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# Meters to Miles Bookmark Page Miles to Meters (Swap Units) Format Accuracy Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result. Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result. Note: For a pure decimal result please select 'decimal' from the options above the result. Show formula mi = m * 0.00062137 Show working Show result in exponential format Meters Miles 0m 0.00mi 1m 0.00mi 2m 0.00mi 3m 0.00mi 4m 0.00mi 5m 0.00mi 6m 0.00mi 7m 0.00mi 8m 0.00mi 9m 0.01mi 10m 0.01mi 11m 0.01mi 12m 0.01mi 13m 0.01mi 14m 0.01mi 15m 0.01mi 16m 0.01mi 17m 0.01mi 18m 0.01mi 19m 0.01mi Meters Miles 20m 0.01mi 21m 0.01mi 22m 0.01mi 23m 0.01mi 24m 0.01mi 25m 0.02mi 26m 0.02mi 27m 0.02mi 28m 0.02mi 29m 0.02mi 30m 0.02mi 31m 0.02mi 32m 0.02mi 33m 0.02mi 34m 0.02mi 35m 0.02mi 36m 0.02mi 37m 0.02mi 38m 0.02mi 39m 0.02mi Meters Miles 40m 0.02mi 41m 0.03mi 42m 0.03mi 43m 0.03mi 44m 0.03mi 45m 0.03mi 46m 0.03mi 47m 0.03mi 48m 0.03mi 49m 0.03mi 50m 0.03mi 51m 0.03mi 52m 0.03mi 53m 0.03mi 54m 0.03mi 55m 0.03mi 56m 0.03mi 57m 0.04mi 58m 0.04mi 59m 0.04mi Start Increments Accuracy Format Print table < Smaller Values Larger Values > ## Meters 1 m is equivalent to 1.0936 yards, or 39.370 inches. Since 1983, the metre has been officially defined as the length of the path travelled by light in a vacuum during a time interval of 1/299,792,458 of a second. mi = m * 0.00062137 ## Miles A unit of length equal to 1760 yards

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# Mile/Hour/Minute to Feet/Second Squared Converter Acceleration Mile/Hour/Minute Feet/Second Squared 1 Mile/Hour/Minute = 0.024444444444444 Feet/Second Squared ## How many Feet/Second Squared are in a Mile/Hour/Minute? The answer is one Mile/Hour/Minute is equal to 0.024444444444444 Feet/Second Squared and that means we can also write it as 1 Mile/Hour/Minute = 0.024444444444444 Feet/Second Squared. Feel free to use our online unit conversion calculator to convert the unit from Mile/Hour/Minute to Feet/Second Squared. Just simply enter value 1 in Mile/Hour/Minute and see the result in Feet/Second Squared. ## How to Convert Mile/Hour/Minute to Feet/Second Squared (mi/h/min to ft/s2) By using our Mile/Hour/Minute to Feet/Second Squared conversion tool, you know that one Mile/Hour/Minute is equivalent to 0.024444444444444 Feet/Second Squared. Hence, to convert Mile/Hour/Minute to Feet/Second Squared, we just need to multiply the number by 0.024444444444444. We are going to use very simple Mile/Hour/Minute to Feet/Second Squared conversion formula for that. Pleas see the calculation example given below. $$\text{1 Mile/Hour/Minute} = 1 \times 0.024444444444444 = \text{0.024444444444444 Feet/Second Squared}$$ ## What is Mile/Hour/Minute Unit of Measure? Mile/Hour/Minute or Mile per Hour per Minute is a unit of measurement for acceleration. If an object accelerates at the rate of 1 mile/hour/minute, that means its speed is increased by 1 mile per hour every minute. ## What is the symbol of Mile/Hour/Minute? The symbol of Mile/Hour/Minute is mi/h/min. This means you can also write one Mile/Hour/Minute as 1 mi/h/min. ## What is Feet/Second Squared Unit of Measure? Feet/Second Squared or Feet per Second Squared is a unit of measurement for acceleration. If an object accelerates at the rate of 1 feet/second squared, that means its speed is increased by 1 feet per second every second. ## What is the symbol of Feet/Second Squared? The symbol of Feet/Second Squared is ft/s2. This means you can also write one Feet/Second Squared as 1 ft/s2. ## Mile/Hour/Minute to Feet/Second Squared Conversion Table Mile/Hour/Minute [mi/h/min]Feet/Second Squared [ft/s2] 10.024444444444444 20.048888888888889 30.073333333333333 40.097777777777778 50.12222222222222 60.14666666666667 70.17111111111111 80.19555555555556 90.22 100.24444444444444 1002.44 100024.44 ## Mile/Hour/Minute to Other Units Conversion Table Mile/Hour/Minute [mi/h/min]Output 1 mile/hour/minute in meter/second squared is equal to0.0074506666666667 1 mile/hour/minute in attometer/second squared is equal to7450666666666700 1 mile/hour/minute in centimeter/second squared is equal to0.74506666666667 1 mile/hour/minute in decimeter/second squared is equal to0.074506666666667 1 mile/hour/minute in dekameter/second squared is equal to0.00074506666666667 1 mile/hour/minute in femtometer/second squared is equal to7450666666666.7 1 mile/hour/minute in hectometer/second squared is equal to0.000074506666666667 1 mile/hour/minute in kilometer/second squared is equal to0.0000074506666666667 1 mile/hour/minute in micrometer/second squared is equal to7450.67 1 mile/hour/minute in millimeter/second squared is equal to7.45 1 mile/hour/minute in nanometer/second squared is equal to7450666.67 1 mile/hour/minute in picometer/second squared is equal to7450666666.67 1 mile/hour/minute in meter/hour squared is equal to96560.64 1 mile/hour/minute in millimeter/hour squared is equal to96560640 1 mile/hour/minute in centimeter/hour squared is equal to9656064 1 mile/hour/minute in kilometer/hour squared is equal to96.56 1 mile/hour/minute in meter/minute squared is equal to26.82 1 mile/hour/minute in millimeter/minute squared is equal to26822.4 1 mile/hour/minute in centimeter/minute squared is equal to2682.24 1 mile/hour/minute in kilometer/minute squared is equal to0.0268224 1 mile/hour/minute in kilometer/hour/second is equal to0.0268224 1 mile/hour/minute in inch/hour/minute is equal to63360 1 mile/hour/minute in inch/hour/second is equal to1056 1 mile/hour/minute in inch/minute/second is equal to17.6 1 mile/hour/minute in inch/hour squared is equal to3801600 1 mile/hour/minute in inch/minute squared is equal to1056 1 mile/hour/minute in inch/second squared is equal to0.29333333333333 1 mile/hour/minute in feet/hour/minute is equal to5280 1 mile/hour/minute in feet/hour/second is equal to88 1 mile/hour/minute in feet/minute/second is equal to1.47 1 mile/hour/minute in feet/hour squared is equal to316800 1 mile/hour/minute in feet/minute squared is equal to88 1 mile/hour/minute in feet/second squared is equal to0.024444444444444 1 mile/hour/minute in knot/hour is equal to52.14 1 mile/hour/minute in knot/minute is equal to0.86897624528 1 mile/hour/minute in knot/second is equal to0.014482937421333 1 mile/hour/minute in knot/millisecond is equal to0.000014482937421333 1 mile/hour/minute in mile/hour/second is equal to0.016666666666667 1 mile/hour/minute in mile/hour squared is equal to60 1 mile/hour/minute in mile/minute squared is equal to0.016666666666667 1 mile/hour/minute in mile/second squared is equal to0.0000046296296296296 1 mile/hour/minute in yard/second squared is equal to0.0081481481481481 1 mile/hour/minute in gal is equal to0.74506666666667 1 mile/hour/minute in galileo is equal to0.74506666666667 1 mile/hour/minute in centigal is equal to74.51 1 mile/hour/minute in decigal is equal to7.45 1 mile/hour/minute in g-unit is equal to0.00075975655974942 1 mile/hour/minute in gn is equal to0.00075975655974942 1 mile/hour/minute in gravity is equal to0.00075975655974942 1 mile/hour/minute in milligal is equal to745.07 1 mile/hour/minute in kilogal is equal to0.00074506666666667 Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority.

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시간 제한메모리 제한제출정답맞힌 사람정답 비율 5 초 512 MB31171458.333% ## 문제 You've invented a slight modification of the run-length encoding (RLE) compression algorithm, called PermRLE. To compress a string, this algorithm chooses some permutation of integers between 1 and k, applies this permutation to the first k letters of the given string, then to the next block of k letters, and so on. The length of the string must be divisible by k. After permuting all blocks, the new string is compressed using RLE, which is described later. To apply the given permutation p to a block of k letters means to place the p[1]-th of these letters in the first position, then p[2]-th of these letters in the second position, and so on. For example, applying the permutation {3,1,4,2} to the block "abcd" yields "cadb". Applying it to the longer string "abcdefghijkl" in blocks yields "cadbgehfkilj". The permuted string is then compressed using run-length encoding. To simplify, we will consider the compressed size of the string to be the number of groups of consecutive equal letters. For example, the compressed size of "aabcaaaa" is 4; the first of the four groups is a group of two letters "a", then two groups "b" and "c" each containing only one letter, and finally a longer group of letters "a". Obviously, the compressed size may depend on the chosen permutation. Since the goal of compression algorithms is to minimize the size of the compressed text, it is your job to choose the permutation that yields the smallest possible compressed size, and output that size. ## 입력 The first line of input gives the number of cases, NN test cases follow. The first line of each case will contain k. The second line will contain S, the string to be compressed. Limits • N = 20 • S will contain only lowercase letters 'a' through 'z' • The length of S will be divisible by k • 2 ≤ k ≤ 16 • 1 ≤ length of S ≤ 50000 ## 출력 For each test case you should output one line containing "Case #XY" (quotes for clarity) where X is the number of the test case and Y is the minimum compressed size of S. 2 4 abcabcabcabc 3 abcabcabcabc Case #1: 7 Case #2: 12 ## 채점 및 기타 정보 • 예제는 채점하지 않는다.

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## 11685 11,685 (eleven thousand six hundred eighty-five) is an odd five-digits composite number following 11684 and preceding 11686. In scientific notation, it is written as 1.1685 × 104. The sum of its digits is 21. It has a total of 4 prime factors and 16 positive divisors. There are 5,760 positive integers (up to 11685) that are relatively prime to 11685. ## Basic properties • Is Prime? No • Number parity Odd • Number length 5 • Sum of Digits 21 • Digital Root 3 ## Name Short name 11 thousand 685 eleven thousand six hundred eighty-five ## Notation Scientific notation 1.1685 × 104 11.685 × 103 ## Prime Factorization of 11685 Prime Factorization 3 × 5 × 19 × 41 Composite number Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 11685 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 11,685 is 3 × 5 × 19 × 41. Since it has a total of 4 prime factors, 11,685 is a composite number. ## Divisors of 11685 1, 3, 5, 15, 19, 41, 57, 95, 123, 205, 285, 615, 779, 2337, 3895, 11685 16 divisors Even divisors 0 16 8 8 Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 20160 Sum of all the positive divisors of n s(n) 8475 Sum of the proper positive divisors of n A(n) 1260 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 108.097 Returns the nth root of the product of n divisors H(n) 9.27381 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 11,685 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 11,685) is 20,160, the average is 1,260. ## Other Arithmetic Functions (n = 11685) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 5760 Total number of positive integers not greater than n that are coprime to n λ(n) 360 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1410 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares There are 5,760 positive integers (less than 11,685) that are coprime with 11,685. And there are approximately 1,410 prime numbers less than or equal to 11,685. ## Divisibility of 11685 m n mod m 2 3 4 5 6 7 8 9 1 0 1 0 3 2 5 3 The number 11,685 is divisible by 3 and 5. ## Classification of 11685 • Arithmetic • Deficient • Polite • Square Free ### Other numbers • LucasCarmichael ## Base conversion (11685) Base System Value 2 Binary 10110110100101 3 Ternary 121000210 4 Quaternary 2312211 5 Quinary 333220 6 Senary 130033 8 Octal 26645 10 Decimal 11685 12 Duodecimal 6919 20 Vigesimal 1945 36 Base36 90l ## Basic calculations (n = 11685) ### Multiplication n×i n×2 23370 35055 46740 58425 ### Division ni n⁄2 5842.5 3895 2921.25 2337 ### Exponentiation ni n2 136539225 1595460844125 18642959963600625 217842987174673303125 ### Nth Root i√n 2√n 108.097 22.6922 10.397 6.50917 ## 11685 as geometric shapes ### Circle Diameter 23370 73419 4.28951e+08 ### Sphere Volume 6.68305e+12 1.7158e+09 73419 ### Square Length = n Perimeter 46740 1.36539e+08 16525.1 ### Cube Length = n Surface area 8.19235e+08 1.59546e+12 20239 ### Equilateral Triangle Length = n Perimeter 35055 5.91232e+07 10119.5 ### Triangular Pyramid Length = n Surface area 2.36493e+08 1.88027e+11 9540.76 ## Cryptographic Hash Functions md5 780261c4b9a55cd803080619d0cc3e11 9c018611e4c9a00320ef5bf9ed3e6a38211f9d49 9607851af30b1f5b2c0d91848030d67d5d3c04159dede6bb004851552bfd2dfd f331149a198d0dfcd09decbbf359f05bb00963a93db956a9e3f49cc98897b5596a7f3829f4490a5e917ce7f91a34a123c53b9153645feaf998bd3d6f9737072f 7afe8b12d7b6cd57be63eb0aae649fda82160670

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A145726 Expansion of q * psi(-q) * psi(-q^15) / (psi(-q^3) * psi(-q^5)) in powers of q where psi() is a Ramanujan theta function. 3 1, -1, 0, 0, -1, 1, 0, -1, 0, 1, 0, 0, 1, -2, 1, 2, -3, 1, 1, -2, 3, 0, -3, 1, 2, -2, 0, 2, -6, 3, 4, -7, 3, 2, -5, 6, 2, -8, 3, 5, -6, 2, 4, -12, 7, 10, -15, 6, 5, -13, 12, 4, -18, 7, 11, -14, 6, 10, -24, 14, 20, -32, 12, 12, -29, 24, 9, -36, 15, 22, -30, 13, 22, -50, 27, 36, -63, 26, 24, -56, 45, 22, -69, 30, 42, -62, 27 (list; graph; refs; listen; history; text; internal format) OFFSET 1,14 COMMENTS Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700). LINKS G. C. Greubel, Table of n, a(n) for n = 1..1000 Michael Somos, Introduction to Ramanujan theta functions Eric Weisstein's World of Mathematics, Ramanujan Theta Functions FORMULA Expansion of eta(q) * eta(q^4) * eta(q^6) * eta(q^10) * eta(q^15) * eta(q^60) / (eta(q^2) * eta(q^3) * eta(q^5) * eta(q^12) * eta(q^20) * eta(q^30)) in powers of q. Euler transform of a period 60 sequence. G.f. is a period 1 Fourier series which satisfies f(-1 / (60 t)) = f(t) where q = exp(2 Pi i t). G.f.: x * Product_{k>0} P(15, x^k) * P(60, x^k) where P(n, x) is the n-th cyclotomic polynomial. a(n) = A145727(n) unless n=0. a(n) = -(-1)^n * A131794(n). a(2*n) = - A094022(n). Convolution inverse of A145725. EXAMPLE q - q^2 - q^5 + q^6 - q^8 + q^10 + q^13 - 2*q^14 + q^15 + 2*q^16 - 3*q^17 + ... PROG (PARI) {a(n) = local(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( eta(x + A) * eta(x^4 + A) * eta(x^6 + A) * eta(x^10 + A) * eta(x^15 + A) * eta(x^60 + A) / (eta(x^2 + A) * eta(x^3 + A) * eta(x^5 + A) * eta(x^12 + A) * eta(x^20 + A) * eta(x^30 + A)), n))} CROSSREFS Cf. A094022, A131794, A145725, A145727. Sequence in context: A145727 A131796 A131794 * A322984 A277822 A327616 Adjacent sequences:  A145723 A145724 A145725 * A145727 A145728 A145729 KEYWORD sign AUTHOR Michael Somos, Oct 23 2008 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified June 13 16:17 EDT 2021. Contains 345008 sequences. (Running on oeis4.)

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# calculation of gradients near wall User Name Remember Me Password Register Blogs Members List Search Today's Posts Mark Forums Read November 5, 2009, 02:51 calculation of gradients near wall #1 New Member   Sumeet Kumar Join Date: Mar 2009 Posts: 21 Rep Power: 8 Hi FOAMers, I want to calculate gradients at the cell centroids connected to a wall patch. The quantities I need to evaluate are :- a) del Utau / del tau b) del Unormal / del normal c) del p / del tau Here, Utau is tangential component of velocity Unormal is normal component of velocity tau is a coordinate tangent to local wall face normal is a coordinate normal to local wall face Can someone assist me calculation of these quantities at the cell centroids. Hints and suggestions in this regard are greatly appreciated Thanks & Regards Sumeet November 5, 2009, 09:54 #2 New Member   Sumeet Kumar Join Date: Mar 2009 Posts: 21 Rep Power: 8 Hi Again I tried to calculate following quanitity.. 1) del p / del tau vectorField n = patch().nf() / mag( patch().nf() ) ; volVectorField gradP(fvc::grad(db().lookupObject( "p"))); const vectorField gradPInternal = gradP.boundaryField()[patchI].patchInternalField(); const scalarField gradPtau = mag(gradPInternal - n * (gradPInternal & n)); The scalarField gradPtau will give me the required derivative However the above method cannot be reiterated for calculation of following derivatives.- a) del U(tau) / del tau b) del U(normal) / del normal because the U (tau) and U (normal) vectors are only defined at the cell centroids of patchInternalField list. How do I calculate such derivatives. Please someone advice. Regards, Sumeet Thread Tools Display Modes Linear Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules Similar Threads Thread Thread Starter Forum Replies Last Post simvun OpenFOAM Other Meshers: ICEM, Star, Ansys, Pointwise, GridPro, Ansa, ... 48 May 14, 2012 05:20 unoder OpenFOAM Installation 11 January 30, 2008 21:30 AB CD-adapco 6 November 15, 2004 05:41 Andrea CFX 2 October 11, 2004 05:12 stephane baralon Main CFD Forum 11 September 2, 1999 04:05 All times are GMT -4. The time now is 13:05. Contact Us - CFD Online - Top

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Community Profile # Dan W ### University of Rochester 3 total contributions since 2015 View details... Contributions in View by Solved Sum all integers from 1 to 2^n Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10. mehr als 4 Jahre ago Solved Find nth maximum Find nth maximum in a vector of integer numbers. Return NaN if no such number exists. x = [2 6 4 9 -10 3 1 5 -10]; So ... mehr als 4 Jahre ago

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https://calspasblog.com/6ne3bb4t/yejkg65.php?page=stirling-approximation-binomial-distribution-601b4d

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stirling approximation binomial distribution # stirling approximation binomial distribution 2−n. Normal approximation to the Binomial In 1733, Abraham de Moivre presented an approximation to the Binomial distribution. (8.3) on p.762 of Boas, f(x) = C(n,x)pxqn−x ∼ 1 √ 2πnpq e−(x−np)2/2npq. Derivation of Gaussian Distribution from Binomial The number of paths that take k steps to the right amongst n total steps is: n! For large values of n, Stirling's approximation may be used: Example:. 2N N+j 2 ! Approximating binomial probabilities with Stirling Posted on September 28, 2012 by markhuber | Comments Off on Approximating binomial probabilities with Stirling Let \(X\) be a binomially distributed random variable with parameters \(n = 1950\) and \(p = 0.342\). When Is the Approximation Appropriate? The factorial N! using Stirling's approximation. If kis in fact constant, then this is the best approximation one can hope for. Using Stirling’s formula we prove one of the most important theorems in probability theory, the DeMoivre-Laplace Theorem. (1) taking the logarithm of both sides, we have lnP j = lnN!−N ln2−ln N +j 2 !−ln N −j 2 ! Find 63! In this section, we present four different proofs of the convergence of binomial b n p( , ) distribution to a limiting normal distribution, as nof. 7. 2. Now, consider … Exponent With Stirling's Approximation For n! Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … How-ever, when k= ! k! (1) (but still k= o(p n)), the k! Stirling's Approximation to n! By using some mathematics it can be shown that there are a few conditions that we need to use a normal approximation to the binomial distribution.The number of observations n must be large enough, and the value of p so that both np and n(1 - p) are greater than or equal to 10.This is a rule of thumb, which is guided by statistical practice. Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770). 3 scaling the Binomial distribution converges to Normal. 1 the gaussian approximation to the binomial we start with the probability of ending up j steps from the origin when taking a total of N steps, given by P j = N! The normal approximation tothe binomial distribution Remarkably, when n, np and nq are large, then the binomial distribution is well approximated by the normal distribution. I kept an “exact” calculation of the binomial distribution for 14 and fewer people dying, and then used Stirling's approximation for the factorial for higher factorials in the binomial … In this next one, I take the piecewise approximation concept even further. 3.1. 12In other words, ntends to in nity. He later appended the derivation of his approximation to the solution of a problem asking ... For positive integers n, the Stirling formula asserts that n! In confronting statistical problems we often encounter factorials of very large numbers. According to eq. We can replace it with an exponential expression by making use of Stirling’s Approximation. is a product N(N-1)(N-2)..(2)(1). term is a little inconvenient. k!(n−k)! (n−k)!, and since each path has probability 1/2n, the total probability of paths with k right steps are: p = n! The statement will be that under the appropriate (and different from the one in the Poisson approximation!) N−j 2!

