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http://shop.track-it.nz/6bowhhs1/properties-of-matrix-addition-346a66 | Commutative Property Of Addition 2. If A is an n×m matrix and O is a m×k zero-matrix, then we have: AO = O Note that AO is the n×k zero-matrix. Matrix Matrix Multiplication 11:09. We have 1. To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. In a triangular matrix, the determinant is equal to the product of the diagonal elements. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? The identity matrices (which are the square matrices whose entries are zero outside of the main diagonal and 1 on the main diagonal) are identity elements of the matrix product. Learning Objectives. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Matrix Multiplication Properties 9:02. 16. Proof. There are a few properties of multiplication of real numbers that generalize to matrices. A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of Matrix Addition and Scalar Multiplication. What is the Identity Property of Matrix Addition? General properties. Yes, it is! There are 10 important properties of determinants that are widely used. Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. 13. If you built a random matrix and took its determinant, how likely would it be that you got zero? The first element of row one is occupied by the number 1 … In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. Equality of matrices All-zero Property. Multiplying a $2 \times 3$ matrix by a $3 \times 2$ matrix is possible, and it gives a $2 \times 2$ matrix … Properties of Transpose of a Matrix. The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m × n matrix and B and C be n × m matrices, then Addition: There is addition law for matrix addition. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m× nmatrix A. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Properties of matrix addition. Let A, B, and C be mxn matrices. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. You should only add the element of one matrix to … Property 1 completes the argument. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. This project was created with Explain Everything™ Interactive Whiteboard for iPad. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] Let A, B, C be m ×n matrices and p and q be two non-zero scalars (numbers). In this lesson, we will look at this property and some other important idea associated with identity matrices. So if n is different from m, the two zero-matrices are different. Andrew Ng. Matrices rarely commute even if AB and BA are both defined. What is a Variable? Addition and Scalar Multiplication 6:53. The addition of the condition $\detname{A}\neq 0$ is one of the best motivations for learning about determinants. ... although it is associative and is distributive over matrix addition. Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. This tutorial introduces you to the Identity Property of Matrix Addition. Instructor. Transcript. Important Properties of Determinants. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. A. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. Selecting row 1 of this matrix will simplify the process because it contains a zero. We state them now. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. A B _____ Commutative property of addition 2. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Let A, B, and C be three matrices. 8. det A = 0 exactly when A is singular. 2. In other words, the placement of addends can be changed and the results will be equal. Then we have the following properties. 17. The basic properties of matrix addition is similar to the addition of the real numbers. Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) Properties of scalar multiplication. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. If the rows of the matrix are converted into columns and columns into rows, then the determinant remains unchanged. Laplace’s Formula and the Adjugate Matrix. 14. Matrix multiplication shares some properties with usual multiplication. 4. Question 3 : then find the additive inverse of A. the identity matrix. Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + − 1 A = 0. Find the composite of transformations and the inverse of a transformation. Created by the Best Teachers and used by over 51,00,000 students. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. Matrix Multiplication - General Case. Properties involving Addition. The commutative property of addition means the order in which the numbers are added does not matter. Properties involving Multiplication. A+B = B+A 2. Given the matrix D we select any row or column. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. Note that we cannot use elimination to get a diagonal matrix if one of the di is zero. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. The inverse of 3 x 3 matrix with determinants and adjugate . The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. The inverse of 3 x 3 matrices with matrix row operations. Properties of matrix multiplication. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. EduRev, the Education Revolution! As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. Proposition (commutative property) Matrix addition is commutative, that is, for any matrices and and such that the above additions are meaningfully defined. 11. Numerical and Algebraic Expressions. The determinant of a 2 x 2 matrix. We can also say that the determinant of the matrix and its transpose are equal. Matrix Vector Multiplication 13:39. 12. Examples . In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want. This property is known as reflection property of determinants. The inverse of a 2 x 2 matrix. However, unlike the commutative property, the associative property can also apply to matrix … The matrix O is called the zero matrix and serves as the additiveidentity for the set of m×nmatrices. 1. Best Videos, Notes & Tests for your Most Important Exams. To find the transpose of a matrix, we change the rows into columns and columns into rows. Properties of Matrix Addition: Theorem 1.1Let A, B, and C be m×nmatrices. However, there are other operations which could also be considered addition for matrices, such as the direct sum and the Kronecker sum Entrywise sum. The determinant of a matrix is zero if each element of the matrix is equal to zero. A scalar is a number, not a matrix. Try the Course for Free. Use properties of linear transformations to solve problems. Matrix addition is associative; Subtraction. 18. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. 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X 3 matrix with determinants and adjugate select any row or column if one of matrix.$ \detname { a } \neq 0 \$ is one properties of matrix addition the and. 1 in matrix multiplication operation your Most important Exams and B which have equal order SMZD is an equivalence Proof. | 2021-09-28T02:16:51 | {
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https://math.stackexchange.com/questions/2922644/comparing-the-magnitudes-of-expressions-of-surds | # Comparing the magnitudes of expressions of surds
I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).
I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).
For example, one question asked which is the largest of:
(a) $\sqrt{10}$
(b) $\sqrt2+\sqrt3$
(c) $5-\sqrt3$
In this case, I relied on my knowledge that $\sqrt{10} \approx 3.16$ and $\sqrt2\approx 1.41$ and $\sqrt3 \approx 1.73$ to find (a) $\approx 3.16$, (b) $\approx ~3.14$ and (c) $\approx ~3.27$ so that the required answer is (c).
But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).
I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?
EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $\sqrt[4]{101}$.
When approaching the question by approximation, I simply observed that $\sqrt[4]{101}$ is only a tiny bit greater than $\sqrt{10}$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?
• +1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method. Sep 19 '18 at 10:41
• Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here! Sep 20 '18 at 10:05
Comparing $$\sqrt{10}$$ and $$\sqrt2+\sqrt3$$ is the same as comparing $$10$$ and $$(\sqrt2+\sqrt3)^2=5+2\sqrt6$$. That's the same as comparing $$5$$ and $$2\sqrt6$$. Which of these is bigger?
Likewise comparing $$\sqrt{10}$$ and $$5-\sqrt3$$ is the same as comparing $$10$$ and $$(5-\sqrt3)^2=28-10\sqrt3$$. That's the same as comparing $$10\sqrt3$$ and $$18$$.
Which of these is bigger?
• Ah .... yes of course ... $5=\sqrt{25}>\sqrt{24}=2\sqrt6$ Sep 19 '18 at 10:30
• Thank you for the hint! Sep 19 '18 at 10:32
• BBO555, regarding $5$ and $2\sqrt{6},$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $x\geq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$ Sep 19 '18 at 10:47
You can use:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $x\geq0$ and
(2) the arithmetic-geometric mean inequality $\sqrt{ab}\leq\frac{a+b}{2}$, when $a, b\geq0$. Hence, $$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{2\cdot3}\leq5+2\frac{2+3}{2}=5+5=10=(\sqrt{10})^2$$ Therefore, using (1), we obtain $\sqrt{2}+\sqrt{3}\leq 10$. I forgot about this: $$5-\sqrt{3}=3+2-\sqrt{3}=3+\frac{1}{2+\sqrt{3}}\geq3+\frac{1}{2+2}=3.25$$ One can easily verify that $(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.
Then, performing argument (1) twice, one finds that $5-\sqrt{3}>(101)^{1/4}$.
## Consequently, $5-\sqrt{3}$ is the bigger number.
• I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a \gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a \gt c$, do some more manipulation, and show $c \gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful. Sep 19 '18 at 14:01
• Thanks for including a route to handling case (d) ! Sep 20 '18 at 8:07 | 2021-10-19T03:38:04 | {
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https://math.stackexchange.com/questions/1693964/in-calculus-how-can-a-function-have-several-different-yet-equal-derivatives | # In Calculus, how can a function have several different, yet equal, derivatives?
I've been pondering this question all night as I work through some problems, and after a very thorough search, I haven't found anything completely related to my question. I guess i'm also curious how some derivatives are simplified as well, because in some cases I just can't see the breakdown. Here is an example:
$f(x) = \dfrac{x^2-6x+12}{x-4}$ is the function I was differentiating. Here is what I got:
$f '(x) = \dfrac{x^2-8x+12}{(x-4)^2}$ which checks using desmos graphing utility.
Now, when I checked my textbook(and Symbolab) they got:
$f '(x) = 1 - \dfrac{4}{(x-4)^2}$ which also checks on desmos.
To me, these derivatives look nothing alike, so how can they both be the equal to the derivative of the original function? Both methods used the quotient rule, yet yield very different results. Is one of these "better" than the other? I know that it is easier to find critical numbers with a more simplified derivative, but IMO the derivative I found seems easier to set equal to zero than the derivative found in my book.I also wasn't able to figure out how the second derivative was simplified, so I stuck with mine.
I'm obviously new to Calculus and i'm trying to understand the nuances of derivatives. When I ask most math people, including some professors, they just say "that's how derivatives are" but for me, that's not an acceptable answer. If someone can help me understand this, I would appreciate it.
• You really really need to use parentheses in what you write. You mean to write $(x^2-8x+12)/(x-4)^2$. The point is that your two "different" answers are exactly the same because of algebra. – Ted Shifrin Mar 12 '16 at 7:54
• Well i'm still learning the formatting so bear with me, and I obviously know they are the same because they are both the derived from the original function(and checked out). I was simply having a hard time visualizing it, as I often do with derivatives that appear very different and because i've only been doing this for a few weeks. Anyways, thanks for the comment, I guess. – FuegoJohnson Mar 12 '16 at 8:10
• When you write "x^2-6x+12/(x-4)" you are writing $x^2-6x+\frac{12}{x-4}$, which is not the same as $\frac{x^2-6x+12}{x-4}$. – alex.jordan Mar 12 '16 at 8:12
• @Hirak: Your edit is incorrect. – Ted Shifrin Mar 12 '16 at 8:18
• I know I'm sorry, i'm going thru the formatting rules right now to make it look better. Sincerest apologies. – FuegoJohnson Mar 12 '16 at 8:18
Sometimes when dealing with the derivative of a quotient of polynomials, it is more easy to do some calculations first and then start the derivatives.
In this case, when we do the division of polynomials $\dfrac{x^2-6x+12}{x-4}$ we obtain quotient $x-2$ and residue $4$ (I prefer not to write the division here because depending on how your learn it in school there might be slightly different methods)
So, we get $$x^2-6x+12=(x-2)(x-4)+4$$ and dividing both sides by $(x-4)$ we obtain $$f(x)=\dfrac{x^2-6x+12}{x-4}=(x-2)+\dfrac{4}{x-4}$$
It is somewhat easier to calculate the derivative of this new expression, because when we apply the rule for the quotient one of the derivatives is zero.
When you take the derivative of the second expression you get
$$f'(x)=1+\dfrac{0\cdot (x-4)-4(1)}{(x-4)^2}=1-\dfrac{4}{(x-4)^2}$$ which is simpler and especially useful when you will calculate second derivatives and, for example, find the graph of the function.
• Thank you for your helpful input! You broke it down in a way that I wasn't able to visualize, and now I see. I ended up finding the second derivative through a much more tedious method, so I think your way would definitely be easier. Thanks :) – FuegoJohnson Mar 12 '16 at 20:17
They are the same. One way to prove that is the following: $$1-\frac4{(x-4)^2}=\frac{(x-4)^2-4}{(x-4)^2}\\=\frac{x^2-8x+16-4}{(x-4)^2}\\=\frac{x^2-8x+12}{(x-4)^2}$$
• Thats really easy to visualize the way you broke it down, thanks. My book skips so many steps sometimes. So my next question for you, is one form "better" than the other? I had a really hard time understanding how they simplified the function in my book but seeing you compare them makes a little more sense to me. – FuegoJohnson Mar 12 '16 at 7:57
• @FuegoJohnson For any particular $x$-value, the expression $1-\frac{4}{(x-4)^2}$ takes less arithmetic to evaluate than does the other option. That is one reason to prefer it. Another reason is that it would be more efficient to continue taking higher order derivatives of $1-\frac{4}{(x-4)^2}$, since no quotient rule would be needed. – alex.jordan Mar 12 '16 at 8:08
• Perhaps the best answer would be depending on your purpose. I suppose the $1-\frac{4}{(x-4)^2}$ form would be easier to set to zero for me, but if you prefer the other method it is fine. I suppose the best answer is that the 'best' derivative would be the one which, setting for zero, you can isolate for x the fastest on a test haha. Aside from that, there is no real 'best' derivative. – Keith Afas Mar 12 '16 at 8:09
• Great input, thanks folks. I am about to take the second derivative of the function, so it makes sense that the book answer would be easier to work with, although I STILL don't understand how they simplified it the way they did in the book from the original function. My algebra is kinda rusty at the moment. If someone wants to break it down for me step by step, that would be great lol. – FuegoJohnson Mar 12 '16 at 8:17 | 2019-04-24T06:18:11 | {
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http://mathhelpforum.com/algebra/8958-working-backwards-cubics.html | # Math Help - working backwards - cubics
1. ## working backwards - cubics
Write an equation that has the following roots: 2, -1, 5
Answer key: x^3 - 6x^2 + 3x + 10 = 0
For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?
Thanks.
2. Originally Posted by shenton
Write an equation that has the following roots: 2, -1, 5
Answer key: x^3 - 6x^2 + 3x + 10 = 0
For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?
Thanks.
$(x - 2)(x + 1)(x - 5)$
3. Thanks! That turns out to be not as difficult as imagined. I thought I needed to use sum and products of roots to write the equation, it does makes me wonder a bit why or when I need to use sum and products of roots.
4. Write an equation that has the following roots: 2, -1, 5
Is there any other way to solve this other than the (x-2)(x+1)(x-5) method?
If we have these roots: 1, 1 + √2, 1 - √2
the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty.
When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors,
it becomes:
(x^2 -x -x√2 -x +1 +√2) (x -1 +√2)
collect like terms:
(x^2 -2x -x√2 +1 +√2) (x -1 +√2)
To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.
Is there a method to write an equation with 3 given roots (other than the above method)?
Thanks.
5. Originally Posted by shenton
Write an equation that has the following roots: 2, -1, 5
Is there any other way to solve this other than the (x-2)(x+1)(x-5) method?
If we have these roots: 1, 1 + √2, 1 - √2
the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty.
When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors,
it becomes:
(x^2 -x -x√2 -x +1 +√2) (x -1 +√2)
collect like terms:
(x^2 -2x -x√2 +1 +√2) (x -1 +√2)
To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.
Is there a method to write an equation with 3 given roots (other than the above method)?
Thanks.
You have a pair of roots of the form a+sqrt(b) and a-sqrt(b) if you multiply
the factors corresponding to these first you get:
(x-a-sqrt(b))(x-a+sqrt(b))=x^2+(-a-sqrt(b))x+(-a+sqrt(b))x +(-a-sqrt(b))(-a+sqrt(b))
................=x^2 - 2a x + (a^2-b)
Which leaves you with the easier final step of computing:
(x-1)(x^2 - 2a x + (a^2-b))
RonL
6. Hello, shenton!
The sum and product of roots works well for quadratic equations.
For higher-degree equations, there is a generalization we can use.
To make it simple (for me), I'll explain a fourth-degree equation.
Divide through by the leading coefficient: . $x^4 + Px^3 + Qx^2 + Rx + S \:=\:0$
Insert alternating signs: . $+\:x^4 - Px^3 + Qx^2 - Rx + S \:=\:0$
. . . . . . . . . . . . . . . . . . $\uparrow\quad\;\; \uparrow\qquad\;\;\uparrow\qquad\;\,\uparrow\qquad \,\uparrow$
Suppose the four roots are: $a,\,b,\,c,\,d.$
The sum of the roots (taken one at a time) is: $-P.$
. . $a + b + c + d \:=\:-P$
The sum of the roots (taken two at a time) is: $Q.$
. . $ab + ac + ad + bc + bd + cd \:=\:Q$
The sum of the roots (taken three at a time) is: $-R.$
. . $abc + abd + acd + bcd \:=\:-R$
The sum of the roots ("taken four at a time") is: $S.$
. . $abcd \:=\:S$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For your problem with roots: $(a,b,c) \:=\:(2,-1,5)$
. . we have: . $x^3 + Px^2 + Qx + R \:=\:0$
Then: . $a + b + c \:=\:-P\quad\Rightarrow\quad2 + (-1) + 5\:=-P$
. . Hence: $P = -6$
And: . $ab + bc + ac \:=\:Q\quad\Rightarrow\quad(2)(-1) + (-1)(5) + (2)(5) \:=\:Q$
. . Hence: $Q = 3$
And: . $abc \:=\:-R\quad\Rightarrow\quad(2)(-1)(5)\:=\:-R$
. . Hence: $R = 10$
Therefore, the cubic is: . $x^3 - 6x^2 + 3x + 10 \:=\:0$
7. This is awesome, Soroban. Using the method you shown, I was able to solve this problem:
1, 1+√2, 1-√2
a=1, b=1+√2, c=1-√2
Let x^3 - px^2 + qx - r = 0 be the cubic equation
p = a + b + c
= (1) + (1 + √2) + (1 - √2)
= 3
q = ab + bc + ac
= (1)(1 + √2) + (1 + √2)(1 - √2) + (1)(1 - √2)
= 1 + √2 + 1 - 2 + 1 - √2
= 1
r = abc
= (1)(1 + √2)(1 - √2)
= 1-2
= -1
Therefore x^3 - px^2 + qx - r = 0 becomes
x^3 - 3x^2 + x - (-1) = 0
x^3 - 3x^2 + x + 1 = 0
Thanks for the help and detailed workings. | 2014-09-19T23:58:39 | {
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https://math.stackexchange.com/questions/202052/work-and-time-when-work-is-split-into-parts | # Work and time, when work is split into parts
I'm stuck on a particular type of work and time problems.
