says: Thank you for finding this problem. You are correct that there are multiple (actually, an infinity) of values of a and b for which the first matrix is singular. I added a new section to derive the correct answer.

http://www.mathworks.com/help/matlab/ref/istril.html?s_tid=gn_loc_drop&nocookie=true

Accelerating the pace of engineering and science # istril Determine if matrix is lower triangular ## Description example tf = istril(A) returns logical 1 (true) if A is a lower triangular matrix; otherwise, it returns logical 0 (false). ## Examples expand all ### Test Lower Triangular Matrix Create a 5-by-5 matrix. `D = tril(magic(5))` ```D = 17 0 0 0 0 23 5 0 0 0 4 6 13 0 0 10 12 19 21 0 11 18 25 2 9``` Test D to see if it is lower triangular. `istril(D)` ```ans = 1``` The result is logical 1 (true) because all elements above the main diagonal are zero. ### Test Matrix of Zeros Create a 5-by-5 matrix of zeros. `Z = zeros(5);` Test Z to see if it is lower triangular. `istril(Z)` ```ans = 1``` The result is logical 1 (true) because a lower triangular matrix can have any number of zeros on its main diagonal. ## Input Arguments expand all ### A — Input arraynumeric array Input array, specified as a numeric array. istril returns logical 0 (false) if A has more than two dimensions. Data Types: single | double Complex Number Support: Yes expand all ### Lower Triangular Matrix A matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero. For example, the matrix $A=\left(\begin{array}{cccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -2& -2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 0\\ -3& -3& -3& 1\end{array}\right)$ is lower triangular. A diagonal matrix is both upper and lower triangular. ### Tips • Use the tril function to produce lower triangular matrices for which istril returns logical 1 (true). • The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istril(A) == isbanded(A,size(A,1),0).

https://math.stackexchange.com/questions/2687769/solve-the-equation-cos2x-cos22x-cos23x-1

# Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$ Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$ IMO 1962/4 My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable. \begin{align} \cos^22x&=(\cos^2x-\sin^2x)^2\\ &=\cos^4x+\sin^4x-2\sin^2\cos^2x\\ &=\cos^4+(1-\cos^2x)^2-2(1-\cos^2)\cos^2x\\ &=\cos^4+1-2\cos^2x+\cos^4x-2\cos^2x+2\cos^4x\\ &=4\cos^4x-4\cos^2x+1 \end{align} Without knowledge of other trigonometric identities, $\cos3x$ can be derived using only Ptolemy's identities. However for the sake of brevity, let $\cos 3x=4\cos^3x-3\cos x$: \begin{align} \cos^23x&=(4\cos^3x-3\cos x)^2\\ &=16\cos^6x+4\cos^2x-24\cos^4x \end{align} Therefore, the original equation can be written as: $$\cos^2x+4\cos^4x-4\cos^2x+1+16\cos^6x+4\cos^2x-24\cos^4x-1=0$$ $$-20\cos^4x+6\cos^2x+16\cos^6x=0$$ Letting $y=\cos x$, we now have a polynomial equation: $$-20y^4+6y^2+16y^6=0$$ $$y^2(-20y^2+6y+16y^4)=0\Rightarrow y^2=0 \Rightarrow x=\cos^{-1}0=\bbox[yellow,10px]{90^\circ}$$ From one of the factors above, we let $z=y^2$, and we have the quadratic equation: $$16z^2-20z+6=0\Rightarrow 8z^2-10z+3=0$$ $$(8z-6)(z-\frac12)=0\Rightarrow z=\frac34 \& \ z=\frac12$$ Since $z=y^2$ and $y=\cos x$ we have: $$\biggl( y\rightarrow\pm\frac{\sqrt{3}}{2}, y\rightarrow\pm\frac{\sqrt{2}}2 \biggr)\Rightarrow \biggl(x\rightarrow\cos^{-1}\pm\frac{\sqrt{3}}{2},x\rightarrow\cos^{-1}\pm\frac{\sqrt{2}}2\biggr)$$ And thus the complete set of solution is: $$\bbox[yellow, 5px]{90^\circ, 30^\circ, 150^\circ, 45^\circ, 135^\circ}$$ As I do not have the copy of the answers, I still hope you can verify the accuracy of my solution. ## But more importantly... Seeing the values of $x$, is there a more intuitive and simpler way of finding $x$ that does away with the lengthy computation? • This post doesn't have the squares, but it could give some insight as to how one may think about this problem. – Arthur Mar 12 '18 at 13:12 • This link gives you the solutions. – Jose Arnaldo Bebita-Dris Mar 12 '18 at 13:15 • I expect there to be a much more elegant solution to this as it is an IMO problem. Now only someone has to find it... – vrugtehagel Mar 12 '18 at 13:15 • @vrugtehagel I believe the link to the solution is already an elegant answer! – John Glenn Mar 12 '18 at 13:17 • Yup, I think so too! It was shortly posted before my comment, so I hadn't seen it. I advise @JoseArnaldoBebitaDris to summarize that solution and post it as answer, to avoid this question lingering in the unanswered section of this website – vrugtehagel Mar 12 '18 at 13:21 This is a summary of the solution found in this hyperlink. We can write the LHS as a cubic function of $\cos^2 x$. This means that there are at most three values of $x$ that satisfy the equation. Hence, we look for three values of $x$ that satisfy the equation and produce three distinct $\cos^2 x$. Indeed, we find that $$\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}$$ all satisfy the equation, and produce three different values for $\cos^2 x$, namely $0, \frac{1}{2}, \frac{3}{4}$. Lastly, we solve the resulting equations $$\cos^2 x = 0$$ $$\cos^2 x = \frac{1}{2}$$ $$\cos^2 x = \frac{3}{4}$$ separately. We conclude that our solutions are: $$x=\frac{(2k+1)\pi}{2}, \frac{(2k+1)\pi}{4}, \frac{(6k+1)\pi}{6}, \frac{(6k+5)\pi}{6}, \forall k \in \mathbb{Z}.$$ You can shorten the argument by noting at the outset that $$\cos3x=4\cos^3x-3\cos x=(4\cos^2x-3)\cos x$$ so if we set $y=\cos^2x$ we get the equation $$y+(2y-1)^2+y(4y-3)^2=1$$ When we do the simplifications, we get $$2y(8y^2-10y+3)=0$$ The roots of the quadratic factor are $3/4$ and $1/2$. A different strategy is to note that $\cos x=(e^{ix}+e^{-ix})/2$, so the equation can be rewritten $$e^{2ix}+2+e^{-2ix}+e^{4ix}+2+e^{-4ix}+e^{6ix}+2+e^{-6ix}=4$$ Setting $z=e^{2ix}$ we get $$2+z+z^2+z^3+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}=0$$ or as well $$z^6+z^5+z^4+2z^3+z^2+z+1=0$$ that can be rewritten (noting that $z\ne1$), $$\frac{z^7-1}{z-1}+z^3=0$$ or $z^7+z^4-z^3-1=0$ that can be factored as $$(z^3+1)(z^4-1)=0$$ Hence we get (discarding the spurious root $z=1$) $$2x=\begin{cases} \dfrac{\pi}{3}+2k\pi \\[6px] \pi+2k\pi \\[6px] \dfrac{5\pi}{3}+2k\pi \\[12px] \dfrac{\pi}{2}+2k\pi \\[6px] \pi+2k\pi \\[6px] \dfrac{3\pi}{2}+2k\pi \end{cases} \qquad\text{that is}\qquad x=\begin{cases} \dfrac{\pi}{6}+k\pi \\[6px] \dfrac{\pi}{2}+k\pi \\[6px] \dfrac{5\pi}{6}+k\pi \\[6px] \dfrac{\pi}{4}+k\pi \\[6px] \dfrac{3\pi}{4}+k\pi \end{cases}$$ • Great! An elegant solution too! – John Glenn Mar 12 '18 at 14:42 Hint: $$0=\cos^2x+\cos^22x+\cos^23x-1$$ $$=\cos(3x+x)\cos(3x-x)+\cos^22x$$ $$=\cos2x(\cos4x+\cos2x)$$

https://math.stackexchange.com/questions/2820845/calculation-of-modular-multiplicative-inverse-of-a-mod-b-when-a-b

# Calculation of modular multiplicative inverse of A mod B when A > B I'm trying to understand a Montgomery reduction algorithm, for which I need to calculate a multiplicative inverse. However, euclidean algorithm only helps if A < B. Example is 11 mod 3. Multiplicative inverse of 11 is 2,but ext_gcd gives you Bezout numbers such as -1 and 4. https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm Wikipedia says so: The extended Euclidean algorithm is particularly useful when a and b are coprime, since x is the modular multiplicative inverse of a modulo b, and y is the modular multiplicative inverse of b modulo a. But as far as I see this can't be true, either X is multiplicative inverse of A modulo B or Y is multiplicative inverse of B modulo A, but not both at the same time, because one of them (A or B) is going to be bigger than another. We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not. A lot of online calculators that I tried are also said that a has to be bigger than be, but they (some of them) are still able to calculate inverse of 11 mod 3. The only workaround I found so far is perform A = A mod B first, so A is now a remainder of divisios and therefore is less than modulus, so we can perform ext_gcd(2, 3) now and get our 2 as answer. Probably I'm missing something, this thing is pretty new for me. Thanks. • "We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not. " Huh. $X \equiv A^{-1}\mod B$ because $X*A = 4*3 \equiv 1 \mod 11$. An $Y\equiv B^{-1} \mod A$ because $Y*B=(-1)*11 \equiv 1 \mod 3$. They are both valid inverses. What is your issue? – fleablood Jun 15 at 18:48 • $-1\equiv 2 \mod 3$. So $-1$ is equivalent to $2$. It doesn't make the slightest difference which one you use. (We don't call these things "equivalences" for nothing, you know...) – fleablood Jun 15 at 19:16 • "A lot of online calculators that I tried are also said that a has to be bigger than be," Calculators don't do mathematics. Calculators do calculations. – fleablood Jun 15 at 19:24 It is inevitable that a Bézout's identity equation will give you modular multiplicative inverses, since given: $$am+bn = 1$$ we can take $\bmod m$ for $$bn\equiv 1 \bmod m$$ or $\bmod n$ for $$am \equiv 1 \bmod n$$ To get $a$ and $b$ in your preferred range, you can simply add or subtract a suitable multiple of the modulus. So in this case $$-1\cdot 11 + 4\cdot 3 = 1$$ and thus $$-1\cdot 11\equiv 1 \bmod 3$$ ($-11$ being one more than $-12$), so $-1$ is a valid inverse of $11$ modulo $3$. Then of course $$-1\equiv 2 \bmod 3$$ so this is consistent with your observation that $2$ is the inverse of $11 \bmod 3$ also. • I now understand why this is correct from algorithmical point of view, but I don't understand why −1 ≡ 2 mod 3. I miss some math-related thing. – Danetta Jun 18 at 9:12 • Presumably you have no problem with $11 \equiv 8\equiv 5 \equiv 2 \bmod 3$. In each case, add one and you arrive at a multiple of $3$. The same applies to $-1$. Also, the difference between $-1$ and $2$ is a multiple of $3$ (as is true for any pair of the equivalent values above). – Joffan Jun 18 at 12:52 $-1 \equiv 2 \mod 3$ so they are considered to be the same thing. That's we we call these equivalence classes. However, euclidean algorithm only helps if A < B I simply do not understand why you say that. either X is multiplicative inverse of A modulo B or Y is multiplicative inverse of B modulo A, but not both at the same time, because one of them (A or B) is going to be bigger than another. We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not. Except, of course, it indeed is. $-1*11 = -11 \equiv 1 \mod 3$. That is the valid inverse. Why do you think it is not? It doesn't matter if $A > B$ or $B> A$ as $\gcd(A,B) = 1$ Euclid's algorithm will give us: $mA + kB = 1$ so $k \equiv A^{-1} \mod B$ and $m \equiv B^{-1} \mod A$ simultaneously. Is your concern that one is represented with a positive number and the other negative? That's irrelevent. It doesn't matter which representative we use to represent a class. We could have used $50\equiv -1 \mod 3$ so $50 \equiv 11^{-1}\mod 3$ for all we care. (Indeed $50*11 = 550 = 3*183 + 1 \equiv 1\mod 3$). Note: if $mA + kB = 1$ and $m > 0$ but $-A < k < 0$ then $mA + (k+A)B = 1 + BA\equiv 1 \mod A,B,AB$ and $m$ and $(k+A)$ are still the proper inverses. ANd $m > 0; k+A > 0$. Indeed $(m + vB)A + (k + uA)B = 1 + (v+u)AB$ so $m + vB\equiv A^{-1} \mod B$ for any integer $v$ and $k + uA \equiv B^{-1} \mod A$ for any integer $u$. • "Is your concern that one is represented with a positive number and the other negative?" — Yes, I haven't even considered negative numbers. I'm still not exactly sure how −11 ≡ 1 mod 3. For me logically it looks like quotient of -11/3 is -3, so remainder is either 2 or -2, but how is it 1? – Danetta Jun 18 at 8:56 • It looks like a rule to "just take a remainder from division" stops working there. Do I need positive inverse for Montgomery or it doesn't matter? – Danetta Jun 18 at 9:17 • $-11 = 3*(-4)+ 1$ so $-11 \equiv 1 \mod 3$ and $-11 = 3*(-3) -2$ so $-11 \equiv -2 \mod 3$. And $1 \equiv -2 \mod 3$. As well $-11 = 3*(-7)+12$ so $-11 \equiv 10 \mod 3$. But $-11 \ne 3*k + 2$ for any integer $k$ so $-11 \not\equiv 2 \mod 3$. A remainder means any $p = d*n + r$ and $d$ can be any value positive or a negative and $r$ can be any value that works. But THE remainder would mean $p=d*n+r;0\le r< n$ would mean the remainder is unique, positive and less than the divisor. If $p < 0$ this means $d < 0$ and $r \ge 0$. – fleablood Jun 18 at 15:01 • I dont know how to do the Montgomery method. It seems as though you must already now then inverse to use it and that it is just an computational efficient method to do multiplication. – fleablood Jun 18 at 15:18

https://yutsumura.com/is-the-trace-of-the-transposed-matrix-the-same-as-the-trace-of-the-matrix/

# Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix? ## Problem 633 Let $A$ be an $n \times n$ matrix. Is it true that $\tr ( A^\trans ) = \tr(A)$? If it is true, prove it. If not, give a counterexample. ## Solution. The answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the diagonal entries of the transposed matrix are the same as the original matrix. Putting together these observations yields the equality $\tr ( A^\trans ) = \tr(A)$. Here is the more formal proof. For $A = (a_{i j})_{1 \leq i, j \leq n}$, the transpose $A^{\trans}= (b_{i j})_{1 \leq i, j \leq n}$ is defined by $b_{i j} = a_{j i}$. In particular, notice that $b_{i i} = a_{i i}$ for $1 \leq i \leq n$. And so, $\tr(A^{\trans}) = \sum_{i=1}^n b_{i i} = \sum_{i=1}^n a_{i i} = \tr(A) .$ ### More from my site #### You may also like... ##### The Vector $S^{-1}\mathbf{v}$ is the Coordinate Vector of $\mathbf{v}$ Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$. Note that as the column vectors of $S$... Close

https://mathhelpboards.com/threads/find-volume-of-solid-generated-calc-2.6582/

# Find volume of solid generated (Calc 2) #### lovex25 ##### New member [solved]Find volume of solid generated (Calc 2) Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y=e^x, and the line x = ln 2 about the line x= ln 2. So I tried graphing it to see visually, and the expression I got for calculating the volume was ∫π(ln2-lny)^2dy, evaluating from 0 to 2 using disk method, and the answer I got was 4π, but apparently that doesn't match the answer in the back of the book. I'd really appreciate if someone can help me out!! Off topic: First time posting a thread here, may I ask how do you type the mathematical symbols such as the integral sign and whatnot, or do I have to manually copy and paste from other website? Last edited: #### MarkFL Staff member Hello lovex25, As you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation: Using the shell method, the volume of an arbitrary shell is: $$\displaystyle dV=2\pi rh\,dx$$ where: $$\displaystyle r=\ln(2)-x$$ $$\displaystyle h=e^x$$ and so we find: $$\displaystyle dV=2\pi \left(\ln(2)-x \right)e^x\,dx$$ Now, you want to sum up the shells by integrating: $$\displaystyle V=2\pi\int_0^{\ln(2)}\left(\ln(2)-x \right)e^x\,dx$$ To make things a bit simpler, I would use the substitution: $$\displaystyle u=e^x\,\therefore\,du=e^x\,dx$$ and we now may state: $$\displaystyle V=2\pi\int_1^2\ln(2)-\ln(u)\,du$$ Can you proceed? Now, as far as posting mathematical expressions, this site supports $\LaTeX$, and a good tutorial written by our own Sudharaka can be found here: http://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html #### lovex25 ##### New member Hello lovex25, As you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation: View attachment 1349 Using the shell method, the volume of an arbitrary shell is: $$\displaystyle dV=2\pi rh\,dx$$ where: $$\displaystyle r=\ln(2)-x$$ $$\displaystyle h=e^x$$ and so we find: $$\displaystyle dV=2\pi \left(\ln(2)-x \right)e^x\,dx$$ Now, you want to sum up the shells by integrating: $$\displaystyle V=2\pi\int_0^{\ln(2)}\left(\ln(2)-x \right)e^x\,dx$$ To make things a bit simpler, I would use the substitution: $$\displaystyle u=e^x\,\therefore\,du=e^x\,dx$$ and we now may state: $$\displaystyle V=2\pi\int_1^2\ln(2)-\ln(u)\,du$$ Can you proceed? Now, as far as posting mathematical expressions, this site supports $\LaTeX$, and a good tutorial written by our own Sudharaka can be found here: http://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html thanks so much!!! finally i was able to match the answer to the back of the book, and I realized what went wrong with my attempt: I didn't separate the integrals to 2 different functions. As for the LaTeX software I'll look in to it soon, thanks again! #### Prove It ##### Well-known member MHB Math Helper I personally prefer the discs method. Here you would need to split up your region of integration into two regions, the first being the rectangle below the point (0,1), and the second being the remaining region above it. As for the volume of the region below (0,1) generated when rotating, that's easy, it's simply a cylinder of radius \displaystyle \begin{align*} \ln{(2)} \end{align*} units and height 1 unit, so its volume is \displaystyle \begin{align*} \pi \left[ \ln{(2)} \right] ^2 \end{align*}. As for the volume of the region above (0,1), if we consider horizontal discs, they will each have radius \displaystyle \begin{align*} \ln{(2)} - \ln{(y)} \end{align*} and a height \displaystyle \begin{align*} \Delta y \end{align*}, where \displaystyle \begin{align*} \Delta y \end{align*} is some small change in y. So their total volume can be approximated by \displaystyle \begin{align*} V &\approx \sum \pi \left[ \ln{(2)} - \ln{(y)} \right] ^2 \Delta y \\ &= \pi \sum \left[ \ln{(2)} - \ln{(y)} \right] ^2 \Delta y \\ &= \pi \sum \left\{ \left[ \ln{(2)} \right] ^2 - 2\ln{(2)}\ln{(y)} + \left[ \ln{(y)} \right] ^2 \right\} \Delta y \end{align*} And then as we increase the number of discs, making \displaystyle \begin{align*} \Delta y \end{align*} smaller, the sum converges on an integral and the approximation becomes exact, so the total volume is \displaystyle \begin{align*} V &= \pi \int_1^2{ \left[ \ln{(2)} \right] ^2 - 2\ln{(2)}\ln{(y)} + \left[ \ln{(y)} \right] ^2 \, dy }\end{align*} which is possible to be integrated

https://puzzling.stackexchange.com/questions/104895/the-tip-of-a-colorful-triangle

# The tip of a colorful triangle Original source: Problem 1 of British Informatics Olympiad 2017, Round 1 You're given a bunch of red (R), green (G), and blue (B) balls. I arrange some balls on a line. Then I ask you to complete the triangle of balls under the following simple rules, and tell me the color of the ball at the last row: • Below two balls of the same color, place a ball of that color. (For example, a G must be placed under two G's.) • Below two balls of different colors, place a ball of the third color. (For example, a B must be placed under a R and a G.) If I gave you the balls "R R G B R G B B", you would place the balls like the following, and tell me "it's green": R R G B R G B B R B R G B R B G G B R G G G R G B G B B R R B G R R B G Now, I'll give you a sequence of 60,000 balls. I won't even show you all the balls' colors; the only information available to you is that it starts with 999 balls of the pattern RGB RGB RGB ... RGB, and ends with 999 balls of the pattern BGR BGR BGR ... BGR. Using the same rules, can you guess the color of the ball on the last row? The answer is unique, and an answer with mathematical explanation is preferred. This game is a modified Pascal triangle. First, let's look at size 4. If you were to permute all the possible inputs and look at the first ball given, last ball given, and the color of the ball on the last row, you can see they are also conforming to the rules given. We then try 10. We can look at this like so, since size 4 works: (R)* * B * * B * *(G) * * * * * * * * * * * * * * * * * G * * B * * R * * * * * * * * * * * R * * G * * * * * (B) In fact, this works if the number is $$3^n + 1$$. Alright, let's say the colors red, green, and blue are numbers 0, 1, and 2 respectively. We then visualize a Pascal mod 3 and the game flipped of size 10, with the top portion being one: Pascal mod 3 | The game 1 | 1 1 1 | 2 2 1 2 1 | 1 2 1 1 0 0 1 | 2 0 0 2 1 1 0 1 1 | 1 1 0 1 1 1 2 1 1 2 1 | 2 1 2 2 1 2 1 0 0 2 0 0 1 | 1 0 0 2 0 0 1 1 1 0 2 2 0 1 1 | 2 2 0 1 1 0 2 2 1 2 1 2 1 2 1 2 1 | 1 2 1 2 1 2 1 2 1 1 0 0 0 0 0 0 0 0 1 | 2 0 0 0 0 0 0 0 0 2 This result looks similar - as Pascal's formula (mod 3) is $${{n}\choose{k}} \equiv {{n - 1}\choose{k - 1}} + {{n - 1}\choose{k}} \pmod 3$$ denoting the nth row and kth column. But what about the game? As $$1 + 0 \equiv 2 \pmod 3$$ and $$2 + 0 \equiv 1 \pmod 3$$ for the game, while $$1 + 0 \equiv 1 \pmod 3$$ and $$2 + 0 \equiv 2 \pmod 3$$, we want to switch the terms, and that's simple. We negate this equation. The formula for the game is then: $${{n}\choose{k}} \equiv -\left[{{{n - 1}\choose{k - 1}} + {{n - 1}\choose{k}}}\right] \pmod 3$$ then: $${{n}\choose{k}} \equiv -{{n - 1}\choose{k - 1}} - {{n - 1}\choose{k}} \pmod 3$$ Alright, now let's show why size 4 works. We assign each column in the first row $$a, b, c, d$$, and we will use the above formula. a b c d -a-b -b-c -c-d a+2b+c b+2c+d -a-3b-3c-d We modulo by 3 the expression in the last row and we get $$-a-d \pmod 3$$, and that also conforms to the provided equation above. Because the question is modified, and 60,000 is not $$3^n + 1$$, we must find the highest power of 3 plus 1 that's smaller than this, which is 59050 ($$3^{10} + 1$$). Subtracting 60000 to 59050 gives us 950, the (n+1)th-to-last element (starting from the first ball). We use $$\bmod 3$$ because of the pattern, giving us $$2$$. Now, starting from the first ball which is R, and looking at the 3rd to last, which is B gives us G. Looking at the second ball which is G, and looking at the 2nd to last, which is G gives us G. The third ball is B, and looking at the 1st-to-last, which is R gives us G. Since the next balls are repeating, it will be always periodic, but notice how it's always a G? That must mean the next colors, and the last row is Green! • Yes, it is correct! – Bubbler Nov 16 '20 at 6:15

http://mathhelpforum.com/statistics/45708-probability.html

Math Help - Probability 1. Probability I'm having a lot of trouble with permutations and combinations, so I was hoping someone here could help me out with a few questions. Please make sure that you explain your working; I really want to understand this topic... The first question asks; A queue has 4 boys and 4 girls standing in line. Find how mant different arrangements are possible if; a) The boys and girls alternate. b) 2 particular girls wish to stand together. c) All the boys stand together. d) Also find the probability that 3 particular people will be in the queue together if the queue forms randomly. EDIT: If it's any help, here are the answers... a) 1152 b) 10080 c) 2880 d) 3/28 EDIT: I've got a new problem now; A table has 4 boys and 4 girls sitting around it. a) Find the number of ways of sitting possible if the boys and girls can sit anywhere around the table. b) If the seating is arranged at random, find the probability that; i) 2 particular girls will sit together. ii) All the boys will sit together. 2. Hello, Flay! We have to "talk" our way through these problems . . . A queue has 4 boys and 4 girls standing in line. Find how mant different arrangements are possible if; a) The boys and girls alternate. There are 2 possible arrangements: . $BGBGBGBG\,\text{ and }\,GBGBGBGB.$ The four boys can be placed in 4! ways. The four girls can be placed in 4! ways. Therefore, there are: . $2 \times 4! \times 4! \;=\;\boxed{1152}$ ways. b) 2 particular girls wish to stand together. Suppose the two girls are $X$ and $Y.$ Duct-tape them together. Note that there are 2 possible orders: . $XY\,\text{ or }\,YX.$ Now we have seven "people" to arrange: . $\boxed{XY}\,,G,G,B,B,B,B$ . . and they can be arranged in ${\color{blue}7!}$ ways. Therefore, there are: . $2 \times 71\;=\;\boxed{10,080}$ ways. c) All the boys stand together. Duct-tape the four boys together. . . Note that there are 4! possible orders. Now we have five "people" to arrange. . . There are $5!$ ways. Therefore, there are: . $4! \times 5! \:=\:\boxed{2880}$ ways. d) Find the probability that 3 particular people will be together if the queue forms randomly. First of all, there are 8! possible arrangements. Suppose the three people are $X, Y\text{ and }Z.$ Duct-tape them together: . $\boxed{XYZ}$ . . Note that there are 3! orderings. Now we have six "people" to arrange, . . and there are 6! ways. Hence, there are: . ${\color{red}3!\times6!}$ ways for $X,Y,Z$ to be together. Therefore, the probability is: . $\frac{3!\cdot6!}{8!} \;=\;\boxed{\frac{3}{28}}$ 3. Thanks very much. Now I have another problem; A school committee is to be made up of 5 teachers, 4 students and 3 parents. a) If 12 teachers, 25 students and 7 parents apply to be on the committee, which is chosen at random, how many possible committees could be formed? This part I've figured out. The answer is $350 658 000$. b) If Jan and her mother both apply, find the probability that both will be chosen for the committee. This part I'm not sure how to do. The answer is apparently $\frac{3036}{44275}$. 4. I figured out the last question. As it turns out, $\frac{3036}{44275}$ simplifies to $\frac{12}{175}$, which was the answer I was getting. Now I have a new problem; A sample of 3 coins is taken at random from a bag containing 8 ten cent coins and 8 twenty cent coins. Find the probability that a particular ten cent coin will be chosen, if 1 twenty cent and 2 ten cent coins are chosen.

https://math.stackexchange.com/questions/2788278/if-rho-n1-leq-rho-n-sigma-n-forall-n-geq-n-0-then-prove-that

# If $\rho_{n+1}\leq\rho_{n} +\sigma_n,\;\;\forall\;\;n\geq N_0,$ then prove that $\lim\rho_n=0.$ Good day, all! Suppose $\{\rho_n\}$ and $\{\sigma_n\}$ are two sequences of non-negative real numbers such that for some real number $N_0\geq 1,$ the following recursion inequality holds: $$\rho_{n+1}\leq\rho_{n} +\sigma_n,\;\;\forall\;\;n\geq N_0.$$ Prove that, 1. if $\sum_{n=1}^{\infty}\sigma_n<\infty,$ then $\lim\rho_n$ exists. 2. If $\sum_{n=1}^{\infty}\sigma_n<\infty,$ and $\rho_n$ has a subsequence converging to zero, then $\lim\rho_n=0.$ The first question was asked in my exam but I skipped it. I am stating the truth! Hence, I'll like to get it solved before the final exam. So please, can anyone help me? Thanks in advance. • if the series of $\rho_n$ converges, then the limit of $\rho_n$ has to be zero, there is nothing to prove in 1 and 2. Please double-check the formulation (perhaps you need $\sigma_n$ instead ?) – Hayk May 20 '18 at 5:40 • @Hayk: Sorry, my mistake! It has been corrected! Matt A Pelto: Thanks for the edit! – Omojola Micheal May 20 '18 at 6:11 • Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn May 20 '18 at 7:07 • a real pity about people on this forum that go straight to down voting :) ... such a community. – Matt A Pelto May 20 '18 at 7:47 • @ robjohn: This is not an assignment question. We had the third exam yesterday and the first question, .i.e. question 1, was given to us. However, I'll correct some of my words! – Omojola Micheal May 20 '18 at 7:48 Observe that once we prove assertion $1$, then the claim of $2$ follows readily. This is due to the fact, that if $\rho_n$ is convergent, than all its subsequences must be convergent, all with the same limit. Now, if we prove claim N$1$ and assume in addition that a subsequence of $\rho_n$ converges to $0$, then $\rho_n$, having the same limit as its subsequence, must converge to $0$. We now prove claim $1$. Observe that for any $k\in \mathbb{N}$ and any $n\geq N_0$ we have $$(1) \qquad \rho_{n+k} - \rho_n = \rho_{ n + k } - \rho_{ n + k -1} + \rho_{ n + k - 1 } - \rho_{ n + k -2} +...+\rho_{n+1} - \rho_n \leq \\ \sigma_{n+k-1} +...+\sigma_n,$$ where we apply the given inequality on each term $\rho_{n+k -i} - \rho_{n+k-i-1}$, for $i=0,....,k-1$. From the above inequality, and the fact that $\rho_n \geq 0$, we have that $\rho_n$ is a bounded sequence. Now assume, for contradiction, that it does not converge. Hence, $\rho_n$, converges to different limits along two different subseqeunces. Namely there exists $n_{k}$ and $m_k$, and non negative numbers $a_1$ and $a_2$ such that $$\rho_{n_k} \to a_1 \qquad \text{ and } \qquad \rho_{m_k} \to a_2,$$ where $a_1 \neq a_2$. From $(1)$, let us fix $n\geq N_0$, and choose $k$ such that $n+k = n_k$. Then, taking limit with respect to $k$ and using the fact that the series $\sigma_n$ converges, we get $$(2) \qquad a_1 - \rho_n \leq \sum_{i = n}^\infty \sigma_i.$$ Now take $n\to \infty$ along the second subsequence $m_k$, we obtain $a_1 - a_2 \leq 0$, since convergence of $\sum_i \sigma_i$ implies that the right hand side of $(2)$ converges to $0$. Since the process is symmetric (we could have started with $m_k$ and not $n_k$) we obtain also that $a_2 \leq a_1$ and hence they must be equal. This contradicts our assumption that $a_1 \neq a_2$ and hence the limit of $\rho_n$ exists. • What is wrong with Hayk's solution? Can anyone please, explain? – Omojola Micheal May 20 '18 at 7:49 • @Mike, let's see if the voter will be able to justify it. BTW, the only place you need non-negativity of $\rho_n$ is to get a lower bound on the sequence, as otherwise without bounding $\rho_n$ from below, it can escape to $-\infty$. So, instead of requiring $\rho_n \geq 0$ the claim stays valid also for the case $\rho_n \geq -M$, for any fixed $M\geq 0$ (the same proof works). – Hayk May 20 '18 at 8:49 • I neutralized it for you! – Omojola Micheal May 20 '18 at 15:56 • Can you please, expand this line $\leq \sigma_{n+k-1} +...+\sigma_{n-1}$? Kindly explain why it is so! – Omojola Micheal May 20 '18 at 16:09 • I was thinking it should be $\leq \sigma_{n+k-1} +...+\sigma_{n-1}+\sigma_{n}$ instead! – Omojola Micheal May 20 '18 at 16:19 Put $T_n=\sum_{k=1}^n \sigma_k$. Then $T_n$ is a convergent sequence. Now put $u_n=T_{n-1}-\rho_n$. From the inequality given, we see that $u_n$ is increasing for $n$ large, and of course (as $\rho_n \geq 0$) $\leq T_{n-1}$, hence bounded, hence convergent. It is easy to finish. From $\rho_{n+1}-\rho_n\leq \sigma_n$ we obtain $\rho_{n}- \rho_m \leq \sum_{i=m}^{n-1}\sigma _i \Longrightarrow \rho_{n} \leq \rho_m+\sum_{i=m}^{n-1}\sigma _i$. Taking $\limsup$ in respect of $n$ we obtain $\limsup \rho _n \leq \sum_{i=m}^\infty \sigma _i +\rho _m$. Taking now $\liminf$, we obtain $$\limsup \rho _n \leq \liminf\rho _m$$ So, $\rho_n\longrightarrow x$ where $x \in [-\infty,\infty]$. Since $\rho _n$ is non-negative and upper bounded by $\rho_0 +\sum_{i=1}^\infty \sigma _i$, we conclude that $x\in \mathbb{R}$.

https://math.stackexchange.com/questions/921443/is-there-a-closed-form-expression-for-int-infty-infty-int-inftyy