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https://blog.listcomp.com/math/2018/05/09/radial-function-radial-basis-function-base-exponent-power

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Radial Function / Radial Basis Function / Base / Exponent / Power Yao Yao on May 9, 2018 In mathematics, a radial function is a function defined on a Euclidean space $\mathbf{R}^n$ whose value at each point depends only on the distance between that point and the origin. “radial” 的意思应该是：”只与 radius (半径长度) 有关“。所以有：$\phi (\mathbf{x} )=\phi (\Vert \mathbf{x} \Vert)$ Radial function 的一个特性是：翻转、旋转这类不改变 $\mathbf{x}$ 向量长度的线性变换不会改变 $\phi (\mathbf{x})$ 的值。 … or alternatively on the distance from some other point $c$, called a center, so that $\phi (\mathbf{x} ,\mathbf{c})=\phi (\Vert \mathbf{x} - \mathbf{c} \Vert)$. Radial basis function 是 radial function 的子类。所谓 Radial basis function 就是它的定义中会涉及到一个 power (幂)，然后 radius ($\Vert \mathbf{x} \Vert$ or $\Vert \mathbf{x} - \mathbf{y} \Vert$) 会是这个幂的 base (基数)。比如： • Gaussian RBF: ${\phi (\mathbf{x} ,\mathbf{y}) = \varphi(r) = e^{-(\varepsilon r)^{2}}}$ • Multiquadric RBF: $\phi (\mathbf{x} ,\mathbf{y}) = \varphi(r) = {\sqrt {1+(\varepsilon r)^{2}}}$ Recap: Base / Exponent / Power • $a$ is the base • $n$ is the exponent • $a^n$ is the power, or precisely the $n^{\text{th}}$ power of $a$ Power is first used for the square. Euclid uses the phrase in power, for example he says that magnitudes are commensurable in power when their squares are commensurable. “magnitudes are commensurable in power when their squares are commensurable” 这句话里的 magnitudes 用 “incommensurable” 修饰一下就更好理解了。举个例子： • $\sqrt 2$ is actually incommensurable. • $\sqrt 2$’s square, $2$, is commensurable. • So we say $\sqrt 2$ is protentially commensurable. Thus, from the Greek dunamis to the Latin potentia and finally to power.

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https://www.aqua-calc.com/one-to-one/density/gram-per-liter/long-ton-per-cubic-foot/1

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# 1 gram per liter [g/l] in long tons per cubic foot ## g/l to long tn/ft³ unit converter of density 1 gram per liter [g/l] = 3 × 10-5 long ton per cubic foot [long tn/ft³] ### grams per liter to long tons per cubic foot density conversion cards • 1 through 25 grams per liter • 1 g/l to long tn/ft³ = 3 × 10-5 long tn/ft³ • 2 g/l to long tn/ft³ = 6 × 10-5 long tn/ft³ • 3 g/l to long tn/ft³ = 8 × 10-5 long tn/ft³ • 4 g/l to long tn/ft³ = 0.00011 long tn/ft³ • 5 g/l to long tn/ft³ = 0.00014 long tn/ft³ • 6 g/l to long tn/ft³ = 0.00017 long tn/ft³ • 7 g/l to long tn/ft³ = 0.0002 long tn/ft³ • 8 g/l to long tn/ft³ = 0.00022 long tn/ft³ • 9 g/l to long tn/ft³ = 0.00025 long tn/ft³ • 10 g/l to long tn/ft³ = 0.00028 long tn/ft³ • 11 g/l to long tn/ft³ = 0.00031 long tn/ft³ • 12 g/l to long tn/ft³ = 0.00033 long tn/ft³ • 13 g/l to long tn/ft³ = 0.00036 long tn/ft³ • 14 g/l to long tn/ft³ = 0.00039 long tn/ft³ • 15 g/l to long tn/ft³ = 0.00042 long tn/ft³ • 16 g/l to long tn/ft³ = 0.00045 long tn/ft³ • 17 g/l to long tn/ft³ = 0.00047 long tn/ft³ • 18 g/l to long tn/ft³ = 0.0005 long tn/ft³ • 19 g/l to long tn/ft³ = 0.00053 long tn/ft³ • 20 g/l to long tn/ft³ = 0.00056 long tn/ft³ • 21 g/l to long tn/ft³ = 0.00059 long tn/ft³ • 22 g/l to long tn/ft³ = 0.00061 long tn/ft³ • 23 g/l to long tn/ft³ = 0.00064 long tn/ft³ • 24 g/l to long tn/ft³ = 0.00067 long tn/ft³ • 25 g/l to long tn/ft³ = 0.0007 long tn/ft³ • 26 through 50 grams per liter • 26 g/l to long tn/ft³ = 0.00072 long tn/ft³ • 27 g/l to long tn/ft³ = 0.00075 long tn/ft³ • 28 g/l to long tn/ft³ = 0.00078 long tn/ft³ • 29 g/l to long tn/ft³ = 0.00081 long tn/ft³ • 30 g/l to long tn/ft³ = 0.00084 long tn/ft³ • 31 g/l to long tn/ft³ = 0.00086 long tn/ft³ • 32 g/l to long tn/ft³ = 0.00089 long tn/ft³ • 33 g/l to long tn/ft³ = 0.00092 long tn/ft³ • 34 g/l to long tn/ft³ = 0.00095 long tn/ft³ • 35 g/l to long tn/ft³ = 0.00098 long tn/ft³ • 36 g/l to long tn/ft³ = 0.001 long tn/ft³ • 37 g/l to long tn/ft³ = 0.00103 long tn/ft³ • 38 g/l to long tn/ft³ = 0.00106 long tn/ft³ • 39 g/l to long tn/ft³ = 0.00109 long tn/ft³ • 40 g/l to long tn/ft³ = 0.00111 long tn/ft³ • 41 g/l to long tn/ft³ = 0.00114 long tn/ft³ • 42 g/l to long tn/ft³ = 0.00117 long tn/ft³ • 43 g/l to long tn/ft³ = 0.0012 long tn/ft³ • 44 g/l to long tn/ft³ = 0.00123 long tn/ft³ • 45 g/l to long tn/ft³ = 0.00125 long tn/ft³ • 46 g/l to long tn/ft³ = 0.00128 long tn/ft³ • 47 g/l to long tn/ft³ = 0.00131 long tn/ft³ • 48 g/l to long tn/ft³ = 0.00134 long tn/ft³ • 49 g/l to long tn/ft³ = 0.00137 long tn/ft³ • 50 g/l to long tn/ft³ = 0.00139 long tn/ft³ • 51 through 75 grams per liter • 51 g/l to long tn/ft³ = 0.00142 long tn/ft³ • 52 g/l to long tn/ft³ = 0.00145 long tn/ft³ • 53 g/l to long tn/ft³ = 0.00148 long tn/ft³ • 54 g/l to long tn/ft³ = 0.0015 long tn/ft³ • 55 g/l to long tn/ft³ = 0.00153 long tn/ft³ • 56 g/l to long tn/ft³ = 0.00156 long tn/ft³ • 57 g/l to long tn/ft³ = 0.00159 long tn/ft³ • 58 g/l to long tn/ft³ = 0.00162 long tn/ft³ • 59 g/l to long tn/ft³ = 0.00164 long tn/ft³ • 60 g/l to long tn/ft³ = 0.00167 long tn/ft³ • 61 g/l to long tn/ft³ = 0.0017 long tn/ft³ • 62 g/l to long tn/ft³ = 0.00173 long tn/ft³ • 63 g/l to long tn/ft³ = 0.00176 long tn/ft³ • 64 g/l to long tn/ft³ = 0.00178 long tn/ft³ • 65 g/l to long tn/ft³ = 0.00181 long tn/ft³ • 66 g/l to long tn/ft³ = 0.00184 long tn/ft³ • 67 g/l to long tn/ft³ = 0.00187 long tn/ft³ • 68 g/l to long tn/ft³ = 0.0019 long tn/ft³ • 69 g/l to long tn/ft³ = 0.00192 long tn/ft³ • 70 g/l to long tn/ft³ = 0.00195 long tn/ft³ • 71 g/l to long tn/ft³ = 0.00198 long tn/ft³ • 72 g/l to long tn/ft³ = 0.00201 long tn/ft³ • 73 g/l to long tn/ft³ = 0.00203 long tn/ft³ • 74 g/l to long tn/ft³ = 0.00206 long tn/ft³ • 75 g/l to long tn/ft³ = 0.00209 long tn/ft³ • 76 through 100 grams per liter • 76 g/l to long tn/ft³ = 0.00212 long tn/ft³ • 77 g/l to long tn/ft³ = 0.00215 long tn/ft³ • 78 g/l to long tn/ft³ = 0.00217 long tn/ft³ • 79 g/l to long tn/ft³ = 0.0022 long tn/ft³ • 80 g/l to long tn/ft³ = 0.00223 long tn/ft³ • 81 g/l to long tn/ft³ = 0.00226 long tn/ft³ • 82 g/l to long tn/ft³ = 0.00229 long tn/ft³ • 83 g/l to long tn/ft³ = 0.00231 long tn/ft³ • 84 g/l to long tn/ft³ = 0.00234 long tn/ft³ • 85 g/l to long tn/ft³ = 0.00237 long tn/ft³ • 86 g/l to long tn/ft³ = 0.0024 long tn/ft³ • 87 g/l to long tn/ft³ = 0.00242 long tn/ft³ • 88 g/l to long tn/ft³ = 0.00245 long tn/ft³ • 89 g/l to long tn/ft³ = 0.00248 long tn/ft³ • 90 g/l to long tn/ft³ = 0.00251 long tn/ft³ • 91 g/l to long tn/ft³ = 0.00254 long tn/ft³ • 92 g/l to long tn/ft³ = 0.00256 long tn/ft³ • 93 g/l to long tn/ft³ = 0.00259 long tn/ft³ • 94 g/l to long tn/ft³ = 0.00262 long tn/ft³ • 95 g/l to long tn/ft³ = 0.00265 long tn/ft³ • 96 g/l to long tn/ft³ = 0.00268 long tn/ft³ • 97 g/l to long tn/ft³ = 0.0027 long tn/ft³ • 98 g/l to long tn/ft³ = 0.00273 long tn/ft³ • 99 g/l to long tn/ft³ = 0.00276 long tn/ft³ • 100 g/l to long tn/ft³ = 0.00279 long tn/ft³ #### Foods, Nutrients and Calories CHEF BOYARDEE, BEEF RAVIOLI IN MEAT SAUCE, UPC: 064144810802 weigh(s) 271.57 gram per (metric cup) or 9.07 ounce per (US cup), and contain(s) 97 calories per 100 grams or ≈3.527 ounces  [ weight to volume | volume to weight | price | density ] HONEY HAM and SWISS CHEESE SLIDERS, UPC: 011110112088 contain(s) 245 calories per 100 grams or ≈3.527 ounces  [ price ] #### Gravels, Substances and Oils Substrate, Eco-Complete weighs 1 538 kg/m³ (96.0142 lb/ft³) with specific gravity of 1.538 relative to pure water.  Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical  or in a rectangular shaped aquarium or pond  [ weight to volume | volume to weight | price ] Nickel carbonate basic tetrahydrate [NiCO3 ⋅ 2Ni(OH)2 ⋅ 4H2O] weighs 2 600 kg/m³ (162.3127 lb/ft³)  [ weight to volume | volume to weight | price | mole to volume and weight | density ] Volume to weightweight to volume and cost conversions for Soybean oil with temperature in the range of 10°C (50°F) to 140°C (284°F) #### Weights and Measurements A dyne per square mile is a unit of pressure where a force of one dyne (dyn) is applied to an area of one square mile. Acceleration (a) of an object measures the object's change in velocity (v) per unit of time (t): a = v / t. dyn/pc² to lbf/dam² conversion table, dyn/pc² to lbf/dam² unit converter or convert between all units of pressure measurement. #### Calculators Calculate volume of a spherical segment and its surface area

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# physics posted by . in every minute a pump draws 6 metric cubed of water from a well and issues it at a speed of 5 meters per second from the nozzle situated 4 meters above the level from which the water was drawn. find the average rate at which the pump is working. (1 metric cubed =1000kg) ## Similar Questions 1. ### physics 1)A fire hose held near the ground shoots water at a speed of6.5 mps At what angle should the nozzle point in order that the water land 2.5 meters away? A tank has the shape of an inverted circular cone with a base radius of 5 meters and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when … 3. ### Math (Calculus) A tank has the shape of an inverted circular cone with a base radius of 5 meters and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when … 4. ### Math (Calculus) A tank has the shape of an inverted circular cone with a base radius of 5 meters and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when … 5. ### physics calculate the pressure needed for the pump of a 3.0 cm radius firehouse to pump water through its nozzle of radius 2.0 cm at an average speed of 4.0 m/s when the nozzle is 15 m above the fire truck. 6. ### chemistry The level of water in an Olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2 meters deep) needs to be lowered 5.0 cm. If the water is pumped out at a rate of 4.7 liters per second, how long will it take to lower … 7. ### chemistry The level of water in an Olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2 meters deep) needs to be lowered 3.2 cm. If the water is pumped out at a rate of 5.0 liters per second, how long will it take to lower … 8. ### physics 1. An electric pump, used to obtain water from 20 metres below the ground, is marked 5000W. a) If the pump operates as rated, how much energy is used to pump water every second? 9. ### Physics A pump of 200 W power is lifting 2 kg water from an average depth of 10 meters per second. Find the velocity of water delivered by the pump? 10. ### Math A pump can raise 900 liters of water per minute to a height of 50 meters caculate the h.p.(metric) of the pump. More Similar Questions

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# Subdividing the sides of a square to approximate a circle I have read that you can approximate a sphere from an icosahedron by subdividing the faces and normalizing the points created by the subdivisions. Please see the accepted answer of this question: Algorithm for creating spheres. Wouldn't it follow that you could create a circle by subdividing the sides of a square and normalizing the points? This doesn't seem to work. Could someone explain to me why? - It does work. If it doesn't work for you, you're doing something wrong, but we can't tell what that is unless you tell us what you're doing and how it "doesn't seem to work". "Correct answers are all alike; every incorrect answer is incorrect in its own way." —not Leo Tolstoy – Rahul Jan 1 '13 at 1:52 Sorry for not being clearer. I drew a square on a piece of graph paper. The vertices are (0, 1), (1, 0), (0, -1), (-1, 0). The midpoint of the segment (0, 1), (1, 0) is (0.5, 0.5). This point is normalized but does not lie on the circle that the four vertices of the square lie on. – bwroga Jan 1 '13 at 2:12 The norm of $(0.5,0.5)$ is $\sqrt{0.5^2+0.5^2}=1/\sqrt2\ne1$, so the point is not normalized. You normalize it by dividing it by its norm, which gives $(1/\sqrt2,1/\sqrt2)$, whose norm is $1$ and which does lie on the unit circle. – Rahul Jan 1 '13 at 2:37 I thought that a coordinate was normalized when its components added up to one. Is that incorrect? – bwroga Jan 1 '13 at 2:52 Yes, that's incorrect. A coordinate is normalized when the squares of its components add up to one. – mjqxxxx Jan 1 '13 at 3:20 ## 1 Answer Short answer: Your error is that the norm of a vector $(x,y)$ is not $x+y$, or $\lvert x\rvert+\lvert y\rvert$, but rather $\sqrt{x^2+y^2}$. A vector is normalized if its norm is $1$. If you start with $(\frac12,\frac12)$, its norm is $\sqrt{(\frac12)^2+(\frac12)^2} = \frac1{\sqrt2}$, so you have to divide it by $\frac1{\sqrt2}$ to get the normalized vector $(\frac1{\sqrt2},\frac1{\sqrt2})$. Long answer: Well, in principle there are different kinds of norms one can use, but only one makes sense here. In a geometry context, when someone says "norm" without further qualification, they invariably mean the Euclidean norm, or the $2$-norm, which for an $n$-dimensional vector $(x_1, x_2, \ldots, x_n)$ is $\sqrt{x_1^2+x_2^2+\cdots+x_n^2}$. Why? Because this is how lengths and distances work. Get a sheet of graph paper and mark the points $(0,0)$ and $(3,4)$. Measure the distance between them with a ruler: it's not $3+4=7$, it's $\sqrt{3^2+4^2}=5$. Or, connect the points $(0,0)$, $(3,0)$, and $(3,4)$: a wild right triangle appears! Cast Pythagoras to defeat the beast. In general, though, a norm is just — roughly speaking — something that tells you how far away from zero a vector is, and there are lots of possible ways of measuring that. For example, you could just add up the absolute values of the components, $\lvert x_1\rvert+\lvert x_2\rvert+\cdots+\lvert x_n\rvert$, like you thought; this is known as the $1$-norm. Or you could take just the largest component, $\max(\lvert x_1\rvert,\lvert x_2\rvert,\ldots,\lvert x_n\rvert)$, which gives the maximum norm, or the $\infty$-norm. The thing is, though, that because different norms measure distance differently, the "unit spheres" made of points at distance $1$ have different shapes. In particular, the unit sphere for the $1$-norm is a square oriented at $45^\circ$ degrees, which is exactly what you started with, so normalizing the points on it doesn't change anything. - Thank you for your wonderful answer! – bwroga Jan 1 '13 at 5:22

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## ››Convert kilowatt hour to megawatthour kilowatt-hour megawatthour How many kilowatt-hour in 1 megawatthour? The answer is 1000. We assume you are converting between kilowatt hour and megawatthour. You can view more details on each measurement unit: kilowatt-hour or megawatthour The SI derived unit for energy is the joule. 1 joule is equal to 2.7777777777778E-7 kilowatt-hour, or 2.7777777777778E-10 megawatthour. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between kilowatt hours and megawatthour. Type in your own numbers in the form to convert the units! ## ››Quick conversion chart of kilowatt-hour to megawatthour 1 kilowatt-hour to megawatthour = 0.001 megawatthour 10 kilowatt-hour to megawatthour = 0.01 megawatthour 50 kilowatt-hour to megawatthour = 0.05 megawatthour 100 kilowatt-hour to megawatthour = 0.1 megawatthour 200 kilowatt-hour to megawatthour = 0.2 megawatthour 500 kilowatt-hour to megawatthour = 0.5 megawatthour 1000 kilowatt-hour to megawatthour = 1 megawatthour ## ››Want other units? You can do the reverse unit conversion from megawatthour to kilowatt-hour, or enter any two units below: ## Enter two units to convert From: To: ## ››Definition: Megawatthour The SI prefix "mega" represents a factor of 106, or in exponential notation, 1E6. So 1 megawatthour = 106 . ## ››Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!