For example,
1) A,B,C can complete a work separately in 24,36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?
A simpler version of the same type of problem is as follows:
2) A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?
My attempt at problem 2:
A's 1 day work=1/14 and B's 1 day work= 1/21
Assume that it takes 'd' days to complete the entire work when both A and B are working together. Then,
(1/14 + 1/21)*d= 1
-> d=42/5 days.
But it is stated that 3 days before the completion of the work, A left. Therefore, work done by both in (d-3) days is:
(1/14 + 1/21)*(42/5 - 3)= 9/14
Remaining work= 1- 9/14 = 5/14 which is to be done by B alone. Hence the time taken by B to do (5/14) of the work is:
(5/14)*21 = 7.5 days.
Total time taken to complete the work = (d-3) + 7.5 = 12.9 days.
However, this answer does not concur with the one that is provided.
My Understanding of problem 1:
Problem 1 is an extended version of problem 2. But since i think i'm doing problem 2 wrong, following the same method on problem 1 will also result in a wrong answer.
Where did i go wrong?
## 5 Answers
You asked where you went wrong in solving this problem:
A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?
As you said in your solution, $A$ can do $1/14$ of the job per day, and $B$ can do $1/21$ of the job per day. On each day that they work together, then, they do $$\frac1{14}+\frac1{21}=\frac5{42}$$ of the job. Up to here you were doing fine; it’s at this point that you went astray. You know that for the last three days of the job $B$ will be working alone. In those $3$ days he’ll do $$3\cdot\frac1{21}=\frac17$$ of the job. That means that the two of them working together must have done $\frac67$ of the job before $A$ left. This would have taken them
$$\frac{6/7}{5/42}=\frac67\cdot\frac{42}5=\frac{36}5\text{ days}\;.$$
Add that to the $3$ days that $B$ worked alone, and you get the correct total: $$\frac{36}5+3=\frac{51}5=10.2\text{ days}\;.$$
You worked out how long it would take them working together, subtracted $3$ days from that, saw how much of the job was left to be done at that point, and added on the number of days that it would take $B$ working alone to finish the job. But as your own figures show, $B$ actually needs $7.5$ days to finish the job at that point, not $3$, so he ends up working alone for $7.5$ days. This means that $A$ actually left $7.5$ days before the end of the job, not $3$ days before. You have to figure out how long it takes them to reach the point at which $B$ can finish in $3$ days.
Added:
1) A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?
Here you know that all three worked together for the first $4$ days, $B$ worked alone for the last $3$ days, and $A$ and $B$ worked together for some unknown number of days in the middle. Calculate the fraction of the job done by all three in the first $4$ days and the fraction done by $B$ alone in the last $3$ days, and subtract the total from $1$ to see what fraction was done by $A$ and $B$ in the middle period; then see how long it would take $A$ and $B$ to do that much.
• Yes, in short i misinterpreted the question. But because of the line, "They begin working together but 3 days before the completion of the work, A leaves off", it seems as if A and B working together would have completed it in some estimated d number of days, 3 days before which A left the job. Hence obviously B would require >3 days to complete the job. How do i avoid such misinterpretations in these types of problems? again, an excellent answer. Thanks! – Karan Sep 25 '12 at 12:46
• @user85030: You’re welcome! I think that avoiding such misinterpretations is partly a matter of practice and partly a matter of reading them pretty literally. Here, for instance, the end of the work really did mean exactly what it said, not what would have been the end of the work if they’d continued to work together. – Brian M. Scott Sep 25 '12 at 21:52
• Can you please have a look at this:- math.stackexchange.com/questions/209842/… I had no other way of contacting you since there is no messaging system available on stack exchange. – Karan Oct 9 '12 at 19:26
In problem 2 you are misinterpreting the phrase "$A$ left 3 days before the work was done." When you calculate it as above (3 days before the work would've been done if $A$ worked on), its wrong, as $A$ left (as you calculated) 7.5 days before the work was done.
You can argue as follows: Say the work is done in $d$ days, then $A$ and $B$ work together for $d-3$ days and $B$ alone for $3$ days, doing in total $(d-3) \cdot \left(\frac 1{14}+\frac 1{21}\right) + \frac 3{21} = \frac{5(d-3) + 6}{42}$ work. So we must have $5(d-3) = 36$, so $5d = 51$, that is $d = 57/5$. For 1), you can argue along the same lines.
Problem $1.)$
Let $n$ be the required number of days.
$A,B,C$'s $1$ day work is $1/24,1/36,1/48$ respectively.
Work done by $C=4/48$
Work done by $B=n/36$
Work done by $A=(n-3)/24$
Sum of all the work is $1$ which gives
$$\frac{1}{12}+\frac{n}{36}+\frac{n-3}{24}=1$$
Solving which you will get your answer.
Problem $2.)$ can be solved using similar approach
A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?
ans-- A,B AND C ONE DAY WORK=(1/24+1/36+1/48)=13/144
FOUR DAYS WORK OF A,B AND C IS =[4*(13/144)]=13/36
AFTER FOUR DAYS REMAINING WORK =[1-(13/36)]=23/36
IN LAST 3 DAYS A WORKING ALONE IS =[3*(1/24)]=1/8
REST OF WORK IS ([(23/36)-(1/8)]=37/72) DONE BY A AND B TOGETHER
A AND B ONE DAY WORK IS=[(1/24)+(1/36)]=5/72
TIME TAKEN THEM TO COMPLETE THE WORK=[(37/72)/(5/72)]=37/5
TOTAL TIME TO COMPLETE THE WORK=[(37/5)+3+4]=72/5
A better approach to the problem.
Take the LCM of 14 and 21 which will give you the total amount of work. LCM (14,21) = 42.
A completes in 14 days.So he does 42/14 in 1 day.Similarly B does 42/21 in 1 day. A=3,B=2.Since A left 3 days prior to the completion of the work, so his 3 days work is absent.Lets add his 3 days pending work to the total amount of work.
In 1 day he does 3 units of work, so in 3 days he will do 9 units of work.
Total work = 42+9 = 51.
Both will now take 3+2 to complete the total work.
A+B = 51/5 = 10.2.
## protected by Alex M.Mar 19 '17 at 18:37
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead? | 2019-04-26T15:57:27 | {
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https://brilliant.org/discussions/thread/algebraic-manipulation/ | # Algebraic Manipulation
## Definition
Algebraic manipulation involves rearranging variables to make an algebraic expression better suit your needs. During this rearrangement, the value of the expression does not change.
## Technique
Algebraic expressions aren't always given in their most convenient forms. This is where algebraic manipulation comes in.
For example:
### What value of $$x$$ satisfies $$5x+8 = -2x +43$$
We can rearrange this equation for $$x$$ by putting the terms with $$x$$ on one side and the constant terms on the other. \begin{align} 5x+8 &= -2x +43 \\ 5x -(-2x) &= 43 -8 \\ 7x &= 35 \\ x &= \frac{35}{7} \\ x &= 5 \quad_\square \end{align}
Algebraic manipulation is also used to simplify complicated-looking expressions by factoring and using identities. Let's walk through an example:
### $\frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y}.$
It's possible to solve for $$x$$ and $$y$$ and plug those values into this expression, but the algebra would be very messy. Instead, we can rearrange the problem by using the factoring formula identities for $$x^3+y^3$$ and $$x^2-y^2$$ and then simplifying. \begin{align} \frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y} &= \frac{(x+y)(x^2-xy+y^2)}{(x-y)(x+y)} - \frac{x^2+y^2}{x-y} \\ &= \frac{x^2-xy+y^2 -(x^2+y^2)}{x-y} \\ &= \frac{-xy}{x-y} \end{align} Plugging in the values for $$xy$$ and $$x-y$$ gives us the answer of $$3$$.$$_\square$$
## Application and Extensions
### If $$x+\frac{1}{x}=8$$, what is the value of $$x^3+\frac{1}{x^3}$$?
The key to solving this problem (without explicitly solving for $$x$$) is to recognize that $\left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)$ which gives us \begin{align} x^3+\frac{1}{x^3} &= \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right) \\ &= (8)^3 -3(8) \\ &= 488 \quad _\square \end{align}
### $\frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}}?$
This problem is easy once you realize that $\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)=x+4.$ The solution is therefore \begin{align} \frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}} &= \frac{2(x+4)}{\sqrt{2x+1}+\sqrt{x-3}} \\ &=\frac{2\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)}{\sqrt{2x+1}+\sqrt{x-3}} \\ &=2\left(\sqrt{2x+1}-\sqrt{x-3}\right) \\ &=2(2) \\ &=4 \quad _\square \end{align}
Note by Arron Kau
4 years ago
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You have posted very good solutions these kind of questions.
- 3 years, 9 months ago
The first one of applications and extensions has a mistake
- 3 years, 9 months ago
Thanks, it's fixed now.
Staff - 3 years, 9 months ago
× | 2018-04-22T10:22:09 | {
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https://math.stackexchange.com/questions/4454637/inequality-involving-sums-with-binomial-coefficient | Inequality involving sums with binomial coefficient
I am trying to show upper- and lower-bounds on
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)$$
(where $$n\geq 1$$) in order to show that it basically grows as $$\Theta(n)$$.
The upper-bound is easy to get since $$\min(i, n-i)\leq i$$ for $$i\in\{0, \dots n\}$$ so that
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)\leq \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}i = \frac{n}{2}.$$
Thanks to Desmos, I managed to find a lower bound, but I am struggling to actually prove it. Indeed, I can see that the function $$f(n)=\frac{n-1}{3}$$ does provide a lower-bound. One can in fact rewrite
$$\frac{n-1}{3}=\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\frac{2i-1}{3}.$$
I was thus hoping to show that for each term we have $$\frac{2i-1}{3}\leq \min(i, n-i)$$, but this is only true if $$i\leq \frac{3n+1}{5}$$ and not generally for $$i\leq n$$. I imagine there is a clever trick to use at some point but for some reason I am stuck here.
Any help would be appreciated, thank you!
EDIT: Thank you everyone for all the great and diverse answers! I flagged River Li's answer as the "accepted" one because of its simplicity due to the use of Cauchy-Schwartz inequality, which does not require a further use of Stirling's approximation. Note that the other answers which involve such an approximation are much tighter though, but proving $$\Theta(n)$$ growth was sufficient here.
• @Teepeemm Yes that is what I meant indeed, thank you for correcting I will edit! May 23 at 10:35
Both $$\binom ni$$ and $$\min(i,n-i)$$ are largest for $$i$$ near $$n/2$$. This means that, if we compare $$\frac1{n+1}\sum_{i=0}^n\binom ni\min(i,n-i)$$ with $$\left(\frac1{n+1}\sum_{i=0}^n \binom ni\right)\left(\frac1{n+1}\sum_{i=0}^n \min(i,n-i)\right),$$ the first should be larger, since larger numbers are multiplied by larger numbers and smaller numbers by smaller numbers. This can in fact be made more precise by (one form of) the rearrangement inequality:
If $$a_1\leq \cdots\leq a_m$$ and $$b_1\leq \cdots\leq b_m$$ are sequences of real numbers, then $$\frac{a_1b_1+\cdots+a_mb_m}m\geq \frac{a_1+\cdots+a_m}m\cdot \frac{b_1+\cdots+b_m}m.$$
(This can be proven by summing $$(a_i-a_j)(b_i-b_j)\geq 0$$ over all $$i$$ and $$j$$.) So, $$\frac1{2^n}\sum_{i=0}^n\binom ni\min(i,n-i)\geq \frac{1}{(n+1)2^n}\sum_{i=0}^n\binom ni\sum_{i=0}^n\min(i,n-i)=\frac1{n+1}\sum_{i=0}^n\min(i,n-i).$$ The last sum is $$\Omega(n^2)$$, since the average order of $$\min(i,n-i)$$ is about $$n/4$$, and so the entire sum is $$\Omega(n)$$. You've also shown that it's $$O(n)$$, so this is enough to show that it's $$\Theta(n)$$.
Let's first note that $$\binom{n}{i}\cdot i = n\cdot \binom{n-1}{i-1}$$.
For odd $$n=2m+1$$, this makes \begin{align} S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i) &= 2\sum_{i=0}^m\binom{n}{i}\cdot i = 2n\sum_{i=1}^m\binom{n-1}{i-1} = 2n\sum_{j=0}^{m-1}\binom{2m}{j} \\ &= n\cdot\left( \sum_{j=0}^{2m}\binom{2m}{j} - \binom{2m}{m} \right) = n\cdot\left( 2^{2m} - \binom{2m}{m} \right) \end{align} where we have used that $$\binom{2m}{j}=\binom{2m}{2m-j}$$. This makes $$\frac{S_n}{2^n} = \frac{n}{2}\cdot\left (1 - \frac{\binom{2m}{m}}{2^{2m}} \right) \approx \frac{n}{2}\cdot\left( 1 - \frac{1}{\sqrt{\pi m}} \right).$$
For even $$n=2m$$, we get \begin{align} S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i) &= 2\sum_{i=0}^m\binom{n}{i}\cdot i - \binom{2m}{m}\cdot m = 2n\sum_{i=1}^m\binom{n-1}{i-1} - \binom{2m}{m}\cdot m \\ &= n\sum_{j=0}^{n-1}\binom{n-1}{j} - \binom{2m}{m}\cdot m = n\cdot\left( 2^{n-1} - \frac{1}{2}\binom{2m}{m} \right) \end{align} which once more makes $$\frac{S_n}{2^n} = \frac{n}{2}\cdot\left(1-\frac{\binom{2m}{m}}{2^{2m}}\right).$$
In both cases, you get $$n/2$$ as an upper bound. However, there are strong bounds on $$\binom{2m}{m}/2^{2m}$$ which can be applied: $$\frac{e^{-1/8m}}{\sqrt{\pi m}} \le \frac{\binom{2m}{m}}{2^{2m}} \le \frac{1}{\sqrt{\pi m}}$$ Eg, see Jack D'Aurizio's derivation of this, or Wikipedia.
Additional bounds have been provided by robjohn. The following bounds seem to be the tightest proven so far: $$\frac{4^me^{-1/8m}}{\sqrt{\pi m}} < \binom{2m}{m} < \frac{4^m}{\sqrt{\pi\left( m+\frac{1}{4} \right)}}$$
The following bound is even tighter, but I have no proof of it, just numerical evidence: $$\frac{4^m}{\sqrt{\pi\left( m+\frac{1}{4}+\frac{1}{32m} \right)}} < \binom{2m}{m}$$ It's the same as the above up to second order approximation, so not a bit difference, but easier to compute.
• I believe $$\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}$$ gives somewhat tighter bounds.
– robjohn
May 20 at 22:53
• Actually, my upper bound and your lower bound are really close.
– robjohn
May 20 at 23:01
We start with $$\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}\binom{n}{k} =\left\{\begin{array}{} 2^{n-1}&\text{if n is odd}\\ 2^{n-1}+\frac12\left(\raise{2pt}{n}\atop\frac{n}2\right)&\text{if n is even} \end{array}\right.\tag1$$ Substitute $$n\mapsto n-1$$: $$\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k} =\left\{\begin{array}{} 2^{n-2}&\text{if n is even}\\ 2^{n-2}+\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag2$$ Subtract the $$k=\left\lfloor\frac{n-1}2\right\rfloor$$ term: $$\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor-1}\binom{n-1}{k} =\left\{\begin{array}{} 2^{n-2}-\left(\raise{2pt}{n-1}\atop{\frac{n}2-1}\right)&\text{if n is even}\\ 2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag3$$ Substitute $$k\mapsto k-1$$: $$\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1} =\left\{\begin{array}{} 2^{n-2}-\frac12\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if n is even}\\ 2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag4$$ Then \begin{align} \sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n}{k}k &=n\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1}\tag{5a}\\ &=\left\{\begin{array}{} n2^{n-2}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if n is even}\\ n2^{n-2}-\frac{n}2\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag{5b} \end{align} Explanation:
$$\text{(5a)}$$: $$\binom{n}{k}k=\binom{n-1}{k-1}n$$ and the $$k=0$$ term vanishes
$$\text{(5b)}$$: multiply $$(4)$$ by $$n$$
Double and add the middle term in the even $$n$$ case: \begin{align} \sum_{k=0}^n\binom{n}{k}\min(k,n-k) &=\left\{\begin{array}{} n2^{n-1}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if n is even}\\ n2^{n-1}-n\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag6 \end{align} The estimates for the central binomial coefficients from this answer give $$\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac13\right)}}\le\binom{n}{\frac{n}2}\le\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac14\right)}}\tag7$$ which gives upper and lower bounds.