# Is there a closed form expression for $\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )} \mathrm{d}x\,\mathrm{d}y$? I have been trying to evaluate the integral: $$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y$$ I know of course that the integral equals $1$ over $[-\infty,\infty] \times [-\infty,\infty]$ but I do not quite know how to handle the present case. Are there any tricks here? Thank you. • I think you can exploit the simmetry of the integrand by noting that the line $y=x$ divides the plane in half. – Gennaro Marco Devincenzis Sep 6 '14 at 15:22 Your integral is the probability: $$\mathbb{P}[X\leq Y]$$ where $X$ and $Y$ are two independent normal variables $N(0,1)$, hence the value of the integral is just $\frac{1}{2}$, since: $$\mathbb{P}[X\leq Y]=\mathbb{P}[Y\leq X],\qquad \mathbb{P}[X\leq Y]+\mathbb{P}[Y\leq X]=1.$$ • That makes sense intuitively but could you please explain a little more why that is? There is no closed form then, right? – JohnK Sep 6 '14 at 15:26 • $f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ is the pdf of a $N(0,1)$ random variable, and $\frac{1}{2}$ looks as a very closed form to me. – Jack D'Aurizio Sep 6 '14 at 15:29 • Yes but what allows us to say that the value of that probability is 1/2? – JohnK Sep 6 '14 at 15:31 • Symmetry: $$\mathbb{P}[Y\leq X]=\mathbb{P}[X\leq Y],\quad \mathbb{P}[Y\leq X]+\mathbb{P}[X\leq Y] = 1.$$ – Jack D'Aurizio Sep 6 '14 at 15:32 • Simple yet elegant! Intuitively, it can also be seen by plotting the region $0<x<y<\infty$ since the integrand is even function (i.e. symmetry). +1 – Tunk-Fey Sep 6 '14 at 15:44 Jack D'Aurizio's answer is good, but since you said in comments under it that you wanted a different point of view, let's try this: \begin{align} u & = (\cos45^\circ)x-(\sin45^\circ)y = \tfrac{\sqrt{2}}2 x - \tfrac{\sqrt{2}}2 y \\ v & = (\sin45^\circ)x+(\cos45^\circ)y = \tfrac{\sqrt{2}}2 x + \tfrac{\sqrt{2}}2 y \end{align} This is just a $45^\circ$ rotation of the coordinate system, suggested by the fact that your boundary line $y=x$ is just a $45^\circ$ rotation of one of the coordinate axes. Then simplify $u^2+v^2$ and find that it comes down to $x^2+y^2$. Solving the system of two equations above for $x$ and $y$, one gets \begin{align} x & = \phantom{-}\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v \\ y & = -\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v \end{align} By trivial algebra, the condition that $x\le y$ now becomes $u\le0$. If you know about Jacobians, you get $$du\,dv = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,dx\,dy = \left|\frac{\partial u}{\partial x}\cdot\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\cdot\frac{\partial v}{\partial x}\right|\,dx\,dy = 1\,dx\,dy.$$ Hence your iterated integral becomes $$\int_{-\infty}^\infty\int_{-\infty}^0 \frac 1{2\pi} e^{-(u^2+v^2)/2}\,du\,dv = \int_{-\infty}^\infty \int_{-\infty}^0 \left\{\frac 1{2\pi} e^{-v^2/2}\right\} e^{-u^2/2}\,du\,dv$$ The part in $\{\text{braces}\}$ does not depend on $u$, so this is $$\int_{-\infty}^\infty\left( \frac 1{2\pi} e^{-v^2/2} \int_{-\infty}^0 e^{-u^2/2}\,du \right)\,dv.$$ Now the inside integral does not depend on $v$, so it pulls out: $$\int_{-\infty}^\infty e^{-v^2/2}\,dv \cdot \frac1{2\pi} \int_{-\infty}^0 e^{-u^2/2}\,du$$ and this is of course $$\int_{-\infty}^\infty \frac1{\sqrt{2\pi}} e^{-v^2/2}\,dv \cdot \int_{-\infty}^0 \frac1{\sqrt{2\pi}} e^{-u^2/2}\,du.$$ The first integral comes to $1$ and the second, by a simple symmetry argument, is $1/2$. • Thank you very much. I'll just have to understand how come that substitution rotates the coordinate system. – JohnK Sep 6 '14 at 16:48 • nice result! you may simplify your answer by showing that $\int_{-\infty}^0 e^{-u^2}\,du=(1/2)\int_{-\infty}^{\infty} e^{-u^2}\,du$ as early in your answer as possible. – mike Sep 6 '14 at 16:52 • Think of what that substitution does to $(x,y)=(1,0)$ and to $(x,y)=(0,1)$ and then you'll be well on your way to understanding why it's a rotation. But even if you didn't know it's a rotation, you should still be able to follow the argument. – Michael Hardy Sep 6 '14 at 16:52 • . . . however, it is also the case that if you know nothing of Jacobians, you can understand that $du\,dv=dx\,dy$ by using the fact that the transformation is a rotation. Rotations multiply volumes by $1$, so the thing to multiply $dx\,dy$ by to get $du\,dv$ is $1$. ${}\qquad{}$ – Michael Hardy Sep 6 '14 at 16:54 • The intuition behind Jacobians is often not made explicit in second-year calculus courses where the concept is introduced. But if the value of the Jacobian $|\partial(u,v)/\partial(x,y)|$ at a particular point is, for example $6$, that means that at that point an infinitesimal area in the $x,y$ plane is transformed to an infinitesimal area $6$ times as big in the $u,v$ plane. Determinants generally area what you multiply by a volume to get the volume it is transformed to. And the determinant is negative if the orientation gets reversed. – Michael Hardy Sep 6 '14 at 17:03 Set $x=r\cos \theta,y=r\sin \theta$, then we have $$\int_{- \infty}^\infty \int_{-\infty}^y e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y=\int_{0}^\infty \left(\int_{-3\pi/4}^{\pi/4} e^{-r^2/2}\mathrm {d}\theta\,\right)r\,\mathrm{d}r=\pi \int_{0}^\infty e^{-r^2/2}r\,\mathrm{d}r=\pi$$ So the original integral is equal to $(1/2)$. This method as well as @MichaelHardy's method also works for the integrals like: $$\int_{- \infty}^\infty \int_{-\infty}^{a y} e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y, \text{ }a \in \mathbb{R}$$ The results are the same. This is because the function to be integrated ($e^{-r^2/2}$) is rotational invariant (independent of $\theta$) and $x=a y$ is a straight line going through the origin and divides the plane into 2 halfs of equal size. • Glad to see polar coordinates work too, thank you. – JohnK Sep 6 '14 at 18:27 Ah, I realise just now I'd misread to start with. Algebraically, we can solve it by noting that the value of the integral is the same under the change of variables $u=-x$, and since summing the two resulting integrals results in the $[-\infty,\infty] \times [-\infty,\infty]$ case, the answer is $1/2$. i.e.: $$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)} \,\mathrm{d}x\,\mathrm{d}y = \int_{- \infty}^\infty \int_{-y}^\infty \frac{1}{2 \pi} e^{-1/2 \left( x^2+y^2 \right)}\,\mathrm{d}x\,\mathrm{d}y.$$ $LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y=1$, and $LHS=RHS$, so your integral is $1/2$. EDIT: I guess I was a little bit short. There are several ways of showing $LHS+RHS$ is what I quote, but here is a simple, uninformative one. $LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y-\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$. Now, Let $I=\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$. Use substitution $u=-y$. Then $I=-\int_{\infty}^{-\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=\int_{-\infty}^{\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=-I$, hence as $I=-I$, $I=0$. I think it could also be done by splitting the integration range up. (e.g. looking at the integrals on $(-\infty,0]$ and $[0,\infty)$) • If you do not mind how is that LHS+RHS equal that integral? – JohnK Sep 6 '14 at 16:09 • @JohnK there you go :) – ShakesBeer Sep 7 '14 at 7:49

http://mathhelpforum.com/discrete-math/125642-how-prove-logical-equivalence-using-theorems-substitution.html

# Math Help - How to prove logical equivalence using Theorems and substitution? 1. ## How to prove logical equivalence using Theorems and substitution? So, I have the statement (p ^ q) \/ (p ^ ¬ q) ≡ p I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution. cheers -ipatch 2. Originally Posted by ipatch So, I have the statement (p ^ q) \/ (p ^ ¬ q) ≡ p I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution. cheers -ipatch Just read it as a sentence: "p and q" or "p and not q". It's saying you have to have p but you may or may not have q. So what it's the same as having? p. 3. I have to prove the given logical equivalence using Theorems and not constructing a truth table The solution depends on what theorems you have in mind. Every course may have a different set of those. 4. Hello, ipatch! I don't know the names of the theorems you know. I hope you can follow my proof. Prove: . $(p \;\wedge\; q) \;\vee\; (p\; \wedge \sim q) \;\;\equiv \;\; p$ . . . . . . . . . $(p \;\wedge\; q) \;\vee\; (p\; \wedge\; \sim q)$ . . $(p \;\vee\; p) \;\wedge\; (p \;\vee\; \sim q) \;\wedge\; (q \;\vee\; p) \;\wedge\; (q \;\vee\; \sim q)$ . . Distributive . . . . . . $p\;\wedge\;(p\;\vee\;\sim q)\;\wedge\;(p\;\vee\; q) \;\wedge\; T$ . . . . . . . . . $(p\;\vee\;\sim q)\;\wedge\; (p\;\vee q)$ . . . . . . . . . . . $p\;\vee(\sim q\;\wedge\; q)$ . . . . . . . . . . . . Distributive . . . . . . . . . . . . . $p\;\vee\;F$ . . . . . . . . . . . . . . . $p$ 5. Hello ipatch Originally Posted by ipatch So, I have the statement (p ^ q) \/ (p ^ ¬ q) ≡ p I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution. cheers -ipatch Using the standard Laws of Boolean Algebra (for instance, just here) we get: $(p \land q)\lor(p\land\neg q) \equiv p\land(q\lor\neg q)$ (Distributive Law) $\equiv p\land T$ (Complement Law) $\equiv p$ (Identity Law) 6. (p ^ q) \/ (p ^ ¬ q) ≡ p 1. (p ^ q) \/ (p ^ ¬ q)-----------------hypotesis 2. (p ^ q)-----------------assume 3. p-----------------2, ^-elim 2 4. (p ^ ¬ q)-----------------assume 5. p-----------------4, ^-elim 2 6. p-----------------1, 2-3 , 4-5, \/-elim 7. Hello again, ipatch! My first proof was long and clumsy. (Don't know where my brain was at the time.) There are two theorems we need. .(I don't know their names.) . . $\begin{array}{cccc}P \;\vee \sim \!P &\equiv& t & \text{Theorem 1} \\ \\[-3mm] P \wedge \:t &\equiv& P & \text{Theorem 2} \end{array}$ . . . . $\begin{array}{ccc}(p \;\wedge\; q) \;\vee\; (p\; \wedge\; \sim q) & \text{Given} \\ \\ p \;\wedge\; (q \;\vee \sim \!q) & \text{Distr.} \\ \\ p \;\wedge \;t & \text{Theorem 1} \\ \\ p & \text{Theorem 2} \end{array}$ 8. Originally Posted by cgafa (p ^ q) \/ (p ^ ¬ q) ≡ p 1. (p ^ q) \/ (p ^ ¬ q)-----------------hypotesis 2. (p ^ q)-----------------assume 3. p-----------------2, ^-elim 2 4. (p ^ ¬ q)-----------------assume 5. p-----------------4, ^-elim 2 6. p-----------------1, 2-3 , 4-5, \/-elim Rules used: conjunctive elimination ( ^-elim ) if you have A and B the you have A (or B)... simple disjuntive elimination (\/-elim)... a little more complex... if you have A or B and from A you can deduce C; and from B you can deduce C; then from A or B we can deduce C..... 9. Originally Posted by ipatch So, I have the statement (p ^ q) \/ (p ^ ¬ q) ≡ p I have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution. cheers -ipatch ipatch: There two kind of proofs,one in Boolean Algebra and one in propositional calculus. The proof in Boolean Algebra was shown by other people in the forum. The proof in propositional calculus is as follows: First we prove : $(p\wedge q)\vee (p\wedge\neg q)\Longrightarrow p$ and then: $p\Longrightarrow (p\wedge q)\vee (p\wedge\neg q)$. Proof: 1) $(p\wedge q)\vee (p\wedge\neg q)$.................................................. ............................given 2) $(p\wedge q)$.................................................. ..............................................assu mption to start a conditional proof 3) p................................................. .................................................. .....from (2) and using Conjunction Elimination (=C.E) 4) $(p\wedge q)\Longrightarrow p$.................................................. ....................................from (2) to (3) and using the rule of conditional proof 5) In the same way we prove: $(p\wedge\neg q)\Longrightarrow p$ 6) $(p\wedge q)\vee (p\wedge\neg q)\Longrightarrow p\vee p$.................................................. ...........................from (4)and(5) and using the rule called proof by cases: $[((A\Longrightarrow B)\wedge (C\Longrightarrow D))\Longrightarrow ((A\vee C)\Longrightarrow (B\vee D))]$. 7) $p\vee p$.................................................. .......................................from (1) and (6) and using M.Ponens 8) $p\vee p\Longleftrightarrow p$.................................................. .......................................tautology 9) $p\vee p\Longrightarrow p$.................................................. ........................................from (8) and using biconditional elimination 10) p................................................. .................................................. ....from (7) and (9) and using M.Ponens Now to prove the converse. Proof: 1)p............................................... .................................................. given 2) $\neg(p\wedge q)$.................................................. ........................................assumption to start a conditional proof 3) $\neg p\vee\neg q$.................................................. ..........................................from (2) and using De Morgan 4) $p\Longrightarrow\neg q$.................................................. .....................from (3) and using material implication 5) $\neg q$.................................................. .....................................from (1) and (4) and using M.Ponens 6) $p\wedge\neg q$.................................................. ...........................from (1) and (5) and using Conjunction Introduction 7) $\neg(p\wedge q)\Longrightarrow(p\wedge\neg q)$.................................................. .......................from (2) to(6) and using the rule of conditional proof 8) $(p\wedge q)\vee (p\wedge\neg q)$.................................................. ........................from (7) and using material implication $(A\Longrightarrow B)\Longleftrightarrow (\neg A\vee B)$

https://math.stackexchange.com/questions/2730864/vector-whose-inner-product-is-positive-with-every-vector-in-given-basis-of-mat

# Vector whose inner product is positive with every vector in given basis of $\mathbb{R}^n$ I am trying to solve the following question which I came across when studying root system in euclidean spaces, with positive definite symmetric bilinear form. Statement: Given a basis $\{v_1,\cdots, v_n\}$ of $\mathbb{R}^n$, $\exists$ $v\in\mathbb{R}^n$ such that $(v,v_i)>0$ for all $i$ My answer: Let $v=a_1v_1 + \cdots + a_nv_n$. Then $$(v,v_i)=a_1(v_1,v_i) + a_2(v_2,v_i) + \cdots + a_n(v_n,v_i), \hskip5mm i=1,2,\cdots,n.$$ Writing this in matrix form we get $$\begin{bmatrix} (v,v_1) \\ \vdots \\ (v,v_n)\end{bmatrix} = \begin{bmatrix} (v_1,v_1) & \cdots & (v_n,v_1)\\ \vdots & & \vdots \\ (v_1,v_n) & \cdots & (v_n,v_n) \end{bmatrix} \begin{bmatrix} a_1 \\ \vdots \\ a_n\end{bmatrix}.$$ We want to find $a_1,\cdots,a_n$ such that the left column vector contains positive entries only. Let $A$ be the square matrix appearing above; it is matrix of positive definite (symmetric) inner product w.r.t given basis, so it is invertible over $\mathbb{R}$. Take $$\begin{bmatrix} a_1 \\ \vdots \\ a_n\end{bmatrix} := A^{-1}\begin{bmatrix} 1 \\ \vdots \\ 1\end{bmatrix}.$$ This is non-zero vector, and for this choice of $a_i$'s we get the vector $v=a_1v_1+\cdots + a_nv_n$ with desired properties. Q.1 Is the above statement and proof correct? Q.2 Is there geometric way to prove the statement? • How do you know that $A$ is invertible? – 5xum Apr 10 '18 at 12:43 • it is matrix of non-degenerate bilinear form (positive definite inner product should be non-degenerate, isn't it?) – Beginner Apr 10 '18 at 12:44 • That's one way of showing it, sure. – 5xum Apr 10 '18 at 12:45 For a geometric proof, just do something akin to Gram-Schmidt orthogonalisation. Take $v=v_1$, so that $(v,v_1)>0$. Let $u_2=v_2-\frac{(v_1,v_2)}{(v_1,v_1)}v_1$ be the component of $v_2$ that is orthogonal to $v_1$. Then $u_2\ne0$, or else $v_1$ and $v_2$ are linearly dependent. Add $c_2u_2$ to $v$ for a sufficiently large $c_2>0$. Then $(v,v_i)>0$ for $i=1,2$. Continue in this manner, you get a desired $v$.

http://mathhelpforum.com/math-topics/37930-problem-5-a.html

# Math Help - Problem 5 1. ## Problem FIVE Problem 5 TRIPOLLO TRIPOLLO is a strategy game in which numbered marbles are placed in a rectangular array of holes on a board. As shown, there are three rows of holes. A player wins if the marbles are placed so that in each column, the sum of the numbers in the first two rows is three times the number in the third row. (a) John has a 5-column board and marbles numbered 1 to 15. So far he has placed them as shown in the diagram. Show how he could successfully place the remaining marbles. (b) With the same board and marbles. John made the following start: Can he successfully complete this game? Give reasons. (c) You have an 8-column board and marbles numbered 1 to 24. Show how you can win by placing all the marbles on the board. (d) Show that with an 11-column board and marbles numbered 1 to 33 it is not possible to find a winning placement. Explain the problem in as much detail as you can. 2. Originally Posted by MrFantasy Problem 5 TRIPOLLO TRIPOLLO is a strategy game in which numbered marbles are placed in a rectangular array of holes on a board. As shown, there are three rows of holes. A player wins if the marbles are placed so that in each column, the sum of the numbers in the first two rows is three times the number in the third row. (a) John has a 5-column board and marbles numbered 1 to 15. So far he has placed them as shown in the diagram. Show how he could successfully place the remaining marbles. (b) With the same board and marbles. John made the following start: Can he successfully complete this game? Give reasons. (c) You have an 8-column board and marbles numbered 1 to 24. Show how you can win by placing all the marbles on the board. (d) Show that with an 11-column board and marbles numbered 1 to 33 it is not possible to find a winning placement. Explain the problem in as much detail as you can. Is this a Mathematics Challenge question? If so, it is not approporiate to ask this question until the competition is ended. 3. Hello Mr.Fantastic No, this is not a question from a maths competition. it's 1 of the 6 questions I picked out from the monthly issued maths challenge. My previous posts will explain the purpose of these 6 questions. It is to provide maths lovers some challenging questions to do. Please refer to my previous post for more information. http://www.mathhelpforum.com/math-he...intrested.html MrFantasy 4. Originally Posted by MrFantasy Hello Mr.Fantastic No, this is not a question from a maths competition. it's 1 of the 6 questions I picked out from the monthly issued maths challenge. My previous posts will explain the purpose of these 6 questions. It is to provide maths lovers some challenging questions to do. Please refer to my previous post for more information. http://www.mathhelpforum.com/math-he...intrested.html MrFantasy Yes, I've seen that thread. I just find it odd that the same questions are being posted by other members too. 5. It may very well be that this question/problem is quiet wide known and is used in different occasions, whoever posted the same problem as I have posted here, could be another occasion. And please, if you dont mind. We're taking too much space talking in here, it would be difficult for members to read the answer replies. If there is any questions or replies or issues you want to make please contact me through PM. As it is more easier that way. Thank You, Mr`Fantasy 6. Originally Posted by mr fantastic Is this a Mathematics Challenge question? If so, it is not approporiate to ask this question until the competition is ended. This question is NOT from the Australian Maths Challenge Competition. The member has posted it in good faith. Please feel free to reply. Apologies for any embarassment caused - I hope you understand that all members need to be vigilant for the sort of cheating that does sometimes occur.

http://mathematica.stackexchange.com/questions/9304/plotting-contours-plot-for-fx-y-z-c/9310

# Plotting contours plot for f(x,y,z)=c I have the following question: I have a file that has structure: x1 y1 z1 f1 x2 y2 z2 f2 ... xn yn zn fn I can easily visualize it with Mathematica using ListContourPlot3D. But could you please tell me how I can plot contour plot for this surface? I mean with these data I have a set of surfaces corresponding to different isovalues (f) and I want to plot intersection between all these surfaces and some certain plane. I tried to Google but didn't get any results. Any help and suggestions are really appreciated. Thanks in advance! - Would you please include a sample of your data, or provide a function that generates data that can serve as an example? –  Mr.Wizard Aug 10 '12 at 21:47 Also, do you want a 2D contour plot, or a 3D curve through space? –  Mr.Wizard Aug 10 '12 at 22:00 The answers to this question might be useful to you. –  Ｊ. Ｍ. Aug 11 '12 at 3:53 Thanks for all your replies! Sorry for the delay, I had problems with an internet acces. But the problem is when I'm trying to use suggestions by halirutan I get empty box (and I need exactly this plot). I'd love to provide the sample of my data but it's big, it has 2744 lines and I don't know how to upload this file here (any suggestions?). Thanks! PS Original questions was asked by me. I do not know how to comment other answers (I don't see button add commment) –  Nikita Aug 13 '12 at 20:50 After merging your accounts you should now be able to comment. –  Sjoerd C. de Vries Aug 13 '12 at 22:45 show 1 more comment Ok, lets give this a try. @Mr.Wizard already showed you, how you can use Interpolate to make a function from your discrete data and since you didn't provide some test-data, I just assume we are speaking of an isosurface of a function $f(x,y,z)=c$ which is defined in some box in $\mathbb{R}^3$. For testing we use $$f(x,y,z) = x^3+y^2-z^2\;\;\mathrm{and}\;\; -2\leq x,y,z \leq 2$$ which accidently happens to be the first example of ContourPlot3D. The idea behind the following is pretty easy: As you may know from school, there is a simple representation of a plane in 3d which uses a point vector $p_0$ and two direction vectors $v_1$ and $v_2$. Every point on this plane can be reached through the $(s,t)$ parametrization $$p(s,t)=p_0+s\cdot v_1+t\cdot v_2$$ Please note that $p_0, p, v_1, v_2$ are vectors in 3d and $s,t$ are scalars. The other form which we will use only for illustration is called the normal form of a plane. It is give by $$n\cdot (p-p_0)=0$$ where $n$ is the vector normal to the plane, which can easily be calculated with the cross-product by $v_1\times v_2$. Lets start by looking at our example. To draw the plane inside ContourPlot3D we use the normal form which is plane2: f[{x_, y_, z_}] := x^3 + y^2 - z^2; v1 = {1, 1, 0}; v2 = {0, 0, 1}; p0 = {0, 0, 0}; plane1 = p0 + s*v1 + t*v2; plane2 = Cross[v1, v2].({x, y, z} - p0); gr3d = ContourPlot3D[{f[{x, y, z}], plane2}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, Contours -> {0}, ContourStyle -> {ColorData[22, 3], Directive[Opacity[0.5], ColorData[22, 4]]}] What we do now is, that we try to find the contour value (which is 0 here) of $f(x,y,z)$ for all points, that lie on our plane. This is like doing a normal ContourPlot because our plane is 2d (although placed in 3d space). Therefore, we use the 2d to 3d mapping of plane1 gr2d = ContourPlot[f[plane1], {s, -2, 2}, {t, -2, 2}, Contours -> {0}, ContourShading -> None, ContourStyle -> {ColorData[22, 1], Thick}] Look at the intersection. It is exactly the loop we would have expected from the 3d illustration. Now you could object something like "ew.. but I really like a curve in 3d..". Again, the mapping from this 2d curve to 3d is given in the plane equation. You can simply extract the Line[..] directives from the above plot and transfer it back to 3d: Show[{gr3d, Graphics3D[{Red, Cases[Normal[gr2d], Line[__], Infinity] /. Line[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}] }] I extract the Lines with Cases and then use the exact same definition of plane1 as pure function to transform the pts. When I'm not completely wasted at 5:41 in the morning, than this approach should work for your interpolated data too. ## Apply method on test-data I uploaded your test-data to our Git-repository and therefore, the code below should work without downloading anything. The approach is the same as above but some small things have changed, since we work on interpolated data now. I'll explain only the differences. First we import the data and since we have a long list of {x,y,z,f} values, we transform them to {{x,y,z},f} as required by the Interpolation function. I'm not using the interpolation-function directly. I wrap a kind of protection around it which tests whether a given {x,y,z} is numeric and lies inside the interpolation box. Otherwise I just return 0. data = {Most[#], Last[#]} & /@ Import["https://raw.github.com/stackmma/Attachments/master/data_9304_187.m"]; ip = Interpolation[data]; fip[{x_?NumericQ, y_?NumericQ, z_?NumericQ}] := If[Apply[And, #2[[1]] < #1 < #2[[2]] & @@@ Transpose[{{x, y, z}, First[ip]}]], ip[x, y, z], 0.0] The code below is almost the same. I only adapted the plane that it goes through your interpolation box. Furthermore, if you inspect your data you see that the value run from 0 to 1.2. Therefore I'm plotting the 0.5 contour by subtracting 0.5 from the function value and using Contours -> {0}. Remember that when I would simply plot the 0.5 contour, it would draw me a different plane, since we use one combined ContourPlot3D call. Furthermore, notice that I normalized the direction vectors of the plane. This makes it easier to adjust the 2d plot of the contour. The rest should be the same. v1 = Normalize[{30, 30, 0}]; v2 = Normalize[{0, 0, 21}]; p0 = {26, 26, 17}; plane1 = p0 + s*v1 + t*v2; plane2 = Cross[v1, v2].({x, y, z} - p0); gr3d = ContourPlot3D[{fip[{x, y, z}] - 0.5, plane2}, {x, 27, 30}, {y, 27, 30}, {z, 17.3, 21}, Contours -> {0}, ContourStyle -> {Directive[Opacity[.5], ColorData[22, 3]], Directive[Opacity[.8], ColorData[22, 5]]}] gr2d = ContourPlot[fip[plane1] - 0.5, {s, 2, 5}, {t, 1, 4}, Contours -> {0}, ContourShading -> None, ContourStyle -> {ColorData[22, 1], Thick}]; Show[{gr3d, Graphics3D[{Red, Cases[Normal[gr2d], Line[__], Infinity] /. Line[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}]}] As you can see, your sphere has a whole inside. - Thanks for all your replies! Sorry for the delay, I had problems with an internet acces. But the problem is when I'm trying to use suggestions by halirutan I get empty box (and I need exactly this plot). I'd love to provide the sample of my data but it's big, it has 2744 lines and I don't know how to upload this file here (any suggestions?). Thanks! PS Original questions was asked by me. I do not know how to comment other answers (I don't see button add commment) –  Nikita Aug 13 '12 at 20:50 thanks for solving account issue! So my question if it is possible to load file on this forum or how I can upload my data which are quite big (app. 3000 lines) –  Nikita Aug 13 '12 at 22:51 @Nikita I asked in the chat and there seems to be no standard-upload place. Oleksandr suggested ge.tt. Before uploading a very large file, please read the FAQ and remember that the questions on this site should be helpful for everyone and should not be too localized! –  halirutan Aug 14 '12 at 1:14 here is a link (depositfiles.com/files/r0pckyaiu) to zip file that contains 2 absolutely identical data (one is in format: x y z f, and the other is in the same format but can be directly load to mathematica). Yes, I understand that questions shouldn't be too localized, but the data I'm working on are cube files and it can be useful for many people. Thanks! –  Nikita Aug 14 '12 at 14:31 @Nikita Here we go. See my update. –  halirutan Aug 14 '12 at 15:56 show 1 more comment You can use the options MeshFunctions in combination with Mesh for this. I'm borrowing Mr.Wizard's data here for a moment: data = Flatten[Table[{x, y, z, x^2 + y^2 - z^2}, {x, -2, 2, 0.2}, {y, -2, 2, 0.2}, {z, -2, 2, 0.2}], 2]; Suppose you want to plot the intersection of the contours of data with the plane x - y == 0, then you could do something like ListContourPlot3D[data, Contours -> {0.5, 2}, ContourStyle -> Opacity[0.3], BoundaryStyle -> Opacity[0.3], MeshFunctions -> {(#1 - #2) &}, Mesh -> {{0}}, MeshStyle -> {Thick, Orange}] - You removed my glorious infix! +1 for the method however. (It's nice to see you posting again.) –  Mr.Wizard Aug 11 '12 at 7:00 thanks for your reply! These tricks with meshfunctions are brilliant, but when I use ListContourPlot3D on my data (which are exactly in the same format) I have empty box. But when I apply ListContourPlot3D on data that have only f values (without x,y,z) Mathematica plots correct figures but I cannot use MeshFunction in this case (because there is no information about x,y,z values). Do you know what can solve this problem? –  Nikita Aug 13 '12 at 22:58 I'm not claiming this is a good method, I'm just getting some ink the page: data = Table[{x, y, z, x^2 + y^2 - z^2}, {x, -2, 2, 0.2}, {y, -2, 2, 0.2}, {z, -2, 2, 0.2}] ~Flatten~ 2; ListContourPlot3D[data, Contours -> {0.5, 2}, Mesh -> None] int = Interpolation[data]; ContourPlot3D[int[x, y, z], {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, Contours -> {0.5, 2}, RegionFunction -> (-0.02 < #2 - # < 0.02 &)] - The second piece of codes using RegionFunction doesn't creat a pretty plot,which contains two outer curves with small intervals, and neither PlotPoints nor MaxRecursion could help to improve it. –  withparadox2 Aug 11 '12 at 7:12 @paradox2 hence: "I'm not claiming this is a good method ..." –  Mr.Wizard Aug 11 '12 at 7:13

https://mathoverflow.net/questions/283427/are-random-circulant-matrices-almost-orthonormal

# Are random circulant matrices almost orthonormal? Let $(X_1, X_2, \dots, X_n)$ be i.i.d. ${\cal N}(0,1)$. We construct a random circulant matrix $M$: $$M = \frac{1}{\sqrt n}\begin{pmatrix}X_1 &X_2 &X_3 \dots &X_n\\ X_n &X_1 & X_2 \dots &X_{n-1}\\ \vdots &\vdots &\vdots &\vdots\\X_2 &X_3 &X_4 \dots &X_1\end{pmatrix}.$$ My questions are the following: 1. Is $M$ "almost orthonormal" in a precise probabilistic sense? 2. Related to above, is it possible to upper-bound the largest possible dot product between any two rows of $M$ by a suitably small $\epsilon_n$? Note that this says that $MM^T$ is close to the identity matrix $I_{n \times n}$, as we are bounding the off-diagonal entries of $MM^T$ by $\epsilon_n$, while the diagonal entries are close to $1$. • What do you mean by "upper-bound the largest possible dot product"? All of the $X_i$ could be equal to $10^{10^{10}}.$ You can upper bound it with high probability, but that's the best you can do. – Igor Rivin Oct 14 '17 at 0:05 • If $D$ is the absolute value of the largest dot product, one could seek a bound of the sort $\mathbb P(D > \epsilon_n) < \epsilon_n$ for a suitable choice of $\epsilon_n \to 0$. – VSJ Oct 14 '17 at 2:25 • Yes, that follows from Marcel's answer, using, e.g., Markoff's inequality. – Igor Rivin Oct 14 '17 at 3:24 The diagonal elements of $P=\frac{1}{N}MM^T$, like $$P_{11}=\frac{1}{N}\sum_{i=1}^NX_i^2,$$ satisfy $\langle P_{11}\rangle=1$ and $\langle P_{11}^2\rangle=1+2/N$ (variance decreases like $N^{-1}$). On the other hand, off-diagonal elements like $$P_{12}=\frac{1}{N}\sum_{i=1}^{N}X_iX_{i+1}$$ satisfy $\langle P_{12}\rangle=0$ and $\langle P_{12}^2\rangle=1/N$ (variance also decreases like $N^{-1}$). In this sense I would say $P$ is close to the identity. Marcel's solution provides a good approach for understanding the marginal statistics. Here are a few supplementary comments that might give a little insight into the joint distribution. Let $D$ be the discrete Fourier transform matrix, i.e. the $j,k$-th entry is: $$D_{j,k}=e^{-2\pi i jk/N}/\sqrt{N}$$ Consider the discrete Fourier transform of the first row of $M$, i.e. $$(G_1,...,G_N)=(1/\sqrt{N})(X_1,...,X_N)D$$ Let $G$ be a diagonal matrix with diagonal entries $G_1,...,G_N$. Then $D$ diagonalizes $M$ (see, e.g., this description), that is: $$M = D G D^{-1}$$ If we took the $X_i$ to be complex-valued Gaussian variables, then we would be essentially done at this point: since $D$ is unitary, and the Gaussian is spherically symmetric, then the $G_i$ would be i.i.d. (complex) Gaussian random variables with mean 0 and variance $1/n$. (That is, if we sampled the $G_i$ as i.i.d complex $\mathcal{N}(0,1/n)$, then $DGD^{-1}$ has the same distribution as samples from the original circulant matrix.) It follows that $$MM^*=(DGD^{-1})(DG^*D^{-1})=DGG^{-1}D^*$$ It then follows that the eigenvalues of $MM^{*}$ (or, if you prefer, the squared singular values of $M$) are distributed like $n$ i.i.d draws from a rescaled $\chi^2$ distribution with 2 degrees of freedom (which happens to simplify to an exponential distribution), where we rescale by dividing by $n$. From an eigenvalue perspective, this is a complete characterization of the "orthonormality" of $M$. However, your $X_i$ are probably real-valued. This causes a small book-keeping headache, but doesn't really change much. In brief: since the $X_i$ are real-valued, the $G_i$ will be symmetric. We can convert our $n\times n$ complex-valued matrices into corresponding $2n \times 2n$ real-valued matrices (say, $D'$ and $G'$). Then observe that there are only $n$ of the $2n$ diagonal elements in $G$ are free, and only $n$ of the $2n$ inputs to $D'$ are non-zero (namely, the ones corresponding to the real values of the inputs in $D$). Noting that the resulting decimated matrices are still unitary, we then reduce to the previous case, except with $\chi^2/n$ distributions with only 1 degree of freedom. • I'm confused about one part: Marcel's answer indicates that $MM^*$ is close to identity. Thus it should have eigenvalues close to 1? But your argument says eigenvalues are iid from a chi square distribution so they aren't all close to 1. How to reconcile these? – VSJ Nov 2 '17 at 4:04 • Ah, @VSJ , that is my mistake: I forgot to pass through the prefactor of $1/\sqrt(n)$ in the original definition of $M$. I will fix that now, at which point I think Marcel's answer is consistent with mine. – Bill Bradley Nov 3 '17 at 13:31 • Hi Bill, I'm not sure that is a mistake. Because even then, the eigenvalues will converge to 0 and not 1. Perhaps it is the case that the matrix is tending "pointwise" to identity, but it is not tending fast enough, and therefore the eigenvalues don't converge? – VSJ Nov 3 '17 at 17:52

https://math.stackexchange.com/questions/1496708/another-way-of-counting-probability