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Definitions # Alternating finite automaton In automata theory, an alternating finite automaton (AFA) is a nondeterministic finite automaton whose transitions are divided into existential and universal transitions. For example, let A be an alternating automaton. • For an existential transition $\left(q, a, q_1 vee q_2\right)$, A nondeterministically chooses to switch the state to either $q_1$ or $q_2$, reading a. Thus, behaving like a regular nondeterministic finite automaton. • For a universal transition $\left(q, a, q_1 wedge q_2\right)$, A moves to $q_1$ and $q_2$, reading a, simulating the behavior of a parallel machine. Note that due to the universal quantification a run is represented by a run tree. A accepts a word w, if there exists a run tree on w such that every path ends in an accepting state. A basic theorem tells that any AFA is equivalent to an non-deterministic finite automaton (NFA) by performing a similar kind of powerset construction as it is used for the transformation of an NFA to a deterministic finite automaton (DFA). This construction converts an AFA with k states to an NFA with up to $2^k$ states. An alternative model which is frequently used is the one where Boolean combinations are represented as clauses. For instance, one could assume the combinations to be in DNF so that $\left\{\left\{q_1\right\}\left\{q_2,q_3\right\}\right\}$ would represent $q_1 vee \left(q_2 wedge q_3\right)$. The state tt (true) is represented by $\left\{\left\{\right\}\right\}$ in this case and ff (false) by $emptyset$. This clause representation is usually more efficient. ## Formal Definition An alternating finite automaton (AFA) is a 6-tuple, $\left(S\left(exists\right), S\left(forall\right), Sigma, delta, P_0, F\right)$, where • $S\left(exists\right)$ is a finite set of existential states. Also commonly represented as $S\left(vee\right)$. • $S\left(forall\right)$ is a finite set of universal states. Also commonly represented as $S\left(wedge\right)$. • $Sigma$ is a finite set of input symbols. • $delta$ is a set of transition functions to next state $\left(S\left(exists\right) cup S\left(forall\right)\right) times \left(Sigma cup \left\{ varepsilon \right\} \right) to 2^\left\{S\left(exists\right) cup S\left(forall\right)\right\}$. • $P_0$ is the initial (start) state, such that $P_0 in S\left(exists\right) cup S\left(forall\right)$. • $F$ is a set of accepting (final) states such that $F subseteq S\left(exists\right) cup S\left(forall\right)$. Search another word or see finite automatonon Dictionary | Thesaurus |Spanish

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Sunday, June 26, 2022 HomePreschoolInstructor Tom: Nearly Definitely Instructor Tom: Nearly Definitely Mark Tobey The infinite monkey theorem states {that a} roomful of monkeys with typewriters, given an infinite period of time, will finally, virtually definitely, produce the whole works of Shakespeare. It is a thought experiment concerning the idea of infinity. The human mind is just not actually geared up to suppose when it comes to infinity, so for many people it is a mind-blowing thought, one that’s laborious to carry in our heads for any various minutes. Intellectually, we are able to grasp infinity, however as a sensible matter, due to our perspective as apes known as Homo sapiens, we simply cannot get it. In fact, there may be nothing to say that you need to consider in infinity, at the same time as math, science, and philosophy all vouch for it. In the long run, the existence of infinity is a matter of perception, at the same time as it’s virtually definitely true. You may assert that the whole universe is proscribed in all instructions and I am unable to show you mistaken, though you continue to need to take care of the nothingness that should then infinitely exist exterior the shell during which our universe exists. You would possibly validly assert that the whole universe rests on the shell of a turtle, however then you need to then contemplate upon what that turtle rests. If the reply is turtles all the best way down, then we’re again to discussing infinity. Like I stated, it is mind-blowing. I do know some individuals who ask me to alter the topic after I discuss it. Possibly these first couple paragraphs make you uncomfortable. Possibly you are scoffing on the complete thought of monkeys writing Shakespeare. Possibly you are pondering, “Inform me one thing I did not know.” They’re all legitimate responses to this virtually definitely true, however frankly, ineffective info. The qualifier “virtually definitely” has already appeared not less than as soon as in each paragraph I’ve written to date. It is a mathematical time period that’s used to debate likelihood. It is a hedge towards the extremely unlikely, but nonetheless not inconceivable, probability that these monkeys will fail to provide even a single web page of Shakespeare, at the same time as the mathematics tells us that not solely will they produce the whole works they’ll accomplish that an infinite variety of instances. Some dismiss this kind of musing as a waste of time. It is going to virtually definitely by no means land me a job. It is going to virtually definitely by no means materially enhance my life, besides to the diploma that I take pleasure in enjoying with the concepts. It’s virtually definitely all simply pondering in circles, spherical and spherical, with no starting center or finish, a churn of psychological vitality that could possibly be higher used fixing the world’s issues. These are the arguments most frequently utilized by those that need kids to cease enjoying and begin working. However there’s something I can state, with out the necessity to insert “virtually definitely.” We’re, as people, as Homo sapiens, a form of ape which might be swapped out for the metaphorical monkeys within the theorem with out altering the important premise. And you recognize what? We apes have already produced the whole works of Shakespeare. Certainly, we have produced it and reproduced it time and again and we’re simply getting began. I’ve three of these reproductions proper right here on my bookshelves. Certainly, the web incorporates the whole works so many instances that the quantity is approaching infinity. And we did not even want typewriters or infinite time as a way to do it. Tune in subsequent week once we play with the thought of time journey. It is virtually definitely inconceivable . . . ****** I put a number of effort and time into this weblog. If you would like to assist me please contemplate a small contribution to the trigger. Thanks! RELATED ARTICLES

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A069728 Number of rooted non-separable Eulerian planar maps with n edges. 2 1, 1, 1, 1, 2, 6, 19, 64, 230, 865, 3364, 13443, 54938, 228749, 967628, 4149024, 18000758, 78905518, 349037335, 1556494270, 6991433386, 31609302688, 143755711433, 657301771172, 3020175361634, 13939605844996, 64604720622719 (list; graph; refs; listen; history; text; internal format) OFFSET 0,5 LINKS Gheorghe Coserea, Table of n, a(n) for n = 0..500 Daniel Birmajer, Juan B. Gil, Michael D. Weiner, A family of Bell transformations, arXiv:1803.07727 [math.CO], 2018. Hsien-Kuei Hwang, Mihyun Kang, Guan-Huei Duh, Asymptotic Expansions for Sub-Critical Lagrangean Forms, LIPIcs Proceedings of Analysis of Algorithms 2018, Vol. 110. Schloss Dagstuhl-Leibniz-Zentrum für Informatik, 2018. V. A. Liskovets and T. R. S. Walsh, Enumeration of Eulerian and unicursal planar maps, Discr. Math., 282 (2004), 209-221. FORMULA G.f.: y = A(x) satisfies 0 = y^5 - y^4 - 12*x*y^3 + x*(16*x + 11)*y^2 - 8*x^2*y + x^2. - Gheorghe Coserea, Apr 12 2018 a(n) ~ 75*sqrt(65)/(4394*sqrt(Pi)) * n^(-5/2) * (128/25)^n. - Gheorghe Coserea, and Vaclav Kotesovec, Apr 12 2018 A(x) = 1 + serreverse((1+x)^2*(1+12*x-(1-4*x)^(3/2))/(2*(4*x+3)^2)); equivalently, it can be rewritten as A(x) = 1 + serreverse((y-1)*(y^2+y-1)^2/(y^3*(3*y-2)^2)), where y = A000108(x). - Gheorghe Coserea, Apr 14 2018 EXAMPLE A(x) = 1 + x + x^2 + x^3 + 2*x^4 + 6*x^5 + 19*x^6 + 64*x^7 + 230*x^8 + ... MATHEMATICA Flatten[{1, Table[(Sum[(-1)^j*Binomial[2*n + j - 1, j] * Sum[(-1)^k*2^(n - j - k - 1)*Binomial[j, k] * Binomial[2*n, n - j - k - 1], {k, 0, Min[j, n - j - 1]}], {j, 0, n - 1}] - 2*Sum[(-1)^j*Binomial[2*n + j - 1, j] * Sum[(-1)^k*2^(n - j - k - 2) * Binomial[j, k]*Binomial[2*n, n - j - k - 2], {k, 0, Min[j, n - j - 2]}], {j, 0, n - 2}])/n, {n, 1, 30}]}] (* Vaclav Kotesovec, Apr 13 2018 *) (* In the article by Liskovets and Walsh, p. 218, E'ns(n), the factor -2*Sum[...] is missing. *) PROG (PARI) seq(N) = {   my(x ='x+O('x^N), y=serreverse(x*(1+x/2-x^2/4)^2/(2*(1+x)^2)));   Vec(1+y/2-y^2/4); }; seq(27) \\ Gheorghe Coserea, Apr 12 2018 CROSSREFS Cf. A000257. Sequence in context: A181315 A181734 A216447 * A150083 A220065 A047016 Adjacent sequences:  A069725 A069726 A069727 * A069729 A069730 A069731 KEYWORD nonn AUTHOR Valery A. Liskovets, Apr 07 2002 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified September 22 00:42 EDT 2020. Contains 337276 sequences. (Running on oeis4.)

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# Which is greater the greatest whole number with 4 digits or the greatest least whole number with 5 digits 1 dari eqaachrishar ## Jawapan 2016-01-08T19:33:16+08:00 The greatest least whole number with 5 digits. For example: The greatest whole number with 4 digits = 9999 The greatest least whole number with 5 digits = 10000 10000 greater than 9999. So, the answer is the greatest least whole number with 5 digits.

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# Algebrator can start solving your homework in the next 5 minutes Get Algebrator Now! only \$49.00   \$35.00 2checkout.com is an authorized reseller of goods provided by Softmath June 24th ## It is easy as 1-2-3! Step 1 : Enter your algebra problem straight from the book: Step 2 : Let Algebrator solve it: Step 3 : Ask for an explanation for the steps you don't understand: • solving a system of two and three linear equations (including Cramer's rule) • simplification of algebraic expressions (polynomials (simplifying, degree, division...), exponential expressions, fractions, radicals (roots), absolute values...) • factoring and expanding expressions • finding LCM and GCF • basic step-by-step arithmetics operations (adding, subtracting, multiplying and dividing) • operations with complex numbers (simplifying, rationalizing complex denominators...) • solving linear, quadratic and many other equations and inequalities (including log. and exponential) • graphing general functions • graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions) • operations with functions (composition, inverse, range, domain...) • simplifying logarithms • sequences (classifying progressions, find the nth term of an arithmetic progression...) • basic geometry and trigonometry (similarity, calculating trig functions, right triangle...) • arithmetic and other pre-algebra topics (ratios, proportions, measurements...) • linear algebra (operations with matrices, inverse matrix, determinants...) • statistics (mean, median, mode, range...) ## System requirements: Any PC running any version of Windows. Any Intel Mac running OS X. Get Algebrator Now! only \$49.00   \$35.00 2checkout.com is an authorized reseller of goods provided by Softmath # Our users: My sister bought your software program for my kids after she saw them doing their homework one night. As a teacher, shed recognized the value in a program that provided step-by-step solutions and good explanations of the work. My kids love it. Marsha Stonewich, TX This algebra software has an exceptional ability to accommodate individual users. While offering help with algebra homework, it also forces the student to learn basic math. The algebra tutor part of the software provides easy to understand explanations for every step of algebra problem solution. John Kattz, WA You guys are GREAT!! It has been 20 years since I have even thought about Algebra, now with my daughter I want to be able to help her. The step-by-step approach is wonderful!!! 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Solve the differential equations xy = a2 Asked by Aaryan | 1 year ago |  102 ##### Solution :- From the question it is given that, xy = a2 … [equation (i)] Now, differentiate the equation (i) with respect x, x $$\dfrac{ dy}{dx}$$ + y(1) = 0 x $$\dfrac{ dy}{dx}$$ + y = 0 Answered by Aaryan | 1 year ago ### Related Questions #### Find the particular solution satisfying the given condition (dy/dx) = (x – 1)dy/dx = 2x3y Find the particular solution satisfying the given condition $$(\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y$$ #### If the interest is compounded continuously at 6% per annum, how much worth Rs. 1000 will be after 10 years? If the interest is compounded continuously at 6% per annum, how much worth Rs. 1000 will be after 10 years? How long will it take to double Rs. 1000? #### In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

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In [ ]: %load_ext autoreload %matplotlib inline %config InlineBackend.figure_format = 'retina' In [ ]: from IPython.display import YouTubeVideo, display # Optimized Learning¶ In this notebook, we will take a look at how to transform our numerical programs into their derivatives. Before they worked on JAX, there was another Python package called autograd that some of the JAX developers worked on. That was where the original idea of building an automatic differentiation system on top of NumPy started. ## Example: Transforming a function into its derivative¶ Just like vmap, grad takes in a function and transforms it into another function. By default, the returned function from grad is the derivative of the function with respect to the first argument. Let's see an example of it in action using the simple math function: $$f(x) = 3x + 1$$ In [ ]: # Example 1: def func(x): return 3 * x + 1 # Pass in any float value of x, you should get back 3.0 as the _gradient_. df(4.0) Here's another example using a polynomial function: $$f(x) = 3x^2 + 4x -3$$ Its derivative function is: $$f'(x) = 6x + 4$$ . In [ ]: # Example 2: def polynomial(x): return 3 * x ** 2 + 4 * x - 3 # pass in any float value of x # the result will be evaluated at 6x + 4, # which is the gradient of the polynomial function. dpolynomial(3.0) ## Using grad to solve minimization problems¶ Once we have access to the derivative function that we can evaluate, we can use it to solve optimization problems. Optimization problems are where one wishes to find the maxima or minima of a function. For example, if we take the polynomial function above, we can calculate its derivative function analytically as: $$f'(x) = 6x + 4$$ At the minima, $f'(x)$ is zero, and solving for the value of $x$, we get $x = -\frac{2}{3}$. In [ ]: # Example: find the minima of the polynomial function. start = 3.0 for i in range(200): start -= dpolynomial(start) * 0.01 start We know from calculus that the sign of the second derivative tells us whether we have a minima or maxima at a point. Analytically, the second derivative of our polynomial is: $$f''(x) = 6$$ We can verify that the point is a minima by calling grad again on the derivative function. In [ ]: ddpolynomial = grad(dpolynomial) ddpolynomial(start) Grad is composable an arbitrary number of times. You can keep calling grad as many times as you like. ## Maximum likelihood estimation¶ In statistics, maximum likelihood estimation is used to estimate the most likely value of a distribution's parameters. Usually, analytical solutions can be found; however, for difficult cases, we can always fall back on grad. Let's see this in action. Say we draw 1000 random numbers from a Gaussian with $\mu=-3$ and $\sigma=2$. Our task is to pretend we don't know the actual $\mu$ and $\sigma$ and instead estimate it from the observed data. In [ ]: from functools import partial import jax.numpy as np from jax import random key = random.PRNGKey(44) real_mu = -3.0 real_log_sigma = np.log(2.0) # the real sigma is 2.0 data = random.normal(key, shape=(1000,)) * np.exp(real_log_sigma) + real_mu Our estimation task will necessitate calculating the total joint log likelihood of our data under a Gaussian model. What we then need to do is to estimate $\mu$ and $\sigma$ that maximizes the log likelihood of observing our data. Since we have been operating in a function minimization paradigm, we can instead minimize the negative log likelihood. In [ ]: from jax.scipy.stats import norm def negloglike(mu, log_sigma, data): return -np.sum(norm.logpdf(data, loc=mu, scale=np.exp(log_sigma))) If you're wondering why we use log_sigma rather than sigma, it is a choice made for practical reasons. When doing optimizations, we can possibly run into negative values, or more generally, values that are "out of bounds" for a parameter. Operating in log-space for a positive-only value allows us to optimize that value in an unbounded space, and we can use the log/exp transformations to bring our parameter into the correct space when necessary. Whenever doing likelihood calculations, it's always good practice to ensure that we have no NaN issues first. Let's check: In [ ]: mu = -6.0 log_sigma = np.log(2.0) negloglike(mu, log_sigma, data) Now, we can create the gradient function of our negative log likelihood. But there's a snag! Doesn't grad take the derivative w.r.t. the first argument? We need it w.r.t. two arguments, mu and log_sigma. Well, grad has an argnums argument that we can use to specify with respect to which arguments of the function we wish to take the derivative for. In [ ]: dnegloglike = grad(negloglike, argnums=(0, 1)) # condition on data dnegloglike = partial(dnegloglike, data=data) dnegloglike(mu, log_sigma) Now, we can do the gradient descent step! In [ ]: # gradient descent for i in range(300): dmu, dlog_sigma = dnegloglike(mu, log_sigma) mu -= dmu * 0.0001 log_sigma -= dlog_sigma * 0.0001 mu, np.exp(log_sigma) And voila! We have gradient descended our way to the maximum likelihood parameters :). ## Exercise: Where is the gold? It's at the minima!¶ We're now going to attempt an exercise. The task here is to program a robot to find the gold in a field that is defined by a math function. In [ ]: from inspect import getsource from dl_workshop.jax_idioms import goldfield print(getsource(goldfield)) It should be evident from here that there are two minima in the function. Let's find out where they are. Firstly, define the gradient function with respect to both x and y. To see how to make grad take a derivative w.r.t. two arguments, see the official tutorial for more information. In [ ]: from typing import Callable pass dgoldfield(3.0, 4.0) Now, implement the optimization loop! In [ ]: # Start somewhere pass ## Exercise: programming a robot that only moves along one axis¶ Our robot has had a malfunction, and it now can only flow along one axis. Can you help it find the minima nonetheless? (This is effectively a problem of finding the partial derivative! You can fix either the x or y to your value of choice.) In [ ]: def grad_ex_3(): pass # Start somewhere and optimize! x = 0.1 for i in range(300): dx = dgoldfield_dx(x) x -= dx * 0.01 x For your reference we have the function plotted below. In [ ]: import matplotlib.pyplot as plt from matplotlib import cm fig, ax = plt.subplots(subplot_kw={"projection": "3d"}) # Change the limits of the x and y plane here if you'd like to see a zoomed out view. X = np.arange(-1.5, 1.5, 0.01) Y = np.arange(-1.5, 1.5, 0.01) X, Y = np.meshgrid(X, Y) Z = goldfield(X, Y) # Plot the surface. surf = ax.plot_surface( X, Y, Z, cmap=cm.coolwarm, linewidth=0, antialiased=False, ) ax.view_init(elev=20.0, azim=20) In [ ]:

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CC-MAIN-2022-49

latest

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https://simplywall.st/stocks/gb/capital-goods/aim-rai/ra-international-group-shares/news/is-ra-international-group-plcs-lonrai-roe-of-16-impressive/

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latest # Is RA International Group PLC’s (LON:RAI) ROE Of 16% Impressive? One of the best investments we can make is in our own knowledge and skill set. With that in mind, this article will work through how we can use Return On Equity (ROE) to better understand a business. By way of learning-by-doing, we’ll look at ROE to gain a better understanding of RA International Group PLC (LON:RAI). RA International Group has a ROE of 16%, based on the last twelve months. That means that for every £1 worth of shareholders’ equity, it generated £0.16 in profit. ### How Do I Calculate Return On Equity? The formula for return on equity is: Return on Equity = Net Profit ÷ Shareholders’ Equity Or for RA International Group: 16% = 8.524372 ÷ US\$54m (Based on the trailing twelve months to June 2018.) It’s easy to understand the ‘net profit’ part of that equation, but ‘shareholders’ equity’ requires further explanation. It is all earnings retained by the company, plus any capital paid in by shareholders. You can calculate shareholders’ equity by subtracting the company’s total liabilities from its total assets. ### What Does Return On Equity Mean? Return on Equity measures a company’s profitability against the profit it has kept for the business (plus any capital injections). The ‘return’ is the amount earned after tax over the last twelve months. A higher profit will lead to a higher ROE. So, all else being equal, a high ROE is better than a low one. That means it can be interesting to compare the ROE of different companies. ### Does RA International Group Have A Good ROE? One simple way to determine if a company has a good return on equity is to compare it to the average for its industry. The limitation of this approach is that some companies are quite different from others, even within the same industry classification. The image below shows that RA International Group has an ROE that is roughly in line with the Construction industry average (15%). That’s not overly surprising. ROE tells us about the quality of the business, but it does not give us much of an idea if the share price is cheap. If you are like me, then you will not want to miss this free list of growing companies that insiders are buying. ### The Importance Of Debt To Return On Equity Companies usually need to invest money to grow their profits. That cash can come from issuing shares, retained earnings, or debt. In the case of the first and second options, the ROE will reflect this use of cash, for growth. In the latter case, the use of debt will improve the returns, but will not change the equity. That will make the ROE look better than if no debt was used. ### Combining RA International Group’s Debt And Its 16% Return On Equity Although RA International Group does use a little debt, its debt to equity ratio of just 0.024 is very low. Its very respectable ROE, combined with only modest debt, suggests the business is in good shape. Conservative use of debt to boost returns is usually a good move for shareholders, though it does leave the company more exposed to interest rate rises. ### The Key Takeaway Return on equity is useful for comparing the quality of different businesses. A company that can achieve a high return on equity without debt could be considered a high quality business. All else being equal, a higher ROE is better. But when a business is high quality, the market often bids it up to a price that reflects this. Profit growth rates, versus the expectations reflected in the price of the stock, are a particularly important to consider. So you might want to check this FREE visualization of analyst forecasts for the company. But note: RA International Group may not be the best stock to buy. So take a peek at this free list of interesting companies with high ROE and low debt. To help readers see past the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price-sensitive company announcements. The author is an independent contributor and at the time of publication had no position in the stocks mentioned. For errors that warrant correction please contact the editor at editorial-team@simplywallst.com.

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CC-MAIN-2019-30

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https://stephanieevergreen.com/visualizing-regression/

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Guest Post: Visualizing Regression Effectively Updated Note from Stephanie: This blog post generated a lot of discussion. Some of that is in the comments here, some of that has been deleted, some of it came from Twitter and via my inbox. Be sure to read the comments to get a sense of the critique. At the bottom of this post, before the comments, I’ve provided some of my reasoning because I think context helps and to explain why I’ve had to delete comments (Hint: Threats!) Note from Stephanie: I outlined a few ways to show regression data in my latest book but they all avoid the regression table itself. This guest post from William Faulkner, João Martinho, and Heather Muntzer illustrates how to improve the simple table and how to take that data even further into something that doesn’t require a PhD to interpret. The world is going to hell in a handbag, and it’s because data viz people haven’t stepped up to the plate. There’s a dark horse of data viz hidden in plain sight, which has for decades made a mess of one of humanity’s most crucial quantitative tools. This villain is the regression table. The world runs on regressions. How many times have you heard “studies show that [blah],” or “it turns out [blah] leads to [blah]”? More often than not, these ‘facts’ are (over)simplified interpretations of a regression. Regressions are THE most common statistical way to determine whether there’s a relationship between two things – like doing yoga and wearing tight pants, or, as we’ll see in a sec, a person’s race and likelihood of being shot by the police. People interpret the results of regressions using regression tables (and little else) A ton of super important decisions get made on the basis of simple statements like “studies show you can reduce [blah] by [blah]%.” And these invariably come from a regression table, which usually looks something like this example analysis of 1974 cars, testing whether those with automatic or manual transmissions are more efficient: (R user? DIY kit available here) Regression tables are TERRIBLE visualization tools. The WORST. Nobody wants to look at that thing! Are you kidding? And worse, even if you’re a quantitative genius really interested in the results, it’s STILL hard to intuit what’s going on. For The Love Of Humanity, Let’s Fix This. WARNING: This middle section is for the nerds. If you don’t run regressions yourself, feel free to skip down to Section III The Tame Tweaks Even without going wild, we can just stop being so careless. With just a few simple tweaks, we can go from this: To this: We’re not claiming perfection, but at least we’re not being as cruel to our audience. Seems a lot easier now to see that the automatic-manual distinction is not as important for efficiency when we account for weight and horsepower. Let’s Go Nuts Think outside the box (ahem, table), when it comes to regressions, maybe we can just graph the coefficients? How changing the horsepower, weight, and transmission type of the average 1974 car seems to affect its mileage: Again, not perfect. But it’s a start towards diagrams that intuitively show what we really care about in most cases: • Size of coefficients • Uncertainty of coefficients (confidence intervals and/or statistical significance) • Explanatory power of the model overall And that’s it! Because It Is Important. Last year, Harvard professor Dr. Fryer released a working paper inspiring some controversial headlines. Did the paper really claim that blacks were 23% less likely than whites to be shot during an encounter with police? The whole hullabaloo boils down to – you guessed it – a regression table which is, as per usual, practically indecipherable: Let’s try our tweaks from before: Add a possible title, like: As we factor in other variables, such as whether the suspect had a weapon, whether the bias is towards blacks or whites becomes a lot less clear and we can’t be nearly as precise about the amount of bias. That’s better. Not perfect, but better. Turns out as we consider more and more aspects of the encounter, that strong bias towards police shooting white suspects gets a lot more muddled. And It’s Not Even That Hard. But you’re right. Who cares about nuance? In a world which constantly steamrolls detail in the name of thumbs-up or thumbs-down now-and-forever conclusions, who’s got time to worry about subtleties? We’d like to think some people do. We’d like to think oh-so-many-more would take interest were it not for these bristling anathemas – regression tables. Regression analysis and data viz experts, let’s give folks a chance: • Be nice to your audience. If you put a regular, white, asterix-splattered regression table in front of them, that’s inconsiderate. So don’t. Instead, • Use accessible labels, translate jargon • Take out extra decimals • Use color, shading, and transparency to express the key info in multiple ways. • Consider what’s important about the analysis – this means both the finding itself and the degree of uncertainty surrounding it! Tools like heatmaps and the coefficient charts above help to put all that detail out there in a quickly digestible format. Big problem. Reeeeeasonably easy solutions. Now the fate of the world is up to you. No pressure. Authors: William Faulkner – Director, Flux RME João Martinho – Evaluation Specialist, C&A Foundation Heather Muntzer – Independent Designer PS: We asked for data from the Harvard team to replicate this study and produce even better visualizations. Despite a few kind replies, they never got around to sharing it. PPS: More materials from this project are available in this Google Drive folder. PPPS: Questions? Email William. Updated Editor’s Note: I’m updating my editor’s note on this post because of how laughably out of hand things have gotten. Statisticians are really unnerved by some of the wording used in this guest post. It’s ok to disagree. I welcome those discussions and comments because they help everyone keep evolving their thinking. But I’m heavily screening all comments posted to this thread from this point forward because now I’m getting threats like this one: If you do not bring that uninformative guest post down, or fix it, I will bring this to the attention of the media and my fellow colleagues at ASA. Trust me, you do not want that kind of attention. Um what? Report me to the American Statistical Association? LOL More than one person took issue not with the content of the blog itself, but with the way that I asked the critics to improve upon what they didn’t like. One person wrote: I am profoundly upset with your and Faulkner’s reaction to the comments. Rule #1, as I’ve stated before, is that this is my blog. I’ll say what I want. I’ll outright and without apology delete any comments that attempt to tell me how to handle commenters or whether to pull a post. If you feel so strongly that it is bad, don’t read it. Start your own blog. That doesn’t mean I defend errors. It means that I’m ok with mistakes – I’ve made plenty of my own – especially when they foster good discussion. But good discussion means generation of new ideas. Only one statistician in all of this mix has agreed to make a better attempt – in a few weeks. Yes, of course, I’m asking critics to do better. Being an armchair critic is easy. To paraphrase Brene Brown, if you aren’t in the arena with me – actively trying to make things better by putting forth efforts that could be wrong or critiqued – I’m not interested in your opinion. Several commenters questioned my intelligence. And then one guy (they were almost all white guys) said: People are just trying to help you. LOL you dudes are so funny. Insulting my intelligence is not help. It might help, actually, to understand a bit more about me and the guest post authors. I have a PhD in interdisciplinary research and evaluation. I met the guest authors at an evaluation conference. (Did you see that the lead author’s name is William Faulkner?? How can you NOT have tequila with this guy?) If you don’t know what evaluation is, it’ll be good for me to explain it to you because there’s a pretty big difference between conducting pure statistics and evaluation. Evaluators are like researchers in that we seek to generate knowledge but we conduct our studies for real organizations who are trying to learn whether they’ve made an impact with their work, or whether new strategies could help them be more efficient. We use anything from observation to a random controlled trial to get at the data. Our methods often have to be creative, since we are collecting data from actual humans, not in clinical settings. Our analyses are always rigorous. And we have to generate explanations of those analyses for real human decision-makers, in time for them to actually make use of it. Our audience is real life, not a journal. Those explanations can be very challenging to compose. It can be difficult to balance statistical jargon fidelity with the need to speak in a plain language for the understanding and action-taking on the table with our clients. Will and team made one of the first attempts I’ve ever seen at making regression more digestible for people. Of course the first attempt will never be perfect. But kudos to them for giving it a shot, instead of just running some stats and wondering why the audience doesn’t get it (or worse, questioning the audience’s intelligence). Twice as many people sent love and support for this post as those statisticians who got furious. And that’s because people are hungry and eager for something better than the way the stats people have been doing it. So keep building. Ever forward, friends. Oh and please please PLEASE report me to the American Statistical Association! I’ve seen some slidedecks from their conferences over the years. You could use my help. Learn something new? Share this helpful info with a friend who needs an extra perk today or post it to your social where your third cousin can benefit, too.

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CC-MAIN-2021-31

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http://www.physicsforums.com/showthread.php?p=3835745

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# 3 balls connected by massless strings by akorn3000 Tags: balls, connected, massless, strings P: 7 Suppose you have 3 balls, all of equal mass, M. They are connected to each other by equal-length (each length L) strings of negligible mass, such that one ball is suspended in the middle and the two balls at the end are themselves suspended from the ceiling at two points, a distance X from one another. Using M, X, and L, how would I go about finding the exact angles the balls make with the horizontal? I tried summing the forces and the moments, but I think I must be doing something wrong. Any help would be appreciated. Attached Thumbnails Sci Advisor P: 2,426 1) Use symmetry to assume that middle mass is equidistant from the other two. (Seems you've done that.) 2) Solve the problem for just one mass hanging from the other two as a function of distance between the later two. Call that distance y. 3) Solve for force between ceiling and the mass attached to it using the solution from above as input. You should treat y as one of the variables you are solving for. Edit: Just realized that there is a much easier way. What can you say about horizontal component of T1 vs horizontal component of T2? P: 7 Quote by K^2 Edit: Just realized that there is a much easier way. What can you say about horizontal component of T1 vs horizontal component of T2? They must be balanced, so T1*cos(theta1) = T2*cos(theta2) P: 7 ## 3 balls connected by massless strings The way I see it, each string wants to reduce its tension by hanging straight down. The strings on top are pulled inward by the middle mass they have to support, and so they pull outward, while the strings on the bottom want to reduce their tension and pull inward. P: 2,426 Quote by akorn3000 They must be balanced, so T1*cos(theta1) = T2*cos(theta2) And you know what the vertical components are. So given theta1, you should have no trouble computing theta2, right? P: 7 I got as far as writing out the force diagrams for each mass: Taking the left mass for instance, balancing the forces in the y direction gives T2*sin(theta2) = mg + T1*sin(theta1), and for the bottom mass, 2*T1*sin(theta1) = mg. Substituting gives 3*T1*sin(theta1) = T2*sin(theta2), and if you divide that by T1*cos(theta1) = T2*cos(theta2) you get 3*tan(theta1) = tan(theta2). This relationship seems right at a cursory glance, since the slope of the upper strings must be steeper than the bottom ones, although I still do not know what the angles are independently and based on X, L, and M. It's quite possible that I'm missing something very obvious. I also know that X = L*cos(theta1) + L*cos(theta2). I forgot to mention before but I do not want either of the tensions in the final answer, which seems to make this problem tricky for me. Sci Advisor P: 2,426 That's the relationship. So theta2=arctan(3*tan(theta1)). Not the most pleasant expression, but there you go. Now, can you express X in terms of theta1, theta2, and L? P: 7 Quote by K^2 That's the relationship. So theta2=arctan(3*tan(theta1)). Not the most pleasant expression, but there you go. Now, can you express X in terms of theta1, theta2, and L? The simplest I can see is X = 2*L*[cos(theta1) + cos(theta2)] So I suppose if you substitute theta2 = arctan(3*tan(theta1)) you'd have (X/2L) = cos(theta1) + cos(arctan(3*tan(theta1))... yeesh. Sci Advisor P: 2,426 Yeah. And as far as I can tell, that's about as good as it gets. You can also build a 4th order polynomial equation for T_x = Tcos(theta), which obviously has a general solution, but it's a bit lengthy to say the least. Point is, I don't think you can get a clean algebraic solution here. But given specific number for X/L, you can easily solve it numerically. P: 7 And if you isolate theta1 instead, I suppose you would have theta1 = arctan((1/3)*tan(theta2)), so in that case you'd have (X/2L) = cos(arctan((1/3)*tan(theta2)) + cos(theta2). Heh, I guess nobody said the answer had to be pretty. P: 7 Thank you for all your help, I appreciate it. Related Discussions Introductory Physics Homework 4 Introductory Physics Homework 6 Introductory Physics Homework 1 Advanced Physics Homework 3 Beyond the Standard Model 10

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CC-MAIN-2013-48

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https://www.accountingweb.com/practice/team/tailor-your-day-for-productivity-focuser-or-juggler-quiz

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# Tailor Your Day For Productivity - Focuser or Juggler Quiz Knowing your work style, and personal patterns is essential in managing your day-to-day activities. How you balance your daily activities is crucial for the upcoming tax season, or as you prepare for entry into the accounting profession. Some people are focusers - they work best when they focus on one project at a time until it is completed. Some people are jugglers - they work best with several projects going at the same time. You can decide which is your style for taking on projects by completing this Self Awareness Quiz! Determine whether each of the statements below is true for you, using the following scoring: 5=almost always 4=usually 3=sometimes 2=not usually 1=almost never 1. I thrive on multiple projects. 2. I like to become intensely, exclusively absorbed with one project at time. 3. In my mind, I divide a project into several distinct stages of development, such as the planning stage, action stage, and the follow-up stage, and like to work on two or more projects that are in various states. 4. I tend to eat and breathe a project, day and night, until it is finished. 5. When I feel blocked, tired, or bored with one project, I can often make progress on a different one. 6. I feel scattered, distracted, and frustrated when I must deal with more than one project, even though they are in various stages of completion. 7. I work well when I am thinking about an upcoming project during the breathing spaces of an active project. 8. When I am forced to switch my attention from the action stage of my primary project to the planning stage of another one, I tend to become consumed by the new one. 9. I like to assign the solution of certain problems on one project to my subconscious mind, while I consciously focus on another project. 10. The only way I can solve problems that arise on a project is to devote myself exclusively to that project. Total your score for the odd-numbered statements, which reflect the juggler style. A score between 18 & 25 means you are a juggler; between 13 & 17, you are probably a juggler and a focuser; between 5 & 12, you are not a juggler. Now total your score for the even-numbered statements, which reflect the focuser style. A score between 18 & 25 means you are a focuser; between 13 & 17, you are probably a focuser and a juggler; between 5 & 12, you are not much of a focuser. Are you a Juggler: If you are an effective juggler, you may be able to get more done during the same time frame than your focuser co-workers. Note: At the task level, almost everyone works best when they focus on the one task at any one point in time. If you are a juggler, you are probably easily bored with one major project that goes on and on. The problem may be your ability to stay focused long enough on one project to make significant headway and to keep going at it until it is completed. Here are some tips to assist you with those long ongoing projects: 1. Divide each project into stages and make a step-by-step plan for each stage. 2. Delegate as many of the tasks as you need to and set a schedule for following up on each task. 3. As soon as one project is underway, look for another. Are you a Focuser: If you are a focuser, you are likely to stick with a major project until it is complete. If your job requires you to juggle more than one project at a time, you may run into barriers. If this is the case, try these tips: 1. If you can choose the one project that promises to be most fruitful and best suited for you, make a plan for completing it, and stick with the project until it is finished, you are in your element 2. If other projects must be started, can you delegate the necessary tasks to someone else? 3. If your position requires you to juggle more than one major project at time, learn to compartmentalize. It is difficult to tear yourself away from your current project, but once you make the transition, it does get easier. Use your focusing skills to zero in on the new project. If you are both a juggler and a focuser this can be a great asset. Being able to totally focus on a task in the moment is essential for success in most positions, but so is juggling several projects. Being aware of these two styles and how to use them most effetively will increase your productivity.