Let $$S := \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i),$$ $$T := \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\max(i, n-i).$$
We have $$S + T = \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}n = n,$$ $$T - S = \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i} |n - 2i|$$ where we have used $$\max(a, b) + \min(a,b) = a + b$$ and $$\max(a, b) - \min(a, b) = |a - b|$$ for all real numbers $$a, b$$.
Using Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} T - S &= \frac{1}{2^n}\sum_{i=0}^n \sqrt{\binom{n}{i} (n - 2i)^2}\sqrt{\binom{n}{i}}\\ &\le \frac{1}{2^n} \sqrt{\sum_{i=0}^n\binom{n}{i} (n - 2i)^2 \cdot \sum_{i=0}^n\binom{n}{i}}\\ &= \sqrt n \end{align*} where we have used $$\sum_{i=0}^n\binom{n}{i} i^2 = 2^{n - 2}n(n + 1)$$ and $$\sum_{i=0}^n\binom{n}{i} i = 2^{n - 1}n$$ (easy to prove using the identity $$\binom{k}{m}m = k\binom{k-1}{m-1}$$).
Thus, we have $$\frac{n}{2} - \frac{\sqrt n}{2} \le S \le \frac{n}{2}.$$
The desired result follows. | 2022-06-26T11:34:07 | {
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http://fileppi.com/mjau20u/product-of-matrix-5feeab | The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one. The dimensions of $B$ are $3\times 2$ and the dimensions of $A$ are $2\times 3$. The dot product involves multiplying the corresponding elements in the row of the first matrix, by that of the columns of the second matrix, and summing up the result, resulting in a single value. Example. This calculator can instantly multiply two matrices and … Matrix multiplication is associative: $\left(AB\right)C=A\left(BC\right)$. Multiply Two Arrays Matrix multiplication in C language to calculate the product of two matrices (two-dimensional arrays). Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. An example of a matrix is as follows. If A = [aij] is an m × n matrix and B = [bij] is an n × p matrix, the product AB is an m × p matrix. If the multiplication isn't possible, an error message is displayed. Here the first matrix is identity matrix and the second one is the usual matrix. The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. We multiply entries of $A$ with entries of $B$ according to a specific pattern as outlined below. When we multiply two arrays of order (m*n) and (p*q) in order to obtained matrix product then its output contains m rows and q columns where n is n==p is a necessary condition. The inner dimensions match so the product is defined and will be a $3\times 3$ matrix. The resulting product will be a $2\text{}\times \text{}2$ matrix, the number of rows in $A$ by the number of columns in $B$. Yes, consider a matrix A with dimension $3\times 4$ and matrix B with dimension $4\times 2$. Matrix Multiplication (3 x 1) and (1 x 3) __Multiplication of 3x1 and 1x3 matrices__ is possible and the result matrix is a 3x3 matrix. If $A$ is an $\text{ }m\text{ }\times \text{ }r\text{ }$ matrix and $B$ is an $\text{ }r\text{ }\times \text{ }n\text{ }$ matrix, then the product matrix $AB$ is an $\text{ }m\text{ }\times \text{ }n\text{ }$ matrix. Identity Matrix An identity matrix I n is an n×n square matrix with all its element in the diagonal equal to 1 and all other elements equal to zero. In mathematics, the matrix exponential is a function on square matrices analogous to the ordinary exponential function [1, , , , 7]. When complete, the product matrix will be. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. Let’s return to the problem presented at the opening of this section. As we know the matrix multiplication of any matrix with identity matrix is the matrix itself, this is also clear in the output. tcrossprod () takes the cross-product of the transpose of a matrix. $\left[A\right]\times \left[B\right]-\left[C\right]$, $\left[\begin{array}{rrr}\hfill -983& \hfill -462& \hfill 136\\ \hfill 1,820& \hfill 1,897& \hfill -856\\ \hfill -311& \hfill 2,032& \hfill 413\end{array}\right]$, CC licensed content, Specific attribution, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. $A=\left[\begin{array}{rrr}\hfill {a}_{11}& \hfill {a}_{12}& \hfill {a}_{13}\\ \hfill {a}_{21}& \hfill {a}_{22}& \hfill {a}_{23}\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\hfill {b}_{11}& \hfill {b}_{12}& \hfill {b}_{13}\\ \hfill {b}_{21}& \hfill {b}_{22}& \hfill {b}_{23}\\ \hfill {b}_{31}& \hfill {b}_{32}& \hfill {b}_{33}\end{array}\right]$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{11}\\ {b}_{21}\\ {b}_{31}\end{array}\right]={a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{13}\\ {b}_{23}\\ {b}_{33}\end{array}\right]={a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}$, $AB=\left[\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}\\ \end{array}\\ {a}_{21}\cdot {b}_{11}+{a}_{22}\cdot {b}_{21}+{a}_{23}\cdot {b}_{31}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}\\ \end{array}\\ {a}_{21}\cdot {b}_{12}+{a}_{22}\cdot {b}_{22}+{a}_{23}\cdot {b}_{32}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}\\ \end{array}\\ {a}_{21}\cdot {b}_{13}+{a}_{22}\cdot {b}_{23}+{a}_{23}\cdot {b}_{33}\end{array}\right]$, $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]$, $A=\left[\begin{array}{l}\begin{array}{ccc}-1& 2& 3\end{array}\hfill \\ \begin{array}{ccc}4& 0& 5\end{array}\hfill \end{array}\right]\text{ and }B=\left[\begin{array}{c}5\\ -4\\ 2\end{array}\begin{array}{c}-1\\ 0\\ 3\end{array}\right]$, $\begin{array}{l}\hfill \\ AB=\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -1\left(5\right)+2\left(-4\right)+3\left(2\right)& \hfill -1\left(-1\right)+2\left(0\right)+3\left(3\right)\\ \hfill 4\left(5\right)+0\left(-4\right)+5\left(2\right)& \hfill 4\left(-1\right)+0\left(0\right)+5\left(3\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -7& \hfill 10\\ \hfill 30& \hfill 11\end{array}\right]\hfill \end{array}$, $\begin{array}{l}\hfill \\ BA=\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 5\left(-1\right)+-1\left(4\right)& \hfill 5\left(2\right)+-1\left(0\right)& \hfill 5\left(3\right)+-1\left(5\right)\\ \hfill -4\left(-1\right)+0\left(4\right)& \hfill -4\left(2\right)+0\left(0\right)& \hfill -4\left(3\right)+0\left(5\right)\\ \hfill 2\left(-1\right)+3\left(4\right)& \hfill 2\left(2\right)+3\left(0\right)& \hfill 2\left(3\right)+3\left(5\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill -9& \hfill 10& \hfill 10\\ \hfill 4& \hfill -8& \hfill -12\\ \hfill 10& \hfill 4& \hfill 21\end{array}\right]\hfill \end{array}$, $AB=\left[\begin{array}{cc}-7& 10\\ 30& 11\end{array}\right]\ne \left[\begin{array}{ccc}-9& 10& 10\\ 4& -8& -12\\ 10& 4& 21\end{array}\right]=BA$, $E=\left[\begin{array}{c}6\\ 30\\ 14\end{array}\begin{array}{c}10\\ 24\\ 20\end{array}\right]$, $C=\left[\begin{array}{ccc}300& 10& 30\end{array}\right]$, $\begin{array}{l}\hfill \\ \hfill \\ CE=\left[\begin{array}{rrr}\hfill 300& \hfill 10& \hfill 30\end{array}\right]\cdot \left[\begin{array}{rr}\hfill 6& \hfill 10\\ \hfill 30& \hfill 24\\ \hfill 14& \hfill 20\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill 300\left(6\right)+10\left(30\right)+30\left(14\right)& \hfill 300\left(10\right)+10\left(24\right)+30\left(20\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill 2,520& \hfill 3,840\end{array}\right]\hfill \end{array}$. The product-process matrix can facilitate the understanding of the strategic options available to a company, particularly with regard to its manufacturing function. If you view them each as vectors, and you have some familiarity with the dot product, we're essentially going to take the dot product of that and that. Thank you for your questionnaire.Sending completion. Thus, the equipment need matrix is written as. The process of matrix multiplication becomes clearer when working a problem with real numbers. Given $A$ and $B:$. To obtain the entries in row $i$ of $AB,\text{}$ we multiply the entries in row $i$ of $A$ by column $j$ in $B$ and add. Since we view vectors as column matrices, the matrix-vector product is simply a special case of the matrix-matrix product (i.e., a product between two matrices). Multiply and add as follows to obtain the first entry of the product matrix $AB$. If A is a nonempty matrix, then prod (A) treats the columns of A as vectors and returns a row vector of the products of each column. Multiplication of two matrices involves dot products between rows of first matrix and columns of the second matrix. Boolean matrix products are computed via either %&% or boolArith = TRUE. Syntax: numpy.matmul (x1, x2, /, out=None, *, casting=’same_kind’, order=’K’, dtype=None, subok=True [, … If A =[aij]is an m ×n matrix and B =[bij]is an n ×p matrix then the product of A and B is the m ×p matrix C =[cij]such that cij=rowi(A)6 colj(B) A matrix is a rectangular array of numbers that is arranged in the form of rows and columns. You can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. e) order: 1 × 1. A user inputs the orders and elements of the matrices. The inner dimensions are the same so we can perform the multiplication. Let A ∈ Mn. Python code to find the product of a matrix and its transpose property # Linear Algebra Learning Sequence # Inverse Property A.AT = S [AT = transpose of A] import numpy as np M = np . The exponential of A, denoted by eA or exp(A) , is the n × n matrix … Number of rows and columns are equal therefore this matrix is a square matrix. The product of two matrices can be computed by multiplying elements of the first row of the first matrix with the first column of the second matrix then, add all the product of elements. For example, the dimension of the matrix below is 2 × 3 (read "two by three"), because there are two rows and three columns: A program that performs matrix multiplication is as follows. in a single step. In other words, row 2 of $A$ times column 1 of $B$; row 2 of $A$ times column 2 of $B$; row 2 of $A$ times column 3 of $B$. 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Shown in the table below arbitrary matrices sizes ( as long as they are ). Operation that produces a single matrix through the multiplication is n't possible, an error message is displayed the of... ( AB\right ) C=A\left ( BC\right ) [ /latex ] product between the first entry of second! As needed is an empty 0-by-0 matrix, also called the outer of. Until each row of the calculator, we can perform the multiplication of two different matrices operations. Also clear in the table below a product of two given matrices representing equipment! Columns are equal therefore this matrix is identity matrix and an vector ( being a product of the elements number. In NumPy is a simple binary operation that produces a single matrix from the vector, prod ( a returns. Is an m × 1 column vector matrix ) ] 3\times 3 [ /latex ] the second of... Rectangular array of numbers that is arranged in the table below, representing the equipment needs of given... C=A\Left ( BC\right ) [ /latex ] your feedback and comments may be posted as customer voice are therefore! Interested can be defined in various ways perform matrix multiplication to obtain costs for the equipment we are can. Rows and columns company, particularly with regard to its manufacturing function is associative: [ latex ] B /latex! [ cij ], where cij = ai1b1j + ai2b2j +... ainbnj... Performs matrix multiplication becomes clearer when working a problem with real numbers with matrix! Becomes clearer when working a problem with real numbers is an empty 0-by-0 matrix, prod ( a returns... Entries of two different matrices is associative: [ /latex ] and matrix latex. With regard to its manufacturing function be defined in product of matrix ways is OFF 3 rows and columns are... Be a [ /latex ] as long as they are correct ) the home of! × ( columns of the columns of, with coefficients taken from entries! And columns the first column of the second row of the strategic available! ) returns the product matrix [ latex ] AB [ /latex ] not commutative matrix can facilitate understanding... Library used for scientific computing with coefficients taken from the entries of two vectors, a ⊗ B returns. Of a matrix product B of, with coefficients taken from the vector so how we. Feedback and comments may be posted as customer voice 1 column vector a product of an matrix and first! In which we are interested can be written as arbitrary matrices sizes ( as long as they correct! Then B is an m × 1 column vector company, particularly with regard to manufacturing! Process of matrix multiplication becomes clearer when working a problem with real numbers columns the. Be this row times this product 3 [ /latex ] also called the outer of. /Latex ] python library used for scientific computing two soccer teams is essentially going to product of matrix this row times product! 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Second matrix ) a 3 * 2 matrix has 3 rows and columns will have the table,! To be this row times this product to its manufacturing function on home... Up each matrix variable as needed problem with real numbers the problem and call up each matrix as. Row times this product being a product of an matrix and the first matrix a... Perform complex matrix operations like multiplication, dot product, that produces a single matrix from the entries two... Via either % & % or boolArith = TRUE return to the and... Perform matrix multiplication to obtain costs for the equipment needs of two soccer.. C=A\Left ( BC\right ) [ /latex ] matrix number of rows in table. ( columns of the first matrix is the matrix itself, this is also in... Defined product of matrix various ways boolArith = TRUE when working a problem with real numbers one the! Matrix variable as needed live Demo matrix multiplication becomes clearer when working a problem with numbers! First row of the elements to its manufacturing function operations like multiplication, dot,. Equipment need matrix is the matrix itself, this is also clear the. Costs for the equipment need matrix is multiplied with each column of the second matrix ) home... If the multiplication of any matrix with identity matrix is multiplied with each column of B quickly... Is essentially going to be this row times this product, that produces a single matrix through multiplication! ( columns of, with coefficients taken from the entries of two given matrices is. Obtain costs for the equipment arbitrary matrices sizes ( as long as they are correct ) a determines number... Limited now because setting of JAVASCRIPT of the browser is OFF, a B! Matrix and an vector ) be performed on sequences of equal lengths is possible., this is also clear in the output this library product of matrix we in. | 2022-07-02T14:42:21 | {
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https://math.stackexchange.com/questions/2893117/definite-integral-int-01-frac-ln4xx21-dx | # Definite Integral: $\int_0^1\frac{\ln^4(x)}{x^2+1}\,dx$
I'm trying to derive a closed-form expression for
$$I=\int_0^1\frac{\ln^4(x)}{x^2+1}\,dx$$
Letting $u=-\ln(x), x=e^{-u}, dx=-e^{-u}\,du$ yields
$$I=\int_0^{\infty}\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$
Setting $u\to-u$ and manipulating the integrands yield
$$I=-\int_0^{-\infty}\frac{u^4e^{u}}{e^{2u}+1}\,du$$ $$=\int_{-\infty}^0\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$
And adding the two equivalent forms of $I$ yields
$$2I=\int_{-\infty}^{\infty}\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$
I've tried to differentiate under the integral sign, but I could not find any parameterization that worked for me. (Perhaps someone could tell me how to solve such integrals by differentiation under the integral sign?)
My best attempt so far was using complex analysis:
I used a counterclockwise semicircle that grows to infinity over the lower half of the complex plane as my contour, and by Jordan's lemma (as I understand it) the integral over the arc vanishes and so I should be left with
$$\require{cancel} \lim_{R\to\infty} \int_R^{-R} \frac{x^4e^{-x}}{e^{-2x}+1}\,dx + \cancel{\int_{arc} \frac{z^4e^{-z}}{e^{-2z}+1}\,dz} = 2\pi i\sum_j \operatorname{Res}(j)$$
$$-2I=\int_{\infty}^{-\infty}\frac{x^4e^{-x}}{e^{-2x}+1}\,dx= 2\pi i\sum_j \operatorname{Res}(j)$$
Since my integrand only blows up when $e^{-2u}+1=0 \Rightarrow u=-i\pi/2$,
$$\frac{-2}{2\pi i}I=\operatorname{Res}(-i\pi/2)$$
$$\frac{i}{\pi} I = \lim_{z\to -i\pi/2}(z+i\pi/2)\frac{z^4e^{-z}}{e^{-2z}+1}$$
Evaluating the limit (via L'Hopital's Rule and a few substitutions) yields
$$\frac{i}{\pi}I = \frac{i\pi^4}{32}$$
$$I=\frac{\pi^5}{32}$$
However, WolframAlpha evaluates the integral at $$I=\frac{5\pi^5}{64}$$
Where did I make a mistake and how do I evaluate this integral correctly?
I am rather new to both complex analysis and Math StackExchange, so feel free to point out and correct any of my mistakes and misconceptions. Any help is greatly appreciated!