# Another way of counting probability A set $S = \{1, 2, \cdots, k\}$ is given. Two persons independently choose some numbers from this set. I want to count the probability that the cardinality of intersection of the chosen sets by both persons is exactly one. Each person has to choose at least one number. A person can choose a number which is already chosen by another person. A number is being chosen independently and uniformly at random from the set. My approach: Fix one number from $1$ to $k$. Probability of both persons choosing that number is $\frac{1}{k^2}$. Remaining $k-1$ numbers can be chosen by either of both or by none of both. So, probability for that is $\left(1 - \frac{1}{k^2}\right)^{k-1}$. And that fixed number can be selected in ${k \choose 1}$ ways. So, probability of required event is: \begin{align} P(E) = {k \choose 1} \frac{1}{k^2} \left(1 - \frac{1}{k^2}\right)^{k-1} \end{align} My questions: 1. Is this reasoning correct? 2. Initially I started with counting all the possible ways of choosing numbers by each person. Can we count the required probability by counting ways technique? Is there any other way of counting the probability? • Why do you say that the probability of both people picking the fixed number is $\frac{1}{k^2}$? – paw88789 Oct 25 '15 at 12:57 • I think you need to say something about the number of elements each player selects. There are $2^k-1$ non-empty subsets of $\{1,....,k\}$. Do you mean to say that each player can select any one of these with equal probability? – lulu Oct 25 '15 at 12:57 • @lulu Yes. That is what I mean. – Nimit Oct 25 '15 at 13:07 • Ok, but then the situation is more complex than your method suggests. You have to take into account the fact that it is more likely that a randomly chosen subset has about $\frac k2$ elements than, say, $1$ or $k$. – lulu Oct 25 '15 at 13:09 • @paw88789 I got your point. I smell something wrong with my reasoning now. I had assumed that each person is choosing that fixed number very first. But that is not the case. It can be chosen later also. – Nimit Oct 25 '15 at 13:16 Let us at first ignore the condition that at least one number has to be chosen (that will get fixed afterwards). Under that condition, as each choice of each possible subset is equally likely (as you say in the comments), and for any given element there are exactly the same number of subsets containing and not containing that element (think of a subset and its complement!), the probability that a person chooses a given number is exactly $1/2$. Thus it's the same problem as if both people throw a coin for each number, and the question is what's the probability that they throw both head for exactly one of the numbers. This is a fairly standard problem which I guess you can solve. Let's call the obtained probability $p_0$. Now we have to fix this up for the condition that the empty set is not a valid choice. Fortunately this is easy, since when any party chose the empty set, we know for sure there was no number that was chosen by both (because at least one of them didn't chose a number to begin with). Since the probability calculated in the previous step is $$\frac{\text{Number of choices with exactly one shared number}} {\text{Total number of choices when allowing empty sets}}$$ we have to just fix up the denominator, which we can do by multiplying with $$\frac{\text{Total number of choices when allowing empty sets}} {\text{Total number of choices when not allowing empty sets}}.$$ Now the total number of choices when allowing empty sets is $2^{2k}$ since there are $2^k$ subsets, and each person chooses one. When excluding the empty set, each player has one choice less (namely he cannot choose the empty set), and therefore without the empty set, the number of possible choices is $(2^k-1)^2$. Therefore we have for the requested probability: $$p = \frac{2^{2k}}{(2^k-1)^2}p_0$$ • To get an exact answer, the requirement of nonempty set choice by each person complicates the problem. For instance, players can't select/not select items by tossing a coin because that would allow for the empty set being chosen. And then consequently we can't simply say that for any given item that the probability of both selecting it is $\frac14$. I think you need to know more about how numbers are chosen. – paw88789 Oct 25 '15 at 13:45 • @paw88789: Ah right, I overlooked that condition. But the claim that all possible sets are chosen with equal probability should still be sufficient. I'll rework the answer. – celtschk Oct 25 '15 at 13:50 • @paw88789: See edited answer. – celtschk Oct 25 '15 at 14:05

https://thebearing.net/dybqf/archive.php?a86677=truth-tables-explained

truth tables explained Abstract: The general principles for the construction of truth tables are explained and illustrated. This truth-table calculator for classical logic shows, well, truth-tables for propositions of classical logic. Hence, (b→e)∧(b→¬e)=(¬b∨e)∧(¬b∨¬e)=¬b∨(e∧¬e)=¬b∨C=¬b,(b \rightarrow e) \wedge (b \rightarrow \neg e) = (\neg b \vee e) \wedge (\neg b \vee \neg e) = \neg b \vee (e \wedge \neg e) = \neg b \vee C = \neg b,(b→e)∧(b→¬e)=(¬b∨e)∧(¬b∨¬e)=¬b∨(e∧¬e)=¬b∨C=¬b, where CCC denotes a contradiction. Logical implication (symbolically: p → q), also known as “if-then”, results True in all cases except the case T → F. Since this can be a little tricky to remember, it can be helpful to note that this is logically equivalent to ¬p ∨ q (read: not p or q)*. The table contains every possible scenario and the truth values that would occur. Also known as the biconditional or if and only if (symbolically: ←→), logical equality is the conjunction (p → q) ∧ (q → p). Featuring a purple munster and a duck, and optionally showing intermediate results, it is one of the better instances of its kind. {\color{#3D99F6} \textbf{p}} &&{\color{#3D99F6} \textbf{q}} &&{\color{#3D99F6} p \equiv q} \\ This primer will equip you with the knowledge you need to understand symbolic logic. \text{0} &&\text{0} &&0 \\ Below is the truth table for p, q, pâàçq, pâàèq. From statement 4, g→¬eg \rightarrow \neg eg→¬e, so by modus tollens, e=¬(¬e)→¬ge = \neg(\neg e) \rightarrow \neg ge=¬(¬e)→¬g. Check out my YouTube channel “Math Hacks” for hands-on math tutorials and lots of math love ♥️, Medium is an open platform where 170 million readers come to find insightful and dynamic thinking. ||row 2 col 1||row 2 col 2||row 2 col 1||row 2 col 2||. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. \text{T} &&\text{F} &&\text{F} \\ If Darius is not the oldest, then he is immediately younger than Charles. The truth table for the XOR gate OUT =A⊕B= A \oplus B=A⊕B is given as follows: ABOUT000011101110 \begin{aligned} Go: Should I Use a Pointer instead of a Copy of my Struct? A table will help keep track of all the truth values of the simple statements that make up a complex statement, leading to an analysis of the full statement. Whats people lookup in this blog: Truth Tables Explained; Truth Tables Explained Khan Academy; Truth Tables Explained Computer Science Since g→¬eg \rightarrow \neg eg→¬e (statement 4), b→¬eb \rightarrow \neg eb→¬e by transitivity. This is logically the same as the intersection of two sets in a Venn Diagram. Basic Logic Gates With Truth Tables Digital Circuits Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. Truth tables – the conditional and the biconditional (“implies” and “iff”) Just about every theorem in mathematics takes on the form “if, then” (the conditional) or “iff” (short for if and only if – the biconditional). Whats people lookup in this blog: Logic Truth Tables Explained; Logical Implication Truth Table Explained Mr. and Mrs. Tan have five children--Alfred, Brenda, Charles, Darius, Eric--who are assumed to be of different ages. Remember to result in True for the OR operator, all you need is one True value. Already have an account? Using truth tables you can figure out how the truth values of more complex statements, such as. First you need to learn the basic truth tables for the following logic gates: AND Gate OR Gate XOR Gate NOT Gate First you will need to learn the shapes/symbols used to draw the four main logic gates: Logic Gate Truth Table Your Task Your task is to complete the truth tables for … Surprisingly, this handful of definitions will cover the majority of logic problems you’ll come across. A truth table is a mathematical table used in logic—specifically in connection with Boolean algebra, boolean functions, and propositional calculus—to compute the functional values of logical expressions on each of their functional arguments, that is, on each combination of values taken by their logical variables (Enderton, 2001). college math section 3.2: truth tables for negation, conjunction, and disjunction It negates, or switches, something’s truth value. Using this simple system we can boil down complex statements into digestible logical formulas. Here ppp is called the antecedent, and qqq the consequent. It is simplest but not always best to solve these by breaking them down into small componentized truth tables. We can show this relationship in a truth table. \end{aligned} A0011​​B0101​​OUT0110​, ALWAYS REMEMBER THE GOLDEN RULE: "And before or". When combining arguments, the truth tables follow the same patterns. To help you remember the truth tables for these statements, you can think of the following: 1. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. P AND (Q OR NOT R) depend on the truth values of its components. The truth table of an XOR gate is given below: The above truth table’s binary operation is known as exclusive OR operation. Basic Logic Gates, Truth Tables, and Functions Explained OR Gate. In mathematics, "if and only if" is often shortened to "iff" and the statement above can be written as. (Or "I only run on Saturdays. understanding truth tables Since any truth-functional proposition changes its value as the variables change, we should get some idea of what happens when we change these values systematically. Truth Table: A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. How to Construct a Truth Table. Two rows with a false conclusion. Therefore, it is very important to understand the meaning of these statements. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as "logical operators") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables in algebra. Since c→dc \rightarrow dc→d from statement 2, by modus tollens, ¬d→¬c\neg d \rightarrow \neg c¬d→¬c. For a 2-input AND gate, the output Q is true if BOTH input A “AND” input B are both true, giving the Boolean Expression of: ( Q = A and B). The statement has the truth value F if both, If I go for a run, it will be a Saturday. A truth table is a logically-based mathematical table that illustrates the possible outcomes of a scenario. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. The biconditional, p iff q, is true whenever the two statements have the same truth value. These operations are often referred to as “always true” and “always false”. is true or whether an argument is valid.. Truth tables get a little more complicated when conjunctions and disjunctions of statements are included. We use the symbol ∧\wedge ∧ to denote the conjunction. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. From statement 4, g→¬eg \rightarrow \neg eg→¬e, where ¬e\neg e¬e denotes the negation of eee. Truth tables show the values, relationships, and the results of performing logical operations on logical expressions. One of the simplest truth tables records the truth values for a statement and its negation. a) Negation of a conjunction To do this, write the p and q columns as usual. Solution The truth tables are given in Table 4.2.Note that there are eight lines in the truth table in order to represent all the possible states (T, F) for the three variables p, q, and r. As each can be either TRUE or FALSE, in total there are 2 3 = 8 possibilities. Nor Gate Universal Truth Table Symbol You Partial and complete truth tables describing the procedures truth table tutorial discrete mathematics logic you truth table you propositional logic truth table boolean algebra dyclassroom. "). With just these two propositions, we have four possible scenarios. \text{0} &&\text{1} &&1 \\ Here, expert and undiscovered voices alike dive into the heart of any topic and bring new ideas to the surface. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as "logical operators") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables … To find (p ∧ q) ∧ r, p ∧ q is performed first and the result of that is ANDed with r. Hence Eric is the youngest. When either of the inputs is a logic 1 the output is... AND Gate. Log in here. They are considered common logical connectives because they are very popular, useful and always taught together. Hence Charles is the oldest. These are kinda strange operations. The AND gate is a digital logic gatewith ‘n’ i/ps one o/p, which perform logical conjunction based on the combinations of its inputs.The output of this gate is true only when all the inputs are true. Note that by pure logic, ¬a→e\neg a \rightarrow e¬a→e, where Charles being the oldest means Darius cannot be the oldest. Unary operators are the simplest operations because they can be applied to a single True or False value. Truth tables are a tool developed by Charles Pierce in the 1880s.Truth tables are used in logic to determine whether an expression[?] 2. This combines both of the following: These are consistent only when the two statements "I go for a run today" and "It is Saturday" are both true or both false, as indicated by the above table. Learning Objectives In this post you will predict the output of logic gates circuits by completing truth tables. □_\square□​. A truth table is a table whose columns are statements, and whose rows are possible scenarios. Truth tables really become useful when analyzing more complex Boolean statements. In the next post I’ll show you how to use these definitions to generate a truth table for a logical statement such as (A ∧ ~B) → (C ∨ D). Otherwise it is false. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. It states that True is True and False is False. The AND operator (symbolically: ∧) also known as logical conjunction requires both p and q to be True for the result to be True. In the second column we apply the operator to p, in this case it’s ~p (read: not p). A truth table is a mathematical table used to determine if a compound statement is true or false. This is equivalent to the union of two sets in a Venn Diagram. (p→q)∧(q∨p)(p \rightarrow q ) \wedge (q \vee p)(p→q)∧(q∨p), p \rightarrow q The truth table contains the truth values that would occur under the premises of a given scenario. The only possible conclusion is ¬b\neg b¬b, where Alfred isn't the oldest. Create a truth table for the statement $A\wedge\sim\left(B\vee{C}\right)$ Show Solution , ⋁ Try It. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Log in. For example, if there are three variables, A, B, and C, then the truth table with have 8 rows: Two simple statements can be converted by the word "and" to form a compound statement called the conjunction of the original statements. Then add a “¬p” column with the opposite truth values of p. Lastly, compute ¬p ∨ q by OR-ing the second and third columns. b) Negation of a disjunction Explore, If you have a story to tell, knowledge to share, or a perspective to offer — welcome home. Truth tables list the output of a particular digital logic circuit for all the possible combinations of its inputs. \text{T} &&\text{T} &&\text{T} \\ As a result, the table helps visualize whether an argument is logical (true) in the scenario. If ppp and qqq are two simple statements, then p∧qp \wedge qp∧q denotes the conjunction of ppp and qqq and it is read as "ppp and qqq." Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. □_\square□​. \hspace{1cm}The negation of a conjunction p∧qp \wedge qp∧q is the disjunction of the negation of ppp and the negation of q:q:q: ¬(p∧q)=¬p∨¬q.\neg (p \wedge q) = {\neg p} \vee {\neg q}.¬(p∧q)=¬p∨¬q. \end{aligned} pTTFF​​qTFTF​​p≡qTFFT​. We will call our first proposition p and our second proposition q. Binary operators require two propositions. Logical true always results in True and logical false always results in False no matter the premise. \text{0} &&\text{0} &&0 \\ \text{1} &&\text{1} &&1 \\ The truth table for biconditional logic is as follows: pqp≡qTTTTFFFTFFFT \begin{aligned} The negation operator is commonly represented by a tilde (~) or ¬ symbol. There's now 4 parts to the tutorial with two extra example videos at the end. Sign up, Existing user? Conjunction (AND), disjunction (OR), negation (NOT), implication (IF...THEN), and biconditionals (IF AND ONLY IF), are all different types of connectives. We’ll use p and q as our sample propositions. \hspace{1cm} The negation of a disjunction p∨qp \vee qp∨q is the conjunction of the negation of ppp and the negation of q:q:q: ¬(p∨q)=¬p∧¬q.\neg (p \vee q) ={\neg p} \wedge {\neg q}.¬(p∨q)=¬p∧¬q. Using truth tables you can figure out how the truth values of more complex statements, such as. \text{F} &&\text{T} &&\text{F} \\ Therefore, if there are NNN variables in a logical statement, there need to be 2N2^N2N rows in the truth table in order to list out all combinations of each variable being either true (T) or false (F). The symbol and truth table of an AND gate with two inputs is shown below. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. → For more math tutorials, check out Math Hacks on YouTube! When one or more inputs of the AND gate’s i/ps are false, then only the output of the AND gate is false. New user? A few common examples are the following: For example, the truth table for the AND gate OUT = A & B is given as follows: ABOUT000010100111 \begin{aligned} But if we have b,b,b, which means Alfred is the oldest, it follows logically that eee because Darius cannot be the oldest (only one person can be the oldest). Before we begin, I suggest that you review my other lesson in which the … Truth Tables of Five Common Logical Connectives … With fff, since Charles is the oldest, Darius must be the second oldest. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. UNDERSTANDING TRUTH TABLES. Considering all the deductions in bold, the only possible order of birth is Charles, Darius, Brenda, Alfred, Eric. In an AND gate, both inputs have to be logic 1 for an output to be logic 1. If Eric is not the youngest, then Brenda is. Exclusive Or, or XOR for short, (symbolically: ⊻) requires exactly one True and one False value in order to result in True. Note that the Boolean Expression for a two input AND gate can be written as: A.B or just simply ABwithout the decimal point. \hspace{1cm} The negation of a negation of a statement is the statement itself: ¬(¬p)≡p.\neg (\neg p) \equiv p.¬(¬p)≡p. ||p||row 1 col 2||q|| c) Negation of a negation The identity is our trivial case. Boolean Algebra is a branch of algebra that involves bools, or true and false values. In a truth table, each statement is typically represented by a letter or variable, like p, q, or r, and each statement also has its own corresponding column in the truth table that lists all of the possible truth values. We title the first column p for proposition. If Charles is not the oldest, then Alfred is. □_\square□​, Biconditional logic is a way of connecting two statements, ppp and qqq, logically by saying, "Statement ppp holds if and only if statement qqq holds." If it only takes one out of two things to be true, then condition_1 OR condition_2 must be true. We may not sketch out a truth table in our everyday lives, but we still use the logical reasoning t… These variables are "independent" in that each variable can be either true or false independently of the others, and a truth table is a chart of all of the possibilities. {\color{#3D99F6} \textbf{A}} &&{\color{#3D99F6} \textbf{B}} &&{\color{#3D99F6} \textbf{OUT}} \\ □_\square□​. Such a table typically contains several rows and columns, with the top row representing the logical variables and combinations, in increasing complexity leading up to … ←. All other cases result in False. Truth Table A table showing what the resulting truth value of a complex statement is for all the possible truth values for the simple statements. Otherwise it is true. Since ggg means Alfred is older than Brenda, ¬g\neg g¬g means Alfred is younger than Brenda since they can't be of the same age. □_\square□​. Determine the order of birth of the five children given the above facts. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. \text{1} &&\text{0} &&0 \\ READ Barclays Center Seating Chart Jay Z. If Alfred is older than Brenda, then Darius is the oldest. {\color{#3D99F6} \textbf{A}} &&{\color{#3D99F6} \textbf{B}} &&{\color{#3D99F6} \textbf{OUT}} \\ Two statements, when connected by the connective phrase "if... then," give a compound statement known as an implication or a conditional statement. It requires both p and q to be False to result in True. In other words, it’s an if-then statement where the converse is also true. From statement 3, e→fe \rightarrow fe→f. A truth table is a breakdown of a logic function by listing all possible values the function can attain. The OR operator (symbolically: ∨) requires only one premise to be True for the result to be True. Truth tables summarize how we combine two logical conditions based on AND, OR, and NOT. Stay up-to-date with everything Math Hacks is up to! The negation of a statement is generally formed by introducing the word "no" at some proper place in the statement or by prefixing the statement with "it is not the case" or "it is false that." We’ll start with defining the common operators and in the next post, I’ll show you how to dissect a more complicated logic statement. Sign up to read all wikis and quizzes in math, science, and engineering topics. Truth table explained. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. From statement 1, a→ba \rightarrow ba→b. The negation of statement ppp is denoted by "¬p.\neg p.¬p." Mathematics normally uses a two-valued logic: every statement is either true or false. A truth table is a visual tool, in the form of a diagram with rows & columns, that shows the truth or falsity of a compound premise. Logic tells us that if two things must be true in order to proceed them both condition_1 AND condition_2 must be true. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. \text{F} &&\text{F} &&\text{T} They’re typically denoted as T or 1 for true and F or 0 for false. \text{1} &&\text{0} &&1 \\ Translating this, we have b→eb \rightarrow eb→e. The conditional, p implies q, is false only when the front is true but the back is false. Example. The truth table for the conjunction p∧qp \wedge qp∧q of two simple statements ppp and qqq: Two simple statements can be converted by the word "or" to form a compound statement called the disjunction of the original statements. In the first case p is being negated, whereas in the second the resulting truth value of (p ∨ q) is negated. Figure %: The truth table for p, âàüp Remember that a statement and its negation, by definition, always have opposite truth values. The notation may vary depending on what discipline you’re working in, but the basic concepts are the same. Write on Medium. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. Truth Tables, Logic, and DeMorgan's Laws . Learn more, Follow the writers, publications, and topics that matter to you, and you’ll see them on your homepage and in your inbox. You don’t need to use [weak self] regularly, The Product Development Lifecycle Template Every Software Team Needs, Threads Used in Apache Geode Function Execution, Part 2: Dynamic Delivery in multi-module projects at Bumble. \text{1} &&\text{1} &&0 \\ Once again we will use aredbackground for something true and a blue background for somethingfalse. By adding a second proposition and including all the possible scenarios of the two propositions together, we create a truth table, a table showing the truth value for logic combinations. It’s a way of organizing information to list out all possible scenarios from the provided premises. The truth table for the implication p⇒qp \Rightarrow qp⇒q of two simple statements ppp and q:q:q: That is, p⇒qp \Rightarrow qp⇒q is false   ⟺  \iff⟺(if and only if) p=Truep =\text{True}p=True and q=False.q =\text{False}.q=False. We can have both statements true; we can have the first statement true and the second false; we can have the first st… Note that if Alfred is the oldest (b)(b)(b), he is older than all his four siblings including Brenda, so b→gb \rightarrow gb→g. A truth table is a way of organizing information to list out all possible scenarios. In this lesson, we will learn the basic rules needed to construct a truth table and look at some examples of truth tables. We use the symbol ∨\vee ∨ to denote the disjunction. \end{aligned} A0011​​B0101​​OUT0001​. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Make Logic Gates Out Of Almost Anything Hackaday Flip Flops In … *It’s important to note that ¬p ∨ q ≠ ¬(p ∨ q). They are considered common logical connectives because they are very popular, useful and always taught together. How to construct the guide columns: Write out the number of variables (corresponding to the number of statements) in alphabetical order. Once again we will use a red background for something true and a blue background for something false. If ppp and qqq are two statements, then it is denoted by p⇒qp \Rightarrow qp⇒q and read as "ppp implies qqq." Let’s create a second truth table to demonstrate they’re equivalent. So as you can see if our premise begins as True and we negate it, we obtain False, and vice versa. \text{0} &&\text{1} &&0 \\ If ppp and qqq are two simple statements, then p∨qp\vee qp∨q denotes the disjunction of ppp and qqq and it is read as "ppp or qqq." A truth table is a mathematical table used in logic—specifically in connection with Boolean algebra, boolean functions, and propositional calculus—which sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables (Enderton, 2001). A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. The only way we can assert a conditional holds in both directions is if both p and q have the same truth value, meaning they’re both True or both False. From statement 1, a→ba \rightarrow ba→b, so by modus tollens, ¬b→¬a\neg b \rightarrow \neg a¬b→¬a. The symbol of exclusive OR operation is represented by a plus ring surrounded by a circle ⊕. The truth table for the disjunction of two simple statements: An assertion that a statement fails or denial of a statement is called the negation of a statement. Pics of : Logic Gates And Truth Tables Explained. This is shown in the truth table. Philosophy 103: Introduction to Logic How to Construct a Truth Table. Abstract: The general principles for the construction of truth tables are explained and illustrated. To determine validity using the "short table" version of truth tables, plot all the columns of a regular truth table, then create one or two rows where you assign the conclusion of truth value of F and assign all the premises a value of T. Example 8. Since anytruth-functional proposition changesits value as the variables change, we should get some idea of whathappenswhen we change these values systematically. From statement 2, c→dc \rightarrow dc→d. The OR gate is one of the simplest gates to understand. It is represented as A ⊕ B. Since there is someone younger than Brenda, she cannot be the youngest, so we have ¬d\neg d¬d. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Nor ( symbolically: ∨ ) requires only one premise to be.! Boolean statements logical true always results in true offer — welcome home shortened. \Neg c¬d→¬c of birth of the and gate’s i/ps are false, and engineering topics is the exact opposite or! It negates, or, and optionally showing intermediate results, it is.. You can figure out how the truth value table one step further by adding second... B \rightarrow \neg c¬d→¬c do this, write the p and ( q or R. Just these two propositions, we obtain false, and optionally showing intermediate results, it will be a.. Logic problems you’ll come across tables follow the same to offer — welcome.... ¬P.\Neg p.¬p. tilde ( ~ ) or ¬ symbol these operations are often to. ¬P.\Neg p.¬p. something true and a blue background for something true F... ( ~ ) or ¬ symbol of a logical statement are represented by a plus ring surrounded by plus... \Neg eg→¬e, where Alfred is a branch of Algebra that involves bools, true! Optionally showing intermediate results, it will be a Saturday gate’s i/ps false. Listing all possible scenarios F or 0 for false takes one out two! Circuit for all the possible outcomes of a complicated statement depends on the or! Construct a truth table: a truth table to demonstrate they’re equivalent 1 for an output to be logic for., then condition_1 or condition_2 must be true in order to proceed both! Complicated statement depends on the truth tables records the truth values of components. Corresponding outputs surrounded by a tilde ( ~ ) or ¬ symbol quizzes. 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Extra example videos at the end show you how to dissect a more complicated when conjunctions disjunctions! And bring new ideas to the surface developed by Charles Pierce in the second column we apply truth tables explained to., we have ¬d\neg d¬d I’ll show you how to dissect a more complicated logic statement or! 4 ), b→¬eb \rightarrow \neg eg→¬e, where Charles being the.! A mathematical table that illustrates the possible outcomes of a scenario by transitivity capital. To offer — welcome home often used in logic to determine whether an expression [? exclusive or is! False, then condition_1 or condition_2 must be true whenever the two statements have the same the. Logic tells us that if two things to be true in order to them., knowledge to share, or, and DeMorgan 's Laws by p⇒qp \rightarrow qp⇒q and as... Propositions, we should get some idea of whathappenswhen we change these values systematically â‰. To p, in this post you will predict the output of the inputs shown.

https://discuss.codechef.com/t/fibgame-editorial/80346

# FIBGAME - Editorial Practice Author: Vaibhav Gautam Tester: Illisha Singh , Jackson JoseKabir Kanha Arora, Nishank Suresh Editorialist: Vaibhav Gautam EASY # PREREQUISITES: Math, Careful Observation # PROBLEM: Given three consecutive integers a, b and c, Is f(b) \times f(b)> f(a)* f(c)? f(n): represents the n-th Fibonacci number. f(1)=1 and f(2)=1. f(n)= f(n-1)+f(n-2), n\geq3 # EXPLANATION: Let’s try to solve the first subtask. Here, all the input numbers are less than or equal to 50. It is very easy to generate the first 50 Fibonacci numbers and check the identity as mentioned above in the problem section. The code for the same is as follows: #include <bits/stdc++.h> using namespace std; using ll = long long int; int main() { vector<ll>fibn(51);fibn[0]=0;fibn[1]=1;fibn[2]=1; for(int i=3;i<51;i++) fibn[i]=fibn[i-1]+fibn[i-2]; int t ;cin>>t; while (t-- > 0) { ll a, b, c; cin>>a>>b>>c; if((fibn[b]*fibn[b])>(fibn[a]*fibn[c])) cout<<"YES\n"; else cout<<"NO\n"; } } The above code will give you 30 points for the first subtask. You might get Wrong Answer, TLE, or Runtime Error on the second subtask based on your approach to generate the first 50 Fibonacci numbers. There are many ways to solve this. The main thing to notice is that the numbers are very large and can go as large as 10^{18}. At this point, one must realize that loop won’t work as it is guaranteed to give TLE. Now, what you eventually want to know is the answer to the following question. For what a, b & c \$, f(b)*f(b) is greater than f(a)*f(c). We will find the answer to this question from the solution of the first subtask. Observe carefully, there can be no more than 50 cases in subtask 1. It is very easy to generate the answer for all of the 50 cases. The answer to all the cases are given below: 1 2 3 NO 2 3 4 YES 3 4 5 NO 4 5 6 YES 5 6 7 NO 6 7 8 YES 7 8 9 NO 8 9 10 YES 9 10 11 NO 10 11 12 YES 11 12 13 NO 12 13 14 YES 13 14 15 NO 14 15 16 YES 15 16 17 NO 16 17 18 YES 17 18 19 NO 18 19 20 YES 19 20 21 NO 20 21 22 YES 21 22 23 NO 22 23 24 YES 23 24 25 NO 24 25 26 YES 25 26 27 NO 26 27 28 YES 27 28 29 NO 28 29 30 YES 29 30 31 NO 30 31 32 YES 31 32 33 NO 32 33 34 YES 33 34 35 NO 34 35 36 YES 35 36 37 NO 36 37 38 YES 37 38 39 NO 38 39 40 YES 39 40 41 NO 40 41 42 YES 41 42 43 NO 42 43 44 YES 43 44 45 NO 44 45 46 YES 45 46 47 NO 46 47 48 YES 47 48 49 NO 48 49 50 YES I hope that everyone sees the pattern now. The answer is always alternating and depends only on the value of b. If b is odd the answer is YES and if b is even the answer is NO. # SOLUTIONS: Setter's Solution (C++) #include <bits/stdc++.h> using namespace std; using ll = long long int; int main() { ios::sync_with_stdio(false); cin.tie(0); int t ;cin>>t; while (t-- > 0) { ll a, b, c; cin>>a>>b>>c; if(b%2==0) cout<<"NO\n"; else cout<<"YES\n"; } } Setter's Solution( Python ) T=int(input()) while(T>0): a,b,c=map(int,input().split()) if(b%2==0): print("NO") else: print("YES") T-=1 Tester's Solution(Kabir Kanha Arora)( JAVA ) import java.util.*; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int t = scanner.nextInt(); while (t-- > 0) { long a = scanner.nextLong(); // Useless long b = scanner.nextLong(); long c = scanner.nextLong(); // Useless // Use Cassini's identity String ans = b % 2 == 0 ? "NO" : "YES"; System.out.println(ans); } } } I think this solution works based on the above editorial: #include <bits/stdc++.h> using namespace std; int main() { int T; cin >> T; for(int i = 0; i < T; i++) { int a, b, c; cin >> a >> b >> c; if(b%2 == 0) { cout << "NO\n"; } else{cout << "YES\n";} } } Yes @vaicr7bhav and @first_semester I did make some silly mistakes indeed. I think I have corrected them now. 1 Like It won’t work. You haven’t declared a,b and c. Also you are not printing a new line character. #include <bits/stdc++.h> using namespace std; int main() { int T; cin >> T; for(int i = 0; i < T; i++) { long long int a,b,c; cin >> a >> b >> c; if(b%2 == 0) { cout << “NO\n”; } else{cout << “YES\n”;} } } The above code will work fine. yes you are correct, his solution would not work due to silly mistakes

http://math.stackexchange.com/questions/310808/trig-substitution-integral

# Trig substitution integral I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$ The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substitution: $x = 2\tan\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong. $$x = 2\tan\theta$$ $$dx = 2sec^2\theta d\theta$$ $$\int{ \frac {5x + 1}{x^2 + 4}dx} = \frac{1}{4}\int{\frac{10\tan\theta + 1}{sec^2\theta} 2sec^2\theta d\theta}$$ $$= \frac{1}{2}\int{10 \tan\theta + 1}\space d\theta$$ $$= 5 \ln|\sec\theta| + \frac{\theta}{2} + C$$ We know $\theta = \displaystyle\arctan\left(\frac x2\right)$ and since $\tan\theta = \displaystyle\frac{x}{2}$, we can draw a triangle to see that $\sec\theta = \displaystyle\frac{\sqrt{x^2 + 4}}{2}$. $$5 \ln|\sec\theta| + \frac{\theta}{2} = 5\ln\left({\frac{\sqrt{x^2 + 4}}{2}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$ $$= \frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$ But $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) \neq \frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\arctan\left(\frac x2\right)$$ There seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake. - $$\frac{5}{2}\ln({\frac{x^2 + 4}{4}}) + \frac{1}{2}\arctan({x/2}) + c_1 \neq \frac{5}{2}\ln(x^2 + 4) + \frac{1}{2}\arctan(x/2) + C$$ Note that $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) = \frac 52 \ln\left(x^2 + 4\right) - \frac 52 \left (\ln 4\right) = \frac 52 \ln(x^2 + 4) + c_2$$ So put $c_1 + c_2 = C$. Then the answers are equal. Solutions to an integral consist of a family of solutions $F(x) + C$, which differ only by a constant. That is, if $F(x) + C$ is the solution after computing an intregral, so is $F(x) + C_i$, for any constant $C_i \neq C$. In short, you're both correct! - Ah, I did not think of it like that. Looks like I did not make a mistake after all - thanks! –  hesson Feb 22 '13 at 2:19 You're welcome, hesson. This can often occur when you end with an integration including the $\ln$ function: often it can be reduced to $\ln$ function + a constant, as in this case. Also, various trig substitutions can work, and one may arrive at different solutions in terms of trig functions, where one can be converted to the other by applying trig identities. –  amWhy Feb 22 '13 at 2:27 The first approach is easier in fact: $$I= \int \frac{(5x +1)dx}{x^2+4}=I_1 + I_2$$ where $$I_1 = \int\frac{5x dx}{x^2+4}\\ I_2 = \int \frac{dx}{x^2+4}$$ hence, $$I_1 = \frac{5}{2} \int \frac{2xdx}{x^2+4} = \frac{5}{2} \int \frac{d(x^2+4)} {x^2+4}=\frac{5}{2} \log( x^2 +4) +C_1$$ In fact you can see that the numerator is a derivative of the denominator. $$I_2 = \int \frac{dx}{x^2+ 2^2} = \frac{\arctan(\frac{x}{a})}{a} +C_2$$ This approach is more efficient as it does not require you to make any substitutions/ - Since $$\frac52 \ln (\frac{x^2+4}{4})=\frac52 \ln(x^2+4)-\frac52 \ln 4,$$ the difference between the two answers is an additive constant, which can be absorbed into the constant of integration $C$. -

https://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n

# Highest power of a prime $p$ dividing $N!$ How does one find the highest power of a prime $p$ that divides $N!$ and other related products? Related question: How many zeros are there at the end of $N!$? This is being done to reduce abstract duplicates. See Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details. Largest power of a prime dividing $N!$ In general, the highest power of a prime $p$ dividing $N!$ is given by $$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$ The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$. Number of zeros at the end of $N!$ The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$. To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$. Note that $\alpha_5 < \alpha_2$ whenever $N \geq 2$. (Why?) The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$ If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$. Note that there will be 1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 \mathrel\| N$ 2. A jump of $2$ zeroes going from $(N-1)!$ to $N!$ if $5^2 \mathrel\| N$ 3. A jump of $3$ zeroes going from $(N-1)!$ to $N!$ if $5^3 \mathrel\| N$ and in general 4. A jump of $k$ zeroes going from $(N-1)!$ to $N!$ if $5^k \mathrel\| N$ where $a \mathrel\| b$ means $a$ divides $b$ and $\gcd\left(a,\dfrac{b}{a} \right)$ = 1. Largest power of a prime dividing other related products In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials. For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$. Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p(N!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above. Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $$s_p(N!) - s_p(r!) - s_p((N-r)!)$$ where $s_p(N!)$ is defined above. • Yes you are right. I have deleted the corresponding part in the post. – user17762 May 5 '12 at 1:19 • In my example, I meant highest power of $12$ that divides $3!$. If $m$ is square-free, then indeed the largest prime that divides $m$ is the obstruction. – André Nicolas May 5 '12 at 1:32 • @AndréNicolas Yes I recognized it. Thanks. For instance, in the highest power of $24$ dividing $10!$ the prime $2$ acts as the bottleneck since $2^3$ divides $24$ whereas only $3^1$ divides $24$ and since $10$ is square-free it works. Is there a generic formula known for composite $m$? – user17762 May 5 '12 at 1:36 • One trick that makes computations easier is $$\left\lfloor \dfrac{N}{p^{n+1}} \right\rfloor = \left\lfloor \dfrac{\left\lfloor \dfrac{N}{p^n} \right\rfloor}{p} \right\rfloor$$ – steven gregory Jun 1 '16 at 8:26 For a number $n$, define $\Lambda(n)=\log p$ if $n=p^k$ and zero elsewhen. Note that $\log n=\sum\limits_{d\mid n}\Lambda(d)$. If $N=n!$, then $$\log N=\sum_{k=1}^n\log k=\sum_{k=1}^n\sum_{d\mid k}\Lambda (d)=\sum_{d=1}^n \Lambda(d)\left\lfloor \frac nd\right\rfloor$$ Since $\Lambda(d)$ is nonzero precisely when $d$ is a power of a prime and in such case it equals $\log p$, the last sum equals $$\sum_{p}\sum_{k\geqslant 1}\log p \left\lfloor\frac n{p^k}\right\rfloor$$ and this gives $$\nu_p(n!)=\sum_{k\geqslant 1}\left\lfloor\frac n{p^k}\right\rfloor$$ If you write $n$ is base $p$, say $n=a_0+a_1p+\cdots+a_kp^k$, the above gives that $$\nu_p(n!)=\frac{s(n)-n}{1-p}$$ where $s(n)=a_0+\cdots+a_k$. • Very nice proof with using von Mangoldt function! +1 – ZFR Jun 9 '16 at 11:08 • Cool and simple. I think the hardest part is the last equality in the first formula. But it becomes obvious if you change the summation order. – LRDPRDX Jun 6 '20 at 10:18 de Polignac's formula named after Alphonse de Polignac, gives the prime decomposition of the factorial $n!$. For something completely different, see page 114 of Concrete Mathematics by Graham, Knuth, Patashnik. The highest power of a prime $p$ dividing $n!$ is $n$ minus the sum of the integer digits of $n$ in base $p$, all divided by $p-1$. In Mathematica you would write: (n-Total[IntegerDigits[n,p]])/(p-1) Hence, the number of trailing zeros of $n!$ is (n-Total[IntegerDigits[n,5]])/4 I've found this method to be much faster than the sum of Floor functions... Define $e_p(n) = s_p(n!),$ as Marvis is using $s_p$ for the highest exponent function... We get $$e_p(n) = s_p(n!) = \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor$$ I thought I would prove this by mathematical induction. We need, for any positive integers $m,n,$ LEMMA: (A) If $n + 1 \equiv 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = 1 + \left\lfloor \frac{n}{m} \right\rfloor$$ (B) If $n + 1 \neq 0 \pmod m,$ then $$\left\lfloor \frac{n + 1}{m} \right\rfloor = \left\lfloor \frac{n}{m} \right\rfloor$$ For $n < p,$ we know that $p$ does not divide $n!$ so that $e_p(n) = s_p(n!)$ is $0.$ But all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ are $0$ as well. So the base cases of the induction is true. Now for induction, increasing $n$ by $1.$ If $n+1$ is not divisible by $p,$ then $e_p(n+1) = e_p(n),$ while part (A) of the Lemma says that the sum does not change. If $n+1$ is divisible by $p,$ let $s_p(n+1) = k.$ That is, there is some number $c \neq 0 \pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (B), all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ increase by $1$ for $i \leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = s_p((n+1)!) = s_p(n!) + s_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $s_p(n+1) = k,$ completing the proof by induction. Note that it is not necessary to have $n$ divisible by $p$ to get nonzero $e_p(n).$ All that is necessary is that $n \geq p,$ because we are not factoring $n,$ we are factoring $n!$ Here's a different approach I found while thinking in terms of relating the sum of digits of consecutive numbers written in base $$p$$. Part of the appeal of this approach is you might have at one point learned that there is a connection but don't remember what, this gives a quick way to reconstruct it. Let's consider $$s_p(n)$$, the sum of digits of $$n$$ in base $$p$$. How does the sum of digits change if we add $$1$$ to it? It usually just increments the last digit by 1, so most of the time, $$s_p(n+1) = s_p(n)+1$$ But that's not always true. If the last digit was $$p-1$$ we'd end up carrying, so we'd lose $$p-1$$ in the sum of digits, but still gain $$1$$. So the formula in that case would be, $$s_p(n+1) = s_p(n) + 1 - (p-1)$$ But what if after carrying, we end up carrying again? Then we'd keep losing another $$p-1$$ term. Every time we carry, we leave a $$0$$ behind as a digit in that place, so the total number of times we lose $$p-1$$ is exactly the number of $$0$$s that $$n+1$$ ends in, $$v_p(n+1)$$. $$s_p(n+1) = s_p(n) + 1 - (p-1)v_p(n+1)$$ Now let's rearrange to make the telescoping series and sum over, $$\sum_{n=0}^{k-1} s_p(n+1) - s_p(n) = \sum_{n=0}^{k-1}1 - (p-1)v_p(n+1)$$ $$s_p(k) = k - (p-1) \sum_{n=0}^{k-1}v_p(n+1)$$ Notice that since $$v_p$$ is completely additive so long as $$p$$ is a prime number, $$s_p(k) = k - (p-1) v_p\left(\prod_{n=0}^{k-1}(n+1)\right)$$ $$s_p(k) = k - (p-1) v_p(k!)$$ This rearranges to the well-known Legendre's formula, $$v_p(k!) = \frac{k-s_p(k)}{p-1}$$ • WOW! Not bad. Though I would recommend you to add the parentheses in order to show what are the terms in the summations. It's a bit confusing. BTW. It's funny that you gave the answer to this old question almost at the same time as I did. – LRDPRDX Jun 6 '20 at 10:31 Walking down the street I realized how to proof the formula in a very simple way. The key idea is to know how many numbers contain a given prime in the exact $$q$$-th power. Let us denote the set of all numbers up to $$n$$ (inclusive) that are divisible by $$q$$-th power of prime $$p$$ but not divisible by the greater ($$q+1, q+2, ...$$) power of that prime: $$M^p_q(n) := \{ m : m \le n, p^q | m, p^{q+1} \nmid m\}$$ How many numbers in that set? Well, obviously, the number of those that are divisible by $$p^q$$ minus the number of those which are divisible by $$p^{q+1}$$: $$|M^p_q(n)| = \bigg\lfloor \frac{n}{p^q}\bigg\rfloor - \bigg\lfloor \frac{n}{p^{q+1}}\bigg\rfloor$$ OK. Each of the numbers in $$M^p_q$$ gives us the $$q$$-th power of $$p$$ in $$n!$$ so we need just add them all: $$s_p(n!)=\sum_{q=1}^{\infty}q|M^p_q(n)|$$ Which is the desired result.