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CC-MAIN-2018-05

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# Re: TC quesions (sic) ```Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br> Ralph Zekelman wrote: > > The ideal voltage gain > > is given by V2max/V1max = sqrt(L2/L1). > > Hi Antonio, > > Would you please go thru a derivation of this? > Begin at the beguine. By the usual lumped model: L1=primary inductance L2=secondary inductance C1=primary capacitance C2=secondary capacitance (self-capacitance of L2 and terminal capacitance) V1=voltage in C1 (input voltage) V2=voltage in C2 (output voltage) Energy initially stored in the primary capacitor: E1=0.5*C1*V1max^2 Maximum energy at the secondary capacitance: E2=0.5*C2*V2max^2 By energy conservation, assuming low losses in the energy transfer: E2=E1 And so: V2max=V1max*sqrt(C1/C2) But the two LC circuits are tuned to the same frequency: L1*C1=L2*C2 and so: C1/C2=L2/L1 and finally: V2max=V1max*sqrt(L2/L1) Antonio Carlos M. de Queiroz ```

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CC-MAIN-2013-20

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https://docs.microsoft.com/es-es/dotnet/api/system.decimal.op_multiply?view=netframework-4.7.2

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# Decimal.Multiply(Decimal, Decimal)Decimal.Multiply(Decimal, Decimal)Decimal.Multiply(Decimal, Decimal)Decimal.Multiply(Decimal, Decimal) Operator ## Definición Multiplica dos valores Decimal especificados.Multiplies two specified Decimal values. ``````public: static System::Decimal operator *(System::Decimal d1, System::Decimal d2);`````` ``public static decimal operator * (decimal d1, decimal d2);`` ``static member ( * ) : decimal * decimal -> decimal`` ``Public Shared Operator * (d1 As Decimal, d2 As Decimal) As Decimal`` #### Parámetros d1 Decimal Decimal Decimal Decimal Primer valor que se va a multiplicar.The first value to multiply. d2 Decimal Decimal Decimal Decimal Segundo valor que se va a multiplicar.The second value to multiply. #### Devoluciones Resultado de multiplicar `d1` por `d2`.The result of multiplying `d1` by `d2`. #### Excepciones El valor devuelto es menor que MinValue o mayor que MaxValue.The return value is less than MinValue or greater than MaxValue. ## Comentarios El Multiply método define la operación del operador de multiplicación para Decimal los valores.The Multiply method defines the operation of the multiplication operator for Decimal values. Permite el código como el siguiente:It enables code such as the following: ``````using System; public class Example { public static void Main() { Decimal number1 = 16.8m; Decimal number2 = 4.1m; Decimal number3 = number1 * number2; Console.WriteLine("{0:N2} x {1:N2} = {2:N2}", number1, number2, number3); } } // The example displays the following output: // 16.80 x 4.10 = 68.88 `````` ``````Module Example Public Sub Main() Dim number1 As Decimal = 16.8d Dim number2 As Decimal = 4.1d Dim number3 As Decimal = number1 * number2 Console.WriteLine("{0:N2} x {1:N2} = {2:N2}", number1, number2, number3) End Sub End Module ' The example displays the following output: ' 16.80 x 4.10 = 68.88 `````` Si el lenguaje que está usando no admite operadores personalizados, llame al Multiply método en su lugar.If the language you're using doesn't support custom operators, call the Multiply method instead. El método equivalente para este operador esDecimal.Multiply(Decimal, Decimal)The equivalent method for this operator is Decimal.Multiply(Decimal, Decimal)

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CC-MAIN-2019-43

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https://www.khanacademy.org/kmap/operations-and-algebraic-thinking-d/more-with-multiplication-division/map-associative-property-of-multiplication/v/associate-property-of-multiplication

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If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. # Associative property of multiplication Sal regroups numbers to simplify multiplication problems. ## Want to join the conversation? • What’s a faster way to do it? • well, Layla Shultz, you can do the standard algorithm way or you can just do it faster in a way of the standard algorithm. Hope this helps! • 40*267*25 evaluate using suitable property • what is the direct simplification of the defenition of what the associative property is? I don't understand this definition. • The associative property of multiplication let's us move / change the placement of grouping symbols. It does not move the numbers. For example: (2 x 4) x 5 can be changed into 2 x (4 x 5) Both expressions create the same result. • What is 12 x 42 x 56? • Does this work for double grouping of polynomials? • is it necessary to use this method for solving problems like these? • This means that (5 X 3) X 4 = (5 X 4) X 3 = (4 X 3) X 5 Is is correct ? • (5x3) x4 = (5x4) x3 = (4x3) x5 The above statement is true. • are there practices for this lesson? • Yes there are practices for this (1 vote) • that is a good example lan pulizzotto • This is to easy, no matter what order, its always the same. Though, the questions are never this easy. Why? ## Video transcript - [Instructor] So, what we're gonna do is get a little bit of practicing multiple numbers together and we're gonna discover some things. So, first I want you to figure out what four times five times two is. Pause the video and try to figure it out on your own. Alright, so whatever your answer is, some of you might have done it this way, some of you might have said hey, what is four times five and then you multiplied it by two, so what you would really have done is you would have done four times five first, so that's why I put parentheses around that and then you would have multiplied by two and what would you have gotten? Weil, the four times five part, that is of course 20 and then you multiply that times two and you would get 40 which of course would be correct, four times five times two is indeed equal to 40. Now, what I want you to do now is as quickly as possible try to figure out what five times two times four is. Really quick, pause the video, try to figure that out. Well, some of you might have tried and you might have done it in a similar way where you tried to figure out five times two is first and you said okay, five times two is equal to 10 and then I'd multiply that times four and then you would say well, gee, this is same thing as I got last time. Is there something interesting going on? And the interesting thing that you might realize is in both cases we're multiplying the same three numbers. We are just doing it in a different order. Here we multiplied four times, we wrote it out in a different order, four times five times two. Here we wrote five times two times four. Here we did the four times five first, here we did the five times two first but notice we got the same result. Now, I'd encourage you, pause this video. Try to multiply these numbers in any order, maybe you do two times four first. In fact, let's just do that. Let's do two times four, two times four and then multiply that by five. What is this going to be equal to? Well, you might notice again this is two times four is eight, you multiply that times five. Well, once again, we got 40, so you might see a pattern here. It doesn't matter which order we multiply these things in. In fact, you could write four times five times two. You could do the four times five first, four times fives times two or you could do four times five times two, so you could do four times five times two. So, it doesn't matter which order you multiply these things in. In every case you are going to get 40. Now, there's a very fancy term for this, the associative property of multiplication but the main realization is and it's not just true with the three numbers, in fact, you've seen something similar with two numbers where it doesn't matter what order you multiply them in but what you see with three numbers and even if you tried it with four or five or really 1,000 numbers being multiplied together, as long as you're just multiplying them all, it doesn't matter what order you're doing it with. It doesn't matter in what order you associate them with. Here we did four times five first, four times five first, here we did five times two first but in either case we got the same result and I'd encourage you, after this video, try to draw it out. Try to think about why that actually makes intuitive sense, why this is true in the world and it's nice because it simplifies our life when we're doing mathematics and not only now but in our future mathematical career.

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http://nasawavelength.org/resource-search?facetSort=1&topicsSubjects%5B%5D=Mathematics&topicsSubjects%5B%5D=Earth+and+space+science%3AEarth+processes%3AEarth%27s+energy+budget&topicsSubjects%5B%5D=The+nature+of+science&instructionalStrategies%5B%5D=Identifying+similarities+and+differences&instructionalStrategies%5B%5D=Computer+simulations%2Fmodels%2Fvisualizations&resourceType%5B%5D=Instructional+materials%3AProblem+set&resourceType%5B%5D=Instructional+materials%3AUnit

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## Narrow Search Audience Topics Earth and space science Earth processes Mathematics The nature of science Resource Type [-] View more... Learning Time Materials Cost Instructional Strategies [-] View more... SMD Forum Filters: Your search found 8 results. Topics/Subjects: Mathematics Earth's energy budget The nature of science Instructional Strategies: Identifying similarities and differences Computer simulations/models/visualizations Resource Type: Problem set Unit Sort by: Per page: Now showing results 1-8 of 8 # Earth's Energy Budget Unit Plan The four lessons in this unit build toward a student understanding of each component of the energy budget formula - and how the contribution of each component changes due to location and time of year. In order, the four lessons consist of: deriving... (View More) Audience: High school, Informal education Materials Cost: Free # Earth System Responses to Natural and Human-Induced Changes (Grades 7-9) This unit consists of five activities, all of which focus on the response of plant life-cycle events to climate change. Students participate in discussions, field observations, data collection and analyses, plant identification, seed dispersal... (View More) # MRC: Where is the Best Place to Measure? (Grades 3-5) This lesson plan teaches how to select the landing site for a planetary surface investigation, using the 5E learning cycle. Students will be able to determine a landing site for their Mars rover; work with their team to summarize information and... (View More) # MRC: Where is The Best Place to Measure? (Grades 6-8) Learners work in teams to determine a landing site for their Mars Rover that best relates to their scientific question. They use technology skills to research Gale Crater through an online interactive module and learn about features of Mars through... (View More) Audience: Middle school # Carbon in the Future and You Unit three of the "Carbon Connections: The Carbon Cycle and the Science of Climate" curriculum examines the role of carbon and the carbon cycle in future climate. Students discover how scientists determine Earth's average temperature and the role of... (View More) # Earth Exploration Toolbook: Annotating Change in Satellite Images This chapter walks users through a technique for documenting change in before-and-after sets of satellite images. The technique can be used for any set of time-series images that are spatially registered to show the exact same area at the same... (View More) # Monitoring the Global Environment In this online, interactive module, students learn how enhanced Earth remote-sensing capabilities are used by dozens of satellites that are continuously collecting data from multiple vantage points. This allows scientists from different countries to... (View More) Audience: Middle school, High school Materials Cost: Free # Urban Sprawl In this activity, learners will use satellite images of the San Francisco Bay Area to evaluate urban development in an earthquake-prone region. They will prepare an executive summary of their findings that incorporates diagrams or images. Links to... (View More) Audience: Middle school Materials Cost: Free 1

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CC-MAIN-2018-51

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https://www.airmilescalculator.com/distance/msp-to-abq/

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crawl-data/CC-MAIN-2020-40/segments/1600402127075.68/warc/CC-MAIN-20200930141310-20200930171310-00627.warc.gz

# Distance between Minneapolis, MN (MSP) and Albuquerque, NM (ABQ) Flight distance from Minneapolis to Albuquerque (Minneapolis–Saint Paul International Airport – Albuquerque International Sunport) is 981 miles / 1578 kilometers / 852 nautical miles. Estimated flight time is 2 hours 21 minutes. Driving distance from Minneapolis (MSP) to Albuquerque (ABQ) is 1230 miles / 1980 kilometers and travel time by car is about 21 hours 48 minutes. ## Map of flight path and driving directions from Minneapolis to Albuquerque. Shortest flight path between Minneapolis–Saint Paul International Airport (MSP) and Albuquerque International Sunport (ABQ). ## How far is Albuquerque from Minneapolis? There are several ways to calculate distances between Minneapolis and Albuquerque. Here are two common methods: Vincenty's formula (applied above) • 980.513 miles • 1577.983 kilometers • 852.043 nautical miles Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth. Haversine formula • 979.924 miles • 1577.035 kilometers • 851.531 nautical miles The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points). ## Airport information A Minneapolis–Saint Paul International Airport City: Minneapolis, MN Country: United States IATA Code: MSP ICAO Code: KMSP Coordinates: 44°52′55″N, 93°13′18″W B Albuquerque International Sunport City: Albuquerque, NM Country: United States IATA Code: ABQ ICAO Code: KABQ Coordinates: 35°2′24″N, 106°36′32″W ## Time difference and current local times The time difference between Minneapolis and Albuquerque is 1 hour. Albuquerque is 1 hour behind Minneapolis. CDT MDT ## Carbon dioxide emissions Estimated CO2 emissions per passenger is 149 kg (330 pounds). ## Frequent Flyer Miles Calculator Minneapolis (MSP) → Albuquerque (ABQ). Distance: 981 Elite level bonus: 0 Booking class bonus: 0 ### In total Total frequent flyer miles: 981 Round trip?

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CC-MAIN-2020-40

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https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-8-section-8-2-perimeter-and-area-of-polygons-exercises-page-371/41

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## Elementary Geometry for College Students (7th Edition) (a) $\frac{P_{EGIK}}{P_{ABCD}} = \frac{\sqrt{5}}{3}$ (b) $\frac{A_{EGIK}}{A_{ABCD}} = \frac{5}{9}$ (a) Let each side of the square $ABCD$ be 3 units. We can find the length of each side of the square $EGIK$: $L = \sqrt{(1)^2+(2)^2}$ $L = \sqrt{1+4}$ $L = \sqrt{5}$ We can find the ratio: $\frac{P_{EGIK}}{P_{ABCD}} = \frac{4\times \sqrt{5}}{4\times 3} = \frac{\sqrt{5}}{3}$ (b) We can find the ratio: $\frac{A_{EGIK}}{A_{ABCD}} = \frac{(\sqrt{5})^2}{(3)^2} = \frac{5}{9}$

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http://www.phy.ntnu.edu.tw/ntnujava/index.php?action=printpage;topic=2315.0

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NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/ Easy Java Simulations (2001- ) => Collaborative Community of EJS => Topic started by: lookang on October 15, 2011, 09:54:27 pm Title: Ejs Open Source Double Slit Diffraction Model Post by: lookang on October 15, 2011, 09:54:27 pm Ejs Open Source Double Slit Diffraction Model by lookang based on the works of Fu-Kwun Hwang found here reference:Ejs Open Source Single Slit Diffraction Model by Fu-Kwun Hwang remixed by lookang http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2314.0 Superposition of several waves http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2082.0 by Fu-Kwun Hwang other good appletsJDK Double slit interference by Fu-Kwun Hwang http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=20.0 Title: Re: Ejs Open Source Double Slit Diffraction Model Post by: lookang on October 16, 2011, 08:22:38 pm changes made:redesign the slits to be 2 instead of the older single slitmade the 2 slits symmetric centered with moved by y slideradd the 2 slits scalarField by introducing the idea of superposition from the top set of secondary sourcesCode:for(int i=0;i yc=yi+i*dy; // lay the source in y yctop=yitop+i*dy; // duplicate for top slit ycc[i] = yi+i*dy; // by lookang for drawing source  ycctop[i] = yitop+i*dy; // by lookang for drawing source for(int j=0;j //  if (x     x=xmin+j*d; //    } //    else if (x>x1){//   x=x1+j*d;  // r=x-x1; // r is distance from array point x to slit position x1//   x2=(x-xc)*(x-xc);//   }  for(int k=0;k   y=ymin+k*d;      if (x      r=x-x1; // r is distance from array point x to slit position x1       zsum[j][k]=Math.sin(kw*r-omega*t)*n; // need n to compensate for the magnitude     }  else if (x>=x1){   r=Math.sqrt((x-xc)*(x-xc)+(y-yc)*(y-yc));   rtop=Math.sqrt((x-xctop)*(x-xctop)+(y-yctop)*(y-yctop));   zsum[j][k]+=Math.sin(kw*r-omega*t)+Math.sin(kw*rtop-omega*t);   }     } }}add path difference tools to allow sense making that when path difference = integer of wavelength, correspond to constructive interference. and 1/2 of integer is destructive interference like in Ejs Open Source Ripple Tank Interference Model java applet http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2067.017 october 2011add d separation = ycentre2*2 to align with teaching and learningthis simulation is made possible by the open source codes shared by Fu-Kwun Hwang, the true master teacher!enjoy!

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https://uknowledge.uky.edu/ce_etds/85/

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# THE USE OF 3-D HIGHWAY DIFFERENTIAL GEOMETRY IN CRASH PREDICTION MODELING 2-14-2019 2019 ## Degree Name Doctor of Philosophy (PhD) ## Document Type Doctoral Dissertation Engineering ## Department/School/Program Civil Engineering ## Abstract The objective of this research is to evaluate and introduce a new methodology regarding rural highway safety. Current practices rely on crash prediction models that utilize specific explanatory variables, whereas the depository of knowledge for past research is the Highway Safety Manual (HSM). Most of the prediction models in the HSM identify the effect of individual geometric elements on crash occurrence and consider their combination in a multiplicative manner, where each effect is multiplied with others to determine their combined influence. The concepts of 3-dimesnional (3-D) representation of the roadway surface have also been explored in the past aiming to model the highway structure and optimize the roadway alignment. The use of differential geometry on utilizing the 3-D roadway surface in order to understand how new metrics can be used to identify and express roadway geometric elements has been recently utilized and indicated that this may be a new approach in representing the combined effects of all geometry features into single variables. This research will further explore this potential and examine the possibility to utilize 3-D differential geometry in representing the roadway surface and utilize its associated metrics to consider the combined effect of roadway features on crashes. It is anticipated that a series of single metrics could be used that would combine horizontal and vertical alignment features and eventually predict roadway crashes in a more robust manner. It should be also noted that that the main purpose of this research is not to simply suggest predictive crash models, but to prove in a statistically concrete manner that 3-D metrics of differential geometry, e.g. Gaussian Curvature and Mean Curvature can assist in analyzing highway design and safety. Therefore, the value of this research is oriented towards the proof of concept of the link between 3-D geometry in highway design and safety. This thesis presents the steps and rationale of the procedure that is followed in order to complete the proposed research. Finally, the results of the suggested methodology are compared with the ones that would be derived from the, state-of-the-art, Interactive Highway Safety Design Model (IHSDM), which is essentially the software that is currently used and based on the findings of the HSM. ## Digital Object Identifier (DOI) https://doi.org/10.13023/etd.2019.219 COinS

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https://staging-pubs.nctm.org/browse?access=all&pageSize=10&sort=datedescending&t=representations&t_0=communication

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# Browse ## You are looking at 1 - 10 of 12items for : • Representations • Communication • Refine by Access: All content Clear All Restricted access ## GPS: From Folding to Dynamic Geometry Environments In this month's Growing Problem Solvers, we focused on supporting students' understanding of congruence and similarity through rigid motions and transformations. Initial understandings of congruence and similarity begin in first grade as students work with shapes in different perspectives and orientations and reflect on similarities and differences. Restricted access ## Problems to Ponder Problems to Ponder provides 28 varied, classroom-ready mathematics problems that span grades PK-12, arranged in order of grade band. Links to the problem answers are available in this department. Restricted access ## Using Covariation Reasoning to Support Mathematical Modeling Table representations of functions allow students to compare rows as well as values in the same row. Restricted access ## Spaghetti Sine Curves: Virtual Environments for Reasoning and Sense Making The Spaghetti Sine Curves activity, which uses GeoGebra applets to enhance student learning, illustrates how technology supports effective use of physical materials. Restricted access ## Develop Reasoning through Pictorial Representations Solutions coupled with drawings can illustrate students' understandings or misunderstandings, particularly in the area of proportional reasoning. Restricted access ## Sorting Out Ideas about Function A card-sorting task can help students extend their understanding of functions and functional relationships. Restricted access A tool used in reading theory is adapted to help mathematics teachers ask good questions that help students interpret displays of information. Restricted access ## Stop Teaching and Let Students Learn Geometry A geometry course for teachers—easily adaptable to a high school geometry class—integrates technology, reasoning, communication, collaboration, reading, writing, and multiple representations. Restricted access ## When pictures just will not do Restricted access ## Delving into Limits of Sequences Through these calculus activities, students reach an understanding of the formal limit concept in a way that enables them to construct the formal symbolic definition on their own.