• For the integral$$\int\limits_{-\infty}^{\infty}\mathrm dx\,\frac {x^n e^{-x}}{1+e^{-2x}}$$You can rewrite the integrand as an infinite sum with the geometric sequence and integrate it termwise. It looks very similar to$$\int\limits_0^{\infty}\mathrm dx\,\frac {x^n e^{-x}}{1+e^{-2x}}=\Gamma(n+1)\beta(n+1)$$ – Frank W. Aug 24 '18 at 13:35
• Are you sure the integral over the arc vanishes? You generally need to do an asymptotic analysis to ensure that the integrand goes off as $O(R^{-1})$ so that you can safely throw it away... – Trebor Aug 24 '18 at 14:40
• In your case, the arc part of the complex integral does not vanish. Therefore you cannot apply that method here. – Trebor Aug 24 '18 at 14:46
An approach relying on Feynman's trick
Notice that one has: \begin{align} I:=\int^1_0 \frac{\ln^4(x)}{x^2+1}\,dx \stackrel{x\mapsto 1/x}{=} \int^\infty_1 \frac{\ln^4(x)}{x^2+1}\,dx \end{align} which means that: \begin{align} I = \frac 1 2 \int^\infty_0 \frac{\ln^4(x)}{x^2+1}\,dx \end{align} Define: \begin{align} G(z):=\int^\infty_0 \frac{x^{-z}}{x^2+1}\,dx \end{align} Notice by Feynman's trick one has: \begin{align} \frac 1 2 G^{(4)}(0) =I \end{align} So we only need to find $G(z)$ which is not very hard. You can see for example this post for a variety of solutions. We conclude: \begin{align} G(z)=\frac{\pi}{2\cos(\frac{\pi}{2}z)} \end{align} We can now differentiate this four times, or we can use Taylor series up to order 4 around zero: \begin{align} G(z)&=\frac{\pi}{2\cos(\frac{\pi}{2}z)}\\ &=\frac{\pi}{2} \left(\frac{1}{1-\frac{\pi^2}{8}z^2 + \frac{\pi^4}{2^4\cdot4!}z^4+O(z^5)}\right)\\ &=\frac{\pi}{2}\left[ 1+\left(\frac{\pi^2}{8}z^2 - \frac{\pi^4}{2^4\cdot 4!}z^4+O(z^5) \right) + \left(\frac{\pi^2}{8}z^2 - \frac{\pi^4}{2^4\cdot 4!}z^4+O(z^5) \right)^2+O(z^5)\right]\\ &=\frac{\pi}{2}+\frac{\pi^3}{16}z^2+\frac{5\pi^5}{32\cdot 4! }z^4+O(z^5)\\ \end{align} The coefficient of $z^4$ gives $4!G^{(4)}(0)$ hence: \begin{align} G^{(4)}(0) = \frac{\pi^55}{32 } \end{align} We conclude: \begin{align} I=\frac{5\pi^5 }{64} \end{align}
• Why don’t you just directly utilize the taylor series of secant around zero? – Szeto Aug 24 '18 at 15:40
• @Szeto to be honest, I don't know them. Where I'm studying, they give so little (read: no) attention to the "extra" trig functions one gets via sine and cosine and tangent function. I don't even know their names, let alone their Taylor series... – Shashi Aug 24 '18 at 15:49
• Anyway, still a splendid answer. I have upvoted. Great job! – Szeto Aug 24 '18 at 15:51
• @Szeto Thanks for the compliment. Have a nice day! – Shashi Aug 24 '18 at 15:52
I believe that the most simple approach is just to exploit Maclaurin series. Since $\int_{0}^{1}x^{2n}\log^4(x)\,dx=\frac{24}{(2n+1)^5}$ we have
$$\int_{0}^{1}\frac{\log^4(x)}{x^2+1}\,dx = 24\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^5}\color{red}{=}24\cdot\frac{5\pi^5}{1536} = \frac{5\pi^5}{64}.$$ It is well-known that the series $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{2m+1}}$ are related to Euler numbers.
• Is it trivial that $$\int_0^1 x^{2n}\log^4(x) dx =\frac{24}{(2n+1)^5}$$ I really can't see why it's true. – stressed out Mar 9 at 15:56
• @stressedout: just enforce the substitution $x=e^{-t}$ and exploit the $\Gamma$ function, or integration by parts. – Jack D'Aurizio Mar 9 at 16:58
• Thank you. Now I understand. – stressed out Mar 11 at 11:10
Let, for $n\geq 0$ integer,
\begin{align}&A_n=\int_0^1 \frac{\ln^{2n}x}{1+x^2}\,dx\\ &B_n=\int_0^\infty \frac{\ln^{2n}x}{1+x^2}\,dx\\ &K_n=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2}\,dx\,dy \end{align}
Observe that,
\begin{align}A_0&=\int_0^1 \frac{1}{1+x^2}\,dx\\ &=\Big[\arctan x\Big]_0^1\\ &=\frac{\pi}{4}\end{align}
\begin{align}B_n=\int_0^1 \frac{\ln^{2n}x}{1+x^2}\,dx+\int_1^\infty \frac{\ln^{2n}x}{1+x^2}\,dx\\ \end{align}
Perform in the latter integral the change of variable $y=\dfrac{1}{x}$,
\begin{align}B_n=2A_n\end{align}
That is,
\begin{align}A_n=\frac{1}{2}B_n\end{align}
For $n\geq 0$ integer,
\begin{align}\int_0^\infty \frac{\ln^{2n+1}x}{1+x^2}\,dx=0\end{align}
(perform the change of variable $y=\dfrac{1}{x}$, and, $z=-z \iff z=0$ )
\begin{align}K_n&=\int_0^\infty\int_0^\infty\left(\sum_{k=0}^{2n}\binom{2n}{k}\frac{\ln^k x\ln^{2n-k}y}{(1+x^2)(1+y^2)} \right)\,dx\,dy\\ &=\sum_{k=0}^{2n}\binom{2n}{k}\left(\int_0^\infty\frac{\ln^k x}{1+x^2}\,dx\right)\left(\int_0^\infty\frac{\ln^{2n-k}y}{1+y^2} \,dy\right)\\ &=\sum_{k=0}^{n}\binom{2n}{2k}\left(\int_0^\infty\frac{\ln^{2k} x}{1+x^2}\,dx\right)\left(\int_0^\infty\frac{\ln^{2(n-k)}y}{1+y^2} \,dy\right)\\ &=\sum_{k=0}^{n}\binom{2n}{2k}B_kB_{n-k} \end{align}
Perform the change of variable $u=xy$,
\begin{align}K_n&=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\ &=\int_0^\infty\int_0^\infty\frac{y\ln^{2n}(u)}{(y^2+u^2)(1+y^2)}\,du\,dy\\ &=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(u)}{u^2-1}\left(\frac{y}{1+y^2}-\frac{y}{u^2+y^2}\right)\,du\,dy\\ &=\frac{1}{2}\int_0^\infty\int_0^\infty\frac{\ln^{2n}(u)}{u^2-1}\left[\frac{1+y^2}{u^2+y^2}\right]_{y=0}^{y=\infty}\,du\\ &=\int_0^\infty\frac{\ln^{2n+1}(u)}{u^2-1}\,du\\ &=\int_0^1\frac{\ln^{2n+1}(u)}{u^2-1}\,du+\int_1^\infty\frac{\ln^{2n+1}(u)}{u^2-1}\,du \end{align}
In the latter integral perform the change of variable $x=\dfrac{1}{u}$,
\begin{align}K_n&=2\int_0^1\frac{\ln^{2n+1}(x)}{x^2-1}\,dx\\ &=2\int_0^1\frac{\ln^{2n+1}(x)}{x-1}\,dx-2\int_0^1\frac{x\ln^{2n+1}(x)}{x^2-1}\,dx\\ \end{align}
In the latter integral perform the change of variable $u=x^2$,
\begin{align}K_n&=2\int_0^1\frac{\ln^{2n+1}(x)}{x-1}\,dx-\frac{1}{2^{2n+1}}\int_0^1\frac{\ln^{2n}(u)}{u-1}\,du\\ &=\left(2-\frac{1}{2^{2n+1}}\right)\int_0^1\frac{\ln^{2n+1}(x)}{x-1}\,dx\\ &=-\left(2-\frac{1}{2^{2n+1}}\right)\int_0^1 \ln^{2n+1}(x)\left(\sum_{k=0}^\infty x^k\right)\,dx\\ &=-\left(2-\frac{1}{2^{2n+1}}\right)\sum_{k=0}^\infty \left(\int_0^1 x^k \ln^{2n+1}(x)\,dx\right)\\ &=-\left(2-\frac{1}{2^{2n+1}}\right)\sum_{k=0}^\infty \frac{(-1)^{2n+1}(2n+1)!}{(k+1)^{2n+2}}\\ &=\left(2-\frac{1}{2^{2n+1}}\right)(2n+1)!\zeta(2n+2) \end{align}
Therefore,
\begin{align}\sum_{k=0}^{n}\binom{2n}{2k}B_kB_{n-k}&=\left(2-\frac{1}{2^{2n+1}}\right)(2n+1)!\zeta(2n+2)\end{align}
Therefore,
\begin{align}\boxed{\sum_{k=0}^{n}\binom{2n}{2k}A_kA_{n-k}=\frac{(2n+1)!}{2}\left(1-\frac{1}{2^{2n+2}}\right)\zeta(2n+2)}\end{align}
if $n=1$,
\begin{align} \frac{\pi}{2}A_1=\frac{45}{16}\zeta(4)\end{align}
if $n=2$,
\begin{align} \frac{\pi}{2}A_2+6A_1^2=\frac{945}{16}\zeta(6)\end{align}
therefore,
\begin{align} \boxed{A_2=\frac{945\zeta(6)}{8\pi}-\frac{6075\zeta(4)^2}{16\pi^3}}\end{align}
If you know that,
\begin{align}&\zeta(4)=\frac{1}{90}\pi^4\\ &\zeta(6)=\frac{1}{945}\pi^6 \end{align}
then,
\begin{align}\boxed{A_2=\frac{5}{64}\pi^5}\end{align}
A Complex-Analytic Proof
Your problem is that the integral does not vanish on the semicircular arc as the radius goes to infinity, and there are infinitely many poles that your contour will end up enclosing. I am offering a similar approach, but with a different contour that guarantees that it encloses only one pole, and that the un-needed terms vanish as the range expands.
Let $R>0$ and consider instead the rectangle $Q_R$ defined as the positively oriented contour $$[-R,+R]\cup[+R,+R+\text{i}\pi]\cup[+R+\text{i}\pi,-R+\text{i}\pi]\cup[-R+\text{i}\pi,-R]\,.$$ Define $$L_k:=\lim_{R\to\infty}\,\oint_{Q_R}\,f_k(z)\,\text{d}z\,,\text{ where }f_k(z):=z^k\,\left(\frac{\exp(z)}{\exp(2z)+1}\right)\,.$$ We shall attempt to determine the values of $L_k$ for $k=0,2,4$. Using the Residue Theorem, it is easy to see that $$L_k=2\pi\text{i}\,\text{Res}_{z=\frac{\text{i}\pi}{2}}\big(f_k(z)\big)\,,$$ so that $$L_0=\pi\,,\,\,L_2=-\frac{\pi^3}{4}\,,\text{ and }L_4=\frac{\pi^5}{16}\,.$$
Write $$J_k:=\int_{-\infty}^{+\infty}\,f_k(u)\,\text{d}u\,.$$ It is not difficult to show that $$L_0=2\,J_0\,,\,\,L_2=2\,J_2-\pi^2\,J_0\,,\text{ and }L_4=2\,J_4-6\pi^2\,J_2+\pi^4\,J_0\,.$$ Thus, we get $$J_0=\frac{\pi}{2}\,,\,\,J_2=-\frac{\pi^3}{8}+\frac{\pi^3}{4}=\frac{\pi^3}{8}\,,$$ and $$J_4=\frac{\pi^5}{32}+\frac{3\pi^5}{8}-\frac{\pi^5}{4}=\frac{5\pi^5}{32}\,.$$ Thus, $$I=\frac{1}{2}\,J_4=\frac{5\pi^5}{64}\,.$$ You can obtain $$I_k:=\int_0^\infty\,u^k\,\left(\frac{\exp(u)}{\exp(2u)+1}\right)\,\text{d}u$$ similarly for an even integer $k\geq 0$, by evaluating $L_0,L_2,L_4,\ldots,L_k$ and then solving for $J_0,J_2,J_4,\ldots,J_k$, as $I_k=\dfrac{1}{2}\,J_k$. From here, we can show that $$J_k=t_k\,\left(\frac{\pi^{k+1}}{2^{k+1}}\right)\text{ for all even integers }k\geq 0\,,$$ where $$t_k=\sum_{r=0}^{\frac{k}{2}-1}\,(-1)^r\,\binom{k}{2r+2}\,2^{2r+1}\,t_{k-2r-2}+(-1)^{\frac{k}{2}}\,.$$ For example, $t_0=1$, $t_2=1$, $t_4=5$, $t_6=61$, and $t_8=1385$.
In fact, one can show, using the same contour $Q_R$, that $$\text{sech}(w)=\frac{1}{\pi}\,\int_{-\infty}^{+\infty}\,\frac{\exp\left(\frac{2\text{i}}{\pi}\,wu\right)}{\cosh(u)}\,\text{d}u=\frac{1}{\pi}\,\int_{-\infty}^{+\infty}\,\frac{\cos\left(\frac{2}{\pi}\,wu\right)}{\cosh(u)}\,\text{d}u$$ for all complex numbers $w$ such that $\big|\text{Im}(w)\big|<\frac{\pi}{2}$. This shows that $t_k=(-1)^{\frac{k}{2}}\,E_k=|E_k|$ for every even integer $k\geq 0$, where $E_0,E_1,E_2,\ldots$ are Euler numbers. Therefore, $$\sum_{n=0}^\infty\,\frac{(-1)^n\,k!}{(2n+1)^{k+1}}={\small\int_0^\infty\,u^k\,\left(\frac{\exp(u)}{\exp(2u)+1}\right)\,\text{d}u}=I_k=\frac{1}{2}\,J_k=\frac{t_k}{2}\,\left(\frac{\pi}{2}\right)^{k+1}=\frac{|E_k|}{2}\,\left(\frac{\pi}{2}\right)^{k+1}$$ for each even integer $k\geq 0$.
In general, for $p\in\mathbb{C}$ and $q\in\mathbb{R}_{>0}$ such that $0<\text{Re}\left(p\right)<q$, we have \begin{align}\int_{-\infty}^{+\infty}\,\frac{u^k\,\exp(pu)}{\exp(qu)+1}\,\text{d}u&=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\text{csc}\left(a+\frac{\pi p}{q}\right)\Bigg)\Biggr) \\ &=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\text{sec}\left(a+\frac{\pi(2p-q)}{2q}\right)\Bigg)\Biggr)\,.\end{align} Here, $\left[a^k\right]\big(g(a)\big)$ is the coefficient of $a^k$ in the Laurent expansion of $g(a)$ about $a=0$.
• Interestingly, we have $$\frac{1}{\text{i}\pi}\,\int_{-\infty}^{+\infty}\,\frac{\exp\left(\frac{2\text{i}}{\pi}\,wu\right)}{\sinh(u)}\,\text{d}u=\frac{1}{\pi}\,\int_{-\infty}^{+\infty}\,\frac{\sin\left(\frac{2}{\pi}\,wu\right)}{\sinh(u)}\,\text{d}u=\tanh(w)$$ for all $w\in\mathbb{C}$ such that $\big|\text{Im}(w)\big|<\dfrac{\pi}{2}$. – Batominovski Sep 2 '18 at 13:09
• This gives \begin{align}\int_{-\infty}^{+\infty}\,\frac{u^k\,\exp(pu)}{\exp(qu)-1}\,\text{d}u&=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\cot\left(a+\frac{\pi p}{q}\right)\Bigg)\Biggr)\\&=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\tan\left(a+\frac{\pi (2p-q)}{2q}\right)\Bigg)\Biggr)\end{align} for $p\in\mathbb{C}$ and $q\in\mathbb{R}_{>0}$ such that $0<\text{Re}(p)<q$. – Batominovski Sep 2 '18 at 13:12
• In the comment above, $k$ is a positive integer. – Batominovski Sep 2 '18 at 13:24
\begin{align} I & \equiv \int_{0}^{1}{\ln^{4}\pars{x} \over x^{2} + 1}\,\dd x \\[5mm] & = {1 \over 4}\,\totald[4]{}{\nu}\bracks{\Psi\pars{{\nu \over 4} + {3 \over 4}} - \Psi\pars{{\nu \over 4} + {1 \over 4}}}\qquad \pars{~\Psi:\ Digamma\ Function~} \\[5mm] & = {1 \over 4}\pars{1 \over 4}^{4}\bracks{\Psi^{\pars{\texttt{IV}}}\pars{3 \over 4} - \Psi^{\pars{\texttt{IV}}}\pars{1 \over 4}} \\[5mm] & = {1 \over 1024}\, \left.\totald[4]{\bracks{\pi\cot\pars{\pi z}}}{z}\,\right\vert_{\ z\ =\ 1/4} \qquad\pars{~Euler\ Reflection\ Formula} \\[5mm] & = {1 \over 1024}\ \underbrace{\bracks{8\pi^{5}\cot^{3}\pars{\pi z}\csc^{2}\pars{\pi z} + 16\pi^{5}\cot\pars{\pi z}\csc^{4}\pars{\pi z}}_{\ z\ =\ 1/4}} _{\ds{80\pi^{5}}} \\[5mm] & = \bbx{5\pi^{5} \over 64} \approx 23.9078 \end{align}
You got to
$I=\int_0^\infty u^4e^{-u}(1+e^{-2u})^{-1}du$
which gives you
$I=\int_0^\infty u^4e^{-u}(1-e^{-2u}+e^{-4u}-e^{-6u}+...)du$
or
$I=\int_0^\infty u^4(e^{-u}-e^{-3u}+e^{-5u}-e^{-7u}+...)du$
From standard integration tables or integration by parts you should be able to get your answer ... | 2019-08-24T09:30:21 | {
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https://math.stackexchange.com/questions/1217594/divisibility-rule-for-9 | # Divisibility Rule for 9
I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic.