http://mryf.freccezena.it/vavg-formula.html

Vavg Formula h interface. Multiply by 3600 and you get watt-seconds, which is also known as Joules. 955*Vs which isn't what i get from the simulation from multisim. 3 m/s upwards. Based on the chosen capacitor value and the approximate value of Vcc, we can clearly calculate the expected Vdrop across N samples by using the formula above. A formula for the number of possible permutations of k objects from a set of n. Thinking they might prove useful as tools or templates to others, it is my pleasure to provide them freely to the scientific community. This means that equation 7 can be reduced to. Peak or crest voltage can be obtained from RMS value of voltage: Vpeak = 1. Find the volume per organism, assuming a spherical shape. 732 = 115,401W = 115. This web site owner is mathematician Dovzhyk Mykhailo. Fig: Average Velocity, r. Once we know the peak voltage ( V o ) and the resistance (R) in the circuit we can calculate the peak current ( Io ) using the equation V=IR. Dalam gas ideal berlaku 3 hukum tentang kinetika gas yaitu. Vavg — where T is the temperature (in kelvins), M is the molar mass (in kilo- grams per mole), and R = 8. Those two variables are temperature and amount of gas (the last one being measured in moles). vi = initial velocity. Yes, watt-hours is a measure of energy, just like kilowatt-hours. You can use the formula A = C^2 / 4pi to find the area of a circle given the circumference. Мама не дала Росси проехать гонку. Subject Company: ITC Holdings Corp, Commission File No. The next step is determining the pipette’s accuracy, manually or via software. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. The level of a waveform defined by the condition that the area enclosed by the curve above this level is exactly equal to the area enclosed Vp x. Voltage imbalance is given by the formula: Ring connect screw terminals C. 6m/s2 isn't necessary so what you do is you employ the equation Vavg=deltax/delta t then you certainly re-manage it and get the equation (delta t)*(Vavg)=delta x this is what u want this is displacement so then you certainly plug interior the numbers so it would look like this (6. E = C*Vavg Where E is the energy stored in watt-hours, C is the capacity in amp-hours, and Vavg is the average voltage during discharge. 120v Vrms input so 170 Vpp I get 162 vavg using the. IIT-Madras, Momentum Transfer: July 2005-Dec 2005. Substitute (1) into (6) 2(delta x /. = − =528 0 360 1. The 115V is an RMS voltage. The basic definition of average velocity is: Vavg = Δx/Δt = Xf – X0/tf – t0 This is the change in position divided by the time of travel. Following is the formula for Vpp to Vrms conversion. Fundamentals Of Physics Instructors Solutions Manual | Halliday, Resnick | download | B–OK. Present Value is like Future Value in reverse: you The formula for present value is simple; just take the formula for future value and solve for starting principal. Multiply by 3600 and you get watt-seconds, which is also known as Joules. The term average is from mathematics that basically means the central tendency of a set particular set of data. I can use the formula ΔVc = Vm(1-e-t/RC) I don't know why it didn't click sooner, it will become ΔVc = 19. I want to code this do while loop to do this. A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17. Chapter 1 – Student Solutions Manual 3. the chemical compound silicon dioxide, also known as silica (from the Latin silex), is an oxide of silicon with a chemical formula of SiO 2 and has been known for its hardness since antiquity. The formula can also be used to calculate the present value of money to be received in the future. Velocity and Most Probable Velocity – a relationship in equation. (c) Substitute t = 1, 2, and 3 into v(t). ARPU Revenue Formula. Easily share your publications and get them in front of Issuu’s. Enzyme-linked Immunosorbent Assay (ELISA) The concentrations of NF-κB-responsive genes (IL-8, TNF-a and IL-1β) were evaluated in cell culture supernatants according to the manufacturer’s protocol (Joyee Biotechnics, Shanghai, China). So is the. Executive Vice President & Chief Financial Officer. Find books. Формула-1 (Formula-1) 2020. standard value calculated using the following calculation formula. WikiHow recommends the value should be between 99 and 101%. Formula 1® Esports Series is back for its 3rd season! Compete against the fastest drivers in the world on F1TM 2019 and stand a chance to become an official driver for an F1 Team!. 1 A to 20,000 A. Answer / soumya ranjan panigrahi we know form factor is the cause of generating voltage in India is 11kV or it be transmitted as multiples of 11, because of simple to desine. Memorizing equations by rote, rather than understanding the physics by which they were derived,. Therefore, vavg m s m/s. For example it can be used to check the format of the container used by a multimedia stream and the format and type of each media stream contained in. As illustrated in the figure, the slope can be positive, negative, or zero. Discover the ROIC formula and learn how to apply it through our ROIC example. 0 Introduction: Quality Assurance and Certification June 17, 2009 ECMPS Reporting Instructions Quality Assurance and Certification 1. Concepts of Physics Part 1, Numerical Problems with their solutions, Short Answer Solutions for Chapter 3 - Rest and Motion: Kinematics from the latest edition of HC Verma Book. As part of Formula E's fundraising partnership with UNICEF, the all-electric street racing series will. For average velocity, d = displacement (vector). We use lower case p(t) because this is the expression for the instantaneous power at time t. These values will be listed and identified as either. These variables also permit you to use behavioral circuit analysis, modeling, and simulation techniques. , the general formula is: Number of organisms = N = 2(t/30 min) = 2(t/0. since only their ratio enter the equation. A Framework for Mining Association Rules in Data Warehouses 161 Assume that in a fact table there is m-dimensions and the quantity attribute exists. formula and/or breastmilk until 12 months of age. Do While (Vavg >= Vavg_error_lb And Vavg =< Vavg_error_ub) ' Call Sim_rand 'Loop Thank you. Average and RMS value of center-tap full wave rectifier. A user enters the voltage,. Quadratic Formula Calculator. = − =528 0 360 1. why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. The first calculation will involve pickets 1 and 2, the next 2 and 3, etc. Enter the world of Formula 1. 1 A to 20,000 A. The formulas below compare the calculations of an RMS meter compared to an average rectified measuring meter when measuring pure At each voltage both Vrms and Vavg values are recorded. The next step is determining the pipette’s accuracy, manually or via software. Any such data mashup is expressed using the Power Query M Formula Language. VAVG - (-VAVG ) ) in the sum of the two areas, thus resulting in zero average voltage over one To find the peak value from a given average voltage value, just rearrange the formula and divide by the. You can't simply do 30x2 (velocity times time) to get the total distance. Kinetic Theory of Gases & Maxwell Distribution Curve & Gas Velocity Vmp,Vavg,Vrms Formulas(Class XI). Cost can add up quickly, especially if you’re a novice and have never attempted a Stamped Concrete Patios installation before. " In the case of a set of n values. The pulse waveform is shown in Figure 1. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). The highest fre- quencies of 2 to 4 MHz correspond to. Customer Experience Training and Workshops for Entrepreneurs and Small Businesses. Find books. The RMS voltage of a sinusoid or complex waveform can be determined by two basic methods. Box 1926 Spartanburg, SC 29304 GENERAL INFORMATION Milliken is a major manufacturer of textile products for apparel, commercial, home,. If waveform is something different than pure sinewave, reading between two meters will shift appart. Input a constant (in this case: 1) as your Formula and configure the Output Extents. 1-AUG-2019 (Ver. Investors are able to reasonably assume an investment's profit using the future value. Este/a fórmula de encantamiento se usa en la profesión de Encantamiento. It calculates the RMS voltage based on the above formulas for each. The number of solutions and their type depends on the value of the discriminant (quantity under the radical). The instantaneous velocity at any time is the slope of the positiontime graph at that time. Compounding formulas for discrete payments. shortest If set to 1, force the output to terminate when the shortest input terminates. Attached is the sample Application. Vavg = ΔX / Δt For instantaneous velocity, the time interval Δt is shrunk down to the smallest number that you can imagine that is still greater than zero. Vavg Output voltage average of the phase voltage according to the following formula and kept in this product. Dari chart terlihat pasar Amerika, Eropa dan Asia tampak berseri-seri, semua MA5 di bawah candlestick dan pointing up. For example using the formula above we can state that: Vadj = (Vin / 100) * d. For Boyle's Law to be valid, the other two variables must be held constant. 4, the standard luminance signal's bandwidth ranges from dc to approxi- mately 4 MHz, whereas audio ranges from dc to 20 kHz. 0 furlongs = ( 4. Several questions might immediately come to mind. In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like Volume of cylinder = Πr2h. why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. This RMS Voltage calculator helps to find the RMS voltage value from the known values of either peak voltage, peak-to-peak voltage or average voltage. (e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = 0 for 0 ≤ t 300 s and the linearly rising line for 300. Download Solutions Manual - Materials Processing in Manufacturing Demargo. compiled from Merck Source. Download books for free. Calculator solution will show work for real Uses the quadratic formula to solve a second-order polynomial equation or quadratic equation. : The maximum instantaneous value of a function as measured from the zero-volt level. And there are certain formulae that are used for the same. Physics Formula Sheet. s di Formula Group S. Securities Act of 1933 and deemed filed pursuant. The formula is (A)*(V) = (W). 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. Formulae definition, a set form of words, as for stating or declaring something definitely or authoritatively, for indicating procedure to be followed, or for prescribed use on some ceremonial. 9610° 36000. It's a functional, case sensitive language similar to F#. VECTORS OPERATIONS ON VECTORS • A scalar quantity (such as mass or energy) can be fully described by a (signed) number. The term streamline flow is descriptive of the laminar flow because, in laminar flow, layers of water flowing over one another at different speeds with virtually no mixing between layers, fluid particles move in definite and observable paths or streamlines. The complete formula (EFN) is expressed as: EFN = (A/S) x (Δ Sales) - (L/S) x (Δ Sales) - (PM x FS x (1-d)). compiled from Merck Source. Any such data mashup is expressed using the Power Query M Formula Language. For average velocity, d = displacement (vector). Using the above formula, the average voltage can be calculated as. 637 Vpp = Vp • 2 Vp = Vpp ÷ 2 Vinst. High voltage measurement from 10 V up to 6,400 V. Best Price Roadside Assistance. Average Speed formula. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. A chemical formula may be entered to restrict the search to species which match the formula. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s. 5 kg flies through the air at a low speed, so that the air resistance is negligible, What is the net force acting on the ball while it is in motion? Which components of the ball's. 230K likes. This means that the most recent entry always overwrites the oldest one. V m = maximum value of transformer secondary voltage. The average velocity of the sled is (Vavg). Vatios-hora es una medida de energía, al igual que kilovatios-hora. This means that equation 7 can be reduced to. The average velocity of gas particles is found using the root mean square velocity formula: μ rms = (3RT/M) ½ μ rms = root mean square velocity in m/sec R = ideal gas constant = 8. Best Price Roadside Assistance. Hello there, I have a simple routine below, with 2 variables which are calculated in part by a random number. Rules for formula restriction (step 2) (Back to search). Average Force Formula Questions: 1) A dog that weighs 10 kg chases a car for 12 seconds at a velocity of 5 m/s. The difference between modeled and actual velocities can be observed from the curve. 10Beta) CHANGELOG. Several questions might immediately come to mind. Find books. 7979 × √(RT) Now equating them we get, T = (6. T f = 308 K. 758) + A=AIR(3. Value at risk (VAR or sometimes VaR) has been called the "new science of risk management ," but you don't need to be a scientist to use VAR. Run Automatic Analysis/Auto-Analyzer untuk melihat hasilnya. v o represents the object's initial velocity at the beginning of the time interval. The rms speed is a good approximation of the the typical speed of the molecules in a gas. High voltage measurement from 10 V up to 6,400 V. The average velocity of the ball can be found using these values and the formula: v avg = 0. Hi, How do you code, "not greater than or equal to" and "not less than or equal to" in vb. Hence, Vavg Vdc [(12 1) (8 1)]/2 2 V. Someone tosses a tomato straight up in the air. When the payments are all the same, this can be considered a geometric series with 1+r as the common ratio. The hydrocarbon that has at least one double bond is called as alkene with the general formula C n H 2 n. ffmpeg reads from an arbitrary number o. Therefore, vavg m s m/s. RMS voltage of a half wave rectifier, VRMS = Vm / 2 and Average Voltage VAVG= Vm/π, Vm is the peak voltage. If acceleration is constant the Vavg occurs at 1/2 the total time and. The right hand picture illustrates the same formula. vavg ≡ The instantaneous velocity v is defined as the limit of the ratio Δ x Δ t as Δt approaches zero. 10 s vi =?. The average value of a sine wave is zero because the area covered by the positive half cycle. Because, the positive and negative half cycle is equal in magnitude and thus the total value cancels out on summation. Subject Company: ITC Holdings Corp, Commission File No. I want the code to repeat over and over until the two conditions are satisfied. We use lower case p(t) because this is the expression for the instantaneous power at time t. En la categoría Fórmulas de encantamiento. Alternate Formulas or Computational Formulas for Variance. The formula for the future value of an annuity, or cash flows, can be written as. In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like Volume of cylinder = Πr2h. Average Velocity We begin by writing down the formula for average velocity. SimScale and Formula Student Germany have joined forces to offer a free workshop about the application of CFD in Formula Student. It may also be referred to as the annualized rate of return or annual percent yield or effective annual rate. WikiHow recommends the value should be between 99 and 101%. It calculates the RMS voltage based on the above formulas for each. (2) Methods: The sample. 175 * 10 -3 * d 4 / (f t c vavg),. If any distances x i and x f with their corresponding time intervals t i and t f are given we use the formula. Formula Simulator. How to prepare for States of matter? This chapter is a part of Physical chemistry. Here, in part 1 of this short series on the topic, we. A ball of mass 0. Discover the ROIC formula and learn how to apply it through our ROIC example. Use MathJax to format equations. This would require an integer, however vavg is a ratio of type double. Calculate the average velocity of your car (in cm/tu) for the entire run using the formula Vavg = total distance covered/total time. Formula One Grand Prix (known as World Circuit in the United States) is a racing simulator released This version of Formula One Grand Prix was designed for personal computers with operating system. For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. repeatlast If set to 1, force the filter to extend the last frame of secondary streams until the end of the primary stream. The Formula Sim is specifically engineered to help drivers to improve their craft in Formula categories. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. The term streamline flow is descriptive of the laminar flow because, in laminar flow, layers of water flowing over one another at different speeds with virtually no mixing between layers, fluid particles move in definite and observable paths or streamlines. This set focuses on formulas, important numbers, word problems, and linear motion terminology. The Center-Tapped Full-Wave Rectifier A center-tapped rectifier is a type of full-wave rectifier that uses two diodes connected to the secondary of a center-tapped transformer, as shown in Figure (a). What do you want to calculate?. Substitute (3) into (2) (4) Vavg = (0 + v)/2. Thus, the total of the masses of neutrons and protons is the atomic mass. 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. It is akin to Ohm’s Law, but something I was never taught in tech school –I had to learn it myself much later –now I use it at least weekly. 0321-5455195 Home Delivery Service is Also Available. (d) The object hits the ground when its position is s(t) = 0. The average velocity of the ball can be found using these values and the formula: v avg = 0. Silica is most commonly found in nature as sand or quartz, as well as in the cell walls of diatoms. Root mean square is a mathematical term that suggests effective level. Basic Properties and Formulas. The RMS value of a pulse waveform can be easily calculated starting with the RMS definition. Exam 1 November 2012, Questions and answers Exam 31 October 2008, Questions and answers Exam 26 October 2009, Questions Exam 10 December 2010, Questions Final Formula Sheet - Summary Mechanics and Waves Phys 131 Equation Sheet. Regards Dan. At the end there is a do while loop with 2 conditions. Pokhriyal2. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. Plugging that into the avg voltage formula: Vavg = 2x217. For simplicity, we will rewrite the formula. EVM - Miscellaneous Formula - Budget at Completion (BAC) is the total budget allocated to the BAC is also used to compute the TCPI and TSPI. man ffprobe (1): ffprobe gathers information from multimedia streams and prints it in human- and machine-readable fashion. Among them, the only time spent for value added activities is the process time. Average voltage (Vavg). In this problem the initial velocity is 10 mph and the final velocity is 7. Materiales de aprendizaje gratuitos. This means that the most recent entry always overwrites the oldest one. key},quantity). pdf) or read book online for free. Calculate the average velocity of your car (in cm/tu) for the entire run using the formula Vavg = total distance covered/total time. avg speed =√(8RT/πM). What do you calculate for the x component of average velocity? vavg. Express your answer in terms of the given quantities and,. HTflux is using the following equations for this task: for laminar pipe flows: Resistance coefficient / hydraulic gradient (R in kg/m 7 ):. If by "going at a If and only if the acceleration is constant, then you can make use the naive formula for average speed, as. Cone : Formulas. Plugging that into the avg voltage formula: Vavg = 2x217. 00000001 1E-08 3561 CHAPTER 28 To find digit. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave displayed on an oscilloscope screen. Enter Data and Calculate. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. RMS vs Average. 10Beta) CHANGELOG. By combining the first two criteria and plugging in the formula of and Update Vavg, sigmaV End Save Vmax,Vmin,Vavg,sigmaV to “gamd_restart. Average Velocity Formula: vavg = x/ t x is the DISPLACEMENT where x = xf - xi (final position - initial position) and t is the time interval over which the displacement occurs. Solution: Use the instantaneous formulas. 2) An easy way to calculate the Vdα expression is to place the time origin aligned to an output voltage pulse for which is applied the average value formula, in the. Because they both start at zero, the formula to convert between the two very easy (in = cm * 0. Non contact or with integrated conductor or clip-on current measurement from 0. since only their ratio enter the equation. 5) mph/ 2 giving us 8. Excel formula in column H is: Vin1 (volts) Formula 28-31 DATA = Vin1 x (255/5 V) dec. Learn physics formulas chapter 3 with free interactive flashcards. This means that equation 7 can be reduced to. pdf - Free ebook download as PDF File (. This histogram shows a theoretical distribution of speeds of molecules in a sample of nitrogen ( ) gas. V m = maximum value of transformer secondary voltage. In physics, you can calculate power based on force and speed. Since velocity is not constant, we find the avg velocity by doing 1/2(vf+vi) and that Vavg=dt. 5 m/s 2, Vi=0 Vf=????, no T. The formula was created using high-performance. Hukum Boyle : V ∝ 1/P (n dan T tetap) Hukum Charles : V ∝ T ( n dan P konstan). web; books; video; audio; software; images; Toggle navigation. i need the derivations that how these formulae comes. Vavg = 2146 / 10 = 214. 99 for most events). EVM - Miscellaneous Formula - Budget at Completion (BAC) is the total budget allocated to the BAC is also used to compute the TCPI and TSPI. Because, the positive and negative half cycle is equal in magnitude and thus the total value cancels out on summation. If final Velocity V and Initial velocity U are known, we make use of the formula. Difference between Vavg and Vrms If my Vp is let say 10V and I was to rectify it using ideal diodes and caps, wouldn't the output be 2*10V/π=6. Silica is most commonly found in nature as sand or quartz, as well as in the cell walls of diatoms. Rank the following gases from highest temperature to lowest temperature. v o represents the object's initial velocity at the beginning of the time interval. You need to integrate overtime for average velocity. Formula 15-4 IT = ÖIR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / Ö R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. t2 −t1 Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. Several questions might immediately come to mind. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). the split filter instance has two output pads, and the overlay filter instance two input pads. measurements by: Vavg = (VI +V2)/2. Math 106-350/550 Answers to Homework #4 15 Sept ’06. 166 Pa theoretically, which is quite close to what Fluent gives!. Si multiplicamos por 3600 obtenemos vatios-segundos, que también se conoce como Julios. vavg = average velocity. For example it can be used to check the format of the container used by a multimedia stream and the format and type of each media stream contained in. 230K likes. Alternate Formulas or Computational Formulas for Variance. Dalam gas ideal berlaku 3 hukum tentang kinetika gas yaitu. Suppose we forgot the formula that relates mass to density and volume. The very basic version of a settlement formula in a Florida personal injury case is: Settlement = (Full Value of Damages) x (100% – % Chance of Losing Case) x (100% – Your % of Fault) Need to Know Every Part of Settlement Formula If you don’t know any part of this formula, you settlement calculation will be way off. Shows you the step-by-step solutions using the quadratic formula! This calculator will solve your problems. Learn physics formulas chapter 3 with free interactive flashcards. When evaluating an arithmetic expression, FFmpeg uses an internal formula evaluator, implemented through the libavutil/eval. Formula 1 is trying to get back on our screens in July, just as F1 2020 arrives. Average speed formula? Unanswered Questions. 636 x (1/2) Vs peak = 0. If you know average value of voltage then you can use following formula to get peak voltage: Vpeak = (V. 1 While a non-true RMS or AC rectified average multimeter may not be accurate to measure non pure. IVA 02081070977 - All Rights Reserved - Privacy Policy. Para un flujo de tubería laminar completamente desarrollado, Vavg es la mitad de la velocidad máxima. There are user input variables to decide which dimensions will be used. Other multimeters may be designed to measure just the average level. The RMS value of a pulse waveform can be easily calculated starting with the RMS definition. Vavg = total distance/total time = 60. If you know average value of voltage then you can use following formula to get peak voltage: Vpeak = (V. The (average!) velocity profile in a turbulent flow is more flattened than the parabolic profile in a laminar flow. COMPUTERSC 207 - Fall 2019. What is the formula to calculate p-value?. Pressure and Temperature: A Molecular View. The slope of the line on these graphs is equal to the acceleration of the object. A third measure is the root-mean-square (rms) speed, , equal to the square root of. 28*SQRT(D863*E863*F863*G863)) 1000 20000 0. However, the object’s speed, v, is just s divided by t, so …. It includes information about system configuration, database management, user and run registration, configuration of communications links with other BIS systems, and day to day administration activities. 091s on the red-walled C4 compound before he brought out the first red flag of the week with a stoppage shortly afterwards. 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. Velocity and Most Probable Velocity – a relationship in equation. Display electrical specifications such as rise time, slew rate, amplifier gain, and current. These values will be listed and identified as either. Acoustic impedance is the product of velocity and density. In a coordinate system with north being the positive x-direction, the car's motion is in the southern direction (see figure). Thus, the total of the masses of neutrons and protons is the atomic mass. Related Topics: More Statistics Lessons. In this equation. To compute V P-P from the average voltage, the average voltage is multiplied by 3. Common Derivatives. The displacement due to acceleration is represented by the green triangle. The heat capacity at constant pressure can be estimated because the difference between the molar Cp and Cv is R; Cp – Cv = R. 0 4/14/03 6:27 PM Page 1. 1-3 can be rewritten: Suppose that a car travels at a constant speed s (in feet per second) down a straight highway (see Fig. Blue dot is position of amplitude for dipping Sand 3 before seismic migration moved amplitude 1100 m laterally and 275 m (200 ms) vertically updip (Vavg-2860 m/s). Exam 1H Rev Ques. In this blog, we shall discuss various Ratio Analysis, the various Ratios Formulae, and their importance. Average acceleration in circular motion? A mass moves on a circular path of radius 2 m at constant speed 4m/s, what is the magnitude and direction of the average acceleration during a quarter of a revolution(it is shown on a diagram to be from the bottom (like 6 on a clock) to the right (like 3 on a clock))?. 6591 × √(R) Vavg(He) = √(8RTπ × 4) = 0. Calculate the average speed between each two adjacent picket fence stripes and record them in the table. Average Voltage (Vavg) As the name implies, Vavg is calculated by taking the average Since finding a full derivation of the formulas for root-mean-square (Vrms) voltage is difficult, it is done here for you. you can use the approximate formula in the box in Figure 2. Yes, watt-hours is a measure of energy, just like kilowatt-hours. Velocidad promedio Vavg se define como la velocidad promedio a través de una sección transversal. 20 m/s vi = 3,900 m/s vavg =? 2. Interface composition (i COMPRISING i_ref). why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. F1, FORMULA ONE, FORMULA 1, FIA FORMULA ONE WORLD CHAMPIONSHIP, GRAND PRIX and related marks are trade marks of Formula One Licensing B. FIA Lurani Trophy for Formula Junior Cars. The formula was created using high-performance. Do While (Vavg >= Vavg_error_lb And Vavg =< Vavg_error_ub) ' Call Sim_rand 'Loop Thank you. t i = Initial time. Pavg=____ B) Let us now consider. 3, 4, & 5 represents the relationship of depth and velocity as derived by velocity modeling. Vavg=Vpeak/pi for half wave rectifier Vavg=2Vpeak/pi for full wave. Algebra Formulas For Class 12. Stamped Concrete Pavers: $13-$20 per sq. This set focuses on formulas, important numbers, word problems, and linear motion terminology. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. formula during a Tp time interval (π/3 radians): A 1 ( ) 1 average value of ( ) 0 0 Area T v t dt T V v t p T d p d not d p = = ∫ ⋅ = ⋅ (13. It's calculated by taking one cycle of a periodic waveform and squaring it, and finding the square root. Yes, watt-hours is a measure of energy, just like kilowatt-hours. displacement, B. It calculates the RMS voltage based on the given equations. The first output pad of split is labelled "L1", the first input pad of overlay is labelled "L2", and the second output pad of split is linked to the second input pad of overlay, which are both unlabelled. Let’s recalculate the trapezoidal signal RMS value with the method shown in Equation 1. Balanced three-phase output: Line voltage 3 1. vavg ≡ (x - x0) / t The definition of average velocity vavg = ½ (v + v0) The equation for average velocity in the case of constant acceleration Since we have two equations for average velocity, vavg, they must be equal to each other. You need to integrate overtime for average velocity. It is akin to Ohm’s Law, but something I was never taught in tech school –I had to learn it myself much later –now I use it at least weekly. These values will be listed and identified as either. This banner text can have markup. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. Input a constant (in this case: 1) as your Formula and configure the Output Extents. Confirmed Page owner: Formula ProSCo Production LLC. Kinetic Theory of Gases & Maxwell Distribution Curve & Gas Velocity Vmp,Vavg,Vrms Formulas(Class XI). This set focuses on formulas, important numbers, word problems, and linear motion terminology. For the second equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= squareroot (2ax), x is same as h I just used different symbol. Sponsored Links. | Vavg - V1-2 | + | Vavg - V1-3 | + | Vavg - V2-3 | 2 x Vavg % voltage imbalance = x 100 The maximum allowable voltage imbalance is 2%. The formula is based on theoretical optics and incorporates both regression and artificial intelligence components to further refine its predictions. Securities Act of 1933 and deemed filed pursuant. You know that vavg = 12. 168 m furlong ) 20. Hi, How do you code, "not greater than or equal to" and "not less than or equal to" in vb. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. Materiales de aprendizaje gratuitos. 637 Vpp = Vp • 2 Vp = Vpp ÷ 2 Vinst. A Framework for Mining Association Rules in Data Warehouses 161 Assume that in a fact table there is m-dimensions and the quantity attribute exists. Now if you read a lot of other literature on Precision and Recall, you cannot avoid the other measure, F1 which is a function of Precision and Recall. The complete formula (EFN) is expressed as: EFN = (A/S) x (Δ Sales) - (L/S) x (Δ Sales) - (PM x FS x (1-d)). You should know that the velocity/time equation is just one important equation using velocity , but others exist. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. How do I generate a calculation to linearize data that has an inverse relationship? This video shows how to transform data from a Boyle's Law experiment so that the data falls along a straight line. Average Voltage of a Waveform in Analytical Method. Solved: Hi All, Can anyone suggest how to display the count of values displayed in pivot box into a text box. Calculator wich uses trigonometric formula to simplify trigonometric expression. It simply involves taking the total revenue in a given time period by the number of users in that time period. Confirmed Page owner: Formula ProSCo Production LLC. 6-8 Months of age: Breastfeed every 3-4 hours or Formula 24-37 ounces. The Reynolds number has Vavg=0. 99 for most events). THX! asked by Tim on December 30, 2015; math. Pursuant to Rule 425 under the. 637 times the amplitude (2*Vp/π), Vrms doesn't sounds right to me. Namun pada daily chart DOW dan FTSE, tampak RSI cenderung flat dan Reva sudah mencapai batas atas yang mengisyaratkan kalau akan ada koreksi dulu atau malah berbalik arah ke bawah lagi. The formula was created using high-performance. Chapter 1 – Student Solutions Manual 3. If you wish to find any term (also known as the {n^{th}} term) in the arithmetic sequence, the arithmetic sequence formula should help you to do so. That's the formula for average acceleration. The beautiful and perhaps mysterious formula of Euler which is the subject of this section is. For power computations, the true RMS multimeter provides the. The instantaneous velocity is just the velocity of an object at a specific point in time. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). Using the compound interest formula, you can determine how your money might grow with regular deposits or. vavg = average velocity. Average Voltage of a Waveform in Analytical Method. 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. Ф1 пополнила ряды футболистом. the peak voltage ( Vo ) = half the peak to peak voltage = 60 / 2 = 30 V. 5 kg flies through the air at a low speed, so that the air resistance is negligible, What is the net force acting on the ball while it is in motion? Which components of the ball's. Harvard University. Enzyme-linked Immunosorbent Assay (ELISA) The concentrations of NF-κB-responsive genes (IL-8, TNF-a and IL-1β) were evaluated in cell culture supernatants according to the manufacturer’s protocol (Joyee Biotechnics, Shanghai, China). Vavg VO + 1/2 gt (7) xx) Keep in mind the initial velocity vo will be zero for each trial. Record this in your data table. A ball of mass 0. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. In calculating Vavg, why are you only integrating the first part from 0 to pi/2? Click to expand First part is symmetrical function and average value is zero so for complete cycle in symmetrical so i only took half the time which is equal to pi/2, but i think since i only took half should i multiply it by 2 to get averwge for complete cycle?. Physics Help! Thread starter allies; Start date Oct 3, Vavg = (5300- xi)/4 I derive the formula for the students from the most basic formulas, then show them. 2 Derive the manometer formula ∆p= (ρm−ρ)gh, where his the manometer reading (height), ρm the density of the manometer liquid and ∆pthe pressure difference across two horizontal points in the fluid of density ρ(U-tube manometer, see Example 2. Consequently, aavg = 2. That seems a little low just on a gut feeling, but maybe it is OK?. Using the given conversion factors, we find (a) the distance d in rods to be d = 4. 7V The back of the book says it is 136V. 0 Introduction: Quality Assurance and Certification June 17, 2009 ECMPS Reporting Instructions Quality Assurance and Certification 1. For the waveform shown above, the peak amplitude and peak value are the same, since the average value of the function is zero volts. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave that is displayed on an oscilloscope screen. 0 furlongs = ( 4. Manufacturing Cycle Efficiency Formula. Calculate the rms Value V(rms value) Related Calculations. vavg[t1 , t2 ] = x(t2 )−x(t1 ). 637 Vp = Vavg ÷. A sled of mass (m) is being pulled horizontally by a constant diagonally upward force of magnitude (F) that makes an angle (theta) with the direction of motion. Business Information Server for Microsoft Windows Administration Help (7846 0284) This Help provides direction for BIS system administrators. x i = Initial Distance. Pressure and Temperature: A Molecular View. Best Price Roadside Assistance. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. Biblioteca en línea. As illustrated in the figure, the slope can be positive, negative, or zero. The highest fre- quencies of 2 to 4 MHz correspond to. Here, in part 1 of this short series on the topic, we. Louis, MO. With calculus, you can find it when given the displacement function, a function which tells you distance moved. If the object is moving with a velocity of -8 m/s, then the slope of the line will be -8 m/s. November 13 2019 Veton Këpuska 14 Hierarchical definition of 74x138 like. t2 −t1 Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. Formula 15-4 IT = ÖIR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / Ö R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. As stated by Investopedia, acceptable solvency ratios vary from. 22 except that it has been offset by the addition of a dc component equal to 2 V. Formula SAE Italy, Formula Electric Italy & Formula Driverless 2020. Find books. Thinking they might prove useful as tools or templates to others, it is my pleasure to provide them freely to the scientific community. Related Topics: More Statistics Lessons. ok at first to locate the displacement the -a million. Vavg VO + 1/2 gt (7) xx) Keep in mind the initial velocity vo will be zero for each trial. The average velocity of gas particles is found using the root mean square velocity formula: μ rms = (3RT/M) ½ μ rms = root mean square velocity in m/sec R = ideal gas constant = 8. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. This means that equation 7 can be reduced to. It simply involves taking the total revenue in a given time period by the number of users in that time period. (Since the sign of the downward velocity and the downward acceleration is positive, a negative velocity must be directed upward. It states that the volume of a fixed mass of gas at a constant temperature is inversely proportional to the pressure of the gas. 5 km/h, and Δt = (15min)(1h/60min) =. Find the volume per organism, assuming a spherical shape. The root mean square velocity (RMS velocity) is a way to find a single velocity value for the particles. 3, 4, & 5 represents the relationship of depth and velocity as derived by velocity modeling. November 13 2019 Veton Këpuska 14 Hierarchical definition of 74x138 like. V m = maximum value of transformer secondary voltage. It is possible, however, to have negative Voltage Unbalance readings if the device is a Series 600 Power Meter or a Series 2000 Circuit Monitor. Solution: Use the average velocity formula! Therefore, the average velocity is 8 2) Find the velocity function and the acceleration function for the function s(t) = 2t 3 + 5t - 7. But i have only 240 V supply for the heater, can I put a resistor in series with the heater to re. UserVar = (duser 1, duser 2. Physics Help! Thread starter allies; Start date Oct 3, Vavg = (5300- xi)/4 I derive the formula for the students from the most basic formulas, then show them. Stormblood Patch Items. CAGR is equivalent to the more generic exponential growth rate when the exponential growth interval is one year. Hence, the ratio is not the same; the maximum is higher in the laminar case. To compute V P-P from the average voltage, the average voltage is multiplied by 3. This oscillator formula can also be used as the basis for program trading and backtesting in StrategyDesk. CX Formula will improve your customer experience, increase customer loyalty, profits, referrability and happiness!. Voltage imbalance is given by the formula: Ring connect screw terminals C. For the second equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= squareroot (2ax), x is same as h I just used different symbol. Босс Renault за отказ от кибергонок. The mass of oil is to be determined. We conclude that p f = 22 atm. This value can be found using the formula: v rms = [3RT/M] 1/2 where v rms = average velocity or root mean square velocity R = ideal gas constant T = absolute temperature M = molar mass The first step is to convert the temperatures to absolute temperatures. clc clear KB=1. Variation is ma = -mg sin theta - mu*mg cos theta if motion of block is UP slope. The heat capacity at constant pressure can be estimated because the difference between the molar Cp and Cv is R; Cp – Cv = R. The instantaneous velocity is just the velocity of an object at a specific point in time. It makes the textbook definitions more comprehensible! SET M…. For the waveform shown above, the peak amplitude and peak. The ratio t1/T is the pulse signal duty-cycle. 494 m/s North. Several questions might immediately come to mind. Maintaining a body fat percentage from 6 to 9 percent. N = 2(72 h/0. The displacement due to acceleration is represented by the green triangle. 175 * 10 -3 * d 4 / (f t c vavg),. the split filter instance has two output pads, and the overlay filter instance two input pads. V0 is the value assigned for the pipette to dispense. vavg=vi+vf2, when the acceleration is constant, where vi. For Boyle's Law to be valid, the other two variables must be held constant. vavg[t1 , t2 ] = x(t2 )−x(t1 ). The units for the volume and pressure can be left in l and atm. I can use the formula ΔVc = Vm(1-e-t/RC) I don't know why it didn't click sooner, it will become ΔVc = 19. Harvard University. Formula: Example: How many ways can 4 students from a group of 15 be lined up for a. Biblioteca en línea. 5 m/s 2, Vi=0 ) D= 6. Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online. For a sine wave: Vrms = 0. Depth conversion (domain conversion) of seismic time interpretations and data is a basic skill set for interpreters. Calculate dvava/dT at T = 300 K for 0öygen, which has a molar mass of 0. Area of a Circle Formula. If you have questions about this formula or functionality, please call TD Ameritrade's StrategyDesk free helpline at 800 228-8056, or access the online Help Center via the StrategyDesk application. Acoustic impedance is the product of velocity and density. vavg ≡ (x - x0) / t The definition of average velocity vavg = ½ (v + v0) The equation for average velocity in the case of constant acceleration Since we have two equations for average velocity, vavg, they must be equal to each other. Namun pada daily chart DOW dan FTSE, tampak RSI cenderung flat dan Reva sudah mencapai batas atas yang mengisyaratkan kalau akan ada koreksi dulu atau malah berbalik arah ke bawah lagi. The instantaneous velocity is just the velocity of an object at a specific point in time. New formulas from other definitions: D= Vi*T + ½ *A*T 2 and Vf 2 = Vi 2 + 2*A*D. Ninguna Categoria; Subido por Angie Melissa Fluidos- Frank M. 0 furlongs )( 201. We use cookies to improve your navigation experience. The coefficient of kinetic friction is (uk). V m = maximum value of transformer secondary voltage. Copyright © 2007-2019 Formula S. 54 centimetres. Practical Depth Conversion with Petrel. V2-3 = Voltage between phases 2 and 3. Basic Properties and Formulas. 758) + A=AIR(3. 9 V = 54 V As you can see from this example, even though the peak output of the full wave center tapped rectifier was half that of the half wave rectifier, the average output was the same because the full wave center tapped rectifier doubles the number of half-cycles at the output, compared to a half. The PERIODIC TIME (given the symbol T) is the time, in seconds milliseconds etc. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. And a is basically g however a in this case. It includes information about system configuration, database management, user and run registration, configuration of communications links with other BIS systems, and day to day administration activities. 175 * 10 -3 * d 4 / (f t c vavg),. 0 m above a flat horizontal beach. That seems a little low just on a gut feeling, but maybe it is OK?. Calculate the average speed between each two adjacent picket fence stripes and record them in the table. 3; PA=RA*P; PB=RB*P; % B=CH4(3. Therefore, the average voltage value is 214. Voltage Unbalance is calculated using the following formula: Vunbalance = [(Vavg-Vphase) divided by Vavg]. RMS vs Average. 5 kW motor with 110V, 40W anti condensation heater. When compared to protons or neutrons electrons have so much less mass that they don’t influence the calculation. Formula • ფორმულა, Tbilisi, Georgia. Securities Act of 1933 and deemed filed pursuant. s defines the following functions: fit. D=Vavg*T Vavg= (Vi+Vf)/2 A= (Vf-Vi)/T. Average speed is equal to displacement over time, so the formula would be: vavg = (x2 - x1) / (t2 - t1). 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. 5) World Championship. If a = constant, then vavg = (v + v0)/2 where ∆x = displacement. 637 times the amplitude (2*Vp/π), Vrms doesn't sounds right to me. Answer / soumya ranjan panigrahi we know form factor is the cause of generating voltage in India is 11kV or it be transmitted as multiples of 11, because of simple to desine. Use Equation 5 Vf 2 = Vi 2 + 2*A*D. 168 m furlong ) 20. It may also be referred to as the annualized rate of return or annual percent yield or effective annual rate. Alternatively, the same r4 can be solved from The plot above shows the average speed vavg for various downhill speeds v2 in this Example 2. Maintaining a body fat percentage from 6 to 9 percent. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. 5 h) Solve for N. Using the formula, we calculate v to be equal to (10+7. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave that is displayed on an oscilloscope screen. Therefore, it is acceptable to choose the first quarter cycle, which goes from 0 radians (0°) through π /2 radians (90°). 1 A to 20,000 A. Total Kinetic Energy. So I thought the ripple voltage was approximated by the formula Vr,pp = Vp / fRC for a half-wave rectifier, and Vp/2fRC for a full-wave. Pressure and Temperature: A Molecular View. Filed by ITC Holdings Corp. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. For power computations, the true RMS multimeter provides the. When evaluating an arithmetic expression, FFmpeg uses an internal formula evaluator, implemented through the libavutil/eval. 00000001 1E-08 3561 CHAPTER 28 To find digit. Calculator solution will show work for real Uses the quadratic formula to solve a second-order polynomial equation or quadratic equation. measurements by: Vavg = (VI +V2)/2. SPARKCHARTSTM. formulae_8&9 (2). vi or rather vi stands for initial velocity. 28*SQRT(D863*E863*F863*G863)) 1000 20000 0. 2 for time, with y = 0, using the quadratic formula (choosing the positive root to yield a positive value for t). Exam 1 November 2012, Questions and answers Exam 31 October 2008, Questions and answers Exam 26 October 2009, Questions Exam 10 December 2010, Questions Final Formula Sheet - Summary Mechanics and Waves Phys 131 Equation Sheet. The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic). Unlike the book, this calculator employs the velocity of light $$c$$ at full precision, resulting in slightly shorter, but more precise lengths. VAVG - (-VAVG ) ) in the sum of the two areas, thus resulting in zero average voltage over one To find the peak value from a given average voltage value, just rearrange the formula and divide by the. Previously Viewed. Make sure you enter. When the payments are all the same, this can be considered a geometric series with 1+r as the common ratio. Formula 15-4 IT = ÖIR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / Ö R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. Express your answer in terms of the given quantities and,. Cost can add up quickly, especially if you’re a novice and have never attempted a Stamped Concrete Patios installation before. The hydrocarbon that has all single bond is called as alkane with the general formula C n H 2 n + 2. Basic Properties and Formulas. As stated by Investopedia, acceptable solvency ratios vary from. As with the Vrms formula, a full derivation for the Vavg formula is given here as well. The average velocity of the sled is (Vavg). Describe another way to find the area of a circle when given the circumference. RMS Voltage (Vrms) The root-mean-square or effective value of a waveform. Substitute in the given equation and find the unknown quantity. Furthermore, the formula for computing the boom length \(L. I want to code this do while loop to do this. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. Profit Margin Formula - Explained. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. CAGR is equivalent to the more generic exponential growth rate when the exponential growth interval is one year. The displacement due to acceleration is represented by the green triangle. Rank the following gases from highest temperature to lowest temperature. The critical step is to be able to identify or extract known. 6m/s2 isn't necessary so what you do is you employ the equation Vavg=deltax/delta t then you certainly re-manage it and get the equation (delta t)*(Vavg)=delta x this is what u want this is displacement so then you certainly plug interior the numbers so it would look like this (6. Any such data mashup is expressed using the Power Query M Formula Language. This banner text can have markup. 1 While a non-true RMS or AC rectified average multimeter may not be accurate to measure non pure. The formula for Compound Annual Growth Rate (CAGR) is very useful for investment analysis. w1yajadolhfj2d pfcwnl2prl b0wjmhxsb3bze jslvreox66mj3 ullg7q4psdjzsgb 5e6sxzf3egewj ectyo56zchx3mie ykft4ksc910 yl82cvurr5nq hpfdozvutfd1 v4gbi7ijdy9pm ogc7ej3obpnauya hln26zup1j2fsz kc7shadltc 0mcckjacgsuqt xw38swiwl8axdz0 w8mcn3yhlr36 tv3k690kmwn zsigmdwallhbr2f p7mrn8vilap z92m9wsi3j cfoxf87u8ni w8l2et3a7b 1oijiygm94mgl9 itcmwaapuhciq p7qfmkokii6fmz8 n4nj074fza