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CC-MAIN-2022-21

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# One Checklist That You Should Keep In Mind Before Attending Simplest Form Improper Fractions | Simplest Form Improper Fractions You abominably accept several problems: So i am activity to accord you a quick arbitrary of the botheration and why I anticipate you accept fabricated the mistake. Simplifying Fractions Is Really Simple, When You Follow The … | simplest form improper fractions (i) In the architect fraction::fraction(), you accept accounting integral, numerator, denominator = 0; This DOES NOT do what you think, it does NOT set basic or numerator. If you capital to do that again you should accept done integral=numerator=denominator=0; . Abundant abundant bigger would accept been to use an initializer account e.g. fraction::fraction() : integral(0),numerator(0),denominator(0) {} .Note: Almost all avant-garde compilers would accept accustomed you a admonishing about that if you had angry your warnings on. No abecedarian should anytime abridge after warnings on. (ii) Ascribe is not absolutely amiss but so ugly, absolutely abnormal fractions would be ok?What about account a band of ascribe and again agreeable it based on e.g. 4/5 or 3 4/5 Putting an Improper Fraction in Simplest Form | simplest form improper fractions (iii) acquaintance atom abettor Never accomplish article a acquaintance that doesn’t charge to be. This aloof doesn’t charge to be a friend. It should additionally accept this form: atom abettor (const fraction&) const; Again it is (a) added able (b) abundant beneath acceptable to go wrong. (iv) gcd and gcdr should be one action and not alarm anniversary added backwards and forwards, or gcd_prep should alarm gcd (which would be the recursive function) already alone [to acquiesce the analysis for abrogating to the outside.] Simplifying Improper Fractions | simplest form improper fractions (v) abettor is accounting in such a way that I anticipate you did not absolutely assignment out what is activity on, on cardboard first. Several things are activity on, and mistakes are made.– Aboriginal abstracted the access of the fraction, accomplish the accomplished about-face to the simplest accessible atom in a abstracted adjustment e.g. const fraction& simplify(); . Write that action so that if you accord it a atom 3 14/4 it can accomplish it 6 1/2.– Second cipher the accession in a simpler way: The absolute aberration that you fabricated in your cipher was accomplishing this:You alarm the action as this Simplifying Fractions Calculator | simplest form improper fractions and you accept uninitialized objects:You would accept capital : However, AGAIN agenda the poor best of name fraction1 is a typo abroad from fraction, and that is activity to advance to HORRIBLE debugging problems. Never use a capricious name that is beneath than three belletrist abroad from a chic name in the acceptance scope. Note: fraction_sum is ONE letter abroad back atom _sum actual accessible to type. Reduce Improper Fractions to simplest form | Improper … | simplest form improper fractions Finally h=h/gcd(h,i); i=i/gcd(h,i); which to assignment requires the OLD h not the new one. However, amuse do not EVER use a set of acting names that meaningless, had you accounting int aI, aD, aN; etc it would accept been actor times easier to read. [Although I still anticipate that is a poor best of names] Sorry for the continued continued list, but it is MUCH bigger to do this amiss once, back abounding of these account chase through in abounding abounding languages, and they absolutely will aftereffect every one of you C programs. ShowMe – 8,8 as an Improper fraction to simplest form | simplest form improper fractions One Checklist That You Should Keep In Mind Before Attending Simplest Form Improper Fractions | Simplest Form Improper Fractions – simplest form improper fractions | Encouraged to my personal website, on this period I am going to teach you with regards to keyword. And after this, this is actually the first photograph: Equations with Mixed Numbers / Improper Fractions – math … | simplest form improper fractions What about photograph preceding? is actually that will awesome???. if you feel so, I’l m explain to you several photograph once again below:

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# Python Operators In this tutorial, you will learn about Python operators with the help of examples. Operators are symbols that are used to perform operations on values and variables. For example, ``print (6 + 5) # 11`` Here, `+` is an operator that adds 6 and 5. ## Types of Python Operators The different types of operators used in Python are: • Arithmetic Operators • Comparison Operators • Logical Operators • Bitwise Operators • Special Operators Let's learn about them in detail. ## Arithmetic Operators Arithmetic operators perform arithmetic operations like addition, multiplication, subtraction etc. List of Arithmetic operators: ``````Addition: + Subtraction: - Multiplication: * Division: / Floor division: // Remainder: % Exponent: **`````` ### Example: Python Arithmetic Operator ``````x = 7 result = x + 5 # subtraction result = x - 5 print('Result from subtraction:', result) # multiplication result = x * 10 print('Result from multiplication:', result) # division result = x / 10 print('Result from division:', result) # floor division quotient = x // 2 print('Quotient:', quotient) # modulus division remainder = x % 2 print('Remainder:', remainder) # exponent result = x ** 2 print('Exponent:', result)`````` Output ```Result from addition: 12 Result from subtraction: 2 Result from multiplication: 70 Result from division: 0.7 Quotient: 3 Remainder: 1 Exponent: 49``` ## Comparison Operators Comparison operators compare values between operands and return `True` or `False`. For example, `>`, `<`, `==` etc. Some of the comparison operators: Operator Example Meaning > x > y x greater than y < x < y x less than y == x == y x is equal to y != x != y x is not equal to y >= x >= y x is greater than or equal to y <= x <= y x is less than or equal to y ### Example: Python Comparison Operator `````` x = 5 y = 6 # greater than print('x is greater than y:', x > y) # less than print('x is less than y:', x `````` Output ``` x is greater than y: False x is less than y: True x is less than or equal to y: True ``` ## Logical Operators Logical Operators perform logical operations and returns `True` or `False`. Some of the logical operators are: • and - returns `True` if both operands are true • or - returns `True` if one of the operands is true • not - returns `True` if both operands are false ### Example: Python Logical Operator ``````x = False y = True # and operation result = x and y print('x and y is', x and y) # or operation result = x or y print('x or y is',x or y) # not operation result = not x print('not x is',not x)`````` Output ```x and y is False x or y is True not x is True``` ## Bitwise Operators Bitwise Operators perform bitwise calculations on integers. It simply performs bit by bit comparison. Some of the bitwise operators are: Operator Example Meaning & x & y Bitwise AND | x | y Bitwise OR << x << y Bitwise left shift >> x >> y Bitwise right shift ~ ~x Bitwise NOT ^ x ^ y Bitwise XOR(exclusive OR) ### Example: Python Bitwise Operator ``````x = 5 # 5 = 0000 0101 in binary y = 6 # 6 = 0000 0110 in binary # Bitwise AND print(x & y) # 4 # Bitwise OR print(x | y) # 7 # Bitwise NOT print(~x) # -6 # Bitwise left shift print(x > y) # 0 # Bitwise XOR print(x ^ y) # 3`````` ## Special Operators In Python, there are special types of operators like: • identity operator • membership operator ### Identity Operators Identity operators are used to check if two values are located on the same part of the memory. The two identity operators are: • `is` - returns `True` if both variables are of same object • `is not` - returns `True` if both variables are not the same object ### Example: Python Identity Operators ``````# integers x1 = 7 y1 = 7 # strings x2 = 'Python' y2 = 'Python' # lists x3 = [5,6,7] y3 = [5,6,7] print(x1 is y1) # True print(x2 is not y2) # False print(x3 is y3) # False`````` Here, • `x1` and `y1` are both equal and identical integers. So, `x1 is y1` returns `True` • `x2` and `y2` are both equal and identical strings. So, `x2 is not y2` returns `False` • `x3` and `y3` are both equal but not identical lists. So, `x3 is y3` returns `False` Note: The interpreter locates lists separately in memory. So they are not identical. ### Membership Operators Membership operators check whether the given value is in the sequence(list, string, set, tuple and dictionary). Membership operators are: • `in` - returns `True` if value is in the sequence • `not in` - returns `False` if value is not in the sequence ### Example: Python Membership Operator ``````# membership operator with string x = 'Python' print('P' in x) # True print('P' not in x) # False # membership operator with dictionaries y = {1:'a',2:'b'} print(1 in y) # True # membership operator with lists z = ['a', 'b'] print('a' in z) # True # membership operator with sets s = {5, 6, 7} print(6 not in s) # False`````` Here, in the dictionary, the membership operator checks whether the key(not the value) is found or not.

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# Moment of Inertia 1. Feb 14, 2014 ### GeneralOJB I read that for a rotating body the kinetic energy $E_k = \sum \frac{1}{2}mv^2 = \frac{1}{2}{\omega}^2∑mr^2 = \frac{1}{2}I{\omega}^2$ where $I$ is the moment of inertia. If we did the same thing for momentum then $P = ∑mv = \omega\sum mr$ So why is angular momentum $I\omega=\omega\sum mr^2$? Shouldn't the momentum just be the sum of the momentum of all the particles, like we did with kinetic energy? Also why should I believe that this quantity $I\omega$ is conserved? Last edited: Feb 14, 2014 2. Feb 14, 2014 Use LaTeX $E_k = \sum\frac{1}{2}mv^2 = \frac{1}{2}\sum m(\omega r)^2 = \frac{1}{2}\omega^2\sum mr^2 = \frac{1}{2}I\omega^2$ 3. Feb 14, 2014 ### GeneralOJB Ah, didn't know we had LaTeX. 4. Feb 14, 2014 ### A.T. That would be the total linear momentum, not the total angular momentum. Note that linear and rotational kinetic energy are both of the same physical scalar quantity. While linear and angular momentum are two different vector quantities. You should look at the vector formulas for momentum to understand it better.

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• :00Days • :00Hours • :00Mins • 00Seconds A new era for learning is coming soon Suggested languages for you: Americas Europe Q1E Expert-verified Found in: Page 180 ### Fundamentals Of Differential Equations And Boundary Value Problems Book edition 9th Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider Pages 616 pages ISBN 9780321977069 # Decide whether or not the method of undetermined coefficients can be applied to find a particular solution to the given equation. ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{t}}}^{-1}{{\mathbf{e}}}^{t}$ No, the method of undetermined coefficients can’t be applied. See the step by step solution ## Step 1: Use the method of undetermined coefficients to find a particular solution to the given differential equation. Given equation, $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}-\mathrm{y}={\mathrm{t}}^{-1}{\mathrm{e}}^{\mathrm{t}}$ Here, the given differential equation is non-homogeneous. According to the method of undetermined coefficients, The method of undetermined coefficients applies only to non-homogeneities that are polynomials, exponentials, sine, or cosine, or products of these functions. ## Step 2: Final conclusion. The R.H.S. of the equation ${\mathbf{t}}^{-1}{\mathbf{e}}^{t}$ is not in the form of polynomials, exponentials, sine or cosine, or the product of these t functions. From the definition of a polynomial, it should have only non-negative integers as powers. And we know that, ${\mathbf{t}}^{\mathbf{-}\mathbf{1}}$ is not a polynomial. Therefore, it is not a product of a polynomial and an exponential function. So, the method of undetermined coefficients cannot be applied.

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# Questions tagged [fft] The fast Fourier transform is an efficient algorithm to compute the discrete Fourier transform (DFT) and its inverse. 3,176 questions Filter by Sorted by Tagged with 1 vote 57 views ### What Are the Units of a Fourier Transform? I know that a Fourier transform (FT) outputs a distribution of how much of each frequency is present in a signal, but what would be the units of that distribution? It's not a proportion or a percent ... 27 views ### Is it possible to reconstruct the continuous frequency spectra from DFT/FFT? Given the output of some FFT (lets call it X) with N bins (lets say 2048) is it possible to recover the values in between the bins in the same way that I can recover the values in between samples of a ... 1 vote 75 views ### Excel FFT of Two Sine Waves Not Giving Expected Results I did an FFT in Excel on a 440 Hz time domain sine wave, $y_1(t) = \sin(2\pi\cdot440\cdot t)$, and got the expected output in the frequency domain: a single-spike distribution with 100% of the signal ... 8 views ### Pixel Shift when using Fourier Interpolation I am working on a Super-Resolution network, and want to try to use Fourier interpolation, different from the normally used bicubic interpolation as it relates to MRI imaging. I have tried using this ... 1k views ### Why doesnt DFT Padding cause sinc like features I'm new to the land of DSP so any incorrect terms please let me know. It seems padding the time domain signal can make the magnitude spectrum look 'nicer', the fact it doesn't gain any more useful ... 62 views ### Peak FFT Frequency Off By One Bin I used Excel's FFT feature to compute an FFT of a 440Hz sine wave that I also generated with Excel. The peak FFT value should occur at 440 Hz. Instead, it occurred one bin later at 528 Hz, so I'm ... 39 views ### Min. sample duration for FFT of low frequent signals obtained from zeroIF receiver I am new here and a beginner with SDR & FFT. For antenna pattern measurements, a flying probe antenna is to be used to sample the electromagnetic near-field in front of an antenna. A lightweight ... 1 vote 37 views ### Discussion on the relationship between the FFT magnitude spectrum and the corresponding actual amplitude This might be a bit long, but it's a question that has troubled me for a long time, and I hope to get everyone's help. When studying signals and systems, we usually only pay attention to the relative ... 58 views ### Jitter model does not behave as expected after FFT So I'm currently trying to understand the relationship between jitter and phase noise. To help my understanding I did write a little simulation in matlab. I used this model in the time domain, relying ... 72 views ### What's Nyquist frequency in DFT? I'm currently studying digital signal processing at university but I can't figure out what the Nyquist frequency means in the DFT coefficients, I know what the Nyquist frequency is in the sampling ... 515 views ### pocketfft delivers wrong values does anyone understand how to use the pocketfft by martin reinecke? Link: https://gitlab.mpcdf.mpg.de/mtr/pocketfft Basically it's just this snipped of code: ... 42 views ### Possible to combine and/or compare a swept Sine and a wide band noise signal? I am working on a problem where two profiles need to be compared. One is a swept sine in $m/s^2$ (Acceleration) and the other is a sine-on-random signal composed of a swept sine $m/s^2$ and a random ... 463 views ### FFT of channel impulse response in python I have this channel impulse response. $h = [1,\,0.5,\,0.3+0.3j]$ The channel is like this by doing a FFT for h. I resample (up-sample) my OFDM symbol to have smaller path delays and convolve it with ... 28 views ### Can a bad quality sound be measured via THD? I'm trying to create an app that measures the audio quality, this is ideally to filter out blown out or cracking speaker who has different types of distortions, what would be the best measurement ... 131 views ### How to find frequency , location and phase of two sinusoid signal buried in noise? There is a data set which has two different sinusoids buried in noise. Data record has following properties: A data records with a total data length of 2048 points. The sampling frequency is 100 Hz. ... 76 views ### FFT Window question - high resolution windows So I am pretty familiar with FFTs and FFT windows. I know that in general there is a tradeoff between mainlobe width and sidelobe levels. However, I recently stumbled across this discussion, where ... 1 vote 45 views 7k views ### Intuition for sidelobes in FFT I was wondering if there's a intuitive way to understand why sidelobes appear when performing an FFT on a signal of fixed length? 44 views ### Is the noise at different sampling points related? [closed] Given the power spectral density of noise, if the values at different sampling points of the same noise are added, what is the power spectral density at this time? Is there any correlation between the ... 1 vote 53 views ### How am I supposed to acquire/measure vertical and horizontal effective resolution from my images? I apologize first for some questions that might be dumb. Unfortunately, I am only a bachelor student (writing my bachelor thesis in physics) and have almost zero image processing experience. And I ... 78 views ### Influence of phase discontinuity on power spectrum density I was wondering about the influence of phase discontinuities on a power spectrum density calculated with f.e. Welch's method. I know you shouldn't have any phase discontinuities in your signal, but I'... 1 vote 430 views ### Is there a FFT algorithm which is both in-place and constant geometry? I work in crypto where the NTT (Number Theoretical Transform) which is just a FFT in finite fields is used. I wanted to know if there exists an implementation of the FFT which is both in-place and ... 181 views ### Using the log power spectrum as features for a Classifier I have a set of time series signals for which I have to develop a anomaly detection algorithm. I am considering using classifiers like SVM to do this. However I am confused about how to properly ... 75 views ### Noise at Corners of MATLAB's ifft2 I'm trying to implement the code in this tutorial in MATLAB using MATLAB's ifft2() function instead of the algorithm used in the shader (looks to be some type of Radix-2 algorithm). It is used to ... 1 vote 64 views ### Debug the difference between two Frft implementations I need help to understand the difference between two fractional Fourier transform implementations. On this website two different implementations for the fractional Fourier transform are presented (...

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# Fibonacci generator and gives the number back as output • June 19th, 2009, 07:59 PM aurora Fibonacci generator and gives the number back as output I have a program that i need to complete. I wrote the first one (FibonacciGenerator) with stated variables - everything worked fine I now want to split it up so that the user can give a number and the program gives the correct number back. (ie, fib(5)=5 AND NOT fib(5)=2) Please help as I know what i need to do but the how escapes me. constructor portion Code : ``` /* Class used to generate a Fibonacci number with a given input */ public class FibonacciGenerator { /** Construct a FibonacciGenerator object to generate a Fibonacci number */ public FibonacciGenerator() { initialize(); }   /** Initializes the FibonacciGenerator object. Allows the object to be reset and reused. */   public void initialize() { fold1 = 1; fold2 = 1; fnew = 1; count = 0; } /** Method used to calculate a fibonacci number @return fnew the fibonacci number */ public int nextNumber() { if(count < 2) { count++; return 1; } else { fnew = fold1 + fold2;   fold2 = fold1; fold1 = fnew;   return fnew; } }   private int fold1; private int fold2; private int count; private int fnew; }``` Test Program (I know i need to return the input to the constructor file but how??) Code : ``` import javax.swing.JOptionPane;   /** Test driver class for FibonacciGenerator class */ public class FibonacciGeneratorTester { public static void main(String[] args) { String input = JOptionPane.showInputDialog( "Enter a Number or hit ''Cancel'' to exit:"); int n = Integer.parseInt(input);   FibonacciGenerator fg = new FibonacciGenerator();   int next = n;   if (n < 2) next = 1; else { for (int i = 3; i <= n; i++) { next = fg.nextNumber(); } }   System.out.println("fib(" + n + ") = " + next);   System.exit(0); } }``` Any help is appreciated • June 20th, 2009, 12:58 AM Freaky Chris Re: help with FibonaccciGenerator Welcome to Java Programming Forums, Aurora. Have a read of this, a previous topic about the fibonacci sequence see if it helps you out. If it doesn't then get back to us and I'll have a proper look over your code. http://www.javaprogrammingforums.com...-sequance.html Regards, Chris

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# Find the Value of K for Which the Roots of the Equation 3x2 - 10x + K = 0 Are Reciprocal of Each Other. - Mathematics Sum Find the value of k for which the roots of the equation 3x2 - 10x + k = 0 are reciprocal of each other. #### Solution The given equation is 3"x"^2 - 10"x" + "k" = 0 Roots of the given equation are reciprocal of each other. Let α and 1/alpha be the roots of the given equation. Product of roots = c/alpha ⇒ α . (1)/α = "k"/(3) ⇒ "k"/(3) = 1 ⇒ "k" = 3 Concept: Nature of Roots Is there an error in this question or solution?