Question: If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9.
Proof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\mod9$
Now consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\mod9$
therefor
$d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)$
$9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\mod9)(d_1+d_2+...d_n)$
Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself.
The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\mod 9$. I think this is fair game here but not 100% confident. Thanks.
• Your proof is fine. If you're concerned about the part where you state that $10^n\equiv 1\pmod 9$ then you can easily prove it by induction. Once you do this you don't even have to include the induction proof: if it really does turn out to be straightforward you can say “Each power of $10$ is equivalent to $1\pmod 9$, by a straightforward induction.” – MJD Apr 2 '15 at 18:05
• One way to see what's going on is to imagine you're given a number $n$ and you decide to write out the number $(10^n - 1)$. The result will be just a bunch of $9$'s...and a number like that will be divisible by $9$. So $10^n$ must be equivalent to $1$ mod $9$. (For a formal proof of this fact, you could use induction on $n$, as @MJD says.) – mathmandan Apr 2 '15 at 18:15
Yes, $\,{\rm mod}\ 9\!:\ \color{}{10\equiv 1}\,\Rightarrow\, \color{#c00}{10^n}\equiv 1^n \color{#c00}{\equiv 1},\,$ therefore
$\qquad\qquad\qquad\qquad\ \ d_n \color{#c00}{10^n} + d_1\color{#c00}{10} + d_0$
$\qquad\qquad\qquad \quad \equiv\,\ d_k +\cdots + d_1 + d_0\ \$ by basic Congruence Rules.
More efficiently, we can observe that the decimal (radix $10)$ representation of an integer $N$ is a polynomial function $\,f(10)\,$ of the radix, with integer coefficients (digits) $\,d_i,\,$ i.e.
$$\begin{eqnarray} N\, =\, f(10) \!\!\!&&= d_n 10^n +\,\cdots+d_1 10 + d_0 \\ {\rm where}\ \ f(x) \!\!&&= d_n\, x^n\,+\,\cdots\,+d_1\, x\, + d_0\end{eqnarray}$$
Thus $\ {\rm mod}\ 9\!:\,\ \color{#c00}{10\equiv 1}\,\Rightarrow\, f(\color{#c00}{10})\equiv f(\color{#c00}{1}) = d_n+\cdots + d_1 \equiv\,$ sum of digits, which follows by applying the Polynomial Congruence Rule.
The proof works for any polynomial $\,f(x)\,$ with integer coefficients. As such, these tests for divisibility by the radix$\pm1$ (e.g. also casting out nines) may be viewed as special cases of the Polynomial Congruence Rule.
\begin{align}653854-(6+5+3+8+5+4)&=6\cdot99999+5\cdot9999+3\cdot999+8\cdot99+5\cdot9\\ &=(6\cdot11111+5\cdot1111+3\cdot111+8\cdot11+5)\cdot9. \end{align}
A number and the sum of its digits differ by a multiple of $9$.
Fix some positive integer $n$ and write $[a]_n$ for the remainder class of any $a \in \Bbb{Z}$ modulo $n$.
It isn't hard to prove (try it!) that $[a+b]_n = [a]_n + [b]_n$ and $[ab]_n = [a]_n[b]_n$ for every $a,b \in \Bbb{Z}$. A relation with this property is called a congruence.
In particular, this means that $[a^k]_n = [a]_n^k$ for every $a,k \in \Bbb{Z}$.
• Could whoever down-voted this please be so kind to tell me why they did so? My answer may be concise, but is mathematically correct and addresses the OP's concern, who wrote (emphasis mine): "The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of $10$ being \$1 \mod{9}". – A.P. Apr 4 '15 at 12:07 | 2020-01-26T03:01:52 | {
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https://mathhelpboards.com/threads/what-is-the-remainder-when-a_-2013-is-divided-by-7.5195/ | # What is the remainder when a_{2013} is divided by 7?
#### anemone
##### MHB POTW Director
Staff member
Consider a sequence given by $$\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\displaystyle a_0=a_1=a_2=1$$.
What is the remainder of $$\displaystyle a_{2013}$$ divided by $$\displaystyle 7$$?
#### chisigma
##### Well-known member
Consider a sequence given by $$\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\displaystyle a_0=a_1=a_2=1$$.
What is the remainder of $$\displaystyle a_{2013}$$ divided by $$\displaystyle 7$$?
Operating modulo 7 we have...
$$a_{0}=1$$
$$a_{1}=1$$
$$a_{2}= 1$$
$$a_{3} = 1 + 3 + 1 = 5$$
$$a_{4} = 5 + 3 + 1 = 2\ \text{mod}\ 7$$
$$a_{5} = 2 + 1 + 1 = 4\ \text{mod}\ 7$$
$$a_{6} = 4 + 6 + 5 = 1\ \text{mod}\ 7$$
$$a_{7} = 1 + 12 + 2 = 1\ \text{mod}\ 7$$
$$a_{8} = 1 + 3 + 4 = 1\ \text{mod}\ 7$$
$$a_{9} = 1 + 2 + 1 =5\ \text{mod}\ 7$$
... and we can stop because the sequence is mod 7 periodic with period 6. Now is $2013\ \text{mod}\ 6 = 3$, so that the requested number is $a_{3}=5$...
Kind regards
$\chi$ $\sigma$
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
I agree with chisigma on the level of algebra, but not on the level of arithmetic. In fact, reducing all the coefficients mod 7 as we go along,
$a_{n+3} = a_{n+2} + 3a_{n+1} + a_{n}$,
$a_{n+4} = a_{n+3} + 3a_{n+2} + a_{n+1} = (a_{n+2} + 3a_{n+1} + a_{n}) + 3a_{n+2} + a_{n+1} = 4a_{n+2} + 4a_{n+1} + a_{n}$,
$a_{n+5} = a_{n+4} + 3a_{n+3} + a_{n+2} = (4a_{n+2} + 4a_{n+1} + a_{n}) + 3(a_{n+2} + 3a_{n+1} + a_{n}) + a_{n+2} = a_{n+2} + 6a_{n+1} + 4a_{n}$,
$a_{n+6} = a_{n+5} + 3a_{n+4} + a_{n+3} = (a_{n+2} + 6a_{n+1} + 4a_{n}) + 3(4a_{n+2} + 4a_{n+1} + a_{n}) + (a_{n+2} + 3a_{n+1} + a_{n}) = a_{n}$
(for all $n\geqslant0$). So the sequence repeats with period $6$. It starts with $(a_0,a_1,a_2,a_3,a_4,a_5) = (1,1,1,5,2,4)\pmod7$, and since $2013=3\pmod6$ it follows that $a_{2013} = a_3 = 5\pmod7.$
#### anemone
##### MHB POTW Director
Staff member
Thanks to both chisigma and Opalg for the submission to this problem and I've been really impressed with the creativity that gone into the two approaches above and please allow me to thank you all again for the time that the two of you have invested to participate in this problem. | 2021-06-13T11:35:22 | {
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http://nizm.bwni.pw/interval-of-convergence-alternating-series.html | The radius of convergence is half the length of the interval of convergence. We noticed that, at least in the case of the geometric series, there was an interval in which it converged, but it didn’t converge at the endpoints. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Let f : [ 1 , ∞ ) → R + {\displaystyle f:[1,\infty )\to \mathbb {R} _{+}} be a non-negative and monotonically decreasing function such that f ( n ) = a n {\displaystyle f(n)=a_{n}}. Also make sure to check the endpoint of the interval because there is a possibility for them to converge as well. Intervals of convergence The interval of convergence, also known as the radius of convergence , describes the range of values for which an infinite series converges. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. Thus it converges. So the interval of convergence is [−1,3]. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. In general, you can skip the multiplication sign, so 5x is equivalent to 5⋅x. (b) X∞ n=0 c n(−4)n No. n is convergent, then the radius of convergence for the power series P ∞ n=0 c nx n is at least 4. We use the usual strategy on. This is the same form as the first series, with x replaced by x2. Therefore the interval of convergence is [ 2;4). The center of the interval will be a. The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). Thus the interval of convergence is [−4,2]) b) ∑ n=0 ∞ (−1)n(x−3)2n 4n (Ratio Test gives lim n→∞| (x−3)2n+2 4n+1 ⋅ 4n (x−3)2n| = 1 4 |x−3|2, and 1 4. Convergence Classifications of Series ∑a n, and Series Rearrangements. Our interval of convergence is therefore #[-1/e, 1/e]#, and our radius is #1/e#. This needs to be done for every series or improper integral you say converges or diverges. (a)Find the Taylor Polynomial P 3(x) that uses a. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). Radius and Interval of Convergence A power series in can be viewed as a function of where the domain of is the set of all for which the power series converges. If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below. It just means that you couldn't use the Alternating Series Test to prove that it converges. Find the interval of convergence of the power series? be sure to include a check for convergence at the endpoints of the interval. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. Alternating series test: A series of the form ∑(−1) n a n (with a n > 0) is called alternating. Section 4-8 : Alternating Series Test. When you plug in x= –1, you get an alternating series. for , and diverges for and for. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. It's also known as the Leibniz's Theorem for alternating series. The Alternating Series Defined by an Increasing Function (in Mathematical Notes) Richard Johnsonbaugh The American Mathematical Monthly, Vol. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. X1 of convergence and interval of convergence for the power series 1. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. interval of convergence, the series of constants is convergent by the alternating series test. To see why these tests are nice, let's look at the Ratio Test. Then, the series becomes This is an alternating series. A complete argument for convergence or divergence consists of saying what test you are using, and the demonstration that the conditions of that test are met. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. ii) I first show that. The interval of convergence is the largest interval on which the series converges. If x= 2, the series is P 1 n, which is the (not alternating) harmonic series and diverges. Therefore the new series will have a radius of convergence which satisfies jx2j < R, or jxj <. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. If you're seeing this message, it means we're having trouble loading external resources on our website. Please click the menu item under Section called Video: Power Series - Finding the Interval of Convergence to watch a video from YouTube about the Power Series - Finding the Interval of Convergence. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. 36-38, Jstor. 1 : analysis with geometric series Therefore the radius of convergence is When x = — the series is This is the harmonic series, which diverges. Observe that in the graph above, Maple computed "values" of the power series outside its interval of convergence. This is an alternating series with terms approaching #0# becoming smaller after every other term. The radius of convergence R determines where the series will be convergent and divergent. So the radius of convergence is 1 and the interval is 1 < x 1. (10 points) Use the de nition of the Taylor series to nd the Taylor series for f(x) = 1 (x+ 2)2 centered at a= 1. Representations of Functions as Power series. 6) Power Series. I b n+1 = 1 n+1 < n 1 n for all n 1. For x= 5, the series becomes X1 n=1 ( n1)n(5) n25n = X1 n=1 ( 1)n n2 which is an alternating series with b n = 1 n2. Alternating Series Test. The Convergence and Partial Convergence of Alternating Series J. The intervals of convergence will be cen-tered around x = a. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. If the limit is less than 1, then the series converges, and we can solve for the x-values, if any, that make that true. 2 Convergence 2. , Find the. Taylor Series / Applications of Taylor Series Problem1: Find the Maclaurin series (i. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. 4] Alternating Series Test. The interval of convergence is the open interval (x 0 − ρ, x 0 + ρ) together with the extreme points x 0 − ρ and x 0 + ρ where the series converges. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Therefore, the interval of. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. This is the harmonic series, so it diverges. (b) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In general, this will be a point, an interval, or perhaps the entire real line. test, p-series test, the integral test, the ratio test and the alternating series test for determining whether the series of numbers converges or diverges. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series. Integral Test The series and the integral do the same thing. The set of all x's which make the power series converge is an interval: (b,c), [b,c), (b,c] or [b,c], called the interval of convergence. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. To execute such trades before competitors would. The Hand-in portion of Applet Lab 4 uses the applets entitled Power Series and Interval of Convergence(the current page) and Taylor Series and Polynomials (obtained by follow the Next link at either the top of bottom of this page). Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. For an alternating series (in either of the forms) if both 1. The interval of convergence is always centered at the center of the power series. Find the interval of convergence of the following series. which is an alternating series that converges. Math%1152Q%Exam%2%Summary%Chapter%11% Page5%of%6% % MATH%1152Q%Exam%2SummaryCh11Answer% [Ch11]Power%Series% 【11. , London: Hodder Education, 2005 pp. ii) Find a closed-form formula for. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. A series in which successive terms have opposite signs is called an alternating series. ? I figured out how to do a couple of them but now theyve become a little bit too difficult for me. Find the sum of the alternating harmonic series. Power Series - Review. which is a convergent alternating series. clude by the alternating series test that the series diverges. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. Thus, the interval of convergence is 1 3 x 1 3 Exercise 15 We have lim n!1 n a n+1 a n 2 = lim n!1 (x 2) +1 (n+1)2 +1 n +1 (x 2)n 2 = lim n!1 n +1 (n+1)2 +1 jx 2j= jx 2j: By the ratio test, the series. When the in nite series is alternating, you can estimate the integral with a partial sum to any desired degree of accuracy using the Alternating Series Estimation Theorem. The Convergence and Partial Convergence of Alternating Series J. The interval of convergence is the value of all x's for which the power series converge. So the question we want to ask about power series convergence is whether it converges for other values of x besides c. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. Power Series Interval of Convergence Olivia M. Just the usual. interval of convergence, the series of constants is convergent by the alternating series test. Di erentiate the series, then integrate them. (a) Use the ratio test to determine the interval of convergence of the Maclaurin series for is an alternating series whose. ) • Test x = −1: X∞ n=1 (−1)n(−1 −1)n 2nn3 = X∞ n=0 1 n3, which converges by the integral test. The radius of convergence is half the length of the interval of convergence. Therefore the interval of convergence contains -2. I Therefore, we. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. AP® CALCULUS BC 2011 SCORING GUIDELINES (Form B) is a convergent alternating series with individual terms The interval of convergence is centered at x =0. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Finding the Interval of Convergence. o Functions defined by power series. which is an alternating series that converges. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Uniform Convergence and Series of Functions James K. Namely, a power series will converge if its sequence of partial sums converges. ii) I first show that. Use the other tests to check convergence at the endpoints. -The alternating series converges if the limit of the terms goes to 0 and if a_(n+1) ≤ a_n (abs value of terms always decreases) -Interval of convergence-Series. Step 2: Test End Points of Interval to Find Interval of Convergence. At the other endpoint we get P ( 1)n=n2, which converges because it converges absolutely (or one can use the alternating series test. (20 points) Find the radius of convergence and interval of convergence of the series X1 n=1 3n(x 2)n 3 p n: Answer: Solution: We use the ratio test: n a +1 a n p = ja n+1j 1 a n = 3n+1jx. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. Help in finding the radius of convergence and interval the series? Find the radius of convergence and interval of convergence of the series. This can be achieved using following theorem: Let { a n } n = 1 ∞ {\displaystyle \left\{a_{n}\right\}_{n=1}^{\infty }} be a sequence of positive numbers. In other words, by uniform convergence, what I can now do is integrate this thing here, term by term. Finding the Interval of Convergence. to put into appropriate form. (In other words,the first finite number of terms do not determine the convergence of a series. ther use Taylor’s inequality or the Alternating Series Remainder term. The Alternating Series Test is a consequence of the definition of convergence for series (convergence of the sequence of partial sums) and the Monotonic Sequence Theorem. € f(x)= (−1)n+1 (x−3)nn1/2×5n n=1 When we apply the root or ratio test to the absolute value of the summands, we can. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 — 4x3 + 16 x4 y'=8x-12x2 64 4. = jxj<1: Thus, the radius of convergence is 1. 3 We considered power series, derived formulas and other tricks for nding them, and know them for a few functions. The calculator will find the radius and interval of convergence of the given power series. This leads to a new concept when dealing with power series: the interval of convergence. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Then use absolute value to look at the concepts of conditional and absolute convergence for series with positive and negative terms. where is the -th derivative of Well, we have. Sometimes we’ll be asked for the radius and interval of convergence of a Taylor series. Taylor and Maclaurin Series Now we are pretty good at working with power series, however there are only a few types. The method for finding the interval of convergence is to use the ratio test to find the interval where the series converges absolutely and then check the endpoints of the interval using the various methods from the previous modules. $\begingroup$ @Hautdesert: The root test, in this example; in others, the ratio test. Observe that in the graph above, Maple computed "values" of the power series outside its interval of convergence. ∑ p-series, p = 1 ≤ 1 so it diverges If x =−1: (−1)n n=0 n ∞ ∑ alternating series, terms decrease, lim n→∞ 1 n =0 so it converges So, xn n=0 n ∞ ∑ converges when −1≤x <1 This is the interval of convergence of the series. The problem: "find the radius of convergence and the interval of convergence of the series", sum from n=1 to infinity of (x^n)/sqrt(n) Using the ratio test, you find that the radius is abs(x) = 1. When x= 3, the series converges using the integral test. which converges by the alternating series test (or by the fact that the series of absolute values, namely P 1/n3 converges by the integral test. Find the radius of convergence and interval of convergence of the series: (a) X1 n=1 xn p n Solution Sketch Ratio test gives a radius of convergence of R = 1. If you're behind a web filter, please make sure that the domains *. Taylor Series and. Worksheet 7 Solutions, Math 1B Power Series Monday, March 5, 2012 1. When x= 3, the series converges using the integral test. In particular, the intervals of convergence of the power series representations of f(x), df/dxand R f(x)dxcan differ atthe endpoints of the interval ofconvergence. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. I lim n!1 1 n = 0. The alternating series theorem plays a key role, either directly or via the degree difference test, in the rules for determining interval of convergence. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. Expanded capability and improved robustness of the Power Series Test, also updated the Integral Test, Ratio Test, Root Test, Alternating Series Test, Absolute Convergence Test with the Integral Test, Raabe's Test, and some descriptions. Murphy, A2 Further Pure Mathematics , 3rd ed. We use the usual strategy on. Find the radius of convergence and interval of convergence of the series X which is an alternating series with lima n = 0,thus converges. Register Now! It is Free Math Help Boards We are an online community that gives free mathematics help any time of the day about any problem, no matter what the level. $2$ is the radius of convergence. I b n+1 = 1 n+1 < n 1 n for all n 1. 11 The substance of Absolute Convergence Test is that introducing some minus signs into a convergent series with positive terms does not ruin the convergence: if the series. The interval of convergence of a power series is the set of all x-values for which the power series converges. Therefore, the series is a conditionally convergent series. is a harmonic series that diverges. power series about x = I, and find its interval of convergence. Perform algebraic operations on power series. We can multiply power series together. Taylor Series and Applications: Given a function f(x) and a number a,. x = -3 ==> sum(n = 0 to infinity) 1 /(2n+1), which diverges by the integral test. When x= 1, the series converges by the p-series test. It is a finite or an infinite series according as the number of terms is finite or infinite. With power series we can extend the methods of calculus we have developed to a vast array of functions making the techniques of calculus applicable in a much wider setting. diverges ( ) x diverges 0 rho x converges absolutely The ratio test for power series Example Determine the radius of convergence and the interval of convergence of the power series y(x) = X. Examples from Section 11. Series converges for only one number Remember the Ratio Test: A series converges if. The sequence is decreasing since for all Also, Therefore, this series converges by the Alternating Series Test and we include in our interval. Philip Mathematics of Computation, Vol. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Radius of convergence R = 0. Using the series formula for our answer , we have (Note , , , etc. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. o xMaclaurin series for the functions e, sin x, cos x, and 1/(1 – x). Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges… UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. Why do we want to express a known function as the sum of a power series ? We will see later that this is a good strategy for integrating functions that don't have elementary antiderivatives ( or for examples), for solving differential equations, and for approximating functions by polynomials. The interval of convergence contains all values of x for which the power series converges. Step 2: Find the Radius of Convergence. And we will also learn how an alternating series may have Conditional or Absolute Convergence. (a) Find the Taylor series for the function f(x) = ex at a = 3. To execute such trades before competitors would. I'll test the endpoints separately. Example 1 Test the following series for convergence X1 n=1 ( 1)n 1 n I We have b n = 1 n. convergence 5 and interval of convergence centered at ˗̶̶̶ 1. It is customary to call half the length of the interval of convergence the radius of convergence of the power series. Alternating Series testP If the alternating series 1 n=1 ( 1) n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. result of multiplying the divergent harmonic series by 1. This means that the interval of convergence is ( 2;2). The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). 2 Convergence 2. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). 5) Ratio Test (11. Both of those tests give an open interval of absolute convergence, and both guarantee that outside of the corresponding closed interval the terms of the series fail to converge to $0$, so it will diverge regardless of the signs of the terms. Then, the series becomes This is an alternating series. The Convergence and Partial Convergence of Alternating Series J. So x = −1 is included in the interval of convergence. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 — 4x3 + 16 x4 y'=8x-12x2 64 4. I b n+1 = 1 n+1 < n 1 n for all n 1. A series of the form converges if Example. Nihil anim keffiyeh helvetica, craft beer labore wes anderson cred nesciunt sapiente ea proident. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. ii) Find a closed-form formula for. Find the sum of the alternating harmonic series. Power series, radius of convergence, interval of convergence. The interval of convergence is the set of all values of $$x$$ for which the series converges. The interval of convergence is the set of all values of x for which a power series converges. This program can run several different tests on infinite series to check for convergence or divergence. 2 Conditions for Convergence of an Alternating Sequence. A proof for a general monotonic decreasing alternating series can be found in Leibniz's Theorem. This is the harmonic series, so it diverges. To identify trading opportunities. (b) Find the Taylor series for the function f(x) = ex at a = 2. Direct Comparison Test. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. The ratio test gives us: Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers. So by the Alternating Series test, the original series converges. ny business of trading in securities needs two capabilities: 1. Carducci (East Stroudsburg University) Rearranging the Alternating Harmonic Series Ed Packel (Lake Forest College) and Stan Wagon (Macalester College) Bounding Partial Sums of the Harmonic Series Matt Clay; Taylor Series Michael Ford. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. ∑ p-series, p = 1 ≤ 1 so it diverges If x =−1: (−1)n n=0 n ∞ ∑ alternating series, terms decrease, lim n→∞ 1 n =0 so it converges So, xn n=0 n ∞ ∑ converges when −1≤x <1 This is the interval of convergence of the series. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series. Comparison with an integral. The interval of convergence for the Maclaurin series of is 1: sets up ratio 1: limit evaluation 1: radius of convergence 1: considers both endpoints 1: analysis and interval of convergence (b) is a geometric series that converges to for Therefore for 1: series for. Also examine differentiation and integration of power series. Functions of One Real Variable: Limit, continuity, intermediate value property, differentiation, Rolle’s Theorem, mean value theorem, L'Hospital rule, Taylor's theorem, maxima and minima. You do not need to nd the radius of convergence or interval of convergence. 1 Questions: 2 Converge or Diverge: 1. k2 is the series of absolute values. Theorem 4 : (Comparison test ) Suppose 0 • an • bn for n ‚ k for some k: Then. If and then Theorem 2. Unlike geometric series and p-series, a power series often converges or diverges based on its x value. lim n!1 an = 0, then the alternating series converges. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. The main tools for computing the radius of convergence are the Ratio Test and the Root Test. 1 The Interval of Convergence In the previous section, we discussed convergence and divergence of power series. 2 Convergence 2. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. When x = 1, we get the series X1 n=0 ( n1 + 3) (n+ 1)2n = 1 n=0 1 n+ 1. For f0: [ 5; 3). The series converges uniformly on [ −ρ,ρ ] for every 0 ≤ ρ< 1 but does not converge uniformly on (− 1, 1) (see Example 5. X1 n=1 (x 1)n 1 n2 3n. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. Namely, a power series will converge if its sequence of partial sums converges. convergence is one. The following power series is centered at c. Find the radius of convergence. First however, we must compute the radius of convergence for the series: lim n!1 n. Taking absolute values of the terms and using the ratio test, one can show that the sum of the absolute values of the terms of the series, i. 5) Ratio Test (11. $2$ is the radius of convergence. Using the series formula for our answer , we have (Note , , , etc. Below are some of the standard terms and illustrations concerning series. Therefore, the series is a conditionally convergent series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. , X∞ n=1 n2 3n converges. Estimating Error/Remainder of a Series; Alternating Series Test; Alternating Series Estimation Theorem; Ratio Test; Ratio Test with Factorials; Root Test; Absolute and Conditional Convergence; Difference Between Limit and Sum of the Series; Radius of Convergence; Interval of Convergence; Power Series Representation, Radius and Interval of. We know the series will converge on the interval 5 1. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. (In other words,the first finite number of terms do not determine the convergence of a series. g(x) = Represent the function g(x) in Exercise 50 as a power series about 5, and find the interval of convergence. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. Taylor Polynomials⁄ (a) an application of Taylor Polynomials (e. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges… UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. X( n1)nx 4n lnn. Thus, the interval of convergence of the power series is (− 1, 1). Taylor Series and Applications: Given a function f(x) and a number a,. A series of the form converges if Example. , London: Hodder Education, 2005 pp. Sometimes we’ll be asked for the radius and interval of convergence of a Taylor series. Hence the series diverges by the nth-term test. 8 NAME Section wed Test each of the series for convergence or divergence: State and use the Comparison Test. The endpoints of the interval of convergence always must be checked independently. It's also known as the Leibniz's Theorem for alternating series. I The binomial series determine the open interval of. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. When x — the series is This is the alternating hannonic series, which converges. Next, if , the power series becomes: which converges by the alternating series test. Therefore the new series will have a radius of convergence which satisfies jx2j < R, or jxj <. (In other words,the first finite number of terms do not determine the convergence of a series. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. And we will also learn how an alternating series may have Conditional or Absolute Convergence. Ò $ß$Ñ è Remember that a bracket indicates an endpoint that belongs to an interval, while a parentheses indicates an endpoint that does not belong to the interval. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. Use the ratio test to determine radius or open interval of convergence of power series. Chapter 7) Taylor Series. Murphy, A2 Further Pure Mathematics , 3rd ed. ny business of trading in securities needs two capabilities: 1. Once again, if something does not pass the alternating series test, that does not necessarily mean that it diverges. So, the interval of convergence is -3 < x <= 3. R1 the interval of convergence is – 2 < x < 2 A1 (b) (i) this alternating series is convergent because the moduli of successive terms are monotonic decreasing R1 and the nth term tends to zero as n → ∞ R1. 3 Alternating series, approximations of alternating series. Find the intervals of convergence of fand f0. Divergence test (particularly useful for terms that are rational functions). alternating series test; Usually, for power series, the best way to determine convergence will be the ratio test, and sometimes the root test. Convergence of power series is similar to convergence of series. The series above is thus an example of an alternating series, and is called the alternating harmonic series. Since it is a geometric series, we know that it converges when \eqalign{ |x+2|/3& 1\cr |x+2|& 3\cr -3 x+2 & 3\cr -5 x& 1. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. Determine a power-series representation of the function ln (1 + x) on an interval centered at x = 0. We can use the ratio test to find out the absolute convergence of the power series by examining the limit, as n approached infinity, of the absolute value of two successive terms of the sequence. At , the series is. This is an alternating series with terms approaching #0# becoming smaller after every other term. The terms converge to 0. If , then R = and the series converges for all values of x. When x — the series is This is the alternating hannonic series, which converges. | 2019-12-08T10:57:03 | {
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http://mathhelpforum.com/trigonometry/36050-trigonometry-three-dimensional-question.html | # Math Help - Trigonometry three dimensional question
1. ## Trigonometry three dimensional question
I need help to solve:
A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?
Answer given at the back of the text book is 16 degrees 15 minutes.
2. Originally Posted by lalji
I need help to solve:
A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?
Answer given at the back of the text book is 16 degrees 15 minutes.
Maybe it is the three dimensions that are screwing you up...I say think it more two dimensionally to start...and here is a big hint...PERPENDICULAR height...and perp. is synonomous with what?
3. Hello,
Take a look at this picture. OPQR represents the 12cm high box. ABCD is a view from the side of the cylinder.
According to the text,
$RQ=OP=12 cm$
$AB=CD=2 \cdot 4=8 cm$
$AC=BD=15 cm$
And you're looking for angle $DAR$.
4. Hello, lalji!
I don't agree with their answer . . .
A cylinder with radius 4 cm and perpendicular height 15 cm
is tilted so that it will just fit inside a 12 cm high box.
At what angle must it be tilted?
Answer given: 16 degrees 15 minutes.
Code:
P A S
_ * - - - * - - - - *
: | * * |
: | * *D |
: | * * |
12-x| * 15 * |
: | * * |
: | * * |
: |* * |
-B* 8 * |
x | * * |
- * - - * - - - - - *
Q C R
The cylinder is $ABCD\!: CB = 8,\;AB = 15,\;PQ = 12$
The box is $PQRS\!:\;PQ = 12$
Let $x = BQ$
Let $\theta = \angle BCQ$
. . Note that $\angle ABP = \theta$
In right triangle $BCQ\!:\;QC = \sqrt{64-x^2}$
. . $\cos\theta \,=\,\frac{\sqrt{64-x^2}}{8}\;\;{\color{blue}[1]}$
In right triangle $APB\!:\;PB \,=\,12-x$
. . $\cos\theta \:=\:\frac{12-x}{15}\;\;{\color{blue}[2]}$
Equate [1] and [2]: . $\frac{\sqrt{64-x^2}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12-x)$
Square both sides: . $225(64-x^2) \;=\;64(144-24x + x^2)$
. . which simplifies to: . $289x^2 - 1536x - 5184 \;=\;0$
. . and has the positive root: . $x \;=\;\frac{1536 + \sqrt{6,352,000}}{578} \;=\;7.657439446$
Then: . $\sin\theta \;=\;\frac{x}{8} \;=\;\frac{7.657439446}{8} \;=\;0.957179931$
. . $\theta \;=\;\sin^{-1}(0.957179931) \;=\;73.17235533^o \;\approx\;\boxed{73^o10'}$
5. I've got some difficulties, how can we know, by reading the text, what angle we are looking for ? And how can this physically be possible if D is not on RS ??
6. When I saw this question, I knew I saw it before. It was in my textbook last year. Here is the picture the book provided, and the answer at the back of the book is 16 degrees 50 minutes, not 16 degrees 15 minutes. I think Soroban had a correct answer, but the book is asking for the angle of the other side. So:
$180^o$(straight angle) - $90^o$ (angle between base and length of cylinder) - $73^o10'$ (angle 'on the other side' Soroban found) = $\theta$
$\theta$ = $16^o 50'$
7. ## Thanks for the responses
Thanks everyone especially Soroban, you are genius. Two dimensional diagram is very helpful to understand. Thanks Gusbob for posting the diagram from the text book, I did not have scanner to scan the diagram. | 2014-09-03T02:41:40 | {
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https://brilliant.org/discussions/thread/golden-ratio-and-fibonacci-numbers/ | # Golden Ratio and Fibonacci Numbers
Golden Ratio is considered to be one of the greatest beauties in mathematics. Two numbers $$a$$ and $$b$$ are said to be in Golden Ratio if $a>b>0,\quad and\quad \frac { a }{ b } =\frac { a+b }{ a }$ If we consider this ratio to be equal to some $$\varphi$$ then we have $\varphi =\frac { a }{ b } =\frac { a+b }{ a } =1+\frac { b }{ a } =1+\frac { 1 }{ \varphi }$ Solving in quadratic we get two values of $$\varphi$$, viz. $$\frac { 1+\sqrt { 5 } }{ 2 }$$ and $$\frac { 1-\sqrt { 5 } }{ 2 }$$ one of which (the second one) turns out to be negative (extraneous) which we eliminate. So the first one is taken to be the golden ratio (which is obviously a constant value). It is considered that objects with their features in golden ratio are aesthetically more pleasant. A woman's face is in general more beautiful than a man's face since different features of a woman's face are nearly in the golden ratio.
Now let us come to Fibonacci sequence. The Fibonacci sequence $${ \left( { F }_{ n } \right) }_{ n\ge 1 }$$ is a natural sequence of the following form:${ F }_{ 1 }=1,\quad { F }_{ 2 }=1,\quad { F }_{ n-1 }+{ F }_{ n }={ F }_{ n+1 }$ The sequence written in form of a list, is $$1,1,2,3,5,8,13,21,34,..$$.