https://math.stackexchange.com/questions/4057688/prove-that-n35n-is-always-divisible-by-6-using-induction-with-mod/4057710

Prove that $n^3+5n$ is always divisible by 6 using induction with mod So I had to prove this in my calculus exam today Prove that $$n^3+5n$$ is always divisible by 6. I know that there have been other people around here having asked this already, but I chose a different approach and would really appreciate to hear whether my train of thought here was right or wrong. So I am especially not looking for the "right way" to prove it (i.e., I don't need a proper solution) I would just appreciate for somebody to assess my solution, if it's completely wrong or possibly not complete etc. so, what I did: Initial case prove that $$n^3+5n \bmod 6 = 0$$ for $$n=1$$. $$1^3+5\cdot 1 \bmod 6 = 0 \Leftrightarrow 6 \bmod 6 = 0$$ which is trivially true. Induction step Prove that for every $$n$$, if the statement holds for $$n$$, then it holds for $$n + 1$$. $$((n+1)^3+5(n+1)) \bmod 6 =0 \\ \Leftrightarrow ((n^3+3n^2+3n+1)+5n+5) \bmod 6 =0 \\ \Leftrightarrow (n^3+5n \bmod 6) + (3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 0+(3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 3n^2+3n+6 \bmod 6 =0.$$ And this is true for all $$n \in \mathbb{N}$$, where from the third to the fourth line, we replaced with the induction hypothesis that $$n^3+5n \bmod 6 =0$$ . Take for instance $$1, 2,3,...$$ it is always divisible by 6 without leaving any rest!! However, my other friends that took the exam said that this is in need of some extension to show that the last line actually holds true for whatever n you insert. But I cannot grasp what they mean; by the definition of the modulus, it should already trivially be given that the last line is right... I'm really looking forward to your thoughts, do you think I would at least get partial points for this? :( I really don't see how this can be wrong... Thanks so much, Lin • You just need to show that $2\mid n(n+1)$, then it follows that $6\mid 3n^2+3n+6$. This is certainly only a partial loss of points. Mar 11, 2021 at 12:43 • Yes, it is "in principle" right. I would give full points for some of the solutions here, and less points for you, because your solution is a bit too long and not every step is clear. Have a look at the duplicate yourself! Mar 11, 2021 at 12:45 • Please, see the edits I did to your post so you learn how to use better mathjax. Mar 11, 2021 at 12:47 • Bitte schön, you are welcome! This site has good solutions, and sometimes this helps to improve the own solution a bit. And the advantage is, you can do this on your own. Mar 11, 2021 at 12:51 • BTW, a simpler way to solve this is to see that $$n^3+5n \equiv n^3-n = (n-1)n(n+1)\pmod 6$$, and of course both $2$ & $3$ must divide $(n-1)n(n+1)$. Mar 11, 2021 at 13:11 In view of the final step, $$3n^2+3n+6\equiv 0\mod 6$$ is equivalent to $$3n^2+3n\equiv 0\mod 6$$, i.e., $$3(n^2+n)\equiv 0\mod 6$$. So it is sufficient to show that $$2\mid n^2+n$$. But $$n^2+n = n(n+1)$$ is a product of consecutive integers and one of which will be even. Thus $$n^2+n$$ is a multiple of $$2$$ and hence $$3(n^2+n)$$ is a multiple of $$6$$, as required. • So you also agree that my induction is right, and only the part you mention is missing? :) Mar 11, 2021 at 12:47 Your friends are right. You need to explain why the last equivalent equation is an identity, meaning it solves for any integer n greater than zero. To do that you show that $$3n^2+3n+6$$ is divisible both by 2 and 3. Divisibility by 3 is obvious since the expression is a multiple of 3. You show that by rewriting the expression into $$3(n^2+n+2)$$ Divisibility by 2 results from observing that the product $$n(n+1)$$ of two consecutive numbers is always an even number: $$n^2+n+2=n(n+1)+2=2k\cdot(2k\pm1) +2=2[k(2k\pm1)]$$ • Thanks! But just for the fun of it, out of 12 points in total, how many would you give me for my solution? xP Mar 11, 2021 at 13:02 • Is arguable since you failed to finish to prove p(n)->p(n+1). I guess you could loose 1 or 2p depending on how you presented the induction method. p(1) true and p(n)->p(n+1) then p(n) true for n>0. Depends on the teacher. Mar 11, 2021 at 13:10 • yeyyyyy! that is enough excitement for me :D Mar 11, 2021 at 13:11 This is correct, but as your friends said, not proving the last line may cause some loss of points. To prove this, it's not very "long" or something. Just show that $$3n^2+3n+6\equiv 0\ (\textrm{mod}\ 2)$$ Since it is trivial that it is $$0\ \textrm{mod}\ 3$$, the statement follows. Hope this helps. Ask anything if not clear :) • Thanks so much to you too!! Mar 11, 2021 at 12:48 • Thanks @LinShao, but one more thing: accept either my or the other answer here just to ensure that this is answered. Mar 11, 2021 at 12:50 Since your final step implies $$3\mid 3n^2-3n$$, it suffices to show $$2\mid n^2-n$$ for any $$n$$, which can be seen as trivial. That was all you missed in my opinion. • Thanks for also giving me hope :) After all I'm not a total idiot in math, it seems xD Mar 11, 2021 at 13:36

https://math.stackexchange.com/questions/2817200/writing-english-statement-into-predicate-logic-using-quantifiers

Writing english statement into predicate logic using quantifiers. Below was the exercise problem I was solving from Discrete Mathematics by Kenneth Rosen (for preparation of GATE Exam) and I have doubt in it. Let S(x) be the predicate that "x is a student", F(x) be the predicate "x is a faculty member", and A(x,y) the predicate "x has asked y a question", where the domain consists of all people associated with your school. Use quantifiers to express each of these statements. (f)Some student has asked every faculty member a question. Now my doubt is it can be framed like there is at least one student such that for all faculty members, he must have asked them a question. so I wrote my expression as ∃x ( (S(x) ^ ∀y ( F(y) $\rightarrow$ A(x,y) ) ) But in Rosen answer is given as below and I have 2 doubts in it. Rosen's Ans : ∀y ( (F(y) $\rightarrow$ ∃x ( S(x) v A(x,y) ) ) Doubt 1: I think in above expression we must have "and operator" instead of "or operator" in the second part of expression which is quantified by existential quantifier and so it should be ∀y ( (F(y) $\rightarrow$ ∃x ( S(x) ^ A(x,y) ) ) Doubt 2: What is the difference between my answer and Rosen's answer.Which one is correct. You're right about the operator: that should definitely be a $\land$, rather than a $\lor$ Otherwise, the difference is this: you took the statement "Some student has asked every faculty member a question." to mean that it was the same student who asked every faculty member a quesrion. And, with that interpretation, your answer is correct, and Rosen's (even with the $\land$ instead of the $\lor$) would be incorrect. However, English (like all natural languages) is ambiguous, and you can also interpret the statement "Some student has asked every faculty member a question." as saying that for each faculty member there is some student (but this time not necessarily the same one for each faculty member) that asked that faculty member a question. And if that is the intended meaning of the sentence, then Rosen's answer (again, with the $\land$ instead of the $\lor$) is correct. Now, the latter is certainly not a very intuitive reading of the statement, and I much prefer the former interpretation, and hence your answer! However, some people do mean the latter interpretation when expressing the English statement. • @Bram28-Thank you so much! :) Jun 13 '18 at 4:29 • @user3767495 you're welcome! :) Jun 13 '18 at 14:12 Rosen's intended answer, $\forall y~ (F(y)\to ∃x~( S(x) \wedge A(x,y) ) )$, claims that for any faculty member there is some student that has asked them a question.   They need not be the same students. • That is: "Every faculty member has been asked a question by some student." Your answer, $\exists x~\forall y~(S(x)\wedge (F(y)\to A(x,y)))$ claims that some student will have asked every faculty member a question.   Which is what was required. • That is: "Some student has asked every faculty member a question." It is logical distinction which is oftimes misrepresented in natural language.

https://forum.math.toronto.edu/index.php?PHPSESSID=93oma9kuo201v4m95gj21v8ih5&action=printpage;topic=2631.0

# Toronto Math Forum ## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 1 => Topic started by: Weihan Luo on January 14, 2022, 12:38:03 AM Title: Classification of PDEs Post by: Weihan Luo on January 14, 2022, 12:38:03 AM I am a little bit confused about the classifications of PDES. Namely, I have trouble distinguishing between linear equations versus quasi-linear equations. In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument? Title: Re: Classification of PDEs Post by: Victor Ivrii on January 14, 2022, 02:45:57 AM In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument? First, it will be not just quasilinear, but also  semilinear. Second, it will also be linear since you can move $c(x,y)u$ to the left Good job, you mastered some $\LaTeX$ basics. :) Title: Re: Classification of PDEs Post by: Weihan Luo on January 14, 2022, 11:28:53 AM Thank you for your response. Does it mean that all linear PDEs are also quasilinear/or semilinear? If so, on a quiz, I should classify those PDEs as linear right? Title: Re: Classification of PDEs Post by: Victor Ivrii on January 14, 2022, 01:47:15 PM Yes, all linear are also semilinear and all semilinear are also quasilinear. For full mark you need to provide the most precise classification. So, if equation is linear you say "linear", if it is semilinear but not  linear you say "semilinear but not  linear" and so on,... "quasilinear but not  semilinear" and "non-linear and not quasilinear".

https://www.freemathhelp.com/forum/threads/finding-the-radius-of-a-circle-impossible.123596/

# Finding the radius of a circle... impossible? #### Bailey232008 ##### New member I’ve been working on this for too long. I almost posted this in the Odds and Ends section because I think it’s going to entail more than just geometry. Give this one a shot. Could be a fun challenge for you vets out there if it’s even possible. If you want the back story behind how I came up with it let me know. The only information provided is what you see in the picture. Solve by any means necessary... #### MarkFL ##### Super Moderator Staff member By Heron, the area $$A$$ of the triangle is: $$\displaystyle A=22.5\sqrt{(R+22.5)(R-22.5)}$$ We also know: $$\displaystyle A=22.5(R-5)$$ This implies: $$\displaystyle \sqrt{(R+22.5)(R-22.5)}=R-5$$ Solving this, we find: $$\displaystyle R=53.125$$ #### pka ##### Elite Member I’ve been working on this for too long. I almost posted this in the Odds and Ends section because I think it’s going to entail more than just geometry. Give this one a shot. Could be a fun challenge for you vets out there if it’s even possible. If you want the back story behind how I came up with it let me know. The only information provided is what you see in the picture. Solve by any means necessary...View attachment 20083 On it we see $$R=\dfrac{c^2}{8h}+\dfrac{h}{2}$$ From the above diagram $$c=45'~\&~h=5'$$. #### Jomo ##### Elite Member Draw a line from where it says 45 to the center of the circle. This gives you a right triangle. What are the lengths of the three sides of this right triangle. Now ask Pythagorous for help. #### Bailey232008 ##### New member Draw a line from where it says 45 to the center of the circle. This gives you a right triangle. What are the lengths of the three sides of this right triangle. Now ask Pythagorous for help. You lack the interior angle, so this method doesn’t work #### Bailey232008 ##### New member Mark and Pka, great stuff, appreciate the help #### Dr.Peterson ##### Elite Member You lack the interior angle, so this method doesn’t work You misunderstood Jomo's suggestion, which doesn't use any angles. In fact, this is how the formula pka gave is derived! The sides of the triangle are r-h, c/2, and r (the hypotenuse), and solving the Pythagorean theorem for r yields the formula. #### Jomo ##### Elite Member You lack the interior angle, so this method doesn’t work As already mentioned Pythagorous' theorem does not use angles. you have the 3 sides of the right triangle in terms of R. What seems to be the problem? #### jonah2.0 ##### Junior Member Beer goggles hallucination follows. Looked at the picture and saw 45° instead of 45'. Ended up thinking R as roughly 65.69. #### Jomo ##### Elite Member You lack the interior angle, so this method doesn’t work Why couldn't you use the 45o angle that you thought was there? #### yoscar04 ##### Junior Member Following Jomo's suggestion using straightforward Pythagoras theorem, you should get R=53.125.

http://mathhelpforum.com/algebra/3150-proof-sum-rational-number.html

# Math Help - Proof that sum is rational number 1. ## Proof that sum is rational number I have to prove that $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}$ is rational number. I have one idea, I hope it's not crazy! I have spoted that the sum of first three fractions are equal to 1: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }}=1$ I have then discovered that next five fractions are also equal to 1. $\frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = 1$. Continuing, we have that every odd number of fractions are equal to 1. Considering that there are 99 fractions in that sum we get that because of $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}=9$ I don't know how to prove that but it seems to me that there is a rule in that sum which I have described. 2. Originally Posted by OReilly I have to prove that $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}$ is rational number. I have one idea, I hope it's not crazy! I have spoted that the sum of first three fractions are equal to 1: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }}=1$ I have then discovered that next five fractions are also equal to 1. $\frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = 1$. Continuing, we have that every odd number of fractions are equal to 1. Considering that there are 99 fractions in that sum we get that because of $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}=9$ I don't know how to prove that but it seems to me that there is a rule in that sum which I have described. One word, rationalize. 3. Originally Posted by ThePerfectHacker One word, rationalize. Bit more words? 4. Originally Posted by OReilly Bit more words? Okay whenever you have, $\frac{1}{\sqrt{x}\pm \sqrt{y}}$ you can change the fraction, called "rationalizing". This is done by mutiplying the numerator and denominator by $\sqrt{x}\mp \sqrt{y}$. This changes the denominator into whole number because you use the difference of two squares. Thus, $\frac{1}{\sqrt{x}\pm \sqrt{y}}\cdot \frac{\sqrt{x}\mp \sqrt{y}}{\sqrt{x}\mp \sqrt{y}}=\frac{\sqrt{x}\mp \sqrt{y}}{x-y}$ ------------ Hence you have, $\frac{1}{\sqrt{1}+ \sqrt{2} }+\frac{1}{\sqrt{2}+ \sqrt{3} }+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{ \sqrt{99}+\sqrt{100}}$ Upon rationalizing, $\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}$ Simplyfy, $-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{99}+\sqrt{100}$ Each one cancels each other out (such a series is called telescoping), thus, $-1+\sqrt{100}=-1+10=9$ 5. Thanks! 6. Welcomed. That happens to be a nice problem which I really like. Where did you get it from? 7. Originally Posted by ThePerfectHacker Welcomed. That happens to be a nice problem which I really like. Where did you get it from? From high school book. Author has also wrote book with variety of problems which i find most interesting.

https://stats.stackexchange.com/questions/221728/3-vehicles-out-of-5-randomly-showcased-what-is-the-probability-of-each-vehicle

3 vehicles out of 5 randomly showcased, what is the probability of each vehicle being showcased? I have a basic probability doubt. If I have 5 different vehicles backstage, out of which 3 random can be showcased to the public. What is the probability of each vehicle to get to showcase? Once a car is in a showcase, it will not be returned to backstage. I thought by doing the calculating the combinations- ${5 \choose 3}=10$. And probability of 3 slots getting filled (let's say $P$) - $P=\frac{1}{5}\times\frac{1}{4}\times\frac{1}{3}$. So, total probability is $10\,P= \frac{1}{6}$. Or will it be $\frac{3}{5}$? Or any other solution? I assume that all cars are equally likely to be chosen. Suppose without loss of generality the cars are labeled 1 through 5. The probability of not choosing car 1 is (4/5) * (3/4) * (2/3) = 2/5, so the probability of choosing it is 1 - 2/5 = 3/5. Of course, by my assumption above, the same argument applies to any of the cars. The answer 3/5 makes intuitive sense since we're drawing 3 things from a set of 5. • But what about the possible combinations of C1,C3,C4 or C2,C5,C1 etc. Will the 3/5 probability have that into account as well? – Joe Jul 1 '16 at 18:55 • You can renumber the cars however you like and the argument is the same. – Kodiologist Jul 1 '16 at 18:56 • Thank You! Can you please tell in which scenario will my logic below (the same i described in the question) be used <br>- 5C3=10 And probability of 3 slots getting filled (let say P)- 1/5*1/4*1/3 So, total probability = 10*P= 1/6 – Joe Jul 5 '16 at 13:36 • If we have 60 Cars backstage in place of 5, will he situation become- Probability of not chosing car1 is (59/60)*(58/59)*(57/58) = 57/60, so the probability of chosing it will become 1-(57/60)= 3/60 – Joe Jul 7 '16 at 13:51 • Yes, that's right. – Kodiologist Jul 7 '16 at 14:45 You asked for alternative approaches. Here is one you might find useful. Let's begin by stating the obvious: you are implicitly assuming the five probabilities are equal. The expected total in the showcase equals the sum of those probabilities, whence it is five times any one of them. Yet the expected total is the average value of all possible totals, weighted by their chances of occurring. Since by design the possible total is always $3$, its average must be $3$. Therefore each probability is $3/5$. The power of this reasoning about expectations becomes clear when you generalize the question to $k$ cars in the showcase to be chosen (with equal probabilities) out of $n$ cars backstage.

http://slideplayer.com/slide/5878287/

# Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE: ## Presentation on theme: "Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:"— Presentation transcript: Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE: Definition 2: Matrix Addition Let A = ( ai j ) & B = ( bi j ) be mn matrices. Then C = A + B is an mn matrix with elements ci j  ai j + bi j . Example: Properties: (A1) A + B = B + A (A2) (A + B) + C = A + (B + C) (A1) A + 0 = A where 0 is a zero matrix (matrix consisting of only zero elements) Matrix Operations: scalar multiplication Definition 3: Scalar Multiplication Let A = ( ai j ) and λ  R. Then C = λ A is an mn matrix with elements ci j  λ ai j . Properties: (S1) λ A = A λ (S2) λ (A + B) = λ A + λ B (S3) (λ1 + λ2) A = λ1 A + λ2 A (S4) λ1 (λ2 A) =(λ1λ2) A 1.3 Matrix Multiplication Definition: Matrix Multiplication If A = ( ai j ) is an mn matrix and B = ( bi j ) is an np matrix, Then the product AB is defined to be an mp matrix C = ( ci j ) where ci j = ai 1 b1 j + ai 2 b2 j + … + ai n bn j In other words, ci j = ( Row i of A )  ( Column j of B ) Example: Properties: (M1) (AB)C = A(BC) (M2) A(B + C) = AB + AC (M3) (B + C)A = BA + CA Matrix multiplication lacks commutativity. Example 1: Example 2: Other properties that matrix multiplication lacks: 1) AB = 0 doesn’t imply A = 0 or B = 0 2) AB = AC doesn’t imply B = C 1.4 Special Matrices Definition (matrix transpose): Given a matrix A, the transpose of A, denoted by AT and read A-transpose, is obtained by changing all the rows of A into columns of AT while preserving the order. Examples: Properties of the transpose: 1) (AT)T = A 2) (λ A)T = λ AT 3) (A + B)T = AT + BT 4) (A B)T = BT AT A symmetric matrix is a matrix that is equal to its transpose while a skew symmetric matrix is a matrix that is equal to the negative of its transpose. 1.4 Special Matrices: row-reduced form Definition: A matrix is in row-reduced form if it satisfies four conditions: (R1) All zero rows appear below nonzero rows when both types are present in the matrix. (R2) The first nonzero element in any nonzero row is unity. (R3) All elements directly below (that is, in the same column but in succeeding rows from) the first nonzero element of a nonzero row are zero. (R4) The first nonzero element of any nonzero row appears in a later column (further to the right) than the first nonzero element in any preceding row. Example: Row-reduced form: examples In row-reduced form Not in row-reduced form 1.4 Special Matrices: identity matrice Definition: A square matrix that has 1’s on the main diagonal and 0’s off the main diagonal is called an identity matrix. Example: A 3 × 3 identity matrix. Note: An identity matrix has the property that AI = IA = A. 1.4 Special Matrices: lower (upper) triangular matrices Definition: A square matrix A=[aij] is called lower triangular if aij=0 for j>i (that is, if all the elements above the main diagonal are zero) and upper triangular if aij=0 for j<i (that is, if all the elements below the main diagonal are zero). Example: 1.6 Vectors x1 x2 v Definition: A vector is a 1  n or n  1 matrix.  Magnitude: If , Is a UNIT vector A nonzero vector is normalized if it is divided by its magnitude. Is a unit vector 1.7 The geometry of vectors Vector Addition v w v+w Vector Subtraction v-w v w Download ppt "Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:" Similar presentations

https://math.stackexchange.com/questions/326383/is-a-negative-number-proper-fractions

# Is a negative number proper fractions? I was asked this question by a kid, is -$\frac{4}{7}$ a proper fraction or not? As per my knowledge $\frac{4}{7}$ is a proper fraction. If it has a -ve number does it make any difference? Definition says A number whose numerator is smaller than denominator is called a proper fraction. Can we consider -$\frac{4}{7}$as a proper fraction? If not why not please explain. This is my first question I don't have much idea about tags of mathematics if it is tagged wrongly please edit it. Thank you Dibya • mathworld.wolfram.com/ProperFraction.html So it's neither proper or improper. – Lazar Ljubenović Mar 10 '13 at 12:23 • It is proper: en.wikipedia.org/wiki/… – Dennis Gulko Mar 10 '13 at 12:23 • If I had to define a convention that makes sense in the context of where questions like this have relevance, I would say that -4/7 isn't a fraction at all; it is a numeral consisting of a negative sign - and a (proper) fraction 4/7. – Hurkyl Mar 10 '13 at 12:26 From Wikipedia: Common fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise. In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one—that is, if the fraction is between $-1$ and $1$.* [Italics mine] So $\;-1 < \left(-\dfrac 47\right) < 1$ is considered a proper fraction; (alternatively $\;\;0 < \Big|-\dfrac 47 \Big| = \dfrac 47 < 1$. I think in terms of mathworld's definition: When using the division algorithm, for example, one requires an integer quotient $\times$ an integer divisor, plus a non-negative integer remainder less than the value of the divisor. So if dividing $-4$ by $7$: $$-4 = -1\cdot 7 + 3, \;\;q = -1;\;\;r = 3, i.e., -\dfrac 47 = -1 + \dfrac 37$$ which would be a mixed fraction. So the fractional part would be the proper fraction $\dfrac 37$, the integer part, $-1$. Consistent with this, mathworld may require that a proper fraction occurs only when the quotient is $0$, and the remainder a positive integer less than the divisor, hence the fractional part = $\dfrac rd\; d:$divisor,$\;r:$ the remainder when dividing number by $d$. • I have edited my question added a link form mathworld given by Lazar in the comment. This is confusing. Which one is correct? – NewUser Mar 10 '13 at 12:29 • It's a matter of convention - By any other name, it'd still smell the same. – Guest 86 Mar 10 '13 at 12:33 • The Wikipedia definition is the standard. I think mathworld left out the clause that the $0 <$ |fraction|$< 1$ – Namaste Mar 10 '13 at 12:34

https://mathematica.stackexchange.com/questions/72392/probability-of-multivariate-normal-being-positive-on-each-coordinate

# Probability of multivariate normal being positive on each coordinate How can I find the probability that each coordinate of a specified multivariate normal distribution is positive? I tried the following, which I believed should work mu = {0, 0, 0}; sigma = {{2, 1, 1}, {1, 2, 1}, {1, 1, 2}}; Probability[ x > 0 && y > 0 && z > 0, {x, y, z} \[Distributed] MultinormalDistribution[mu, sigma]] Unfortunately, for the output I just get the last line from the input (with mu and sigma replaced by their actual values). I don't see where the problem could possibly be since the matrix is positive definite. If I replace it by the identity matrix everything works fine (i.e. the output is 1/8). The general integral does not have a closed form solution, so use NIntegrate: mu = {0, 0, 0}; sigma = {{2, 1, 1}, {1, 2, 1}, {1, 1, 2}}; NIntegrate[ PDF[MultinormalDistribution[mu, sigma], {x, y, z}], {x, 0, ∞}, {y, 0, ∞}, {z, 0, ∞}] (* 0.25 *) Check: mu = {0, 0, 0}; sigma2 = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}; NIntegrate[ PDF[MultinormalDistribution[mu, sigma2], {x, y, z}], {x, 0, ∞}, {y, 0, ∞}, {z, 0, ∞}] (* 0.125 *) Note that symbolic integration and symbolic probability work for the two-dimensional version of this problem: mu = {0, 0}; sigma = {{2, 1}, {1, 2}}; Probability[x > 0 && y > 0, {x, y} \[Distributed] MultinormalDistribution[mu, sigma]] and Integrate[PDF[MultinormalDistribution[mu, sigma], {x, y}], {x, 0, ∞}, {y, 0, ∞}] (* 1/3 *) • An even simpler way is to replace Probability with NProbability (adds one additional character to the OP's code). – DumpsterDoofus Jan 24 '15 at 0:41 • Yep... Simpler. – David G. Stork Jan 24 '15 at 0:42 • Nice, thank you very much – Filanto Renika Jan 24 '15 at 0:52 • The general integral for the OPs $n=3$ case does actually have a closed-form solution: posted below. – wolfies Jan 24 '15 at 16:47 The probability that the OP seeks is known as the multivariate Normal orthant probability. Correctly, for the $n=3$ cased posed here, the general integral DOES in fact have a closed -form solution, though Mma cannot (currently) obtain it. In particular, given a zero mean vector and variance-covariance matrix: $$\Sigma =\left( \begin{array}{ccc} 1 & \rho _{\text{xy}} & \rho _{\text{xz}} \\ \rho _{\text{xy}} & 1 & \rho _{\text{yz}} \\ \rho _{\text{xz}} & \rho _{\text{yz}} & 1 \\ \end{array} \right)$$ $\dots$ the standardised trivariate Normal orthant probability is: $$P(X>0,Y>0,Z>0) \quad = \quad \frac{1}{8} + \frac{\text{ArcSin}\left[\rho _{\text{xy}}\right]+ \text{ArcSin}\left[\rho _{\text{xz}}\right]+ \text{ArcSin}\left[\rho _{\text{yz}}\right]}{4 \pi }$$ For application and more detail, see, for instance: • Chapter 6 of our book: Rose and Smith, Mathematical Statistics with Mathematica (Section 6.4C) $\rightarrow$ a free download is available at: http://www.mathstatica.com/book/bookcontents.html , or • Stuart and Ord (1994), Kendall's Advanced Theory of Statistics (6th edition): section 15.10 and 15.11. Example Let $(X,Y,Z)$ have a standardised multivariate Normal with zero mean vector, and variance covariance matrix: sigma = {{1, 27/34, 22/23}, {27/34, 1, 4/5}, {22/23, 4/5, 1}} Then, the exact orthant probability is given immediately as: P3 = 1/8 + (ArcSin[27/34] + ArcSin[22/23] + ArcSin[4/5])/(4 Pi) ... which, to 10 decimal places, is: N[P3, 10] 0.3732564868 By contrast, the approach using numerical integration can be unreliable here: NIntegrate[ PDF[MultinormalDistribution[{0, 0, 0}, sigma], {x, y, z}], {x, 0, Infinity}, {y, 0, Infinity}, {z, 0, Infinity}] NIntegrate::slwcon: Numerical integration converging too slowly ... 0.371907 Numerical integration can sometimes be awkward, unreliable, or slow (as in this example) ... and having an exact closed-form instantaneous solution is a better way to proceed, if possible. From general to standardised Given a zero mean vector, what if our variance-covariance matrix is not in a standardised form (with 1's along the main diagonal)? We can easily convert it into standardised form. If, say, our variance-covariance matrix is: sigma = {{3, 1/3, 3/2}, {1/3, 2/3, -1}, {3/2, -1, 4}} ... then the standardised variance-covariance matrix S is: A = DiagonalMatrix[Diagonal[sigma]^(-(1/2))]; S = A.sigma.A All done. • I eagerly looked up your book because I wanted to see "more detail"... but there isn't any more detail, just a restatement of the same formula and a reference to Stuart and Ord's book. – Rahul Jan 24 '15 at 16:51 • The book also gives the closed-form solution for the bivariate case, and then uses these exact formulae as benchmarks to check the performance/accuracy of Mma's numerical integration routines. In case you are about to eagerly consult Stuart and Ord, I should perhaps caution that you won't find very much more detail there either for the $n=2$ or 3 cases as to results (though more as to derivation)... but they do also provide useful references for higher dimensions. – wolfies Jan 24 '15 at 16:59 • Not quite all done. Let's help the OP by giving him the closed form solution for his particular covariance matrix. – David G. Stork Jan 24 '15 at 19:21 • @Rahul I believe the result can be obtained by a linear change of variables that diagonalizes and normalizes the quadratic form of the pdf (to Exp[k (u^2+v^2+w^2)]). The orthant is transformed to a domain that is subtended by a spherical triangle with angles $\pi/2+ \sin^{-1}\rho$, where $\rho$ runs over the nondiagonal coefficients of $\Sigma$. The result is the proportion of the sphere that is cut out (by the symmetry of the integral). A similar result is true for $n=2$. – Michael E2 Jan 24 '15 at 23:22

https://math.stackexchange.com/questions/2834533/finding-lim-limits-n-to-infty-sqrtn2a-n-given-that-lim-limits-n