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# physics A positively charged glass rod is brought close to a neutral sphere that is supported on a nonconducting plastic stand. Sketch the distribution of charges on the sphere if it is made of: a) Aluminum b) Non-conducting plastic 1. 👍 1 2. 👎 0 3. 👁 1,204 1. a)Aluminum: the electrons in the sphere would rise towards the positively charged glass rod making the top part of the sphere negative and the bottom part positive b) Non conducting matierial: nothing would happen the sphere would still be neutral 1. 👍 1 2. 👎 0 ## Similar Questions 1. ### PHYSICS A rod 14.0 cm long is uniformly charged and has a total charge of -20.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 36.0 cm from its center. N/C toward the rod away from 2. ### Physics In which case would there be an electrostatic force between two objects? two positively charged objects**** two neutral objects a positively charged object and a neutral object a negatively charged object and a neutral object 3. ### Physics Neutral metal sphere A, of mass 0.10kg, hangs from an insulating wire 2.0m long. An identical metal sphere B, with charge -q is brought into contact with sphere A. Sphere A goes 12 degrees away from Sphere B. Calculate the initial 4. ### science IF WE KEEP A GLASS ROD IN HOT WATER WE CAN STILL HOLD THE GLASS ROD WITH OUR HAND.IT SHOWS THAT GLASS IS A GOOD CODUCTOR OF HEAT. WHETHER TRUE OR FALSE 1. ### Physics (Newton's and Coulomb's Laws) **ASAP** 1. Which law gives the force between two objects that is related to their mass and distance? A. Newton's law of gravitation B. Coulomb's law C. Kepler's law D. Newton's second law of motion 2. What is the property that allows a 2. ### Physical Science Consider three objects: A, B, and C. If A is positively charged, B repels A, and C is attracted by A, what can you say about the charge on B and C? A. B is positively charged and C is negatively charged. B. B is negatively charged 3. ### Physics There are two conducting concentric hollow spheres of outer radii and ( ). The thickness of the material of both spheres is . The inner sphere is negatively charged with charge density . The larger sphere is positively charged 4. ### physics Two similar sphere having charge -2C and +4C respectively are separated by distance "r" exert force (F=16N) third similar neutral sphere is first touched with sphere B and then sphere A. then force between spheres decreases by Two small insulating spheres with radius 7.00×10^−2m are separated by a large center-to-center distance of 0.565m . One sphere is negatively charged, with net charge -2.05μC , and the other sphere is positively charged, with 2. ### physics A negatively charged balloon sticks to a wooden door. However, an uncharged balloon does not stick to a wooden door. What is the nature of the charge on the wooden door? A. Electrically neutral B. Positively charged C. Conductor 3. ### physics An object is formed by attaching a uniform, thin rod with a mass of mr = 6.54 kg and length L = 5.12 m to a uniform sphere with mass ms = 32.7 kg and radius R = 1.28 m. Note ms = 5mr and L = 4R. 1)What is the moment of inertia of 4. ### Natural science What will happen if a positively charged glass rod is brought near another positively charged glass rod that is suspended by a silk thread

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Monday May 2, 2016 # Homework Help: math Posted by jamie on Thursday, January 31, 2013 at 6:23pm. the product of three consecutive integers is 22 less than the cube of the middle integer. Find the integer. • math - Reiny, Thursday, January 31, 2013 at 7:08pm let the 3 consecutive integers be x-1, x, and x+1 ( I wanted to cube x , not x+1 ) x(x-1)(x+1) = x^3 - 22 x^3 - x = x^3 - 22 x = 22 the 3 integers are 21 , 22, and 23 check: product of 3 numbers = 21x22x23 = 10626 cube of 22 = 10648 is the product 22 less than the cube? YES

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# What is 3 2 1 12 ## The sum of all natural numbers One of the most remarkable equations in science is that the sum of all natural numbers - the sum of 1, 2, 3, and so on to infinity - is assigned the value -1/12. This is not a joke and even plays a role in physics. But how can the equation be correct? The two mathematicians in the worth seeing clip from Brady Haran's Numberphile series demonstrate so vividly how one arrives at this result that one can hardly resist their powers of persuasion. And yet: Quite a few other mathematicians have their hair on end. So what is it about when you dig a little deeper than the video? The sum of 1, 2, 3 to infinity is a divergent series: With each addend it gets bigger and bigger, and there is never a shortage of further addends. As soon as such a series appears in a calculation, it prevents mathematicians and physicists from calculating further. In the 18th and 19th centuries, respectively, Leonhard Euler and Bernhard Riemann tackled the problem of divergent series through the trick of "analytical continuation". They cleverly defined new series that are convergent for some values โ€‹โ€‹- that is, despite an infinite number of summands, they do not become infinitely large, but rather reach a defined limit value. For other values, on the other hand, their terms are just the terms of the original divergent sum. One of these new series is defined as the sum 1 + 1/2s + 1 / 3s + ..., where s is a so-called complex number. For certain values โ€‹โ€‹of s it is identical to the (otherwise remarkable) Riemann zeta function (this is how the zeta function is even defined). The highlight: This function also has a clearly defined value there - namely at s = -1 - where the associated series is identical to the sum of 1, 2, 3 ... Namely the value -1/12. The result continues to contradict our common sense. But we don't need to be impressed by this: Even mathematicians are worried about divergent series - infinite numbers are just hard to grasp.

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# 最小公倍数 ## 与最大公因数之关系 ${\displaystyle \operatorname {lcm} (a,b)={\frac {|a\cdot b|}{\operatorname {gcd} (a,b)}}}$ ## 计算方法 ${\displaystyle 216=2^{3}\times 3^{3}}$${\displaystyle 384=2^{7}\times 3^{1}}$${\displaystyle 210=2^{1}\times 3^{1}\times 5^{1}\times 7^{1}}$ ${\displaystyle [216,384,210]=2^{7}\times 3^{3}\times 5^{1}\times 7^{1}=120960}$ ### 递归计算多个整数的最小公倍数 ${\displaystyle a_{1},a_{2},a_{3}}$ 的质因数分解分别为${\displaystyle \prod _{i=1}^{n}p_{i}^{e_{1i}},\prod _{i=1}^{n}p_{i}^{e_{2i}},\prod _{i=1}^{n}p_{i}^{e_{3i}}}$，其中 ${\displaystyle p_{i}}$ 是第 ${\displaystyle i}$ 个质数。 ${\displaystyle \operatorname {lcm} (\operatorname {lcm} (a_{1},a_{2}),a_{3})=\operatorname {lcm} (\prod _{i=1}^{n}p_{i}^{\max(e_{1i},e_{2i})},a_{3})=\prod _{i=1}^{n}p_{i}^{\max(\max(e_{1i},e_{2i}),e_{3i})}=\prod _{i=1}^{n}p_{i}^{\max(e_{1i},e_{2i},e_{3i})}}$ ## 程式代码 ### C# int GCD(int a, int b) { return a % b == 0 ? b : GCD(b, a % b); } int LCM(int a, int b) { return a * b / GCD(a, b); } ### C int GCD(int a, int b) { if(b) while((a %= b) && (b %= a)); return a + b; } int LCM(int a, int b) { return a * b / GCD(a, b); } ### C++ template < typename T > T GCD(T a, T b) { if(b) while((a %= b) && (b %= a)); return a + b; } template < typename T > T LCM(T a, T b) { return a * b / GCD(a, b); } ### PASCAL 1、 var a,b,ans:integer; function gcd(a,b:integer):longint; begin if b=0 then gcd:=a else gcd:=gcd(b,a mod b ) ; end; 2、 var a,b,ans:integer; function lcm(a,b:integer):longint; begin ans:=(a*b) div gcd(a,b); write(ans); end; ### JAVA int GCD(int a, int b) { return a % b == 0 ? b : GCD(b, a % b); } int LCM(int a, int b) { return a * b / GCD(a, b); } ### RUBY def gcd(a, b) b.zero? ? a : gcd(b, a % b) end def lcm(a, b) a * b / gcd(a, b) end ### Python def gcd(a, b): return a if b == 0 else gcd(b, a % b) def lcm(a, b): return a * b / gcd(a, b) ### Golang func GDC(a, b int) int { if a%b == 0 { return b } return GDC(b, a%b) } func LCM(a, b int) int { return a * b / GDC(a, b) } ## 应用 ${\displaystyle {2 \over 21}+{1 \over 6}={4 \over 42}+{7 \over 42}={11 \over 42}}$ ## 参考来源 • 柯召，孙绮，孙琦. 《数论讲义》. 高等教育出版社. 2005. ISBN 753205473X. • 阿尔伯特﹒H﹒贝勒著 谈祥柏译. 《数论妙趣：数学女王的盛情款待》. 上海教育出版社. 1998. ISBN 7040091909.

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Variable Cells Final Reduced Objective Allowable Cell Name Value Cost # Variable cells final reduced objective allowable cell This preview shows page 3 - 8 out of 8 pages. Variable Cells Final Reduced Objective Allowable Cell Name Value Cost Coefficient Increase \$C\$16 X1 0.0994475138 0 0.75 0.1875 \$D\$16 X2 0.2127071823 0 0.9 1.00000E+030 \$E\$16 X3 0.0883977901 0 0.8 0.1166666667 \$F\$16 X4 0.0994475138 0 0.7 0.0924369748 Constraints Final Shadow Constraint Allowable Cell Name Value Price R.H. Side Increase \$C\$23 A Amount Used (LHS) 0.5 0.2635359116 0.5 0.0989010989 \$C\$24 C Amount Used (LHS) 6 0.0135359116 6 0.7619047619 \$C\$25 B Amount Used (LHS) 5 0.0303867403 5 0.3025210084 \$C\$26 D Amount Used (LHS) 5 0.0082872928 5 2.25 Allowable Decrease 0.1774193548 0.1065217391 0.1153846154 1.00000E+030 Allowable Decrease 0.0248962656 1.6739130435 0.5806451613 1.2307692308 Microsoft Excel 14.4 Limits Report Worksheet: [Workbook18]Sheet1 Report Created: 4/21/2015 11:50:27 AM Objective Cell Name Value \$C\$19 Objective Function (Min) X1 0.41 Variable Lower Objective Upper Objective Cell Name Value Limit Result Limit Result \$C\$16 X1 0.10 0.10 0.41 0.10 0.41 \$D\$16 X2 0.21 0.21 0.41 0.21 0.41 \$E\$16 X3 0.09 0.09 0.41 0.09 0.41 \$F\$16 X4 0.10 0.10 0.41 0.10 0.41 Material X1 X2 X3 X4 A 1 1 1 1 C 9 16 8 10 B 12 10 10 8 D 0 14 15 7 E 0 0 0 0 E 0 0 0 0 Obj. Func. Coefficient 0.75 0.9 0.8 0.7 Decision Variables ( Set values equal to 1) Tons Produced X1 X2 X3 X4 0.10 0.21 0.09 0.10 Objective Function 0.41 Constraints Amount Used (LHS) A 0.50 0.5 C 6.00 >= 6 B 5.00 >= 5 D 5.00 >= 5 E - <= 0 E - <= 0 XE 1.00 <= 0 YA 1.00 <= 0 YB 1.00 <= 0 YC 1.00 <= 0 YD 1.00 <= 0 YE 1.00 <= 0 Material Requirements P2 P3 YB YC YD YE mount Availabl 0 0 0 0 0 0 0.5 0 0 0 0 0 0 6 0 0 0 0 0 0 5 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 P2 P3 YB YC YD YE 1.00 1.00 1.00 1.00 1.00 1.00 le #### You've reached the end of your free preview. Want to read all 8 pages? • Fall '14 • objective function, Report Created, LHS)

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• 0 Vote(s) - 0 Average • 1 • 2 • 3 • 4 • 5 "game" - to calculate 1.00...001 ^^ 0.5 = ? nuninho1980 Fellow Posts: 95 Threads: 6 Joined: Apr 2009   04/17/2009, 12:35 PM (This post was last modified: 04/17/2009, 12:37 PM by nuninho1980.) Hi! I already investigated to know the parts of cases decimais of number. to remember - 1.000000000000000000001^0.5 =~ 1.000000000000000000000499999999999999999999875\ 00000000000000000006249999999999999999996093750\ 00000000000000273437499999999999999794921875000\ 000000000161132812499999999999869079589843750000000109100342 1.00000000000000000001 ^^ 0.5 = ?? (in 100 cases decimal) 1.000000000000000000001 ^^ 0.5 = ?? (in 100 cases decimal) you do calcuate. good luck! but I already got results with almost exact. !!warning: x ^^ 0.5 =/= ssqrt(x). « Next Oldest | Next Newest » Messages In This Thread "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/17/2009, 12:35 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by bo198214 - 04/18/2009, 12:42 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/18/2009, 11:55 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by bo198214 - 04/18/2009, 12:19 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/18/2009, 01:49 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by bo198214 - 04/18/2009, 02:28 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/18/2009, 03:26 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by bo198214 - 04/18/2009, 03:57 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/18/2009, 04:37 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by bo198214 - 04/18/2009, 05:16 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/18/2009, 06:15 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by bo198214 - 04/18/2009, 09:33 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 04/20/2009, 04:13 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/20/2009, 11:59 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 04/21/2009, 06:35 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/21/2009, 07:46 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 04/22/2009, 09:43 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/22/2009, 03:45 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 04/23/2009, 07:04 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 04/23/2009, 10:15 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by bo198214 - 04/20/2009, 12:31 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 05/01/2009, 02:42 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 05/01/2009, 08:34 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 05/01/2009, 11:07 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 05/01/2009, 11:53 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 05/02/2009, 12:44 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 05/02/2009, 01:55 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 05/02/2009, 12:54 PM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by andydude - 05/03/2009, 08:31 AM RE: "game" - to calculate 1.00...001 ^^ 0.5 = ? - by nuninho1980 - 05/03/2009, 09:12 PM Users browsing this thread: 1 Guest(s)

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× Didn't find what you were looking for? Ask a question Top Posters Since Sunday 6 A 5 b 4 3 s 3 s 3 L 3 i 3 b 3 d 3 b 3 j 3 # Probability Distribution wrote... Posts: 264 Rep: 7 months ago Probability Distribution Instructions: a) Identify the type of probability distribution shown in the problem: binomial, hypergeometric, poisson etc. b) Identify the given in the problem. c) Solve for the probability.2) The average number of relief goods your household receives during a pandemic is 3 per month. What is the probability that exactly 4 relief goods will be received by your household next month? Read 61 times 1 Reply ### Related Topics Replies Anonymous wrote... 7 months ago For this, you must use Poisson's model.$$f\left(x\right)=\frac{e^{-\mu }\mu ^x}{x!};\ \ \ x=0,1,2,\ ...$$x = number of success per unit time or space$$\mu$$ = mean number of successesQuoteThe average number of relief goods your household receives during a pandemic is 3 per month. What is the probability that exactly 4 relief goods will be received by your household next month?$$\mu = 3$$x = 4$$f\left(4\right)=\frac{e^{-3}3^4}{4!}=0.168\ \Rightarrow \ 16.8\%$$ Explore 214 People Browsing 188 Signed Up Today Related Images 1912 157 192

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7 lutego 2020 7 lutego 2020 # The following informative article will be to answer that the question, how what is mathematics idea? Let us have a look in this question. Let us specify what it is before we can answer this particular question. It’s is but one of the absolute most confusing words in the English terminology. People should produce distinct definitions and answers. So, to get matters easier, let us stick paramountessays on the fundamentals. It’s defined as a procedure, a procedure of accomplishing an objective through the use of analytic and numerical reasoning. To learn one’s hand-eye coordination, you have to make use of paper and a pencil to do mathematical computations. Since you are able to see, using it as a technique to accomplish a goal. Put simply, it could be translated since the word math into different languages like German and French. Moreover, the term in the definition is an synonym of the mathematics, meaning using math. In conclusion, what is math? Could be the simplest problem to solution should people consider the write an essay online terms and definitions. The fact there are many definitions to your phrase, what is math theory, demonstrates that there is just a broad range of significance to your term. It may indicate many things. To get things less complicated, why don’t we take a look at its definitions and utilize these to define it accurately. The definition of this phrase, what is math concept isn’t much difficult to get. You’d know this is the significance supplied in geometry, if you examine elementary school. It would mean the shortest path between two things if you’re wondering exactly what is a plane. What’s really a world is easily the most frequently encountered definition of the term. If you inquire what is the diameter of a circle, then it would mean the circle’s outer region where its own radius falls on some certain stage. The definition of the sentence, what’s math concept, generally comprises those mathematical theories which could be employed to fix a problem. Cases will be the addition, subtraction, multiplication, https://catalog.gmu.edu/courses/biol/ division, trigonometry, algebra, calculus, and geometry. Some examples are ratios, fractions, graphing, linear equations, exponents, square roots, polynomials, inequalities, and inequalities of the second sort. As we have seen, what’s math concept has many definitions and utilize situations. These definitions usually need some abilities or plausible reasoning. However, there are some people who genuinely believe that there is only one definition for that which exactly is math concept. Those who believe in this idea claim that what is mathematics concept is the definition that is possible. It is said that it is definitely better to consider about exactly everything exactly is math theory as a”rule” in the place of complex and so straightforward method. I’ll also add that this theory just applies to how people as well as it is defined by civilizations that are distinct.