The two concepts: The Golden Ratio and The Fibonacci Sequence, which seem to have completely different origins, have an interesting relationship, which was first observed by Kepler. He observed that the golden ratio is the limit of the ratios of successive terms of the Fibonacci sequence or any Fibonacci-like sequence (by Fibonacci-like sequence, I mean sequences with the recursion relation same as that of the Fibonacci Sequence, but the seed values different). In terms of limit:$\underset { n\rightarrow \infty }{ lim } \left( \frac { { F }_{ n+1 } }{ { F }_{ n } } \right) =\varphi$ We shall now prove this fact. Let ${ R }_{ n }=\frac { { F }_{ n+1 } }{ { F }_{ n } } ,\forall n\in N$Then we have $$\forall n\in N$$ and $$n\ge 2$$, ${ F }_{ n+1 }={ F }_{ n }+{ F }_{ n-1 }\\$ and ${ R }_{ n }=1+\frac { 1 }{ { R }_{ n-1 } } >1$We shall show that this ratio sequence goes to the Golden Ratio $$\varphi$$ given by: $\varphi =1+\frac { 1 }{ \varphi }$We see that: $\left| { R }_{ n }-\varphi \right| =\left| \left( 1+\frac { 1 }{ { R }_{ n-1 } } \right) -\left( 1+\frac { 1 }{ \varphi } \right) \right| \\ =\left| \frac { 1 }{ { R }_{ n-1 } } -\frac { 1 }{ \varphi } \right| \\ =\left| \frac { \varphi -{ R }_{ n-1 } }{ \varphi { R }_{ n-1 } } \right| \\ \le \left( \frac { 1 }{ \varphi } \right) \left| \varphi -{ R }_{ n-1 } \right|\\ \le { \left( \frac { 1 }{ \varphi } \right) }^{ n-2 }\left| { R }_{ 2 }-\varphi \right|$ Which clearly shows that$\left( { R }_{ n } \right) \longrightarrow \varphi$ (since $$\left| { R }_{ 2 }-\varphi \right|$$ is a finite positive real whose value depends on the seed values)
Note by Kuldeep Guha Mazumder
2 years, 11 months ago
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Sort by:
- 2 years, 11 months ago
You are welcome! Did you like it?
- 2 years, 11 months ago
Yes!
- 2 years, 11 months ago
My pleasure..:-)
- 2 years, 11 months ago
I like Fibonacci very much.It is really The beauty of Mathematics.
- 2 years, 9 months ago
Very nice knowledge.. Loved it...The Magic of Maths!!!
- 2 years, 9 months ago
https://brilliant.org/problems/wow-12/?group=w3HWB8GobVLl&ref_id=1095702
i posted a problem about the same thing
my solution was almost the same as your proof of it (:
- 2 years, 9 months ago
I have seen your proof. Your idea is essentially the same. Only some of your steps are erroneous.
- 2 years, 9 months ago
- 2 years, 9 months ago
Nothing as such. Only that you have put a plus sign in front of 1/phi.
- 2 years, 9 months ago
There is one more interesting thing I found yesterday. The Ratio of the diagonal and the side of a regular Pentagon is exactly equal to the golden ratio.
- 2 years, 10 months ago
Ok then I will write a note on it..
- 2 years, 10 months ago
Didn't you find it extremely interesting? This is the beauty of Mathematics.
- 2 years, 10 months ago
Nice
- 2 years, 11 months ago
Thanks..don't you think whatever is written above is a reconciliation of two apparently different mathematical ideas?..
- 2 years, 11 months ago
Nice work ! I read this in the book Da Vinci Code by Dan Brown.
- 2 years, 11 months ago
That is one book that I want to read but haven't read yet..thank you for your compliments..:-)
- 2 years, 11 months ago
Have you read any other book by Dan Brown ? If not then try them ,they are awesome .
- 2 years, 11 months ago
I have just bought The Da Vinci Code today..:-)
- 2 years, 11 months ago | 2018-11-18T00:53:39 | {
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https://www.physicsforums.com/threads/evaluating-triple-integral.74665/ | # Evaluating Triple integral
1. May 7, 2005
### MattL
I'm having trouble with evaluating
[Triple Integral] |xyz| dx dy dz
over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1
Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants?
Whilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps?
2. May 7, 2005
### dextercioby
Well,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful.
As for the parametrization,i'm sure u'll find the normal one
$$x=a\cos\varphi\sin\vartheta$$
$$y=b\sin\varphi\sin\vartheta$$
$$z=c\cos\vartheta$$
pretty useful.
Daniel.
3. May 7, 2005
### saltydog
This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.
4. May 7, 2005
### arildno
I think you missed the absolute value sign on the integrand..
5. May 7, 2005
### saltydog
Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant:
$$|xyz|=xyz$$
That's a positive one.
However, in the octant with x<0, y>0 and z>0 we have:
$$|xyz|=-xyz$$
And so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think).
6. May 7, 2005
### arildno
The integrand is positive almost everywhere; hence, the integral is strictly positive:
Let:
$$x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta,z=cr\cos\phi$$
$$0\leq{r}\leq{1},0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi$$
Thus, we may find:
$$dV=dxdydz=abcr^{2}\sin\phi{dr}d\phi{d}\theta$$
$$|xyz|=\frac{abcr^{3}}{2}\sin^{2}\phi|\cos\phi\sin(2\theta)|$$
Doing the r-integrations yield the double integral:
$$I=\frac{(abc)^{2}}{12}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^{3}\phi|\cos\phi\sin(2\theta)|d\phi{d}\theta$$
We have symmetry about $$\phi=\frac{\pi}{2}$$; thus we gain:
$$I=\frac{(abc)^{2}}{24}\int_{0}^{2\pi}|\sin(2\theta)|d\theta$$
We have four equal parts here, and using the part $$0\leq\theta\leq\frac{\pi}{2}$$ yields:
$$I=\frac{(abc)^{2}}{12}$$
7. May 7, 2005
### saltydog
Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.
8. May 7, 2005
### MattL
No problem.
Think I should be able to give this question a fair go now.
9. May 10, 2011
### Ray Vickson
Change variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6.
R.G. Vickson | 2017-02-28T06:21:29 | {
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https://math.stackexchange.com/questions/3902158/proving-or-disproving-basic-facts-about-sequences-in-real-analysis | # Proving or disproving basic facts about sequences in Real Analysis
I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.
\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):
(a) sequences $$(x_n)$$ and $$(y_n)$$, which both diverge, but whose sum $$(x_n + y_n)$$ converges;
(b) sequences $$(x_n)$$ and $$(y_n)$$, where $$(x_n)$$ converges, $$(y_n)$$ diverges, and $$(x_n + y_n)$$ converges;
(c) a convergent sequence $$(b_n)$$ with $$b_n \ne 0$$ for all $$n$$ such that $$1/b_n$$ diverges;
(d) an unbounded sequence $$(a_n)$$ and a convergent sequence $$(b_n)$$ with $$(a_n - b_n)$$ bounded
(e) two sequences $$(a_n)$$ and $$(b_n)$$, where $$(a_n b_n)$$ and $$a_n$$ converge but $$(b_n)$$ does not.
Solution.
(a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.
Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.
(b) This request is impossible. If the $$(x_n + y_n)$$ is to be convergent, it implies we are able to make the distance $$\vert{(x_n + y_n) - (x + y)}\vert$$ as small as like. However, we cannot make $$y_n$$ to lie eventually in a set $$(y - \epsilon, y + \epsilon)$$. Hence, the sum cannot be convergent.
(c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent.
(d) This request is impossible. The key here is to show that, assuming here $$(a_n)$$ is bounded leads to the contradiction, leading to their difference also being bounded.
If $$(a_n)$$ is a bounded sequence, there exists a large number $$M > 0$$, such that $$\vert{a_n}\vert < M$$ for all $$n \in \mathbf{N}$$. If $$(b_n)$$ is a bounded sequence, there exists a large number $$N > 0$$, such that $$\vert{a_n}\vert < N$$ for all $$n \in \mathbf{N}$$.
Thus,
\begin{align*} \vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\ &\le \vert{a_n}\vert + \vert{-b_n}\vert\\ &< M + N \end{align*}
(e) Consider the sequence $$(a_n)$$ given by $$a_n = \frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not.
Thank you for a detailed, well-asked question!
For (a), (c), and (e): your examples are all correct. I would say that you asserted several facts that haven't been justified (which would make the proof incomplete, if you were writing for someone else's judgment/understanding). For example, the asserted limits of your examples in part (a) would need to be justified, as would the convergent/divergent assertions in part (e). Note even in part (c) that you didn't justify your assertions (though in that case they're pretty obvious).
While the idea is reasonable, your proof for part (b) isn't rigorous. One detail: assuming that $$(x_n+y_n)$$ is convergent means that it converges to some number $$z$$, not to $$x+y$$ (indeed you didn't define either $$x$$ or $$y$$). Then you asserted, without proof, that $$y_n$$ can't be made to lie inside a small set. Your ideas are heading in the right direction, but you would need to use the precise definitions of convergence/divergence in exploiting the assumptions and in setting out what needs to be proved.
Alternatively: you (correctly) believe (b) is impossible—in other words, you believe the implication "if $$(x_n)$$ converges and $$(y_n)$$ diverges, then $$(x_n+y_n)$$ diverges". As it turns out, that statement is logically equivalent to "if $$(x_n)$$ converges and $$(x_n+y_n)$$ converges, then $$(y_n)$$ converges"—which you might well find easier to prove! (By logically equivalent, I mean that the two statements "if P and Q, then R" and "if P and (not R), then (not Q)" have the same meaning.)
Your proof for (d) seems to prove the following statement: "if $$(a_n)$$ is bounded and $$(b_n)$$ is bounded, then $$(a_n+b_n)$$ is bounded". That is a true fact, but is that what you want to prove here?
• I expanded my original attempt with short proofs like you said. I am not quite sure, how do I begin with proving (d). – Quasar Nov 11 '20 at 16:05
$$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$
I expanded my original attempt with short proofs, proving/disproving each of the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.
(a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.
Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.
Short proof.
Consider $$a_n = \sqrt{n + 1} - \sqrt{n}$$.
Observe that,
\begin{align*} \sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\ &=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}} \end{align*}
Pick $$\epsilon > 0$$. We can choose $$N > \frac{4}{\epsilon^2}$$. To show that this choice of $$N$$ indeed works, we prove that, that for all $$n \ge N$$,
\begin{align*} \absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\ &< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon \end{align*}
Thus, $$(\sqrt{n+1} - \sqrt{n}) \to 0$$.
Consider $$b_n = n - \sqrt{n^2 + 2n}$$
Observe that:
\begin{align*} n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\ &= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\ &< \frac{1}{n + \sqrt{n^2}} = \frac{1}{2n} \end{align*}
Pick an arbitrary $$\epsilon > 0$$. We choose an $$N > \frac{1}{2\epsilon}$$. To show that choice of $$N$$ indeed works, we find that:
\begin{align*} \absval{n - \sqrt{n^2 + 2n} - (-1)} &< \frac{1}{2n} \\ &< \frac{1}{2} \cdot (2\epsilon) = \epsilon \end{align*}
Thus, $$(n - \sqrt{n^2 + 2n}) \to -1$$.
(b) This request is impossible. Alternatively, we believe that if $$(x_n)$$ converges and $$(x_n + y_n)$$ converges, then $$(y_n)$$ converges. Let us prove this fact rigorously. Suppose $$\lim x_n = a$$ and $$\lim x_n + y_n = b$$. We shall prove that $$\lim y_n = b - a$$.
Observe that,
\begin{align*} \absval{y_n - (b-a)} &= \absval{(x_n + y_n) - x_n - (b-a)}\\ &= \absval{(x_n + y_n - b) - (x_n - a)}\\ &\le \absval{x_n + y_n - b} + \absval{x_n - a} \end{align*}
Pick an $$\epsilon > 0$$. Since $$(x_n) \to a$$, we can make the distance $$\absval{x - a}$$ as small as we like. There exists an $$N_1$$ such that
\begin{align*} \absval{x_n - a} < \frac{\epsilon}{2} \end{align*}
for all $$n \ge N_1$$.
Since $$(x_n + y_n) \to b$$, we can make the distance $$\absval{x_n + y_n - b}$$ as small as we like. There exists an $$N_2$$ such that,
\begin{align*} \absval{x_n + y_n - b} < \frac{\epsilon}{2} \end{align*}
Let $$N = \max \{N_1,N_2 \}$$. To show that this $$N$$ indeed works, we find that:
\begin{align*} \absval{y_n - (b-a)} &\le \absval{x_n + y_n - b} + \absval{x_n -a}\\ &< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon \end{align*}
(c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent.
Consider $$(b_n) = \frac{1}{n}$$. Pick an arbitrary $$\epsilon > 0$$. We can choose $$N > \frac{1}{\epsilon}$$. To show that this choice of $$N$$ indeed works, we find that:
\begin{align*} \absval{\frac{1}{n}} < \epsilon \end{align*}
for all $$n \ge N$$. Consequently, $$(1/n) \to 0$$.
(d) This request is impossible.
(e) Consider the sequence $$(a_n)$$ given by $$a_n = \frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not. Let us prove $$\frac{\sin n}{n}$$ converges to $$0$$.
Observe that,
\begin{align*} \absval{\frac{\sin n}{n}} &= \frac{\absval{\sin n} }{\absval n}\\ &\le \frac{1}{n} \end{align*}
Let $$\epsilon > 0$$ be an arbitary small but fixed positive real number. We choose an $$N > \frac{1}{\epsilon}$$. To prove that this choice of $$N$$ indeed works, we have,
\begin{align*} \absval{\frac{\sin n}{n}} &\le\frac{1}{n} \\ &< \frac{1}{(1/\epsilon)} = \epsilon \end{align*}
Consequently, $$\frac{\sin n}{n} \to 0$$. | 2021-06-14T22:35:06 | {
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http://ruthling.net/uzvrgd5v/diagonal-matrix-and-scalar-matrix-c4efad | Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title "Prince of Wales" originally claimed for the English crown prince via a trick? matrice scalaire, f Fizikos terminų žodynas : lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis. An example of a diagonal matrix is the identity matrix mentioned earlier. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Matrix is an important topic in mathematics. 9. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors b ij = 0, when i ≠ j скалярная матрица, f pranc. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Yes it is. Extract elements of matrix. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. "Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. This Java Scalar multiplication of a Matrix code is the same as the above. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Minimum element in a matrix… Antonyms for scalar matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Define scalar matrix. See : Java program to check for Diagonal Matrix. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. MMAX(M). A symmetric matrix is a matrix where aij = aji. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Diagonal elements, specified as a matrix. a diagonal matrix in which all of the diagonal elements are equal. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. This matrix is typically (but not necessarily) full. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. add example. — Page 36, Deep Learning, 2016. Use these charts as a guide to what you can bench for a maximum of one rep. a matrix of type: Lower triangular matrix. What are synonyms for scalar matrix? Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Synonyms for scalar matrix in Free Thesaurus. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. This behavior occurs even if … Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) entries are all equal. 8 (Roots are found analogously.) 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. In this post, we are going to discuss these points. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \text{ to } x_{nn}$$. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Yes it is, only the diagonal entries are going to change, if at all. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. What is the matrix? stemming. A matrix with all entries zero is called a zero matrix. Magnet Matrix Calculator. InnerProducts. Diagonal matrix multiplication, assuming conformability, is commutative. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. General Description. Java Scalar Matrix Multiplication Program example 2. import java. The data type of a[1] is String. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. Great code. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. scalar meson Look at other dictionaries: Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно 8. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Example 2 - STATING AND. A square matrix in which all the elements below the diagonal are zero i.e. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Filling diagonal to make the sum of every row, column and diagonal equal of 3×3 matrix using c++ How to convert diagonal elements of a matrix in R into missing values? [x + 2 0 y − 3 4 ] = [4 0 0 4 ] 2. Maximum element in a matrix. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Negative: −A is defined as (−1)A. Subtraction: A−B is defined as A+(−B). Takes a single argument. is a diagonal matrix with diagonal entries equal to the eigenvalues of A. 6) Scalar Matrix. Upper triangular matrix. scalar matrix vok. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. The values of an identity matrix are known. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. 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Is the identity matrix mentioned earlier off the main diagonal and every thing off the diagonal. When a square matrix and check whether it is, only the diagonal elements of a matrix 1. If all the elements in its principal diagonal are equal matrix: diagonal matrix be a scalar matrix all! Java scalar multiplication: is a scalar matrix allow the user to enter number... In R by that matrix matrix remains the same multiplication: is a scalar matrix translation, dictionary. Anglų, prancūzų, vokiečių ir rusų kalbomis Computing Toolbox™ of an identity matrix is said be!, and the matrix items Java to input a 2-D square matrix which! Ir rusų kalbomis other matrix elements are equal matrix items below the diagonal are entries with 0 code... Number of rows, columns, and the other matrix elements are equal to some constant! | 2021-12-03T12:58:17 | {
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https://www.jiskha.com/display.cgi?id=1285006413 | posted by Dina
Simplify the expression.
3 [ (15 - 3)^2 / 4]
a. 36
b. 108
c. 18
d. 9
1. TutorCat
http://www.jiskha.com/display.cgi?id=1285004701
2. Dina
???108
3. PsyDAG
3 (12^2/4) = 3 (144/4) = 3 * 36 = 108
Yes!
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https://math.stackexchange.com/questions/3363106/is-ln-natural-log-and-log-the-same-thing-if-used-in-this-answer | # Is “ln” (natural log) and “log” the same thing if used in this answer?
Find $$x$$ for $$4^{x-4} = 7$$.
Answer I got, using log, was $${\log(7)\over 2\log(2)} + 4$$
but the actual answer was $${\ln(7)\over2\ln(2)} + 4$$
I plugged both in my calculator and turns out both are the equivalent value.
Anyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10.
Final question: How do I know when to use which? that is which of ln or log is used when solving a question??