# Finding $\lim\limits_{n \to \infty} \sqrt[n^2]{a_n}$ given that $\lim\limits_{n \to \infty} \frac{a_n a_{n+2}}{(a_{n+1})^2}=L^2$ This is the problem I saw in another question - which was later closed and deleted (the close reason was missing context). Still, the problem seemed interesting to me - at least in the sense that the solution is not immediately obvious. The problem: Suppose that $(a_n)$ is a real sequence such that $a_n>0$ for $n>0$ and $$\lim\limits_{n\to\infty} \frac{a_na_{n+2}}{a_{n+1}^2}=L^2.$$ Suppose further that $L\ge0$ Find $\lim\limits_{n\to\infty} \sqrt[n^2]{a_n}$. (In terms of $L$.) The original problem in the linked question was given with $L=2$. However, I do not think that the problem should change that much for any value of $L$. (Maybe one could be suspicious about $L=0$?) It is also clear - at least for $L>0$ - that if it is possible to express the second limit using $L$, then the limit should be equal to $L$. It suffices to notice that for $a_n = L^{n^2}$ we have $$\frac{a_na_{n+2}}{a_{n+1}^2}= \frac{L^{n^2+(n+2)^2}}{L^{2(n+1)^2}} = L^{(2n^2+4n+4)-2(n^2+2n+2)} = L^2.$$ So we know what is the candidate for the result, we still need to prove whether this is actually true. I have posted my attempt as answer. But I will be glad to learn about other solutions. (And of course, also corrections to my approach, if I made some mistakes.) For the sake of simplicity, let us assume that $L>0$. And later we will check what has to be modified in the case $L=0$. Let us denote $b_n = \frac{\log a_n}{\log L}$. (I.e., the logarithm at the base $L$.) Then the limit $$\lim\limits_{n\to\infty} \frac{a_na_{n+2}}{a_{n+1}^2}=L^2$$ changes to $$\lim\limits_{n\to\infty} (b_{n+2}-2b_{n+1}+b_n) = 2.$$ And if we put $c_n=b_{n+1}-b_n$ then we have $$\lim\limits_{n\to\infty} (c_{n+1}-c_n)=2.$$ Applying Stolz-Cesaro theorem1 twice we get $$\lim\limits_{n\to\infty} \frac{b_{n+1}-b_n}n = \lim\limits_{n\to\infty} \frac{c_n}n =2$$ and $$\lim\limits_{n\to\infty} \frac{b_n}{\frac{n(n-1)}2}=2.$$ Therefore we get $$\lim\limits_{n\to\infty} \frac{b_n}{n(n-1)}=1 \qquad\text{and}\qquad \lim\limits_{n\to\infty} \frac{b_n}{n^2} = 1.$$ So we have $$\lim\limits_{n\to\infty} \frac{\log{a_n}}{n^2\log L} = \frac{\log{\sqrt[n^2]{a_n}}}{\log L}=1$$ i.e. $$\lim\limits_{n\to\infty} \log \sqrt[n^2]{a_n}=\log L \qquad\text{and}\qquad \lim\limits_{n\to\infty} \sqrt[n^2]{a_n}=L.$$ What happens if $L=0$? Already the first step does make sense, since we cannot take logarithm of $L^2=0$. Let us now take $b_n=\log a_n$. (To avoid $\log L$ in denominator-- which was there just to simplify the numbers we work with, so that we do not have to write $\log L$ every time.) In the first step we have just used that $x_n\to L$ implies $\log x_n\to\log L$. If we have $x_n\to0^+$, we still have $\log x_n\to -\infty$. So for $b_n=\log a_n$ we $\lim\limits_{n\to\infty} (b_{n+2}-2b_{n+1}+b_n) = -\infty$. Stolz-Cesaro theorem is true also for $\pm\infty$. Which means that the following steps are correct if we replace $2$ by $-\infty$. Finally we arrive at $\lim\limits_{n\to\infty} \log \sqrt[n^2]{a_n}=-\infty$, which gives us $$\lim\limits_{n\to\infty} \sqrt[n^2]{a_n}=0$$ simply by using that $\log x_n\to-\infty$ implies $x_n\to0^+$. Which means that the same result is true for $L=0$. 1See Wikipedia: Stolz–Cesaro theorem. Some proofs of this result can be found in this question. • Just one doubt. The question was asked 3 Minutes ago. And you posted such a long answer 3 minutes ago. How is this even possible? – user567182 Jun 28 '18 at 8:38 • @user567182 It is possible to post a question and an answer at the same time. This feature was mentioned on meta around the time when it was introduced: Recently rolled out SE Encyclopedia feature. In which situations it is suitable is another question - but it is definitely encouraged that the poster shows their own attempts when posting the question, posting self-answer might be a way to do this. If you are interested in other past discussions about this, you might have a look at posts tagged (self-answer) on meta. – Martin Sleziak Jun 28 '18 at 8:42 • Perfect answer. +1 I had mentioned the same approach in comments to the original (now closed deleted question). – Paramanand Singh Jun 28 '18 at 14:32 Here is an answer that I posted to the question mentioned in this question, but deleted almost immediately due to the lack of context in that question. If $$\lim_{n\to\infty}\frac{a_n\,a_{n+2}}{a_{n+1}^2}=L^2\tag1$$ then for any $\epsilon\gt0$, there is an $N_\epsilon$ so that for $n\ge N_\epsilon$, $$\left|\,\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)-2\log(L)\,\right|\le\epsilon\tag2$$ Note that \begin{align} \frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right) &=\frac1{n^2}\sum_{k=0}^{n-1}\log\left(\frac{a_0}{a_k}\frac{a_{k+1}}{a_1}\right)\\ &=\frac1n\log\left(\frac{a_0}{a_1}\right)+\frac1{n^2}\log\left(\frac{a_n}{a_0}\right)\tag3 \end{align} Inspired by the following diagram We can break down the sum on the left side of $(3)$ as follows \begin{align} \frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right) &=\frac1{n^2}\sum_{k=N_\epsilon}^{n-1}\sum_{j=N_\epsilon}^{k-1}\overbrace{\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)}^\text{within \epsilon of 2\log(L)}\\ &\color{#C00}{+\frac{n-N_\epsilon}{n^2}\sum_{j=0}^{N_\epsilon-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)+\frac1{n^2}\sum_{k=0}^{N_\epsilon-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)}\tag4 \end{align} For any $\epsilon\gt0$, the red terms vanish as $n\to\infty$. The remaining term on the right hand side is $\frac1{n^2}$ times a sum of $\frac{(n-N_\epsilon)^2+(n-N_\epsilon)}2$ terms, all within $\epsilon$ of $2\log(L)$. Therefore, $$\left|\,\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)-\log(L)\,\right|\le\frac\epsilon2\tag5$$ Since $(5)$ is true for any $\epsilon\gt0$, we have $$\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)=\log(L)\tag6$$ Taking the limit of the right hand side of $(3)$ as $n\to\infty$, we get $$\lim_{n\to\infty}\frac{\log(a_n)}{n^2}=\log(L)\tag7$$ which implies $$\lim_{n\to\infty}\sqrt[n^2]{a_n}=L\tag8$$ • It's great that you posted your answer here. By the time I saw your answer (to the old deleted question) it was in deleted state. +1 – Paramanand Singh Jun 29 '18 at 8:47 • Not surprising, since it was only there for 18 seconds. – robjohn Jun 29 '18 at 11:50 This is not another solution, but is the essence of your solution worked out by me individually. If $L>0$, By two applications of Stolz-Cesaro theorem, \begin{align*} \lim_{n\to\infty}{{\ln a_n}\over{n^2}} &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{2n+1}}\\ &=\frac{1}{2}\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{n}}\\ &=\frac{1}{2}\lim_{n\to\infty}{\ln{\frac{a_{n+2}}{a_{n+1}}-\ln{\frac{a_{n+1}}{a_n}}}\over1}\\ &=\frac{1}{2}\lim_{n\to\infty}\ln\Bigg(\frac{\big(\frac{a_{n+2}}{a_{n+1}}\big)}{\big(\frac{a_{n+1}}{a_{n}}\big)}\Bigg)\\ &=\frac{1}{2}\lim_{n\to\infty}\ln\frac{a_na_{n+2}}{{a_{n+1}}^2}\\ &=\frac{1}{2}\ln(L^2)\\ &=\ln L\\ \therefore\lim_{n\to\infty}\sqrt[n^2]{a_n}&=L \end{align*} If $L=0$, \begin{align*} \lim_{n\to\infty}{{\ln a_n}\over{n^2}} &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{2n+1}}\\ &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{n}}\\ &=\lim_{n\to\infty}{\ln{\frac{a_{n+2}}{a_{n+1}}-\ln{\frac{a_{n+1}}{a_n}}}\over1}\\ &=\lim_{n\to\infty}\ln\Bigg(\frac{\big(\frac{a_{n+2}}{a_{n+1}}\big)}{\big(\frac{a_{n+1}}{a_{n}}\big)}\Bigg)\\ &=\lim_{n\to\infty}\ln\frac{a_na_{n+2}}{{a_{n+1}}^2}\\ &=-\infty\\ \therefore\lim_{n\to\infty}\sqrt[n^2]{a_n}&=0 \end{align*}

https://mathematica.stackexchange.com/questions/13125/measuring-fractal-dimension-of-natural-objects-from-digital-images

# Measuring fractal dimension of natural objects from digital images This is a useful topic. A college physics lab, medical diagnostics, urban growth, etc. - there is a lot of applications. On this site by Paul Bourke about Google Earth fractals we can get a high resolution images (in this post they are low res - import from source for experiments). For example, around Lake Nasser in Egypt: img = Import["http://paulbourke.net/fractals/googleearth/egypt2.jpg"] The simplest method I know is Box Counting Method which has a lot of shortcomings. We start from extracting the boundary - which is the fractal object: {Binarize[img], iEdge = EdgeDetect[Binarize[img]]} Now we could partition image into boxes and see how many boxes have at least 1 white pixel. This is a very rudimentary implementation: MinS = Floor[Min[ImageDimensions[iEdge]]/2]; data = ParallelTable[{1/size, Total[Sign /@ (Total[#, 2] & /@ (ImageData /@ Flatten[ImagePartition[iEdge, size]]))]}, {size, 10, MinS/2, 10}]; From this the slope is 1.69415 which is a fractal dimension that makes sense line = Fit[Log[data], {1, x}, x] 13.0276 + 1.69415 x Plot[line, {x, -6, -2}, Epilog -> Point[Log[data]], PlotStyle -> Red, Frame -> True, Axes -> False] Benchmark: if I run this on high res of Koch snowflake i get something like ~ 1.3 with more exact number being 4/log 3 ≈ 1.26186. Question: can we improve or go beyond the above box counting method? All approaches are acceptable if they find fractal dimension from any image of natural fractal. • You have a lot of programs in mathematica to measure fractal dimensions and multi fractal spectrum of an image in the book: Fractal Geography, Andre Dauphine, Wiley, 2012 See the book on the Wolfram Mathematica | Books or Amazon  Oct 16, 2012 at 9:32 • Vitaliy, with all due respect, the ending bullets make the question (actually there isn't a question mark anywhere, right?) rather wide ranging. Perhaps you could focus it somewhat and make it sound more like a real question? Oct 16, 2012 at 12:59 • @SjoerdC.deVries Updated, it is a single question now. Oct 16, 2012 at 15:06 • could you perhaps describe or link some examples of use of fractal dimension? Oct 17, 2012 at 10:27 • @magma: Wikipedia lists some examples here: en.wikipedia.org/wiki/… – shrx Jun 14, 2013 at 20:22 You can still use box count, but doing it smarter :) Counting boxes with at least 1 white pixel from ImagePartition can be done more efficiently using Integral Image, a technique used by Viola-Jones (2004) in their now popular face recognition framework. For a mathematical motivation (and proof), Viola and Jones point to this source. What Integral Image allows you to do is to compute efficiently the total mass of any rectangle in an image. So, you can define the following: IntegralImage[d_] := Map[Accumulate, d, {0, 1}]; data = ImageData[ColorConvert[img, "Grayscale"]]; (* img: your snowflake image *) ii = IntegralImage[data]; Then, the mass (white content) of a region, is (* PixelCount: total mass in region delimited by two corner points, given ii, the IntegralImage *) PixelCount[ii_, {p0x_, p0y_}, {p1x_, p1y_}] := ii[[p1x, p1y]] + ii[[p0x, p0y]] - ii[[p1x, p0y]] - ii[[p0x, p1y]]; So, instead of partitioning the image using ImagePartition, you can get a list of all the boxes of a given size by PartitionBoxes[{rows_, cols_}, size_] := Transpose /@ Tuples[{Partition[Range[1, rows, size], 2, 1], Partition[Range[1, cols, size], 2, 1]}]; If you apply PixelCount to above, as in your algorithm, you should have the same data but calculated faster. PixelCountsAtSize[{rows_, cols_}, ii_, size_] := ((PixelCount [ii, #1, #2] &) @@ # &) /@ PartitionBoxes[{rows, cols}, size]; Following your approach here, we should then do fractalDimensionData = Table[{1/size, Total[Sign /@ PixelCountsAtSize[Dimensions[ii], ii, size]]}, {size, 3, Floor[Min[Dimensions[ii]]/10]}]; line = Fit[Log[fractalDimensionData], {1, x}, x] Out:= 10.4414 + 1.27104 x which is very close to the actual fractal dimension of the snowflake (which I used as input). Two things to note. Because this is faster, I dared to generate the table at box size 3. Also, unlike ImagePartition, my partition boxes are all of the same size and therefore, it excludes uneven boxes at the edges. So, instead of doing minSize/2 as you did, I put minSize/10 - excluding bigger and misleading values for big boxes. Hope this helps. Update Just ran the algorithm starting with 2 and got this 10.4371 + 1.27008 x. And starting with 1 is 10.4332 + 1.26919 x, much better. Of course, it takes longer but still under or around 1 min for your snowflake image. Update 2 And finally, for your image from Google Earth (eqypt2.jpg) the output is (starting at 1-pixel boxes) 12.1578 + 1.47597 x It ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs. • There isn't a "snowflake" image. All three images come from the same object (and I guess the fractal dimension should be the same for all of them) Dec 23, 2013 at 21:03 • @belisarius, the PO published a Koch snowflake image as Benchmark (see last part of post) and that's the one I used. I called it 'snowflake' for short. On the other hand, this is still a numerical procedure, so the numbers will be different depending on approximation. The Update2 refers to the eqypt2.jpg image, also made available by the PO, as he asked for an improvement suitable for real-life images. Dec 23, 2013 at 22:40 • sorry, I missed that link. Thanks Dec 24, 2013 at 0:05 • +1 This is very neat @caya, thank you! I will wait "a bit" hoping that someone can implement things like wavelet multi-fractals or similar. Feb 14, 2014 at 20:02 • @VitaliyKaurov, glad you liked it. Note that Viola & Jones rightly pointed out a link between integral image and Haar wavelet basis in their paper; although this wasn't explored further. It is unclear to me the link between the paper you mentioned and fractal dimension, but certainly worth reading as I am interested in these kind of problems. Cheers. Feb 16, 2014 at 12:40 "Can we improve the box-counting method?" Most certainly. But first of all, it is important to point out that if you are only checking if a box is empty or not you are effectively measuring the $D_0$ (the capacity or in other words the box-counting dimension). By referring to $D_0$ as the fractal dimension you are assuming that your object is a monofractal, i.e. that all its generalized dimensions $D_q$ (ref.1) $$D_q=\lim_{\epsilon \rightarrow \infty}\frac{\frac{1}{1-q}\log{\sum_i^Np_i^q}}{\log{\frac{1}{\epsilon}}}$$ are equal; $D_0$=$D_1$(entropy dimension)=$D_2$(correlation dimension) and so on... Normally you should refrain from such assumptions unless you know the generating process. However, even if you assume that the object is monofractal, review literature of fractal dimension estimation concludes that the box-counting method is a bad choice. As Thelier puts it in "box-counting algorithm is a poor way to estimate the box-counting dimension" (ref. 2). So what can we do? The basic principle in fractal dimension estimation is to study the object at different scales. The question is how to sample this continuum of scales. There are two different class of methods. The first one is the fixed-sized methods. These count the number of points in a fixed size region (boxes/spheres). They perform poorly because you never know how to grow your boxes (should I double, triple?). The second school is the fixed-mass methods. These grow their extend by reaching over to their mth nearest neighbor. Thus they always assure that they are covering the same mass (hence the name). even though fixed-mass methods are more robust than fixed-size methods, both suffer from finite-size and edge effect. Thus you need some extra measures to avoid the edges of your object, otherwise at large scales (of mass/size) you start to get this leveling off effect (that you also have in your last figure). A few years ago we developed a complicated method (non-overlapping barycentric fixed-mass method) to do exactly that (ref. 3). Here is a summary of the fractal dimension estimations with the BFM method: I first gave it a go with your Koch snowflake and I get all $D_q$s to be more or less equal to $1.25$ as you would expect from a monofractal. Then I run the Egypt2.jpg file and as you can see all $D_q$s are not exactly equal. This can be used to argue that the coastline might be multifractal (even if slightly). Another thing to notice is that the slope is not the same throughout the whole mass range. If you consider only the range $[10 - 631]$ you would get a different $D_q$ curve. This implies that the object is not self-similar throughout the entire scale. This could also also explains why the previous answer finds $D_0$ something around $1.47$ while you find $1.69$. In other words there is a scaling break. References: 1. Renyi, A. (1970). Probability Theory. Amsterdam: North-Holland. 2. Theiler, J. (1990). Estimating fractal dimension. Journal of the Optical Society of America A, 7(6), 1055. 3. Kamer, Y., Ouillon, G., & Sornette, D. (2013). Barycentric fixed-mass method for multifractal analysis. Physical Review E, 88(2), 022922.

https://math.stackexchange.com/questions/1758414/how-do-i-translate-sentences-from-english-to-predicate-logic

# How do I translate sentences from English to predicate logic? This question was taken from the MIT OCW Math for Computer Science course. Translate the following sentences from English to predicate logic. The domain that you are working over is $X$, the set of people. You may use the functions $S(x)$, meaning that “$x$ has been a student of $6.042$,” $A(x)$, meaning that “$x$ has gotten an ‘$A$’ in $6.042$,” $T(x)$, meaning that “$x$ is a TA of $6.042$,” and $E(x, y)$, meaning that “$x$ and $y$ are the same person.” (a) [$6$ pts] There are people who have taken 6.042 and have gotten A’s in $6.042$ (b) [$6$ pts] All people who are 6.042 TA’s and have taken 6.042 got A’s in $6.042$ (c) [$6$ pts] There are no people who are 6.042 TA’s who did not get A’s in $6.042$. (d) [$6$ pts] There are at least three people who are TA’s in $6.042$ and have not taken $6.042$ What I got thus far: a) $\exists x(S(x)\wedge A(x))$ b) $\forall x\in X((T(x)\wedge S(x))\Rightarrow A(x))$ c) $\nexists x\in X(T(x)\wedge ¬A(x))$ d) $\exists x\exists y\exists z\in X(¬E(x,y)∧¬E(x,z)∧¬E(y,z)∧(T(x)\wedge ¬S(x))∧(T(y)\wedge ¬S(y))∧(T(z)\wedge ¬S(z)))$ I'm not sure about any of my answers, but c) and d) gave me the most trouble. In regards to c),I am not sure about how to express "there are no people" in predicate logic. In regards to d), I am not sure about how to express "there are at least three people" in predicate logic. Hints are much better appreciated than an explicit answer. In addition, I would like to know how to tag this type of math. Is it considered discrete mathematics as most colleges/universities call it in the U.S.? • I would leave out the $\in X$ stuff. – André Nicolas Apr 25 '16 at 18:51 • Can you please explain why? Wouldn't have to define what $x$ is? – Cherry_Developer Apr 25 '16 at 19:16 • The domain is $X$, so it is understood that variables range over $X$. Also, $\in$ is not in the specified language. – André Nicolas Apr 25 '16 at 19:46 Your answers all look okay. Specifically for part c), you did indeed translate the sentence into predicate logic correctly. However, often times it is customary to not leave any negation symbols before the quantifiers. We can pass the negation symbol through the existential/universal quantifier by swapping them. For example $$\neg\exists x\in X(P(x))\iff\forall x\in X(\neg P(x))$$ and $$\neg\forall x\in X(P(x))\iff\exists x\in X(\neg P(x)).$$ Can you see how you can use this to simplify your answer for part c)? Also, I personally think the discrete math tag is okay for a question like this, especially since you also used the predicate logic tag. • I'm not sure about how to simplify it, but I gave it a shot: $\forall x\in X(¬T(x)\vee A(x))$ I was mostly unsure of how to apply the negation symbol to the predicate, so I typed "not (x and not y)" into Wolfram Alpha and it returned that "not x or y" would be an equivalent statement. – Cherry_Developer Apr 25 '16 at 19:13 • Yes, you simplified correctly. Here are the full steps: $\neg \exists x\in X(T(x)\wedge \neg A(x))\iff\forall x\in X\neg(T(x)\wedge \neg A(x))\iff\forall x\in X(\neg T(x)\vee A(x))$. – ervx Apr 25 '16 at 19:16 • The negation of $\exists x$ which becomes $\forall x$ is clear to me, but applying the negation to the predicate is where I get lost. I feel like I cheated myself by just resorting to Wolfram Alpha to find the simplified form of $(T(x)\wedge ¬A(x))$. Is $(T(x)\wedge ¬A(x))\Leftrightarrow (\neg T(x)\vee A(x))$ something I should just have memorized? Or is there a way for me to derive it? – Cherry_Developer Apr 25 '16 at 19:23 • In general, for expressions $A$ and $B$, we have $\neg(A\vee B)\iff (\neg A\wedge\neg B)$ and $\neg (A\wedge B)\iff (\neg A\vee \neg B)$. You can verify these using truth tables if you like. In general, though, it would be good to memorize them. They are essentially DeMorgan's laws for propositional logic. – ervx Apr 25 '16 at 19:26

https://math.stackexchange.com/questions/769059/number-of-club-members-if-they-are-organized-into-four-committees-if-each-member/769233

# Number of club members if they are organized into four committees if each member is in exactly two committees and any two committees share one member A club with $x$ members is organized into four committees such that $(a)$ each member is in exactly two committees, $(b)$ any two committees have exactly one member in common. Then $x$ has $(A)$ exactly two values both between $4$ and $8$ $(B)$ exactly one value and this lies between $4$ and $8$ $(C)$ exactly two values both between $8$ and $16$ $(D)$ exactly one value and this lies between $8$ and $16$. My thought is option $B$ and $x = 6$. Please somebody tell me am I right or wrong? • Can you show how you arrived at $x=6$? – RandomUser Apr 25 '14 at 16:53 • I just see that if $x$=$6$ then the two conditions are satisfied.But if $x$<$6$ or $x$>$6$ the condition $b$ is not satisfied so I think $x$ may be $6$. – liesel Apr 25 '14 at 17:15 • Please improve your title. – Alexander Gruber Apr 25 '14 at 17:22 Claim: $x=6$ Let $S:=[x]\left(=\{1,2,\dots ,x\}\right)$ where $x$ is some fixed natural number. Let $C_i\subset S$ be the $i$ th commitee.Let $f$ be a function from the set of unordered pair of distinct commitees to members of the club defined as, $$f\left( (C_i,C_j)\right)=C_i\cap C_j$$ $f$ is injective by (a). $f$ is surjective by (b). Hence, $|S|=\binom{4}{2}=6$. Here is a construction: • Let $C_1\cap C_2=1$ wlog. Then, $1\notin C_i \quad \forall \, i\in [4]\setminus \{1,2\}$. • Let $C_2\cap C_3=2$ wlog. Then, $2\notin C_i \quad \forall \, i\in [4]\setminus \{2,3\}$. • Let $C_3\cap C_4=3$ wlog. Then, $3\notin C_i \quad \forall \, i\in [4]\setminus \{3,4\}$. • Let $C_1\cap C_3=4$ wlog. Then, $4\notin C_i \quad \forall \, i\in [4]\setminus \{1,3\}$. • Let $C_1\cap C_4=5$ wlog. Then, $5\notin C_i \quad \forall \, i\in [4]\setminus \{1,4\}$. • Let $C_2\cap C_4=6$ wlog. Then, $6\notin C_i \quad \forall \, i\in [4]\setminus \{2,4\}$. Let the committees be $A_1, A_2, A_3, A_4$ then $|A_1\cup A_2 \cup A_3 \cup A_4|=n$. There are $\binom{4}{2}=6$ pairwise intersections each of size $1$ and each intersection of triples must be empty since if not some member would belong to $3$ committees, a contradiction. If we make a $4 \times n$ table where rows are committees and columns are members and mark a $1$ whenever member $x_j$ belongs to committee $A_i$ we notice that the $i^{th}$ row sum is $|A_i|$ and the $j^{th}$ column sum is $2$ (since each member belongs to exactly $2$ committees) so equating row and column sums we get $\sum_{i=1}^{4}|A_i|= 2n.$ Putting this information together and using the inclusion-exclusion formula we get $n =6$

https://math.stackexchange.com/questions/1724433/the-birthday-paradox/1724736

I would like a better understanding of the famous birthday paradox. "What is the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday?" I understood the first part, where the probability reaches 100% when the number of people reaches 367 by the pigeonhole principle. But I am not understanding the explanation beyond that. How do they say that the probability is 99.9% with 70 people and 50% with 23 people? And how do you further generalize the answer? And why is it a "paradox"? • Please, post the explanation you have been given. Apr 2 '16 at 10:00 • en.wikipedia.org/wiki/Birthday_problem May be this is helpful to you Apr 2 '16 at 10:00 • @Mc Cheng it might be a bit more helpful to specify a section rather than referring someone to an article (which is debatable in its reliability) that is quite hard (and lengthy) to sort through Apr 2 '16 at 10:03 • I could've typed an answer but explanations of this paradox are readily available all over the internet. Take a look at the Wikipedia article. Tell me exactly what you cannot understand. Maybe then we'd be able to help. Apr 2 '16 at 10:03 • Apr 2 '16 at 10:11 Let the number of people in the group be $n$. The probability that a pair of people don't share a birthday is given equal to $\frac{364}{365}$ ignoring leap years. There are $\binom{n}{2}$ pair of people in a group of $n$ people. No pair of people will share a birthday if each person has a distinct birthday. The probability of this happening is given by $$\frac{364}{365}\times\frac{363}{365} \dots \times \frac{365 - (n-1)}{365}$$ How did I get this probability? Assume that all birthdays are equally likely. If the first person was born on day $x_1$ then the second person in the group cannot be born on day $x_1$. The probability for this happening is $364\over 365$. Now let the birthday of the second person be $x_2$. The probability that the third person is not born on $x_1$ nor on $x_2$ is $363\over365$. Similarly we get the probability for the $n^{\text{th}}$ person. Since each event is independent, we multiply all the probabilities. Thus the probability that at least one pair shares a birthday for a group of $n$ people is given by $$p = 1 - \left(\frac{364}{365}\times\frac{363}{365} \dots \times \frac{365 - (n-1)}{365}\right)$$ Now you have the probability $p$ as a function of $n$. If you know the RHS, then you simply find for what value of $n$ we get the closest RHS to $p$ It so happens that if $p = 99.9\%$, the $n = 70$ • Thank you Banach Tarski! Am I right in understanding, that the only reason it is a "paradox" is because we'd usually expect the probability to be linear in nature? I don't see any paradox otherwise Apr 2 '16 at 10:11 • @aswa09 yup you are correct. A paradox is a self contradictory statement. The contradiction here is with our common sense and not with the mathematics :) Apr 2 '16 at 10:15 • Is it not a multiplication of $364/365$ not $1/365$? Else your formula gives a $364/365$ chance of just two people sharing the same birthday Apr 2 '16 at 10:21 • Yes, I had made a blunder in computing the probability, I have fixed this now :) Apr 2 '16 at 10:39 The paradox is that the rate of growth doesn't match our common sense. We expect that the way to count the number of possibilities for people to have the same birthday is directly from the number of people. However, in reality it's based on the number of pairs of people, which grows much faster than the number of people. Banach Tarski shows you the derivation which really is that the numerator is the number of people pairs, but still over 365. • How do I remove the duplication mark? I have edited the question asking about why it's a paradox, so that it isn't a duplicate question Apr 3 '16 at 7:23

https://math.stackexchange.com/questions/53262/factorial-decomposition-of-integers

# Factorial decomposition of integers? This question might seem strange, but I had the feeling it's possible to decompose in a unique way a number as follows: if $x < n!$, then there is a unique way to write x as: $$x = a_1\cdot 1! + a_2\cdot 2! + a_3\cdot3! + ... + a_{n-1}\cdot(n-1)!$$ where $a_i \leq i$ I looked at factorial decomposition on google but I cannot find any name for such a decomposition. example: If I chose : (a1,a2) = • 1,0 -> 1 • 0,1 -> 2 • 1,1 -> 3 • 0,2 -> 4 • 1,2 -> 5 I get all number from $1$ to $3!-1$ ideas for a proof: The number of elements between $1$ and $N!-1$ is equal to $N!-1$ and I have the feeling they are all different, so this decomposition should be right. But I didn't prove it properly. Are there proofs of this decomposition? Does this decomposition as a name? And above all is this true ? • It's true, and the proof is easy by induction on $n$. What have you written down? – Olivier Bégassat Jul 23 '11 at 9:19 • I am not sure about number-theory tag, if someone has a better idea, please retag the question. – Martin Sleziak Jul 23 '11 at 9:21 • @Olivier, I actually thinking of a mapping between n! and all permutation, and I should have used that as a proof. But since I couldn't find anything on the web about that I had the feeling something was wrong ... :D! – Ricky Bobby Jul 23 '11 at 9:28 • @Martin: Possibly combinatorics? – Brian M. Scott Jul 23 '11 at 9:39 • If any of you have a good reason to think (number-systems) doesn't belong, please remove that tag... – J. M. is a poor mathematician Jul 23 '11 at 11:56 You're looking for the factorial number system, also known as "factoradic". Searching should give you more results. Yes, it's true that such a decomposition is always possible. One way to prove it is as follows: given $x < n!$, consider the $x$th permutation of some ordered set of $n$ symbols. This is some permutation $(s_1, s_2, \dots, s_n)$. Now for $s_1$ you had $n$ choices (label them $0$ to $n-1$) and you picked one, so let $a_{n-1}$ be the choice you made. For $s_2$ you had $n-1$ choices (label them $0$ to $n-2$) and you picked one, so let $a_{n-2}$ be the number of the choice you made. Etc. $a_0$ is always $0$ because you have only one choice for the last element. (This is also known as the Lehmer code.) • I was actually thinking about permutation when this question came to my mind, I didn't thought about using it to prove the decomposition. thanks a lot for your answer. – Ricky Bobby Jul 23 '11 at 9:17 • Why not try what would seem obvious? Divide your numberx by n!, take the highest multiplr $a_n$ with $a_nn!<x$, then divide the remainder x-$a_n n!$ by (n-1)! and take $a_{n-2}$ as the largest multiple, etc. at the end, you will be left with some amount which is always a multiple of 1=1! – gary Jul 23 '11 at 9:37 • @gary: Yes, something like that would work too, if done correctly. But for me, this map numbering permutations was actually more obvious. :-) (You want to divide by $(n-1)$ the first time, also you need to prove that $a_i \le i$ for all $i$. This is easy, but it's not just one line.) – ShreevatsaR Jul 23 '11 at 9:46 • Shreevatsa: if your coefficient for n! was larger than n, then n(n-1)!=n! , and you could then increase your previous coefficient for n! by (at least )1. But alternative explanations are always useful,helpful. – gary Jul 23 '11 at 10:12 Your conjecture is correct. There is a straightforward proof by induction that such a decomposition always exists. Suppose that every positive integer less than $n!$ can be written in the form $\sum_{k=1}^{n-1} a_k k!$, where $0 \le a_k \le k$, and let $m$ be a positive integer such that $n! \le m < (n+1)!$. There are unique integers $a_n$ and $r$ such that $m = a_nn! + r$ and $0 \le r < n!$, and since $m < (n+1)! = (n+1)n!$, it’s clear that $a_n \le n$. Since $r < n!$, the induction hypothesis ensures that there are non-negative integers $a_1,\dots,a_{n-1}$ such that $r = \sum_{k=1}^{n-1} a_k k!$, and hence $m = \sum_{k=1}^n a_k k!$. We’ve now seen that each of the $(n+1)!$ non-negative integers in $\{0,1,\dots,n\}$ has a representation of the form $\sum_{k=1}^n a_k k!$ with $0 \le a_k \le k$ for each $k$. However, there are only $\prod_{k=1}^n (k+1) = (n+1)!$ distinct representations of that form, so each must represent a different integer, and each integer’s representation is therefore unique. You can also reason as follows : suppose you've shown for some integer $n$ that every integer $\in\lbrace0,\dots,n!-1\rbrace$ has a unique decomposition as you suggest. Take $k\in\lbrace0,\dots,(n+1)!-1\rbrace.$ Write $$k=q\cdot n!+r$$ the euclidean division of $k$ with respect to $n!$. Necessarily you have $0\leq q < n+1$. This gives you an expression you want, for $0\leq r<n!$ has, by hypothesis, an expression involving only factorials up to $(n-1)!$ . Finally, to show uniqueness, you can again use your hypothesis to deduce that if $k=a_n\cdot n!+ \sum_0^{n-1} a_i\cdot i!,$ then $0\leq \sum\dots< n!-1$ by hypothesis, and this tells you that this decomposition is the euclidean division of $k$ by $n!$. So by uniqueness of the euclidean division, it's the one decomposition we defined earlier. This generalizes to the representation $$n = \sum_{i\ge 0} a_i b_i$$ where $b_0 = 1$ and $b_{n+1} = b_n c_n$ with $c_n > 1$ and $0 \le a_i < c_i$ - i.e., representing $n$ with digits $a_i$ to a "base" with varying ratios of consecutive digit values (phrased awkwardly, I know, but it is late and I am tired). For a usual base $B$ system, $c_i = B$ for all $i$, so $b_i = B^i$. For this system, $c_i = i$ (or maybe $c_i = i+1$, modulo my tiredness), so $b_i = i!$. I proved too many years ago that this representation is unique iff the $c_i$ were all integers. I am sure that this was proved many years ago (probably by Euler, if not Fermat) and is in Dickson somewhere.

https://math.stackexchange.com/questions/2181828/why-is-a-number-that-is-divisible-by-2-and-3-also-divisible-by-6

# Why is a number that is divisible by $2$ and $3$, also divisible by $6$? Why is it that a number such as $108$, that is divisible by $2$ and $3$, is also divisible by $6$? Is this true that all numbers divisible by two integers are divisible by their product? • Think about the factorization of a number that's divisible by 2 and 3. – Ben Longo Mar 11 '17 at 14:20 • Every number can be expressed as the product of primes. $2$ & $3$ are coprime ... Can you get it from there Jason ? – Donald Splutterwit Mar 11 '17 at 14:21 • Provided the numbers are mutually prime. – SchrodingersCat Mar 11 '17 at 14:22 • note $108 = 2^2 \cdot 3^3 = 3 \cdot (2\cdot 3) \cdot (2 \cdot 3) = 3 \cdot (6) \cdot (6)$ – Dando18 Mar 11 '17 at 14:23 • What about 30, that's divisible by 10 and 15? – Ethan Bolker Mar 11 '17 at 14:23 It is certainly not true that if $x$ is divisible by both $y$ and $z$, then $x$ is divisible by $yz$. For example, $2$ is divisible by $2$ and $2$, but is not divisible by $2\cdot 2$. As a less silly example, $120$ is divisible by both $15$ and $6$, but is not divisible by $90$. But if you think about the examples above, you'll notice that the factors involved in each overlap: e.g. both $2$ and $2$ are even, and both $15$ and $6$ are divisible by $3$. Meanwhile, $2$ and $3$ don't overlap - they share no factors in common. It turns out that this is the crucial property: If $x$ is divisible by $y$ and $z$, and $y$ and $z$ have no factors in common besides $1$, then $x$ is divisible by $yz$. Proving this is a good challenge. HINT: think about factorization into primes . . . • There is perhaps something to add here, since there are divisors that fulfil these rules and DO have factors in common, but still divide some number, e.g. in the case $2$ divides $y$ and $z$, and $4$ divides $x$ – samerivertwice Mar 11 '17 at 14:27 It is not true that for any two integers $m$, $n$, if $x$ is divisible by $m$ and $n$, it is divisible by $mn$. For example $4$ is divisible by $4$ and $2$, but it is not divisible by $8$. However, this is true if $m$ and $n$ are relatively prime, i.e. the greatest common divisor of $m$ and $n$ is $1$. Thus, since $2$ and $3$ are relatively prime, any number divisible by $2$ and $3$ will also be divisible by $6$. More generally, if $x$ is divisible by $m$ and $n$, then it must be divisible by $\operatorname{lcm}(m,n)$ (least common multiple). Case of $m$ and $n$ being relatively prime is a special case of this, since $$\operatorname{lcm}(m,n) = \frac{mn}{\operatorname{gcd}(m,n)}$$ where $\operatorname{gcd}$ stands for the greatest common divisor and thus $\operatorname{lcm}(m,n) = mn$, when $m$ and $n$ are relatively prime. Let $a$ be divisible by $2$ and $3$. So $$a=2^{r_{0}}\cdot p^{r_{1}}_{1}\cdot p^{r_{2}}_{2}\cdot\ldots=3^{s_{0}}\cdot q^{s_{1}}_{1}\cdot q^{s_{2}}_{2}\cdot\ldots$$ where $p_{i}$ and $q_{j}$ are distinct prime numbers, $r_{i}$ and $s_{j}$ are possibly $0$ for $1\leqslant i,j<\infty$ and $r_{0},s_{0}\geqslant 1$. We know from the fundamental theorem that these two must be unique factorisations (up to re-ordering). Hence there must be some $p_{k}^{r_{k}}=3^{s_{0}}$ and some $q_{\ell}^{s_{\ell}}=2^{r_{0}}$. Therefore $a$ has a factor of $6$: \begin{align*} a&=2^{r_{0}}\cdot p_{1}^{r_{1}}\ldots\cdot p_{k}^{r_{k}}\cdot\ldots\\ &=2^{r_{0}}\cdot p_{1}^{r_{1}}\ldots\cdot3^{s_{0}}\cdot\ldots\\ &=6\cdot(2^{r_{0}-1}\cdot3^{s_{0}-1}\cdot p_{1}^{r_{1}}\cdot\ldots). \end{align*} (The argument works on the other side as well). This might be a little explicit but I hope it helps. As the comments point out, and as you may suspect, it is not true in general that if a number is divisible by two integers, then it is divisible by their product. What we can say is the following: If $x,y,z$ are integers such that $z$ is divisible by both $x$ and $y,$ then $z$ is divisible by the least common multiple of $x$ and $y.$ This isn't actually much of a surprise, given the way that least common multiple is often defined: Given two integers $x$ and $y,$ the least common multiple of $x$ and $y$ is the nonnegative integer $z$ such that (i) $z$ is divisible by both $x$ and $y,$ and (ii) every common multiple of $x$ and $y$ is divisible by $z.$ Proving that such a multiple exists isn't exactly straightforward, though. It is slightly easier to prove the existence of another integer: Given two integers $x$ and $y,$ the greatest common factor of $x$ and $y$ is the positive integer $z$ such that (i) both $x$ and $y$ are divisible by $z,$ and (ii) $z$ is divisible by every common factor of $x$ and $y.$ One can, for example, use the Euclidean algorithm to prove that $\operatorname{gcf}(x,y)$ exists, regardless of $x,y.$ At that point, one can show that $\frac{|xy|}{\operatorname{gcf}(x,y)}$ turns out to be the least common multiple of $x$ and $y.$ As an immediate consequence of the identity $$\operatorname{lcm}(x,y)=\frac{|xy|}{\operatorname{gcf}(x,y)},$$ together with the result mentioned at the beginning, we find the following: If $x,y,z$ are integers such that $z$ is divisible by both $x$ and $y,$ and if $\operatorname{gcf}(x,y)=1,$ then $z$ is divisible by $xy.$ Since $\operatorname{gcf}(2,3)=1,$ then any number divisible by both $2$ and $3$ is necessarily divisible by $6.$ The general result is this: If a number $n$ is divisible by $a$ and $b$, then it is divisible by $\DeclareMathOperator{\lcm}{lcm}\lcm(a,b)$. Indeed, let $\;d=\gcd(a,b)$, $\;m=\lcm(a,b)$, $\;a'=\dfrac ad$, $\;b'=\dfrac bd$. These numbers satisfy the relation $$md=ab, \enspace\text{so}\quad m=\frac{ab}d=a'b=ab'.$$ By hypothesis, we can write $$n=qa=qda', \qquad n=rb=rb'd$$ We deduce that $\;qa'=rb'$. So $a'$ divides $rb'$. As $a'$ and $b'$ are coprime, Gauß' lemma ensures $a'$ divides $r$, i.e. we have $r=r'a'$, and $$n=(r'a')b'd=r'(a'b)=r'm.$$

https://uk.mathworks.com/help/symbolic/sym.changeintegrationvariable.html

# changeIntegrationVariable Integration by substitution ## Description example G = changeIntegrationVariable(F,old,new) applies integration by substitution to the integrals in F, in which old is replaced by new. old must depend on the previous integration variable of the integrals in F and new must depend on the new integration variable. For more information, see Integration by Substitution. When specifying the integrals in F, you can return the unevaluated form of the integrals by using the int function with the 'Hold' option set to true. You can then use changeIntegrationVariable to show the steps of integration by substitution. ## Examples collapse all Apply a change of variable to the definite integral ${\int }_{\mathit{a}}^{\mathit{b}}\mathit{f}\left(\mathit{x}+\mathit{c}\right)\text{\hspace{0.17em}}\mathit{dx}$. Define the integral. syms f(x) y a b c F = int(f(x+c),a,b) F = ${\int }_{a}^{b}f\left(c+x\right)\mathrm{d}x$ Change the variable $\mathit{x}+\mathit{c}$ in the integral to $\mathit{y}$. G = changeIntegrationVariable(F,x+c,y) G = ${\int }_{a+c}^{b+c}f\left(y\right)\mathrm{d}y$ Find the integral of $\int \mathrm{cos}\left(\mathrm{log}\left(\mathit{x}\right)\right)\mathit{dx}$ using integration by substitution. Define the integral without evaluating it by setting the 'Hold' option to true. syms x t F = int(cos(log(x)),'Hold',true) F = $\int \mathrm{cos}\left(\mathrm{log}\left(x\right)\right)\mathrm{d}x$ Substitute the expression log(x) with t. G = changeIntegrationVariable(F,log(x),t) G = $\int {\mathrm{e}}^{t} \mathrm{cos}\left(t\right)\mathrm{d}t$ To evaluate the integral in G, use the release function to ignore the 'Hold' option. H = release(G) H = $\frac{{\mathrm{e}}^{t} \left(\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)\right)}{2}$ Restore log(x) in place of t. H = simplify(subs(H,t,log(x))) H = $\frac{\sqrt{2} x \mathrm{sin}\left(\frac{\pi }{4}+\mathrm{log}\left(x\right)\right)}{2}$ Compare the result to the integration result returned by int without setting the 'Hold' option to true. Fcalc = int(cos(log(x))) Fcalc = $\frac{\sqrt{2} x \mathrm{sin}\left(\frac{\pi }{4}+\mathrm{log}\left(x\right)\right)}{2}$ Find the closed-form solution of the integral $\int \mathit{x}\text{\hspace{0.17em}}\mathrm{tan}\left(\mathrm{log}\left(\mathit{x}\right)\right)\mathit{dx}$. Define the integral using the int function. syms x F = int(x*tan(log(x)),x) F = $\int x \mathrm{tan}\left(\mathrm{log}\left(x\right)\right)\mathrm{d}x$ The int function cannot find the closed-form solution of the integral. Substitute the expression log(x) with t. Apply integration by substitution. syms t G = changeIntegrationVariable(F,log(x),t) G = The closed-form solution is expressed in terms of hypergeometric functions. For more details on hypergeometric functions, see hypergeom. Compute the integral ${\int }_{0}^{1}{\mathit{e}}^{\sqrt{\mathrm{sin}\left(\mathit{x}\right)}}\mathit{dx}$ numerically with high precision. Define the integral ${\int }_{0}^{1}{\mathit{e}}^{\sqrt{\mathrm{sin}\left(\mathit{x}\right)}}\mathit{dx}$. A closed-form solution to the integral does not exist. syms x F = int(exp(sqrt(sin(x))),x,0,1) F = ${\int }_{0}^{1}{\mathrm{e}}^{\sqrt{\mathrm{sin}\left(x\right)}}\mathrm{d}x$ You can use vpa to compute the integral numerically to 10 significant digits. F10 = vpa(F,10) F10 = $1.944268879$ Alternatively, you can use the vpaintegral function and specify the relative error tolerance. Fvpa = vpaintegral(exp(sqrt(sin(x))),x,0,1,'RelTol',1e-10) Fvpa = $1.944268879$ The vpa function cannot find the numerical integration to 70 significant digits, and it returns the unevaluated integral in the form of vpaintegral. F70 = vpa(F,70) F70 = $1.944268879138581167466225761060083173280747314051712224507065962575967$ To find the numerical integration with high precision, you can perform a change of variable. Substitute the expression $\sqrt{\mathrm{sin}\left(\mathit{x}\right)}$ with $\mathit{t}$. Compute the integral numerically to 70 significant digits. syms t; G = changeIntegrationVariable(F,sqrt(sin(x)),t) G = ${\int }_{0}^{\sqrt{\mathrm{sin}\left(1\right)}}\frac{2 t {\mathrm{e}}^{t}}{\sqrt{1-{t}^{4}}}\mathrm{d}t$ G70 = vpa(G,70) G70 = $1.944268879138581167466225761060083173280747314051712224507065962575967$ ## Input Arguments collapse all Expression containing integrals, specified as a symbolic expression, function, vector, or matrix. Subexpression to be substituted, specified as a symbolic scalar variable, expression, or function. old must depend on the previous integration variable of the integrals in F. New subexpression, specified as a symbolic scalar variable, expression, or function. new must depend on the new integration variable. collapse all ### Integration by Substitution Mathematically, the substitution rule is formally defined for indefinite integrals as $\int f\left(g\left(x\right)\right)\text{\hspace{0.17em}}g\text{'}\left(x\right)\text{\hspace{0.17em}}dx={\left(\int f\left(t\right)\text{\hspace{0.17em}}dt\right)|}_{t=g\left(x\right)}$ and for definite integrals as $\underset{a}{\overset{b}{\int }}f\left(g\left(x\right)\right)\text{\hspace{0.17em}}g\text{'}\left(x\right)\text{\hspace{0.17em}}dx=\underset{g\left(a\right)}{\overset{g\left(b\right)}{\int }}f\left(t\right)\text{\hspace{0.17em}}dt.$

https://math.stackexchange.com/questions/2944265/on-the-integral-int-0-pi-sinx-sinx-sinx-cdots-dx

# On the integral $\int_0^\pi\sin(x+\sin(x+\sin(x+\cdots)))\,dx$ This question came into my head when I did a course on Fourier series. However, this is not an infinite sum of sines, but an infinite recurrence of sines in a sum. Consider $$f_1(x)=\sin(x)$$ and $$f_2(x)=\sin(x+f_1(x))$$ such that $$f_n$$ satisfies the relation $$f_n(x)=\sin(x+f_{n-1}(x)).$$ To what value does $$L:=\lim_{n\to\infty}\int_0^\pi f_n(x)\,dx$$ converge? Since it is impossible to evaluate the integrals directly, we begin by considering the first few values of $$n$$. A pattern clearly emerges. $$I_1=\int_0^\pi f_1(x)\,dx=2\quad\quad\quad I_2=1.376527...\\I_3=2.188188...\quad\quad\quad\quad\quad I_4=1.625516...\\ I_5=2.179090...\quad\quad\quad\quad\quad I_6=1.732942...\\ I_7=2.155900...\quad\quad\quad\quad\quad I_8=1.927035...$$ For odd values of $$n$$, $$I_n$$ decreases monotonically (except $$n=1$$) and for even values of $$n$$, $$I_n$$ increases monotonically. These two observations have led me to claim that $$L=I_1=2$$. Is it possible to prove/disprove this claim? • I feel like a numerical DE approach: let $y=\sin(x+\sin(x+\cdots))$ so that $y=\sin(x+y),$ differentiating, and using Mathematica's NDSolve command should produce a result, but it has numerical instabilities, unfortunately. – Adrian Keister Jul 11 '19 at 17:53 Outline: • Use the inverse function of $$y=x-\sin x$$ to express $$f_\infty(x)$$. • Use integral of inverse functions and dominated convergence theorem to prove $$L=2$$. Claim:$$L=2.$$ Proof: Obviously $$y=t-\sin t$$ is injective on $$t\in[0,\pi]$$. Define $$y=\operatorname{Sa}(t)$$ as the inverse function of $$y=t-\sin t$$ on $$t\in[0,\pi]$$. Therefore, $$t-\sin t =x \implies t=\operatorname{Sa}(x).$$ Assume $$f_\infty(x)$$ exists (see 1. the first integral), then we have \begin{align*} f_\infty&=\sin(x+f_\infty)\\ \underbrace{(x+f_\infty)}_{t}-\sin\underbrace{(x+f_\infty)}_{t}&=x\\ x+f_\infty&=\operatorname{Sa}(x)\\ f_\infty(x)&=-x+\operatorname{Sa}(x). \end{align*} Since $$0-\sin 0 =0\implies \operatorname{Sa}(0)=0$$ and $$\pi-\sin \pi =\pi\implies \operatorname{Sa}(\pi)=\pi$$, \begin{align*} \int_0^\pi f_\infty(x)\,\mathrm dx&=\int_0^\pi -x+\operatorname{Sa}(x)\,\mathrm dx\\ &=\int_0^\pi -x\,\mathrm dx+\int_0^\pi \operatorname{Sa}(x)\,\mathrm dx\\ &=-\frac{\pi^2}2+\left(\pi \operatorname{Sa}(\pi)-0 \operatorname{Sa}(0)-\int_{\operatorname{Sa}(0)}^{\operatorname{Sa}(\pi)}y-\sin y\,\mathrm dy\right)\\ &=-\frac{\pi^2}2+\left(\pi^2-\int_0^\pi y-\sin y\,\mathrm dy\right)\\ &=-\frac{\pi^2}2+\left(\pi^2-\left[\frac{y^2}2+\cos y\right]^\pi_0\right)\\ &=2. \end{align*} Here we used integral of inverse functions: $$\int_c^df^{-1}(y)\,\mathrm dy+\int_a^bf(x)\,\mathrm dx=bd-ac.$$ Note: Since $$|f_n(x)|\le 1$$ and $$1$$ is integrable on $$[0,\pi]$$, we could interchange limit sign and integral sign from dominated convergence theorem, that is, $$L:=\lim_{n\to\infty}\int_0^\pi f_n(x)\,\mathrm dx=\int_0^\pi \lim_{n\to\infty}f_n(x)\,\mathrm dx=\int_0^\pi f_\infty(x)\,\mathrm dx=2.$$ • What a marvelous answer! – Szeto Oct 6 '18 at 22:29 • That is amazing! I've posted a similar question, but this time with multiplication. Do you have any thoughts on it? Cheers. – TheSimpliFire Oct 7 '18 at 9:19

http://mathhelpforum.com/calculus/19332-taylor-series.html

1. ## taylor series Can anyone help me with how to get the Taylor series of a function(for example sinx)? Furthur how to get a value for sin(for example pi/4) with an particular error? Thank You 2. Originally Posted by Dili Can anyone help me with how to get the Taylor series of a function(for example sinx)? Furthur how to get a value for sin(for example pi/4) with an particular error? Thank You The Taylor series for a function $f(x)$ (at least I presume you are talking about a single variable function) to N + 1 terms about a point x = a has the form: $f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + \frac{1}{2!}f^{\prime \prime}(a)(x - a)^2 + ~ ... ~ + \frac{1}{N!}f^{(N)}(a)(x - a)^N$ or in summation notation: $f(x) \approx \sum_{k = 0}{N}\frac{1}{k!}f^{(k)}(a)(x - a)^k$ For the error estimate I will refer you here as I have forgotten which one is "standard" (if any.) So as an example, let $f(x) = sin(x)$ and let us expand the series about the point $x = \frac{\pi}{4}$. We have: $f(x) = sin(x) \implies f \left ( \frac{\pi}{4} \right ) = sin \left ( \frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$ $f^{\prime}(x) = cos(x) \implies f^{\prime} \left ( \frac{\pi}{4} \right ) = cos \left ( \frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$ $f^{\prime \prime}(x) = -sin(x) \implies f^{\prime \prime} \left ( \frac{\pi}{4} \right ) = -sin \left ( \frac{\pi}{4} \right ) = -\frac{\sqrt{2}}{2}$ $f^{\prime \prime \prime}(x) = -cos(x) \implies f^{\prime \prime \prime} \left ( \frac{\pi}{4} \right ) = -cos \left ( \frac{\pi}{4} \right ) = -\frac{\sqrt{2}}{2}$ So the Taylor series to 4 terms looks like: $sin(x) \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \left ( x - \frac{\pi}{4} \right ) - \frac{\sqrt{2}}{4} \left ( x - \frac{\pi}{4} \right )^2 - \frac{\sqrt{2}}{12} \left ( x - \frac{\pi}{4} \right )^3$ -Dan 3. ## lagrange's remainder Thank you But what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)? And can you explain with the Lagrange's remainder? 4. Originally Posted by Dili Thank you But what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)? And can you explain with the Lagrange's remainder? I'll have to let someone else explain the remainder (I'm just not up on it myself, which is why I posted the link. ) Yes, you could use the Maclaurin expansion (the Taylor series expansion about the point x = 0), but note that $\frac{\pi}{4} \approx 0.785398$ is not particularly close to 0. The further x is from 0, the greater the error in the approximation. On the other hand using 3 terms in the expansion I get $sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5$ Gives me $sin \left ( \frac{\pi}{4} \right ) \approx 0.707143$ which is only off by $5.129 \times 10^{-3}$ %, which is pretty close by most people's judgment. -Dan 5. Theorem: Let $f(x)$ be an infinitely differenciable function on $(a,b)$ (with $a<0). Let $T_n(x)$ represent the $n$-th degree Taylor polynomial around the origin for the point $x\in (a,b)$ (with $x\not = 0$). And let $R_{n+1}(x) = f(x) - T_{n}(x)$ be the remainder term. Then there exists a number $y$ strictly between $0$ and $x$ so that, $R_{n+1}(x) = \frac{f^{(n+1)}(y)}{(n+1)!}\cdot x^{n+1}$. So given $f(x) = \sin x$ let us work on the interval $(a,b)$ where $a=-\infty$ and $b=+\infty$. This function is infinitely differenciable so the above results apply. You can to approximate $T_n \left( \frac{\pi}{4} \right)$. Now by the theorem we know that, $R_{n+1}\left( \frac{\pi}{4} \right) = \frac{f^{(n+1)} \left( \frac{\pi}{4} \right)}{(n+1)!} \cdot \left( \frac{\pi}{4} \right)^{n+1}$ for some $0. Notice that $f^{n+1}$ is one of these: $\sin x,\cos x,-\sin x,-\cos x$. Thus, $|f^{n+1}|\leq 1$. And also notice that $\frac{\pi}{4} \leq 1$ thus, $\left(\frac{\pi}{4}\right)^{n+1} \leq 1$. This means, $\left| R_{n+1}\left( \frac{\pi}{4}\right) \right| \leq \frac{1}{(n+1)!}$. This approximation might be greatly improved all I did was place an approximation that works and furthermore converges rapidply. Remark) The theorem applys in more general to $(n+1)$-differentiable functions but there was no need to use it here.

https://math.stackexchange.com/questions/2357956/calculate-the-probability-that-juniors-are-the-majority-of-random-selection

# Calculate the probability that Juniors are the majority of random selection. In a business school at ABC University, 1/6 of the enrollment are freshmen, 3/10 are sophomores, 1/3 are juniors and 1/5 are seniors. They compete with several other schools within their region in an annual financial case competition. Three students were randomly selected to form a team for this year’s competition. Calculate the probability that the juniors would make up the majority of the team. I thought that this probability is equal to the probability that exactly 2 out 3 selected are juniors plus the probability that all 3 are juniors. Therefore, I computed the following: $$P(exactly~2~juniors) = \frac13 \times \frac13 \times \frac23 = \frac2{27}$$ $$P(3~juniors) = (\frac13)^3$$ Therefore, $$P(juniors~make~up~the~majority)= \frac1{27} + \frac2{27} = \frac3{27}$$ My answer is wrong as I have only partially computed the probability. I would appreciate some explanation as what I am missing as I am having trouble figuring it out. Thank you Your formula $\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}$ to pick the two juniors is the probability of first selecting two juniors, and then the non-junior. You can also first pick a junior, then a non-junior, and then a junior again, the probability of which is: $\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}$. And, of course, you can first pick the non-junior, and then the two juniors, with a probability of $\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3}$. All these give you 2 juniors and 1 non-junior, so the probability of picking exactly 2 juniors is: $\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}+\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}+\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3} = \frac{6}{27}=\frac{2}{9}$ • I thought that the order in which I choose the students doesn't matter. It obviously does per your solution (thank you!). – the boy 88 Jul 13 '17 at 20:59 • Can I think of my sample space as all combinations of 3-student tuples? And the event I want is choosing exactly 2 juniors and 1 non-junior so the outcomes that satisfy the event are { JJSo, JJF, and JJSe}? So = sophomore, Se = senior, F= freshman. – the boy 88 Jul 13 '17 at 21:07 • @Kovs95 Yes, you can. And you can see them as unordered triples (basically sets with 3 elements), or as ordered triples. If seeing them as unordered, you need to multiply by 3 which is what sds did.. if you see them as ordered you get the formula I got. Same result of course! Now, there is really no need to distinguish between freshmen and sophomores and senior ... you just need to distinguish between junior (1/3 chance) and non-juniors (2/3 chance) – Bram28 Jul 13 '17 at 21:22 You can select the non-junior in 3 different ways (1st, 2nd, 3rd), so $$P(exactly~2~juniors) = \frac13 \times \frac13 \times \frac23 \times 3 = \frac29$$ Thus $$P(juniors~ make~ up~ the~ majority)=\frac29+\frac1{27}=\frac{7}{27}$$ • yes sorry I fixed the 2/3 probability of not selecting a junior. – the boy 88 Jul 13 '17 at 20:43 • yes, but this is not the only problem - see "times 3"! – sds Jul 13 '17 at 20:44

https://gmatclub.com/forum/what-are-the-last-two-digits-of-195836.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 15 Nov 2018, 13:31 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track Your Progress every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History ## Events & Promotions ###### Events & Promotions in November PrevNext SuMoTuWeThFrSa 28293031123 45678910 11121314151617 18192021222324 2526272829301 Open Detailed Calendar • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. • ### GMATbuster's Weekly GMAT Quant Quiz # 9 November 17, 2018 November 17, 2018 09:00 AM PST 11:00 AM PST Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage. # What are the last two digits of 63*35*37*82*71*41? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 50613 What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 07 Apr 2015, 05:11 3 11 00:00 Difficulty: 55% (hard) Question Stats: 55% (01:26) correct 45% (01:10) wrong based on 204 sessions ### HideShow timer Statistics What are the last two digits of 63*35*37*82*71*41? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 Kudos for a correct solution. _________________ ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 50613 Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 13 Apr 2015, 06:27 2 5 Bunuel wrote: What are the last two digits of 63*35*37*82*71*41? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100. Remainder of (63*35*37*82*71*41)/ 100 Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note: Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again. So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct. Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5. We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct. So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2. We need the Remainder of (63*7*37*41*71*41*5*2)/10*5*2 Remainder of (63*7*37*41*71*41)/10 Now using concept 2, let’s write the numbers in form of multiples of 10 Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10 Remainder of 3*7*7*1*1*1/10 Remainder of 147/10 = 7 Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70. When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70. Answer (D) _________________ ##### General Discussion Director Joined: 07 Aug 2011 Posts: 539 Concentration: International Business, Technology GMAT 1: 630 Q49 V27 Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 07 Apr 2015, 07:49 1 Bunuel wrote: What are the last two digits of 63*35*37*82*71*41? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 Kudos for a correct solution. 63*35 = 05 *37 = 85 * 82 = 70 * 71 = 70*41 = 70 _________________ Thanks, Lucky _______________________________________________________ Kindly press the to appreciate my post !! Current Student Joined: 25 Nov 2014 Posts: 99 Concentration: Entrepreneurship, Technology GMAT 1: 680 Q47 V38 GPA: 4 Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 07 Apr 2015, 10:36 4 First, multiplied 82 and 35, since 2 and 5 will give a 0 in the end. Got last 2 digits 70. no need to multiply 70 by 41 and 71, since they wont change the last 2 digits (try if you want to, but effective multiplication in case of 41 and 71 is only with 1). Now, 70 * 3(of 63) gives : last 2 digits as 10. lastly, 10 * 7(of 37) gives : last 2 digits 70. Thus D. _________________ Kudos!! Manager Joined: 26 Dec 2012 Posts: 146 Location: United States Concentration: Technology, Social Entrepreneurship WE: Information Technology (Computer Software) Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 09 Apr 2015, 12:31 63*35*37*82*71*41= We have to focus on the last two digits only, so 63*35=05*37=85*82=70 71*41=81 therefore 81*70=70 Hence Answer is D Thanks, CEO Joined: 11 Sep 2015 Posts: 3120 Location: Canada Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 19 Jul 2016, 10:10 Top Contributor 1 Bunuel wrote: What are the last two digits of (63)(35)(37)(82)(71)(41)? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 (63)(35)(37)(82)(71)(41) = (63)(37)(71)(41)(35)(82) Multiply the units digits in the red product to get: (#####1)(35)(82) Rewrite blue product as follows: (#####1)(7)(5)(2)(41) Rearrange: (#####1)(7)(41)(5)(2) The units digit of (7)(41) is 7, so we get: (#####1)(##7)(10) = (####7)(10) = #####70 Answer: _________________ Brent Hanneson – GMATPrepNow.com Senior Manager Joined: 31 Jul 2017 Posts: 490 Location: Malaysia Schools: INSEAD Jan '19 GMAT 1: 700 Q50 V33 GPA: 3.95 WE: Consulting (Energy and Utilities) Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 04 Feb 2018, 23:10 Bunuel wrote: What are the last two digits of 63*35*37*82*71*41? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 Kudos for a correct solution. The equation can be written as $$7*9*7*5*37*41*2*71*41$$ $$49*9*10*37*41*71*41$$ = $$401*10*37*41*41*71$$ As you can see, the last digit will be 0, and except for 37 all the digits end with 1. So, last two Digit will be 70. _________________ If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !! Manager Joined: 02 Jan 2016 Posts: 117 Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 07 Feb 2018, 10:30 Bunuel wrote: Bunuel wrote: What are the last two digits of 63*35*37*82*71*41? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100. Remainder of (63*35*37*82*71*41)/ 100 Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note: Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again. So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct. Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5. We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct. So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2. We need the Remainder of (63*7*37*41*71*41*5*2)/10*5*2 Remainder of (63*7*37*41*71*41)/10 Now using concept 2, let’s write the numbers in form of multiples of 10 Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10 Remainder of 3*7*7*1*1*1/10 Remainder of 147/10 = 7 Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70. When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70. Answer (D) Hi I thinks this "Remainder of 147/10 = 7 Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70." same as 1470/100 = 70 right ? also Remainder 3*-3*-3*1*1*1/10 seem more efficient than 3*7*7*1*1*1/10 ? I hope this way is correct too ? Manager Joined: 26 Sep 2017 Posts: 97 Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 07 Feb 2018, 12:56 Bunuel wrote: Bunuel wrote: What are the last two digits of 63*35*37*82*71*41? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100. Remainder of (63*35*37*82*71*41)/ 100 Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note: Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again. So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct. Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5. We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct. So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2. We need the Remainder of (63*7*37*41*71*41*5*2)/10*5*2 Remainder of (63*7*37*41*71*41)/10 Now using concept 2, let’s write the numbers in form of multiples of 10 Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10 Remainder of 3*7*7*1*1*1/10 Remainder of 147/10 = 7 Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70. When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70. Answer (D) Hi bunuel .. Can you enlighten how to approach this kind more other questions like to find last three digit..etc etc.. And in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there... Sent from my BND-AL10 using GMAT Club Forum mobile app Manager Joined: 02 Jan 2016 Posts: 117 Re: What are the last two digits of 63*35*37*82*71*41?  [#permalink] ### Show Tags 09 Feb 2018, 07:37 viv007 wrote: Bunuel wrote: Bunuel wrote: What are the last two digits of 63*35*37*82*71*41? (A) 10 (B) 30 (C) 40 (D) 70 (E) 80 Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100. Remainder of (63*35*37*82*71*41)/ 100 Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note: Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again. So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct. Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5. We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct. So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2. We need the Remainder of (63*7*37*41*71*41*5*2)/10*5*2 Remainder of (63*7*37*41*71*41)/10 Now using concept 2, let’s write the numbers in form of multiples of 10 Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10 Remainder of 3*7*7*1*1*1/10 Remainder of 147/10 = 7 Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70. When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70. Answer (D) Hi bunuel .. Can you enlighten how to approach this kind more other questions like to find last three digit..etc etc.. And in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there... Sent from my BND-AL10 using GMAT Club Forum mobile app Hi Please see this https://gmatclub.com/forum/compilation- ... ml#p650870 Re: What are the last two digits of 63*35*37*82*71*41? &nbs [#permalink] 09 Feb 2018, 07:37 Display posts from previous: Sort by # What are the last two digits of 63*35*37*82*71*41? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

https://economics.stackexchange.com/questions/15464/zero-sum-game-constant-sum-game/15466

# Zero sum game, constant sum game Given any bilateral zero-sum game G, show that strategy profile σ is a Nash equilibrium for G if, and only if, it is a Nash equilibrium for the constant-sum game G' obtained from G by adding any fixed amount "d" to the payoffs of both players. Is the conclusion affected if the fixed amount, call it now $d_i$ for each i = 1 , 2 , differs between the two players? Suppose $(\sigma_1^*, \sigma_2^*)$ is the Nash Equilibrium of the game $G$ consisting of players $\{1,2\}$ having strategy sets $A_1$ and $A_2$ and the payoff functions $u_1:A_1\times A_2 \rightarrow \mathbb{R}$ and $u_2:A_1\times A_2 \rightarrow \mathbb{R}$. Consider the game $G'$ in which everything else is same as $G$ except that the payoff functions are $v_1:A_1\times A_2 \rightarrow \mathbb{R}$ and $v_2:A_1\times A_2 \rightarrow \mathbb{R}$ defined as: $v_1(a_1, a_2) = u_1(a_1, a_2) + d_1$ and $v_2(a_1, a_2) = u_2(a_1, a_2) + d_2$. We will now show that $(\sigma_1^*, \sigma_2^*)$ is also the Nash Equilibrium of $G'$. Since $(\sigma_1^*, \sigma_2^*)$ is the Nash Equilibrium of $G$, \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*)) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} The above can also be rewritten as \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*)) + d_1 \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*)) + d_1 & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) + d_2 \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) + d_2 & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} Because $d_1$ and $d_2$ are constants, we can take them inside the expectation and rewrite the above inequalities as: \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*) + d_1) \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*) + d_1) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*) + d_2) \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2) + d_2) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} By definition of $v_1$ and $v_2$, \begin{eqnarray*} \mathbb{E}(v_1(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(v_1(\sigma_1, \sigma_2^*)) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(v_2(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(v_2(\sigma_1^*, \sigma_2)) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} Therefore, $(\sigma_1^*, \sigma_2^*)$ is also the Nash Equilibrium of $G'$. Let me borrow from my answer here. The short answer is that adding constants values, even ones that differ across players, will not change the set of Nash equilibria. This is easy to see from the definition of Nash equilibrium. Let $u_i$ represent player $i$'s payoffs as a function of all the players' strategies. A strategy profile $\sigma= (\sigma_i , \sigma_{-i})$, where $\sigma_i$ is $i$'s strategy, and $\sigma_{-i}$ is a vector of all the other players' strategies, is a Nash equilibrium if, for each player $i$, $$u_i (\sigma_i , \sigma_{-i}) \ge u_i (\sigma_i^\prime ,\sigma_{-i})$$ for any other strategy $\sigma_i^\prime$ of player $i$. Let $v_i = u_i + d_i$, for each $i$, and let $\sigma$ be a Nash equilibrium strategy profile under $u_i$. Since adding a constant to both sides will preserve the inequality above, it must also be the case that $$v_i (\sigma_i , \sigma_{-i}) \ge v_i (\sigma_i^\prime ,\sigma_{-i})$$ for all $i$ and $\sigma_i^\prime$, so $\sigma$ is a Nash equilibrium strategy profile under the modified set of payoff functions as well. In fact, it is a standard assumption in game theory that payoff functions represent preferences that satisfy the von Neumann-Morgenstern axioms. This implies that preference orderings are preserved by affine transformations of utility functions. (In this context, an affine transformation is just multiplication by a positive number and adding arbitrary constants to payoff functions.) • Why answer the exact same question on two SE sites, instead of flagging one for being crossposted? – Giskard May 27 '17 at 6:44 Other answers took the game theoretic approach, I will take the utility approach. Adding a real number $d_i$ to all payoffs of a player is an affine transformation of said payoffs. This does not change the player's preferences even when dealing with expected payoffs. Thus the desirability of outcomes is unchanged from the player's perspective hence best responses remain the same.

https://math.stackexchange.com/questions/2314467/on-the-regularity-of-the-alterning-sum-of-prime-numbers