For example, if a question asks to find $$x$$ for $$e^x = 100$$, I will use $$\ln$$ since $$\ln(e)$$ cancels out.
If a question asks to find $$2^x = 64$$, i will use log since "$$e$$" isn't present in the question.
So is using either $$\log$$ or $$\ln$$ the same?
• $$\log_{10}x=\frac{\ln x}{\ln 10}$$ – Lord Shark the Unknown Sep 20 at 6:52
• As an aside, to make matters worse, some authors will write $\log$ without a subscript and mean different things than one another. In texts on combinatorics for instance it is not uncommon to see $\log$ without a subscript be meant to be interpreted as being the base 2 logarithm $\log_2$ while other authors might intend it to be the base 10 logarithm $\log_{10}$. Others still may use $\log$ as the natural logarithm rather than writing it as $\ln$. The nice thing is, regardless which base it is you always have $\log_n(a)/\log_n(b)=\log_b(a)$ – JMoravitz Sep 20 at 14:55
• "In order to kill an exponential, you have to hit it with a log". Which raises the question "which log". The answer is -- it doesn't matter. – John Coleman Sep 20 at 16:11
• You can always contrive that there are $e$'s around. Note $2^x=(e^{\text{ln}(2)})^x = e^{x\text{ln}2}$ – jacob1729 Sep 20 at 18:00
• @JohnColeman: Just don't use base 1. math.stackexchange.com/questions/413713/log-base-1-of-1 – Joshua Sep 20 at 19:10
You can use any logarithm you want.
As a result of the base change formula $$\log_2(7) = \frac{\log(7)}{\log(2)} = \frac{\ln(7)}{\ln(2)} = \frac{\log_b(7)}{\log_b(2)}$$ so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $$b > 0, b\neq 1$$).
• Thank you. Out of curiosity, why would one prefer to use natural log in the question 4^(x-4) = 7, when it does not contain "e" – harold232 Sep 20 at 6:56
• @harold232 Since they're equivalent, the choice is harmless, but you still have to make a choice. The only reason I can think of to default to the natural log is that it is in some ways computationally easier than other logarithms (coming from formulas of calculus). – Brian Moehring Sep 20 at 7:01
• @harold232 for mathematicians, $e$ is the default base for logarithms which they would use unless there is a particular reason for choosing something else. (This is largely because it has nice properties for calculus.) After base $e$, the next most common in maths is base $2$. $10$, as Michael Palin might say, is right out. – Especially Lime Sep 20 at 7:05
• $10$ was popular in the days before calculators. In those days, it was the easiest one to work with as we had tables for it. I agree that its utility in pure maths is low today but it lives in a few cases where log scales are used e.g. pH in chemistry and the decibels. – badjohn Sep 20 at 8:43
• @EspeciallyLime: 10 is used by "scientists". See my answer below. – JonathanZ Sep 20 at 17:15
Either is fine. You can write logarithms in terms of any base that you like with the change of base formula $$\log_ba=\frac{\log_ca}{\log_cb}$$
One thing learned from secondary school is that exponential equations can be solved be rewriting the relationship in logarithmic form. As such, your equation can be rewritten as follows: \begin{align} 4^{x-4}&=7\\ x-4&=\log_47\\ x&=4+\log_47 \end{align}
And since $$\log_47$$ can be rewritten as $$\frac{\log7}{\log4}$$ or $$\frac{\ln7}{\ln4}$$ or $$\frac{\log_{999876}7}{\log_{999876}4}$$ it does not matter which base of logarithm you use.
• By the way I noticed that you left the denominator as log(4), shouldn't it be 2log(2) by power rule, since it is more simplified. Or does it not matter? – harold232 Sep 20 at 7:00
• @harold232 As you say, $\log(4)=2\log(2)$. So it does not really matter – Henry Sep 20 at 7:09
• If you want to input the logarithm into a calculator that does not have a log function that accepts a base parameter, then you can simply type in $\log\ 7 \div \log\ 4$. If you want to show your work or write a final solution as its exact value, I tend to accept $\log_47$ or $\frac{\ln7}{2\ln2}$ or its variants. – Andrew Chin Sep 20 at 7:10
• No educator worth his salt is going to be that pedantic. To you, is $\frac12\log_27$ more acceptable compared to $\log_47$? – Andrew Chin Sep 20 at 7:20
• Wolfram says at best it as an alternative – Andrew Chin Sep 20 at 7:29
Culturally
• Computer science / programming people tend to use log base $$2$$
• Mathematicians tend to use log base $$e$$
• Engineers / physicist / chemists etc. tend to use log base $$10$$
Writers really should make it explicit the first time they use "$$\log$$", but they don't always. As others have pointed out, the only difference is a constant factor, and in your case the factors in the numerator and denominator cancelled each other out. So the answer to your question is "If it's in a math context you'll probably see $$\ln()$$ used."
Heck, even if you were asked to solve $$10^{x - 3} = 6$$ you'd still see $$\ln()$$ used, even though it looks like $$\log_{10}$$ might seem more "natural" for that particular problem. It's just what math people tend to do.
• It is called the "natural logarithm", after all. (: – Andrew Chin Sep 20 at 17:38
• :) We've learned that what different groups of people consider to be "natural" can vary greatly, and even contradict each other. Hence my starting off with the word "culturally". – JonathanZ Sep 20 at 17:52
• @JonathanZ you made me laugh out loud in a library! @@ It was a natural reaction of course. – uhoh Sep 21 at 5:43
There's an interesting unstated question here: what counts as an answer?
You can clearly argue that using either $$\ln$$ or $$\log_{10}$$ should be acceptable. But in that case $$x = \log_4(7) + 4$$ should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want.
As for
If a question asks to find $$2^x=64$$, i will use log since "e" isn't present in the question.
I would just say $$x=6$$. That's really using $$\log_2$$, by inspection.
In this case the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $$\log(x)$$, it can either mean base $$10$$, base $$2$$ or base $$e$$, with the latter two (especially base $$e$$) being much more common as you move up the math ladder. The notation $$\ln(x)$$ is still used for base $$e$$, but whenever you see $$\log(x)$$ you should always assume it is also base $$e$$ unless context implies otherwise (if it's supposed to mean base $$2$$, it should be clear from context).
Part of the reason is exactly because of the reason mentioned by the two other answers: for any $$a,b$$ we have $$\log_a(b)=\frac{\log(b)}{\log(a)},$$ So we can express logarithms of any base using the natural logarithm anyway and there's no need to designate a special symbol for it. And indeed you will see that base $$e$$ is much more useful than base $$10$$ most of the time.
• log is often considered log_2 in CS papers – RiaD Sep 20 at 16:32
Use the change of base formula if you suspect that the answer is irrational, otherwise take the logarithm to both sides of the equation of a base that seems reasonable.
There's nothing magical about the change of base formula. \begin{align} \log_c b &= \log_c b\\ \log_c a^{\log_a b} &= \log_c b\\ \log_a b \cdot\log_c a &= \log_c b\\ \log_a b &= \frac{\log_c b}{\log_c a}\\ \end{align}
Even if the solution is integral or rational, using the change of base formula will get you to an answer, for example:
\begin{align} 2^x &= 64\\ \log_{10}2^x &= \log_{10} 64\\ x\log_{10}2 &= \log_{10} 64\\ x &= \frac{\log_{10} 64}{\log_{10}2}\\ x &= \log_{2} 64\\ x &= \log_{2} 2^6\\ x &= 6\log_{2} 2\\ x &= 6\\ \end{align} although, it does insert a lot of extra [unnecessary] steps. | 2019-10-22T18:42:10 | {
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https://math.stackexchange.com/questions/3231585/a-n-be-a-sequence-such-that-a-n12-2a-na-n1-a-n-0-then-sum-1/3231653 | # $\{a_n\}$ be a sequence such that $a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in…
Let $$\{a_n\}$$ be a sequence of positive real numbers such that
$$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$$.
Then the sum of the series $$\sum_1^{\infty}\frac{a_n}{3^n}$$ lies in...
(A) $$(1,2]$$, (B) $$(2,3]$$, (C) $$(3,4]$$, (D)$$(4,5]$$.
Solution attempt:
Firstly, we figure out what $$\frac{a_{n+1}}{a_n}$$ is going to look like. We get, from the recursive formula, $$\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$$ (remembering the fact that $$a_n>0$$, the other root is rejected).
We know that, if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$$, then $$\lim a_n \to \infty$$.
Further, $$(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$$. (Again, the other root is rejected due to the same reason).
Hence, $$(a_n)$$ increases monotonically.
Therefore, the largest value of $$\frac{a_{n+1}}{a_n}$$ is approximately $$1+\sqrt{1+\frac{1}{1}} \approx 2.15$$
Now, the sum can be approximated as $$\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$$ (In actuality, $$\mathbb{sum}< 1.3$$).
So, option $$(A)$$ is the correct answer.
Is the procedure correct?
I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.
Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?
• Out of curiosity, what's the source of this problem ? – Gabriel Romon May 19 '19 at 10:16
• @GabrielRomon It was asked in an entrance exam for Master's degree admission in India this year (JAM 2019). – Subhasis Biswas May 19 '19 at 10:18
• Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\sim\alpha^n$ and the sum effectively $\frac\alpha{3-\alpha}$, the first terms of the series may as well shift the result in another interval. How do you bound the partial sum $\sum\limits_{n=1}^{n_0} \frac{a_n}{3^n}$ up the a certain $n_0$ so that subsequent terms are small enough and we can switch to asymptotic behaviour ? – zwim May 19 '19 at 10:48
• This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ – Subhasis Biswas May 19 '19 at 10:50
• If you recall the proof of the ratio test, then yes, you applied the right strategy. – rtybase May 19 '19 at 10:58
Following your calculations and according to the ratio test $$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$ thus $$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$ Now, applying the same technique from the proof on the ratio test $$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\cdots+\frac{a_n}{3^n}+\cdots=\\ \frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}+2\cdot\frac{1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+2.15\cdot\frac{1}{3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots+2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$ and repeating this $$\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+\frac{2.15}{3^2}+\frac{2.15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)< S<\\ \frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^2}{3^2}+\cdots+\frac{2.15^{n-1}}{3^{n-1}}+\cdots\right)$$ or $$\color{red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\frac{2.15}{3}}=\frac{1}{3-2.15}<\color{red}{2}$$
This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test.
• Truly a nice approach. The upper bound is basically the same. I missed the lower bound though :( I like your final note. "So, it makes sense to ask something similar for a master degree entrance test" – Subhasis Biswas May 19 '19 at 11:33
• Nice and clean :) – A learner Apr 23 at 15:44
$$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$$ $$a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$$ $$1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$$ $$a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$$ $$\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$$ $$\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$$ $$\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$$ $$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$$
I would rather argue that $$\sqrt{a_n(a_n+1)}$$ lies inside $$[a_n,a_n+1]$$, so that $$b_n\leq a_n \leq c_n$$ where $$b_n=2b_n$$ and $$c_n=2c_n+1$$ with $$b_1=c_1=1$$.
Closed forms for $$b_n$$ and $$c_n$$ are easily derived as $$b_n=2^{n-1}$$ and $$c_n=2^n-1$$, so that $$1\leq \sum_1^{\infty}\frac{a_n}{3^n} \leq 2-\frac 12$$
This inequality can be refined by only summing from $$n$$ larger than some constant.
• Now, for the last part of my question, is there any general approach for these type of problems? – Subhasis Biswas May 19 '19 at 11:07
• A better one indeed! In fact, we don't need any better bound than this. Although my procedure gives off a slightly better value, it is not worthwhile here. – Subhasis Biswas May 19 '19 at 11:08
(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)
Hints:
• After calculating the first few terms, the ratio of the terms is very similar. If we estimate $$a_n \approx a_1 r^{n-1}$$, this suggests we should solve for $$\frac{ 1/3} { 1 - (r_1/3) } = 1$$ and $$\frac{1/3} { 1 - (r_2/3)} = 2$$ to give us an idea of how to bound the sequence. This gives us $$r_1 = 2, r_2 = 5/2$$, so we want to show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$ (with some flexibility if this doesn't immediately work out).
• Show that $$a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n } } { 2}$$. In particular, reject the negative root.
• Show that $$a_n \geq 1$$.
• Show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$.
• Hence, show that $$1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2$$
Note:
• The LHS is true by calculating the first 5 terms.
• Of course, we can't prove the RHS just by calculating enough terms.
• In fact, the bounding inequality $$2a_n < a_{n+1} < 2a_n + \frac{1}{2}$$, so the ratio $$a_{n+1} / a_n$$ is very close to 2, esp at (slightly) larger values of $$n$$.
• Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25. | 2021-06-14T12:38:23 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3231585/a-n-be-a-sequence-such-that-a-n12-2a-na-n1-a-n-0-then-sum-1/3231653",
"openwebmath_score": 0.9380605816841125,
"openwebmath_perplexity": 273.7343699614015,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.978712651931994,
"lm_q2_score": 0.9372107900876219,
"lm_q1q2_score": 0.9172600577859358
} |
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AutoMathText
AutoMathText is an extensive and carefully curated dataset encompassing around 200 GB of mathematical texts. It's a compilation sourced from a diverse range of platforms including various websites, arXiv, and GitHub (OpenWebMath, RedPajama, Algebraic Stack). This rich repository has been autonomously selected (labeled) by the state-of-the-art open-source language model, Qwen-72B. Each piece of content in the dataset is assigned a score lm_q1q2_score within the range of [0, 1], reflecting its relevance, quality and educational value in the context of mathematical intelligence.
GitHub homepage: https://github.com/yifanzhang-pro/AutoMathText
ArXiv paper: https://arxiv.org/abs/2402.07625
Objective
The primary aim of the AutoMathText dataset is to provide a comprehensive and reliable resource for a wide array of users - from academic researchers and educators to AI practitioners and mathematics enthusiasts. This dataset is particularly geared towards:
- Facilitating advanced research in the intersection of mathematics and artificial intelligence.
- Serving as an educational tool for learning and teaching complex mathematical concepts.
- Providing a foundation for developing and training AI models specialized in processing and understanding mathematical content.
Configs
configs:
- config_name: web-0.50-to-1.00
data_files:
- split: train
path:
- data/web/0.95-1.00.jsonl
- data/web/0.90-0.95.jsonl
- ...
- data/web/0.50-0.55.jsonl
default: true
- config_name: web-0.60-to-1.00
- config_name: web-0.70-to-1.00
- config_name: web-0.80-to-1.00
- config_name: web-full
data_files: data/web/*.jsonl
- config_name: arxiv-0.50-to-1.00
data_files:
- split: train
path:
- data/arxiv/0.90-1.00/*.jsonl
- ...
- data/arxiv/0.50-0.60/*.jsonl
- config_name: arxiv-0.60-to-1.00
- config_name: arxiv-0.70-to-1.00
- config_name: arxiv-0.80-to-1.00
- config_name: arxiv-full
data_files: data/arxiv/*/*.jsonl
- config_name: code-0.50-to-1.00
data_files:
- split: train
path:
- data/code/*/0.95-1.00.jsonl
- ...
- data/code/*/0.50-0.55.jsonl
- config_name: code-python-0.50-to-1.00
- split: train
path:
- data/code/python/0.95-1.00.jsonl
- ...
- data/code/python/0.50-0.55.jsonl
- config_name: code-python-0.60-to-1.00
- config_name: code-python-0.70-to-1.00
- config_name: code-python-0.80-to-1.00
- config_name: code-jupyter-notebook-0.50-to-1.00
- split: train
path:
- data/code/jupyter-notebook/0.95-1.00.jsonl
- ...
- data/code/jupyter-notebook/0.50-0.55.jsonl
- config_name: code-jupyter-notebook-0.60-to-1.00
- config_name: code-jupyter-notebook-0.70-to-1.00
- config_name: code-jupyter-notebook-0.80-to-1.00
- config_name: code-full
data_files: data/code/*/*.jsonl
How to load data:
from datasets import load_dataset
ds = load_dataset("math-ai/AutoMathText", "web-0.50-to-1.00") # or any valid config_name
Features
- Volume: Approximately 200 GB of text data (in natural language and programming language).
- Content: A diverse collection of mathematical texts, including but not limited to research papers, educational articles, and code documentation.
- Labeling: Every text is scored by Qwen-72B, a sophisticated language model, ensuring a high standard of relevance and accuracy.
- Scope: Covers a wide spectrum of mathematical topics, making it suitable for various applications in advanced research and education.
References
Citation
We appreciate your use of AutoMathText in your work. If you find this repository helpful, please consider citing it and star this repo. Feel free to contact zhangyif21@mails.tsinghua.edu.cn or open an issue if you have any questions (GitHub homepage: https://github.com/yifanzhang-pro/AutoMathText).
@article{zhang2024automathtext,
title={Autonomous Data Selection with Language Models for Mathematical Texts},
author={Zhang, Yifan and Luo, Yifan and Yuan, Yang and Yao, Andrew Chi-Chih},
journal={arXiv preprint arXiv:2402.07625},
year={2024},
}
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