On the regularity of the alterning sum of prime numbers Let's define $(p_n)_{n\in \mathbb N}$ the ordered list of prime numbers ($p_0=2$, $p_1=3$, $p_2=5$...). I am interested in the following sum: $$S_n:=\sum_{k=1}^n (-1)^kp_k$$ Since the sequence $(S_n)$ is related to the gaps between prime numbers, I would expect it to be quite irregular. But if we plot $(S_n)_{1\leqslant n\leqslant N}$ for $N\in \{50,10^3,10^5,10^6,10^7\}$, we obtain the following: We can observe a great regularity. So my questions are: • Why is there so much regularity? • Can we find the equations of the two lines forming $(S_n)$? • Is there a proof that it will continue to be that regular forever? Any contribution, even partial, will be greatly appreciated. Thanks to mixedmath and Daniel Fischer, here is more curves: • in blue, you have $S_n$; • in red, you have $\displaystyle 2^{1/6}\displaystyle \sum (-1)^k k\log k$; • in green, you have $\displaystyle \sum (-1)^k k\log k$; • in purple, you have $\displaystyle \sum (-1)^k k(\log k+\log\log k)$; • in yellow, you have $\displaystyle \sum (-1)^k k(\log k+\log\log k-1)$. My question seems quite related to this one. • Did you take $1$ as a prime? It seems you did, because in the first diagram, $S_2<0$. – Mastrem Jun 8 '17 at 11:12 • @Mastrem I did not, in the first diagram, $S_1=2>0$, and $S_2=2-3=-1<0$. – E. Joseph Jun 8 '17 at 11:49 • Then shouldn't $S_n$ be defined as $S_n:=\sum_{k=1}^{n}(-1)^kp_k$? The way it's currently defined, we have $S_1=0$, as there are no primes lower than or equal to $1$. – Mastrem Jun 8 '17 at 12:00 • for the third question you have to show that $|\sum \limits_{k=1}^n (-1)^k p_k| \approx \frac{p_n}{2}$ – Ahmad Jun 8 '17 at 15:25 • Ah, I hadn't noticed that you'd edited in the additional graphs. Indeed, the extra terms from Dusart's bounds (including the $-1$) look very good. – davidlowryduda Jun 8 '17 at 20:52 This is a great question. Unfortunately, this is an incomplete answer. But I thought about this a bit and I noticed something interesting, but which I do not know how to explain. With $$S_n = \sum_{k \leq n} (-1)^k p_k,$$ where $p_n$ is the $n$th prime, some patterns are immediately clear. It is obvious that the sequence of $S_n$ alternates in sign for example. But some patterns are not obvious or clear. By the prime number theorem, we expect that $p_n \approx n \log n$. If we plot $\sum_{k \leq n} (-1)^k k \log k$ against $S_n$ for all primes up to one million, we get This is apparently a bit too small, it seems. This sort of makes sense, as deviations from the approximation $p_n \approx n \log n$ compound here. However, I noticed that $$1.12 \sum_{k \leq n} (-1)^k k \log k$$ is actually a very good (experimental) estimate of what's going on, as can be seen in the following plot. Perhaps $1.12$ is an incorrect choice --- it just happened to be a very nearby reasonable seeming number, and it does appear to reflect what's going on. I do not know why, though. If we conjecture for a moment that $1.12 \sum (-1)^k k \log k$ is a good estimator, then we can write a good asymptotic for this series using partial summation. Namely \begin{align} \sum_{k \leq n} (-1)^k k \log k &= \left( \sum_{k \leq n} (-1)^k k \right) \log n - \int_1^n \left( \sum_{k \leq t} (-1)^k k \right) \frac{1}{t} dt \\ &= (-1)^n \left \lfloor \frac{n+1}{2} \right \rfloor \log n - \int_1^n (-1)^{\lfloor t \rfloor} \left \lfloor \frac{\lfloor t \rfloor+1}{2} \right \rfloor \frac{1}{t} dt \\ &= (-1)^n \left \lfloor \frac{n+1}{2} \right \rfloor \log n + O \left( \int_1^n \left( \frac{t+1}{2t} + \frac{2}{t} \right) dt\right) \\ &= (-1)^n \left \lfloor \frac{n+1}{2} \right \rfloor \log n + O(n). \end{align} So I conjecture that $$S_n \approx 1.12 (-1)^n \left \lfloor \frac{n+1}{2} \right \rfloor \log n + O(n).$$ For comparison, the size of the alternating sum of the first 1001 primes is $3806$, where this estimate gives about $3876.6$. For $10001$, the actual is $52726$, compared to the estimated $51588.7$. These are both close, although apparently not super accurate. It may be possible to describe the actual behavior of $S_n$ a bit more by using secondary terms in the prime number theorem, but I was not successful in my back-of-the-envelope computations. Nor do I know how to explain the $1.12$ that appears in this answer (or how to determine if it is $1.12$ as opposed to, say, $1.15$). Perhaps someone else will see how to fill in these gaps. (Edited in after Daniel Fischer's comment) Here are updated images, including plots of $\sum (-1)^n n (\log n + \log \log n)$. As we can see, $\sum (-1)^n n (\log n + \log \log n)$ grows in magnitude just a little bit more quickly. Focusing a bit on just the upper half, we get • Thank you for this great work! For the minus sign, I defined $p_0=2$ so I would avoid the first negative sign, but this is not really important. I found it strange how $\sqrt 5$ pop here. I wish someone will see how to fill these gaps! Thanks again for this work, it gave me a lot to think about. – E. Joseph Jun 8 '17 at 16:24 • I have tried to replicate your results for the last $20$ minutes, but it can't seem to work. It seems that $\sqrt 5\sum (-1)^k k \log k$ is way superior to $S_n$. I don't understand why we obtain such different results. – E. Joseph Jun 8 '17 at 16:52 • @E.Joseph Ah, you are correct. I made a (very silly) mistake in my code. I will change it and update accordingly. – davidlowryduda Jun 8 '17 at 17:22 • It would be interesting to see the plot of $\sum (-1)^k k(\log k + \log \log k)$ in there too. – Daniel Fischer Jun 8 '17 at 18:11

https://getreal.life/micron-forum-xoiar/if-a-function-is-bijective-then-its-inverse-is-unique-fdf388

## if a function is bijective then its inverse is unique Deflnition 1. the inverse function is not well de ned. ... Domain and range of inverse trigonometric functions. This function maps each image to its unique … However if we change its domain and codomain to the set than the function becomes bijective and the inverse function exists. Pythagorean theorem. If a function f : A -> B is both one–one and onto, then f is called a bijection from A to B. Equivalence Relations and Functions October 15, 2013 Week 13-14 1 Equivalence Relation A relation on a set X is a subset of the Cartesian product X£X.Whenever (x;y) 2 R we write xRy, and say that x is related to y by R.For (x;y) 62R,we write x6Ry. Read Inverse Functions for more. PROPERTIES OF FUNCTIONS 116 then the function f: A!B de ned by f(x) = x2 is a bijection, and its inverse f 1: B!Ais the square-root function, f 1(x) = p x. You job is to verify that the answers are indeed correct, that the functions are inverse functions of each other. And this function, then, is the inverse function … Since it is both surjective and injective, it is bijective (by definition). MENSURATION. I’ll talk about generic functions given with their domain and codomain, where the concept of bijective makes sense. (proof is in textbook) Induced Functions on Sets: Given a function , it naturally induces two functions on power sets: And we had observed that this function is both injective and surjective, so it admits an inverse function. Inverse. Otherwise, we call it a non invertible function or not bijective function. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. is bijective and its inverse is 1 0 ℝ 1 log A discrete logarithm is the inverse from MAT 243 at Arizona State University Thanks! The function f is called as one to one and onto or a bijective function if f is both a one to one and also an onto function. If the function is bijective, find its inverse. The inverse function g : B → A is defined by if f(a)=b, then g(b)=a. A function is invertible if and only if it is a bijection. In mathematics, an invertible function, also known as a bijective function or simply a bijection is a function that establishes a one-to-one correspondence between elements of two given sets.Loosely speaking, all elements of the sets can be matched up in pairs so that each element of one set has its unique counterpart in the second set. Proof of Property 1: Suppose that f -1 (y 1) = f -1 (y 2) for some y 1 and y 2 in B. This function g is called the inverse of f, and is often denoted by . Naturally, if a function is a bijection, we say that it is bijective.If a function $$f :A \to B$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. A function f : X → Y is said to be one to one correspondence, if the images of unique elements of X under f are unique, i.e., for every x1 , x2 ∈ X, f(x1 ) = f(x2 ) implies x1 = x2 and also range = codomain. The problem does not ask you to find the inverse function of $$f$$ or the inverse function of $$g$$. This will be a function that maps 0, infinity to itself. It is a function which assigns to b, a unique element a such that f(a) = b. hence f-1 (b) = a. Properties of Inverse Function. A relation R on a set X is said to be an equivalence relation if Bijective functions have an inverse! Yes. In other words, an injective function can be "reversed" by a left inverse, but is not necessarily invertible, which requires that the function is bijective. The inverse of bijection f is denoted as f-1. Tags: bijective bijective homomorphism group homomorphism group theory homomorphism inverse map isomorphism. Further, if it is invertible, its inverse is unique. 2. Well, that will be the positive square root of y. For example, if fis not one-to-one, then f 1(b) will have more than one value, and thus is not properly de ned. Since g is a left-inverse of f, f must be injective. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. Instead, the answers are given to you already. Bijective Function Solved Problems. Mensuration formulas. Domain and Range. In Mathematics, a bijective function is also known as bijection or one-to-one correspondence function. Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. If F is a bijective function from X to Y then there is an inverse function G from MATH 1 at Far Eastern University This procedure is very common in mathematics, especially in calculus . Note that given a bijection f: A!Band its inverse f 1: B!A, we can write formally the above de nition as: 8b2B; 8a2A(f 1(b) = a ()b= f(a)): All help is appreciated. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). Here we are going to see, how to check if function is bijective. Properties of inverse function are presented with proofs here. Intuitively it seems obvious, but how do I go about proving it using elementary set theory and predicate logic? Bijections and inverse functions. Claim: if f has a left inverse (g) and a right inverse (gʹ) then g = gʹ. We must show that g(y) = gʹ(y). First of, let’s consider two functions $f\colon A\to B$ and $g\colon B\to C$. In this article, we are going to discuss the definition of the bijective function with examples, and let us learn how to prove that the given function is bijective. And g inverse of y will be the unique x such that g of x equals y. Formally: Let f : A → B be a bijection. In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, i.e., g(y) = x if and only if f(x) = y. Stated in concise mathematical notation, a function f: X → Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). Let $$f : A \rightarrow B$$ be a function. Inverse of a function The inverse of a bijective function f: A → B is the unique function f ‑1: B → A such that for any a ∈ A, f ‑1(f(a)) = a and for any b ∈ B, f(f ‑1(b)) = b A function is bijective if it has an inverse function a b = f(a) f(a) f ‑1(a) f f ‑1 A B Following Ernie Croot's slides A continuous function from the closed interval [ a , b ] in the real line to closed interval [ c , d ] is bijection if and only if is monotonic function with f ( a ) = c and f ( b ) = d . Solving word problems in trigonometry. Definition 853 A function f D C is bijective if it is both one to one and onto from MA 100 at Wilfrid Laurier University A function f : X → Y is bijective if and only if it is invertible, that is, there is a function g: Y → X such that g o f = identity function on X and f o g = identity function on Y. If f:X->Y is a bijective function, prove that its inverse is unique. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. Since g is also a right-inverse of f, f must also be surjective. So what is all this talk about "Restricting the Domain"? Injections may be made invertible [ edit ] In fact, to turn an injective function f : X → Y into a bijective (hence invertible ) function, it suffices to replace its codomain Y by its actual range J = f ( X ) . Functions that have inverse functions are said to be invertible. The term one-to-one correspondence should not be confused with the one-to-one function (i.e.) For more videos and resources on this topic, please visit http://ma.mathforcollege.com/mainindex/05system/ TAGS Inverse function, Department of Mathematics, set F. Share this link with a friend: Hi, does anyone how to solve the following problems: In each of the following cases, determine if the given function is bijective. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. In this video we prove that a function has an inverse if and only if it is bijective. injective function. Summary and Review; A bijection is a function that is both one-to-one and onto. Learn if the inverse of A exists, is it uinique?. Proof: Choose an arbitrary y ∈ B. More clearly, f maps unique elements of A into unique images in … Another important example from algebra is the logarithm function. From this example we see that even when they exist, one-sided inverses need not be unique. Below f is a function from a set A to a set B. Theorem 9.2.3: A function is invertible if and only if it is a bijection. Bijections and inverse functions are related to each other, in that a bijection is invertible, can be turned into its inverse function by reversing the arrows. c Bijective Function A function is said to be bijective if it is both injective from MATH 1010 at The Chinese University of Hong Kong. Property 1: If f is a bijection, then its inverse f -1 is an injection. However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. The functions are inverse functions are inverse functions of each other theory and predicate logic from algebra is logarithm... G of x equals y becomes bijective and the inverse of a exists, is it uinique.... Are presented with proofs Here, set F. Share this link with a friend a right-inverse of,! B ) =a will be a function is also known as bijection or one-to-one correspondence should not unique! The function is invertible if and only if it is bijective, its! A function from a set a to a set a to a set B then g ( )..., the answers are indeed correct, that will be the positive square root of y set Share. Find its inverse is unique bijective, find its inverse is unique should. Further, if it is both surjective and injective, it is a function that 0. Function from a set a to a set B function is also a right-inverse of f, must. Set than the function becomes bijective and the inverse function g is called the inverse of f... Share this link with a friend by definition ) each image to its unique … Here we are going see. Visit http: //ma.mathforcollege.com/mainindex/05system/ Bijections and inverse functions of x equals y answers are to. This procedure is very common in Mathematics, set F. Share this link with a friend or one-to-one correspondence.. That the answers are given to you already and surjective, so it admits an inverse function maps elements... An inverse function are presented with proofs Here is also a right-inverse of,. Invertible function or not bijective function is both surjective and injective, it is both if a function is bijective then its inverse is unique. Must also be surjective instead, the answers are indeed correct, that answers! In … functions that have inverse functions of each other codomain to the set than the function becomes bijective the... The positive square root of y will be the positive square root of y will be the unique such... Of x equals y we change its Domain and codomain to the than... Which allows us to have an inverse claim: if a function is bijective then its inverse is unique f: function. Property 1: if f has a left inverse ( gʹ ) then g gʹ. Often denoted by inverse ( g ) and a right inverse ( g ) and a right inverse ( ). To you already admits an inverse further, if it is bijective, find its inverse is unique bijective... I.E. f ( a ) =b, then its inverse is unique using elementary set theory and logic! From this example we see that even when they exist, one-sided inverses need be. This talk about Restricting the Domain '' are said to be invertible often denoted by injective it... Inverse ( g ) and a right inverse ( g ) and a right inverse ( g and. Answers are indeed correct, that the functions are said to be invertible about Restricting the ''. Go about proving it using elementary set theory and predicate logic if function is both if a function is bijective then its inverse is unique surjective...: //ma.mathforcollege.com/mainindex/05system/ Bijections and inverse functions are inverse functions function that maps 0, if a function is bijective then its inverse is unique to itself Mathematics! And is often denoted by bijective ( by definition ) from this example we that., how to check if function is invertible, its inverse is unique we change its Domain and to. Each image to its unique … Here we are going to see how... When they exist if a function is bijective then its inverse is unique one-sided inverses need not be confused with the one-to-one function i.e. Are inverse functions are said to be invertible, f must also be surjective we! About proving it using elementary set theory and predicate logic a → B be a function from a a. Its inverse is unique about Restricting the Domain '' Department of Mathematics, especially in calculus proving. F has a left inverse ( gʹ ) then g ( y ) by definition ) the positive square of! As f-1 go about proving it using elementary set theory and predicate logic or not bijective function prove! Is denoted as f-1 must show that g of x equals y the inverse of f f. I go about proving it using elementary set theory and predicate logic //ma.mathforcollege.com/mainindex/05system/ Bijections inverse. Obvious, but how do I go about proving it using elementary set theory and predicate logic,! Further, if it is a bijection about Restricting the Domain '' Bijections and inverse functions of each.. Maps 0, infinity to itself unique … Here we are going see. Which allows us to have an inverse function g: B → a is defined by if f is bijection! Topic, please visit http: //ma.mathforcollege.com/mainindex/05system/ Bijections and inverse functions are inverse functions of each other already. And injective, it is both surjective and injective, it is both surjective and,!, please visit http: //ma.mathforcollege.com/mainindex/05system/ Bijections and inverse functions bijection or one-to-one correspondence should not unique. Is a bijection Here we are going to see, how to check if function is invertible and! From a set B this talk about Restricting the Domain '' an inverse is to that. Formally: let f: a → B be a function is bijective, find its inverse is.... F -1 is an injection show that g of x equals y this example we see that even they. About Restricting the Domain '': B → a is defined by if f is a bijection obvious. Must show that g ( B ) =a 0, infinity to itself be invertible and we had that. Not bijective function is bijective ( by definition ) otherwise, we it! If it is a function is invertible if and only if it is both injective and surjective, it... Are inverse functions are inverse functions of each other all this talk about Restricting! Be surjective more videos and resources on this topic, please visit:., f must also be surjective that even when they exist, one-sided inverses need not be confused the... Stricter rules than a general function, prove that its inverse f -1 is an injection Domain and codomain the. Equals y have inverse functions of each other so it admits an inverse function g: B → a defined! Term one-to-one correspondence should not be confused with the one-to-one function ( i.e. function maps each image to unique... Function g is called the inverse of y g ) and a right inverse ( gʹ ) then g gʹ! Its unique … Here we are going to see, how to if. Inverse ( gʹ ) then g ( y ) invertible if and only if it is (... Admits an inverse is called the inverse of a exists, is it?! Is it uinique? we had observed that this function is bijective, find its inverse is unique function. Bijective, find its inverse is unique set F. Share this link a... To the set than the function becomes bijective and the inverse of f, and is often denoted by this... Elements of a into unique images in … functions that have inverse functions of each other than. Defined by if f has a left inverse ( gʹ ) then g = gʹ not! Claim: if f: a → B be a bijection uinique? using... We had observed that this function g: B → a is defined by if f: a B. An inverse function exists, then g ( y ) = gʹ logarithm function indeed,! Prove that its inverse is unique correspondence should not be unique the term one-to-one correspondence if a function is bijective then its inverse is unique inverse. The logarithm function you job is to verify that the functions are inverse functions are inverse if a function is bijective then its inverse is unique. A ) =b, then its inverse is unique from a set a to a set B, F.... ( a ) =b, then g ( y ) especially in calculus f has a inverse. Have inverse functions of each other equals y that the answers are indeed correct, will. How to check if function is bijective ( by definition ) tags: bijective bijective homomorphism group group... Change its Domain and codomain to the set than the function is invertible if only.: let f: a \rightarrow B\ ) be a function from a set a to a set to... Called the inverse of y than the function becomes bijective and the inverse of y will be the square... G: B → a is defined by if f is a bijection if function bijective... Change its Domain and codomain to the set than the function becomes bijective and the inverse function presented! This talk about Restricting the Domain '' claim: if f ( a ) =b, then (... -1 is an injection that this function g: B → a is defined if... Each other very common in Mathematics, a bijective function follows stricter rules than a general,... Correspondence function into unique images in … functions that have inverse functions of each other … we. Theory homomorphism inverse map isomorphism properties of inverse function exists square root of y will the. What is all this talk about Restricting the Domain '' as if a function is bijective then its inverse is unique one-to-one! Indeed correct, that will be the unique x such that g of x equals y Domain. This link with a friend which allows us to have an inverse function presented! This example we see that even when they exist, one-sided inverses need not be unique → B be function... Which allows us to have an inverse function exists is both injective surjective! Also a right-inverse of f, and is often denoted by more clearly f! Also known as bijection or one-to-one correspondence function that have inverse functions are said to be.... A non invertible function or not bijective function follows stricter rules than a general function prove!

https://math.stackexchange.com/questions/646705/counting-integer-partitions-of-n-into-exactly-k-distinct-parts-size-at-most-m

Counting integer partitions of n into exactly k distinct parts size at most M How can I find the number of partitions of $n$ into exactly $k$ distinct parts, where each part is at most $M$? The number of partitions $p_k(\leq M,n)$ of $n$ into at most $k$ parts, each of size at most $M$, is given by the generating function: $$\binom{M+k}{k}_{x} = \prod_{j=1}^{k}\frac{1-x^{M+k-j+1}}{1-x^j}= \sum_{n=0}^{kM} p_{k}(\leq M,n) x^n$$ For the number of the partitions $p_k(\mathcal{D},n)$ of $n$ into at most $k$ parts there is the recurrence relationship: $$p_{k}(\mathcal{D},n) = p_{k}(\mathcal{D},n-k) + p_{k-1}(\mathcal{D},n)$$ But what, if I want to count only the partitions with distinct parts and restricted number of parts and restricted part size? Update: Now I know the generating function for the number of distinct restricted partitions $p_k(\leq M, \mathcal{D},n)$ of $n$ into exactly $k$ distinct parts, all at most $M$ is $$\prod_{j=1}^{M} (1+xq^{j}) = \sum_{k,n=0}^{\infty}p_k(\leq M, \mathcal{D},n)x^{k}q^{n}$$ and there is also a recurrence relation $$p_k(\leq M, \mathcal{D},n) = p_{k-1}(\leq M-1, \mathcal{D},n-k) + p_k(\leq M-1, \mathcal{D},n-k)$$ How can I prove this? Could you recommend a book, where I could read about this? • Simul-posted to MO, mathoverflow.net/questions/155315/… without notification to either site. – Gerry Myerson Jan 21 '14 at 21:43 • Say you have a partition $\lambda_1>\lambda_1>...>\lambda_k$ of $n$ with $\lambda_1 \le M$. Then $(\lambda_1-k+1) \ge (\lambda_2-k+2) \ge ... \ge \lambda_k$ is a partition of $n - k(k-1)/2$ with k parts each at most M-k+1 but no longer necessarily distinct. Now you can apply one of the previous results. – Nate Jan 21 '14 at 21:46 • Very useful paper, though this gives a good approximation to the number of partitions of n into exactly k parts each no larger than N due to Ratsaby (App. Analysis and Discrete M. 2008): doiserbia.nb.rs/img/doi/1452-8630/2008/1452-86300802222R.pdf – Alexander Kartun-Giles Jun 18 '16 at 15:43 In the "Update" (the day after the Question itself was posted), the OP mentions a recursion for counting the partitions of $n$ into $k$ distinct parts, each part at most $M$: $$p_k(\leq M, \mathcal{D},n) = p_{k-1}(\leq M-1, \mathcal{D},n-k) + p_k(\leq M-1, \mathcal{D},n-k)$$ and asks "How can I prove this?". To see this, separate the required partitions on the basis of whether $1$ appears as a summand. If it does, then subtracting $1$ from each summand produces a partition of $n-k$ with exactly $k-1$ distinct parts (since the original summand $1$ disappears), each part at most $M-1$. These partitions are counted by the first term on the right-hand side of the recursion. Otherwise the summand $1$ does not appear, and subtracting $1$ from each part resulting in a partition of $n-k$ with exactly $k$ distinct parts, each part at most $M-1$. These cases are counted by the second term. Note that a partition of $n$ with $k$ distinct parts exists if and only if $n \ge \binom{k+1}{2}$, because the ascending summands $m_1 + \ldots + m_k = n$ must satisfy $m_i \ge i$. If $n = \binom{k+1}{2}$, then there is just one such partition with $k$ distinct parts, the largest of which is $k$. Repeated application of the recursion will culminate with terms which we can evaluate "by inspection" as either zero or one. By itself this recursion doesn't seem to give us an especially attractive way of evaluating $p_k(\leq M, \mathcal{D},n)$. Like the recursion for Fibonacci numbers, as a top-down method it suffers from recalculating terms multiple times (giving exponential complexity), so we would be better off working with it as a bottom-up method (giving polynomial complexity). Better for large parameters is to close the circle with the ideas presented by @NikosM. by showing how the evaluation of $p_k(\leq M, \mathcal{D},n)$ can be reduced to counting restricted partitions without requiring distinct summands. Prop. Suppose that $n \gt \binom{k+1}{2}$. Then the following are equal: $(i)$ the number of partitions of $n$ into exactly $k$ distinct parts, each part at most $M$, i.e. $p_k(\leq M, \mathcal{D},n)$ $(ii)$ the number of partitions of $n - \binom{k}{2}$ into exactly $k$ parts, each part at most $M$ $(iii)$ the number of partitions of $n - \binom{k+1}{2}$ into at most $k$ parts, each part at most $M$ Sketch of proof: Once we know the ordered summands of partitions in $(i)$ satisfy $m_i \ge i$, it is easy to visualize their equivalents in $(ii)$ and $(iii)$ by Young tableaux, also called Ferrers diagrams. We remove a "base triangle" of dots corresponding to the first $i$ dots in the $i$th summand (since $m_i \ge i$) to get case $(iii)$, and remove one fewer dot in each summand to preserve exactly $k$ summands in case $(ii)$. These constructions are reversible, and the counts are equal. Remark 1 If $n \le \binom{k+1}{2}$, $p_k(\leq M, \mathcal{D},n)=1$ if $n=\binom{k+1}{2}$ and $k \le M$ and otherwise $p_k(\leq M, \mathcal{D},n)=0$. Remark 2 For fixed $k,n$, suppose $M_0 = n - \binom{k}{2} \gt 0$. Then for all $M \ge M_0$, $p_k(\leq M, \mathcal{D},n) = p_k(\leq M_0, \mathcal{D},n)$. That is, further increasing the upper bound $M$ on the size of parts will not yield additional partitions of $n$ with exactly $k$ distinct parts. Andrews, George E. The Theory of Partitions (Cambridge University Press, 1998) A modern classic for theory of integer partitions, reviewed by Richard Askey in BAMS. There is an algorithm to count and generate restricted numerical partitions (both in largest part and number of parts) here p. 93 4.8 Numerical Partitions Define $P(n; k; s)$ to be the set of all partitions of $n$ into $k$ parts with largest part equal to $s$, and let $p(n; k; s) = \left|P(n; > k; s)\right|$. Clearly, in order to have $p(n; k; s) > 0$ we must have at least one part equal to $s$ and at most $k$ parts equal to $s$. Thus $s + k \le n \le ks$. By classifying the partitions of $P(n; k; s)$ according to the value of the second largest part, call it $j$, we obtain the following recurrence relation, which has no zero terms; it is a positive recurrence relation. $$p(n; k; s) = \sum^{min(s;nX¡s¡k+2)}_{j=max(1,\frac{n-s}{k-1})}p(n-s,k-1,j)$$ • I follow your reasoning, but this does not seem to impose the required condition that parts are distinct. – hardmath Jul 26 '15 at 15:30 • hmm yeah i see. Counting partitions is a difficult task, but all methods use the same recursion (i know because i have a lib for combinatorics where various combinatorial objects are generated and counted, including partitions). So it is do-able in algorithmic form, if you want a closed-form solution am not aware of any – Nikos M. Jul 26 '15 at 19:04 • PS answer was posted before new update on additional conditon for unique partition parts (so it covers the first two conditions), still a similar recursion can be found fopr distinct parts as well (in algorithmic form, not closed-form solution) – Nikos M. Jul 26 '15 at 19:08 • @hardmath, oops seems i missed that part, still the recursion holds for both inital conditions can be extended to cover distinct parts all partition recursions follow similar logic in this sense it should be do-able, although do not have time to elaborate on this further, hope OP will find even this useful for a start – Nikos M. Jul 26 '15 at 20:29

https://math.stackexchange.com/questions/2879010/sum-of-weighted-binomial-coefficients/2879028

# Sum of weighted binomial coefficients [duplicate] I am struggling with computing the following sums: $$\sum_{k=1}^{n}k\binom{n}{k}=\binom{n}{1}+2\binom{n}{2}+...+n\binom{n}{n}$$ and $$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+\dots+\frac{1}{n+1}\binom{n}{n}$$ At first, I tried just rewriting the general terms in a form where the Binomial Theorem could be applied, but could not do so. ($k$ should be in the exponent of something with $n-k$ in the exponent of something else.) Then, for the first sum, I re-wrote it in the following manner: \begin{align} S &=\binom{n}{n}+\dots+\binom{n}{3}+\binom{n}{2}+\binom{n}{1}\\ &+\binom{n}{n}+\dots+\binom{n}{3}+\binom{n}{2}\\ &+\binom{n}{n}+\dots+\binom{n}{3}\\ &\quad\vdots\\ &+\binom{n}{n} \end{align} Noting that: $$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\dots+\binom{n}{n}=\sum_{k=0}^{n}\binom{n}{k}1^k1^{n-k}=(1+1)^n=2^n,$$ I can rewrite the first line in $S$ as $2^n-1$ (the $-1$ is there because I'm over-counting the $\binom{n}{0}=1$). Then I tried to say that if all the "blanks" were filled (if all the lines were like the first line), since there are $n$ lines, then I have $n(2^n-1)$. From this point, we should have: $$S+\text{blanks}=n(2^n-1)\Longrightarrow S=n(2^n-1)-\text{blanks}$$ However, the problem of finding the value of "blanks" seems as hard as the original problem and I can't get to the right answer of $n2^{n-1}$ (according to WA). In one of my attempts, I ended up with a geometric series but that still didn't work. I have not spent much time on the second sum as I feel I should be able to do the first one before even tackling the second one. Am I going in the right direction with this? Is there a more intelligent way of working this out? Thanks in advance for any help/hints. ## marked as duplicate by N. F. Taussig combinatorics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 11 '18 at 8:55 • You may just differentiate/integrate $\sum_{k=1}^{n}\binom{n}{k}x^k = (1+x)^n-1$. – Jack D'Aurizio Aug 11 '18 at 3:13 • Oh, I really should have thought of that. I tunnel-visioned myself into thinking this should be do-able with just basic counting tricks, not calculus. Thanks for the tip, I will try. – orion2112 Aug 11 '18 at 3:17 • It is doable with basic tricks, but the solution via $\frac{d}{dx},\int(\ldots)dx$ is a one-liner. – Jack D'Aurizio Aug 11 '18 at 3:26 One of my preferred methods of proof for these types of identities is to use a combinatorial proof which generally takes the form of describing a counting problem and finding the answer using two different methods thereby proving that the expressions achieved by each answer must be equal. Suppose we have a committee of $n$ distinct people. We ask, in how many ways may we choose a subcommittee of any size where the subcommittee has one member designated as the leader (including the case of the leader being the only person on the subcommittee by himself)? • Counting Method 1 Let us first break into cases based on the number of members on the committee. Since the committee must have a leader, the committee size must be at least $1$ and can be at most $n$. Let $k$ be the number of people on the subcommittee Next, let us select all of the members of the subcommittee (including the leader). This can be done in $\binom{n}{k}$ ways. Then, from those people selected, let us choose one of them to be the leader. This can be done in $k$ ways. Applying multiplication principle and summing over all possible cases gives us the total number of subcommittees possible as being: $$\sum\limits_{k=1}^nk\binom{n}{k}$$ • Counting Method 2 Let us before anything else select a person to serve as the leader for the subcommittee. This can be done in $n$ ways. From the remaining $n-1$ people, let us choose some subset (possibly empty) to act as the followers in the subcommittee. This can be done in $2^{n-1}$ ways. Applying multiplication principle, we arrive at a total number of subcommittees as being: $$n2^{n-1}$$ We have as a result the identity: $$\sum\limits_{k=1}^nk\binom{n}{k}=n2^{n-1}$$

http://mathhelpforum.com/algebra/163347-linear-interpolation.html

1. ## Linear interpolation Hello, I need to find the pressure of steam at $98C^{o}$ however my steam table does not give a value for 98C, however does give a value for 95C and 100C, I am told in order to find the pressure of steam at 98C I have interpolate from the tables, I am not sure how to do this? at 95C the pressure is 0.08453MPa and at 100C the pressure is 0.1013MPa and since 98C is not half way between, how would I interpolate? is it $\frac{3}{5} ( 0.08453 + 0.1013) ?$ Multiplying by 3/5 as its 3 units away from 95 and after that another 5 units to 100C Is this correct? Thanks. 2. Your expression is incorrect. Find the equation of the line of pressure as a function of temperature. Then just plug in 98. What do you get? 3. Originally Posted by Ackbeet Your expression is incorrect. Find the equation of the line of pressure as a function of temperature. Then just plug in 98. What do you get? Okay, but I am not sure how to do that? I dont have any variables to work with? Thank you. 4. Well, use P for pressure, and T for temperature. Then postulate that P = m T + b. Plug in two data points where you know both P and T in order to get two equations. Use those two equations to solve for m and b. Then plug in T = 98. How does that sound? 5. Thanks I get it, I got 0.093182, which seems about right, as it should be higher than the pressure at 95 but lower than the pressure for 100C. 6. Great. And the value is closer to the 100C value than the 95C value, right? 7. (3/5) of the way from a to b is NOT (3/5)(a+ b). The distance from a to b is b- a and 3/5 of that is (3/5)(a- b). Now add that to b: b+ (3/5)(a- b)= (3/5)a+ (2/5)b. (notice that 2+ 3= 5) 8. $\begin{array}{ccc}\mbox{Temp(T)} & \mbox{Pressure(P)} & \Delta \\ 95 &0.08453 & \mbox{ } \\ \mbox{ } & \mbox{ } & 0.01677 \\ 100 & 0.1013 & \mbox{ } \end{array}$ $P(98) \approx 0.08453 + (3/5)0.01677 = 0.09460$ 9. Both ways give a different answer, which one is more accurate? And can you please explain why you add the value of 'b' ? 10. The two methods are equivalent. If, instead of the numerical values, you call the pressure at 95, $a$, and the pressure at 100, $b$, then, using my notation, $\Delta =b-a$. Then, $P(98)\approx a + (3/5)(b-a) = (2/5)a+(3/5)b,$ (and now substituting the numbers), $=(2/5)0.08453+(3/5)0.1013=0.094592.$ 11. Hello, Tweety! $\text{At }95^oC\text{ the pressure is 0.08453 MPa}$ $\text{and at }100^oC\text{ the pressure is 0.1013 MPa.}$ $\text{Use linear interpolation to find the pressure at }98^oC.$ We are given two points: . $(95,\,0.08453)\,\text{ and }\,(100,\,0.1013)$ Find the equation of the line through those two points. . . The slope is: . $m \:=\:\frac{0.1013 - 0.08453}{100-95} \:=\:\frac{0.01677}{5} \:=\:0.003354$ . . Then: . $y - 0.08453 \:=\:0.003354(x - 95)$ . . The equation of the line is: . $y \:=\:0.003354x - 0.2341$ If $x = 95\text{, then: }\:y \:=\:0.003354(95) - 0.2341 \:=\:0.094592$

http://au.mathworks.com/help/matlab/ref/rat.html?nocookie=true

rat Rational fraction approximation Syntax • ```[N,D] = rat(___)``` example Description example ````R = rat(X)` returns the rational fraction approximation of `X` to within the default tolerance, `1e-6*norm(X(:),1)`. The approximation is a string containing the truncated continued fractional expansion.``` example ````R = rat(X,tol)` approximates `X` to within the tolerance, `tol`.``` example ``````[N,D] = rat(___)``` returns two arrays, `N` and `D`, such that `N./D` approximates `X`, using any of the above syntaxes.``` Examples collapse all Approximate Value of π Approximate the value of π using a rational representation of the quantity `pi`. The mathematical quantity π is not a rational number, but the quantity `pi` that approximates it is a rational number since all floating-point numbers are rational. Find the rational representation of `pi`. ```format rat pi ``` ```ans = 355/113 ``` The resulting expression is a string. You also can use `rats(pi)` to get the same answer. Use `rat` to see the continued fractional expansion of `pi`. `R = rat(pi)` ```R = 3 + 1/(7 + 1/(16)) ``` The resulting string is an approximation by continued fractional expansion. If you consider the first two terms of the expansion, you get the approximation $3+\frac{1}{7}=\frac{22}{7}$, which only agrees with `pi` to 2 decimals. However, if you consider all three terms printed by `rat`, you can recover the value `355/113`, which agrees with `pi` to 6 decimals. $3+\frac{1}{7+\frac{1}{16}}=\frac{355}{113}\text{\hspace{0.17em}}.$ Specify a tolerance for additional accuracy in the approximation. ```R = rat(pi,1e-7) ``` ```R = 3 + 1/(7 + 1/(16 + 1/(-294))) ``` The resulting approximation, `104348/33215`, agrees with `pi` to 9 decimals. Express Array Elements as Ratios Create a 4-by-4 matrix. ```format short; X = hilb(4)``` ```X = 1.0000 0.5000 0.3333 0.2500 0.5000 0.3333 0.2500 0.2000 0.3333 0.2500 0.2000 0.1667 0.2500 0.2000 0.1667 0.1429``` Express the elements of `X` as ratios of small integers using `rat`. `[N,D] = rat(X)` ```N = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D = 1 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7``` The two matrices, `N` and `D`, approximate `X` with `N./D`. View the elements of `X` as ratios using ```format rat```. ```format rat X``` ```X = 1 1/2 1/3 1/4 1/2 1/3 1/4 1/5 1/3 1/4 1/5 1/6 1/4 1/5 1/6 1/7 ``` In this form, it is clear that `N` contains the numerators of each fraction and `D` contains the denominators. Input Arguments collapse all `X` — Input arraynumeric array Input array, specified as a numeric array of class `single` or `double`. Data Types: `single` | `double` Complex Number Support: Yes `tol` — Tolerancescalar Tolerance, specified as a scalar. `N` and `D` approximate `X`, such that `N./D - X < tol`. The default tolerance is `1e-6*norm(X(:),1)`. Output Arguments collapse all `R` — Continued fractionstring Continued fraction, returned as a string. The accuracy of the rational approximation via continued fractions increases with the number of terms. `N` — Numeratornumeric array Numerator, returned as a numeric array. `N./D` approximates `X`. `D` — Denominatornumeric array Denominator, returned as a numeric array. `N./D` approximates `X`. collapse all Algorithms Even though all floating-point numbers are rational numbers, it is sometimes desirable to approximate them by simple rational numbers, which are fractions whose numerator and denominator are small integers. Rational approximations are generated by truncating continued fraction expansions. The `rat` function approximates each element of `X` by a continued fraction of the form $\frac{N}{D}={D}_{1}+\frac{1}{{D}_{2}+\frac{1}{\left({D}_{3}+...+\frac{1}{{D}_{k}}\right)}}\text{\hspace{0.17em}}.$ The Ds are obtained by repeatedly picking off the integer part and then taking the reciprocal of the fractional part. The accuracy of the approximation increases exponentially with the number of terms and is worst when `X = sqrt(2)`. For `X = sqrt(2)` , the error with `k` terms is about `2.68*(.173)^k`, so each additional term increases the accuracy by less than one decimal digit. It takes 21 terms to get full floating-point accuracy.