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# nLab category algebra category theory ## Applications #### Algebra higher algebra universal algebra # Contents ## Idea The space of functions $\mathcal{C}_1 \to R$ on the space of morphisms $\mathcal{C}_1$ of a small category $\mathcal{C}_\bullet$ (with coefficients in some ring $R$) naturally inherits a convolution algebra structure from the composition operation on morphisms. This is called the category convolution algebra or just category algebra for short. Often this is considered specifically for groupoids and hence accordingly called groupoid convolution algebra or just groupoid algebra for short. (For one-object delooping groupoids of groups, groupoid algebras reduce to group algebras.) The inversion operation of the groupoid naturally makes its groupoid algebra into a star-algebra (this is generally the case for category algebras of dagger categories) and accordingly groupoid algebras play a role in C-star-algebra theory. More generally, if the groupoid carries a line 2-bundle then (in its incarnation as a bundle gerbe-like transition bundle) the space of morphisms carries a line bundle (satisfying some compatibility conditions) and one can consider convolution algebras not just of functions, but of sections of this line bundle. The resulting algebra is called the twisted groupoid convolution algebra, twisted by the characteristic class of the line 2-bundle (TXLG). For “bare” categories/groupoids (i.e.: internal to Set) these constructions are canonical. But under mild conditions or else when equipped some suitable extra structure, it generalizes to internal categories/internal groupoids in geometric contexts, notably in topology (topological groupoids) differential geometry (Lie groupoids) and algebraic geometry. In such geometric situations a groupoid convolution algebra equipped with its canonical coalgebra structure over the functions on its canonical atlas is also called a Hopf algebroid and may be used to characterize the geometric groupoid. Therefore to some extent one may think of the relation between groupoids/categories and their groupoid/category algebras as an incarnation of the general duality between geometry and algebra. Since category/groupoid algebras are generically non-commutative, this relation identifies groupoids/categories as certain spaces in noncommutative geometry. From this point of view groupoid convolution algebras have been highlighted and developed notably in (Connes 94). Due to this relation the groupoid convolution product is also referred to as a star product and denoted ”$\star$”. Groupoid C*-algebras form a rich sub-class of all C*-algebras, including crossed product C*-algebras, Cuntz algebras. Groupoid convolution algebras may also be understood as generalizations of matrix algebras, to which they reduce for the case of the pair groupoid. In (Connes 94, 1.1) it was famously argued that when Werner Heisenberg (re-)discovered (infinite-dimensional) matrix algebras as algebras of observables in quantum mechanics, conceptually he rather considered groupoid convolution algebras. This perspective has since been fully developed: in (EH 06) strict deformation quantization is given fairly generally by twisted groupoid convolution algebras. See at geometric quantization of symplectic groupoids for more on this. For the discrete but higher geometry of infinity-Dijkgraaf-Witten theory quantization by higher groupoid convolution algebras is indicated in (FHLT 09). ## Definition ### For bare categories (discrete geometry) ###### Definition Let $\mathcal{C}$ be a small category and let $R$ be a ring. The category algebra or convolution algebra $R[\mathcal{C}]$ of $\mathcal{C}$ over $R$ is the $R$-algebra • whose underlying $R$-module is the free module $R[\mathcal{C}_1]$ over the set of morphisms of $\mathcal{C}$; • whose product operation is defined on basis-elements $f,g \in \mathcal{C}_1 \hookrightarrow R[\mathcal{C}]$ to be their composition if they are composable and zero otherwise: $f \cdot g := \left\lbrace \array{ g \circ f & if\;composable \\ 0 & otherwise } \right. \,.$ ###### Remark We may identify elements in $R[\mathcal{C}_1]$ with functions $\mathcal{C}_1 \to R$ with the property that they are non-vanishing only on finitely many elements. Under this identification for $\phi_1, \phi_2$ two such functions, their product in $R[\mathcal{C}]$ is given by the formula $\phi_1 \cdot \phi_2 \; \colon \; f \mapsto \sum_{f_2 \circ f_1 = f} \phi_2(f_2) \cdot \phi_1(f_1) \,,$ where $f,f_1,f_2 \in \mathcal{C}_1$. In particular if $\mathcal{C}$ is a groupoid so that every morphisms $f$ has an inverse $f^{-1}$ then this is equivalently $\phi_1 \cdot \phi_2 \; \colon \; f \mapsto = \sum_{g \in \mathcal{C}_1} \phi_2(f \circ g^{-1}) \cdot \phi_1(g) \,.$ This expresses convolution of functions. ###### Remark If the small category $\mathcal{C}_\bullet$ is a groupoid, hence equipped with an inversion map $inv \colon \mathcal{C}_1 \to \mathcal{C}_1$ then pullback of functions along this map makes is an involution of the convolution algebra of $\mathcal{C}$ and hence makes it into a star-algebra. More generally for $\mathcal{C}$ equipped with the structure of a dagger-category, pullback along the dagger-fuctor $\dagger \colon \mathcal{C}_1 \to \mathcal{C}_1$ makes the convolution algebra a star-algebra. ### For Lie groupoids ###### Remark If $\mathcal{C}$ is a groupoid with extra geometric structure, then there are natural variants of the above definition. Notably if $\mathcal{C}$ is a Lie groupoid then there is a variant where the functions in remark 1 are taken to be smooth functions and where the convolution sum is replaced by an integration. In order for this to make sense one needs to consider in fact functions with values in half-densities over the manifold $\mathcal{C}_1$. More generally, for a bundle gerbe over a Lie groupoid $\mathcal{C}$, hence a multiplicative line bundle over $\mathcal{C}_1$, one can consider a convolution product on sections of this line bundle tensored with half-densities. ###### Definition Let $\mathcal{G}_\bullet$ be a Lie groupoid. Write $C^\infty_c(\mathcal{G}_1, \sqrt{\Omega})$ for the space of smooth half-densities in $T_{d s = 0}\Gamma_1 \oplus T_{d t = 0}\Gamma_1$ of compact support on its manifold $\mathcal{G}_1$ of morphisms. Let the convolution product $\star \;\colon\; C_c(\mathcal{G}_1, \sqrt{\Omega}) \times C_c(\mathcal{G}_1, \sqrt{\Omega}) \to C_c(\mathcal{G}_1, \sqrt{\Omega})$ be given on elements $f,g \in C_c(\mathcal{G}_1, \sqrt{\Omega})$ over any element $\gamma \in \mathcal{G}_1$ by the integral $(f \star g) \colon \gamma \mapsto \int_{\gamma_2\circ \gamma_1 = \gamma} f(\gamma_1) \cdot f(\gamma_2) \,.$ (Here we regard the integrand naturally as taking values in actual densities tensored with the pullback of $\sqrt{Omega}$ along the composition map. This defines the integration of density-factor which then takes values in $\sqrt{\Omega}$.) The algebra $(C_c^\infty(\mathcal{G}_1, \sqrt{\Omega}), \star)$ is the groupoid convolution algebra of smooth compactly supported functions. As in remark 1, this is naturally a star-algebra with involution $inv^\ast$. This construction originates around (Connes 82). ###### Proposition/Definition For $\mathcal{G}_\bullet$ a Lie groupoid and for $x \in \mathcal{G}_0$ any point in the manifold of objects there is an involutive representation $\pi_x$ of the convolution algebra $(C_c^\infty(\mathcal{G}_1, \sqrt{\Omega}), \star, inv^\ast)$ of def. 3 on the canonical Hilbert space of half-densities $L^2(s^{-1}(x))$ of the source fiber of $x$ given on any $\xi \in L^2(s^{-1}(x))$ by $\pi_x(f) \xi \colon \gamma \mapsto \int_{\gamma_1 \in t^{-1}(\gamma)} f(\gamma_1) \xi(\gamma_1^{-1}\gamma) \,.$ This defines a norm $|{\Vert \Vert}$ on the vector space $C_c^\infty(\mathcal{G}_1, \Omega)$ given by the supremum of the norms in $L^2(s^{-1}(x))$ over all points $x$: ${\Vert f\Vert} \coloneqq Sup_{x \in \mathcal{G}_0} {\Vert \pi_x(f)\Vert} \,.$ The Cauchy completion of the star algebra $(C_c^\infty(\mathcal{G}_1, \sqrt{\Omega}), \star, inv^\ast)$ with respect to this norm is a C-star-algebra, the convolution $C^\ast$-algebra of the Lie groupoid $\mathcal{G}_\bullet$. This is recalled at (Connes 94, prop. 3 on p. 106). ###### Remark With suitable definitions, this construction constitutes something at least close to a 2-functor from differentiable stacks to C-star-algebras and Hilbert bimodules between them: In (Mrčun 99) the convolution algebra construction for étale Lie groupoids is extended to groupoid bibundles and shown to produce a functor to C-star-algebras with (isomorphism classes of) bimodules between them. In (Kališnik-Mrčun 07) it is shown that if one remembers the additional Hopf algebroid structure on the convolution $C^\ast$-alegras (the algebraic analog of remebering the atlas of the differentiable stack of a Lie groupoid) then this construction becomes a full subcategory inclusion of étale Lie groupoids into their convolution Hopf algebroids. In (Muhly-Renault-Williams 87, Landsman 00) the generalization of the construction of a $C^\ast$-bimodule from a groupoid bibundle to general Lie groupoids is discussed (not necessarily étale), but only equivalence-bibundles are considered and are shown to yield Morita equivalence bimodules (no discussion of composition and functoriality here). ## Equivalent characterizations We discuss equivalent characterizations of category algebras/groupoid algebras that are useful in certain context ### As a weak colimit over a constant $2Vect$-valued functor Apparently for $\mathcal{C}$ a groupoid the category algebra of $C$ is the weak colimit over $\mathcal{C}$ of the functor $\mathcal{C} \to Vect\text{-}Mod$ constant on the ground field algebra. This statement is for instance in (FHLT, section 8.4). The 2-cell in the universal co-cone corresponding to the morphism $f \in C$ is the $k\text{-}k[C]$-bimodule homomorphism $f \cdot (-) : A \to A$ that multiplies by $f \in k[C]$ from the left. $\array{ x &&\stackrel{f}{\to}&& y \\ k &&\stackrel{k}{\to}&& k \\ & {k[C]}_{\mathllap{}}\searrow &\swArrow_{f \cdot (-)}& \swarrow_{\mathrlap{k[C]}} \\ && k[C] }$ This description should be compared with the analogous description of the action groupoid by a weak colimit. One sees that the groupoid algebra is a linear incarnation of the action groupoid in some sense. ### In terms of composition of spans The category algebra of a category $C$ is a special case of a general construction of spans (see also at bi-brane). In order not to get distracted by inessential technicalities, consider the case of a finite category $C$, i.e. an internal category in FinSet. This is a span $\array{ && C_1 \\ & {}^{s}\swarrow && \searrow^{t} \\ C_0 &&&& C_0 }$ equipped with a composition operation: a morphism of spans from the composite span $\array{ &&&& C_1 \times_{t,s} C_1 \\ &&& \swarrow && \searrow \\ && C_1 &&&& C_1 \\ & {}^{s}\swarrow && \searrow^{t} && {}^{s}\swarrow && \searrow^{t} \\ C_0 &&&& C_0 &&&& C_0 }$ to the original one, i.e. a morphism $comp : C_1 \times_{t,s} C_1 \to C_1$ which respects source and target morphisms. Given this, consider the trivial vector bundle on the set of objects $C_0$. This is nothing but an assignment $I : C_0 \to Vect_k$ of the ground field $k$ to each element of $C_0$. There are two different ways to pull this vector bundle on objects back to a vector bundle on morphisms, once along the source, once along the target map. Then notice that the set of natural transformations between these two vector bundles $Hom_{[Sets,Vect_k]}(s^* I , t^* I)$ whose elements are 2-arrows of the form $\array{ && C_1 \\ & {}^{s}\swarrow && \searrow^{t} \\ C_0 &&\stackrel{f}{\Rightarrow}&& C_0 \\ & {}_I \searrow && \swarrow_I \\ && Vect_k }$ are canonically in bijection with $k$-calued functions on $C_1$, hence with the vector space spanned by $C_1$, hence with the vector space underlying the category algebra $Hom_{[Sets,Vect_k]}(s^* I , t^* I) \simeq k[C] \,.$ The algebra structure on $k[C]$ is similarly encoded in the diagrammatics: given two elements $\array{ && C_1 \\ & {}^{s}\swarrow && \searrow^{t} \\ C_0 &&\stackrel{f}{\Rightarrow}&& C_0 \\ & {}_I \searrow && \swarrow_I \\ && Vect_k } \;\;\;\; and \;\;\;\; \array{ && C_1 \\ & {}^{s}\swarrow && \searrow^{t} \\ C_0 &&\stackrel{g}{\Rightarrow}&& C_0 \\ & {}_I \searrow && \swarrow_I \\ && Vect_k }$ their pre-composite is the diagram $\array{ &&&& C_1 \times_{t,s} C_1 \\ &&& \swarrow && \searrow \\ && C_1 &&&& C_1 \\ & {}^{s}\swarrow && \searrow^{t} && {}^{s}\swarrow && \searrow^{t} \\ C_0 &&\stackrel{f}{\Rightarrow}&& C_0 &&\stackrel{g}{\Rightarrow}&& C_0 \\ & \searrow &&& \downarrow &&& \swarrow \\ && \to &&Vect_k&& \leftarrow } \,.$ This is a composite transformation between three trivial vector bundles on the set $C_1 \times_{t,s} C_1$ of composable morphisms in $C$. As such, it is a function, which on the element consisting of the composable pair $\stackrel{r}{\to}\stackrel{s}{\to}$ takes the value $f(r)\cdot g(s)$. In order to get back a transformation between vector bundles on $C_1$, hence a transformation between vector bundles on $C_1$, we push forward along the composition map $comp: C_1 \times_{t,s} C_1 \to C_1$. This just means that we add up the values on the fibers of this map. The result is the convolution product $(f\star g) : t \mapsto \sum_{s\circ r = t} f(r)\cdot g(s) \,.$ This is indeed the product in the category algebra. Looking at category algebras realizes them as a puny special case of a bigger story which involves bi-branes as morphisms between $n$-bundles/$(n-1)$-gerbes which live on spaces connected by correspondence spaces. This is related to a bunch of things, such as T-duality, Fourier-Mukai transformations and other issues of quantization. A description of this perspective is in This is related to observations such as described here: ## Examples ### Basic examples ###### Example The convolution algebra of a set/manifold $X$ regarded as a discrete groupoid/Lie groupoid with only identity morphisms is the ordinary function algebra of $X$. ###### Example For $X$ a set the convolution algebra of the pair groupoid $Pair(X)_\bullet$ is the matrix algebra of ${\vert X\vert} \times {\vert X\vert}$ matrices. For $X$ a smooth manifold and $Pair(X)_\bullet$ its pair groupoid regarded as a Lie groupoid its smooth convolution algebra is the algebra of smoothing kernels? on $X$. ###### Example $\mathcal{C} = \mathbf{B}G$ is the delooping groupoid of a discrete group $G$ (the groupoid with a single object and $G$ as its set of morphisms), then def. 1 reduces to that of the group algebra of $G$: $R[\mathbf{B}G] \simeq R[G] \,.$ ### Higher groupoid convolution algebras and n-vector spaces/n-modules under construction We discuss here a natural generalization of the notion of groupoid convolution algebras to higher algebras for higher groupoids. There may be several sensible such generalizations. The one discussed now follows the principle of iterated internalization and naturally matches to the concept of n-modules (n-vector spaces) as they appear in extended prequantum field theory. In order to disentangle conceptual from technical aspects, we first discuss geometrically discrete higher groupoids. The results of this discussion then in particular help to suggest what the right definition of “higher Lie groupoid” in the context of higher convolution algebras should be in the first place. The considerations are based on the following ###### Remark By the discussion at 2-module we may think of the 2-category $k Alg_b$ of $k$-associative algebras and bimodules between them as a model for the 2-category 2Mod of $k$-2-modules that admit a 2-basis (2-vector spaces). Hence the groupoid convolution algebra constructiuon is a 2-functor $C \;\colon\; Grpd \to 2 Mod \,.$ There is then the following systematic refinement of this to higher groupoids and higher algebra: by the discussion at n-module, 3-modules are algebra objects in 2Mod and maps between them are bimodule objects in there. An algebra object in $k Alg_b$ is equivalently a sesquialgebra, an algebra equipped with a second algebra structure up to coherent homotopy, that is exhibited by structure bimodules. Special cases of this are bialgebras, for which these structure bimodules come from actual algebra homomorphisms. Examples of these in turn are Hopf algebras. These we naturally re-discover as special higher groupoid convolution higher algebras in example 4 below. This iterated internalization on the codomain of the groupoid convolution algebra functor has a natural analog on its domain: a 2-groupoid we may present by a double groupoid, namely a groupoid object in an (∞,1)-category in Grpd which is 3-coskeletal as a simplicial object in Grpd. ###### Remark Given a groupoid object $\mathcal{G}_\bullet$ in the (2,1)-topos Grpd hence a double groupoid, applying the groupoid convolution algebra $(2,1)$-functor $C$ to the corresponding simplicial object $\mathcal{G}_\bullet \in Grpd^{\Delta^{op}}$ yields: • groupoid convolution algebras $C(\mathcal{G}_0)$ and $C(\mathcal{G}_1)$, • a $C(\mathcal{G}_1) \otimes_{C(\mathcal{G}_{0,1})} C(\mathcal{G}_1)-C(\mathcal{G}_{0})$-bimodule, assigned to the composition functor $\partial_1 \colon \mathcal{G}_1 \underset{\mathcal{G}_0}{\times} \mathcal{G}_1 \to \mathcal{G}_1$. Under the 2-functoriality of $C$, the Segal conditions satisfied by $\mathcal{G}_\bullet$ make this bimodule exhibi a sesquialgebra structure over $C(\mathcal{G}_{0,1})$. This sesquialgebra we call the the double groupoid convolution 2-algebra of $\mathcal{G}_\bullet$. (Here we make invariant sense of the tensor product by evaluating on a Reedy fibrant representative.) ###### Example Let $G$ be a finite group. Write $\mathbf{B}G$ for its delooping groupoid (the connected groupoid with $\pi_1 = G$). Since this is just a 1-groupoid, there are two natural ways to present $\mathbf{B}G$ as a double groupoid: 1. $\underset{\longrightarrow}{\lim}(\cdots \mathbf{B}G\stackrel{\to}{\stackrel{\to}{\to}} \mathbf{B}G \stackrel{\overset{id}{\to}}{\underset{id}{\to}} \mathbf{B}G) \simeq \mathbf{B}G$; 2. $\underset{\longrightarrow}{\lim}(\cdots G \times G \stackrel{\to}{\stackrel{\to}{\to}} G \stackrel{\to}{\to} *) \simeq \mathbf{B}G$. (The first is “vertically constant”, the second is “horizontally constant”). Applying the groupoid convolution algebra functor to the first presentation yields the groupoid convolution algebra $C(\mathbf{B}G)$ equipped with a trivial multiplication bimodule, hence just the group convolution algebra $C(\mathbf{B}G) \simeq C_{conv}(G)$. Applying however the groupoid convolution algebra functor to the second presentation yields the commutative algebra of functions $C(G)$ equipped with the multiplication bimodule which is $C(G \times G)$ regarded as a $(C(G\times G), C(G))$-bimdodule, where the right action is induced by pullback along the group product map $G \times G \to G$. This bimodule is in the image of the functor $Alg \to Alg_b$ that sends algebra homomorphisms to their induced bimodules, by sending $f \colon A \to B$ to $A$ regarded as an $(A,B)$-bimodule with the canonical left action on itself and the right action induced by $f$. Namely this bimdoule correspondonds to the map $\Delta \colon C(G) \to C(G \times G) \simeq C(G) \otimes C(G)$ given on $\phi \in C(G)$ and $g_1, g_2 \in G$ by $\Delta \phi \colon (g_1, g_2) \mapsto \phi(g_1 \cdot g_2) \,.$ This is that standard coproduct on the standard dual Hopf algebra associated with $G$. In summary this means that (for $G$ a finite group): 1. If we regard $\mathbf{B}G$ as presented as a double groupoid constant on $\mathbf{B}G$, then the corresponding groupoid convolution sesquialgebra (basis for a 3-module) is the convolution algebra of $G$; 2. If instead we regard $\mathbf{B}G$ as presented as the double groupoid which is degreewise constant as a groupoid, then the corresponding groupoid convolution sesquialgebra is the standard (“dual”) Hopf algebra structure on the commutative pointwise product algebra of functions on $G$. ## Properties ### Relation to (twisted) K-theory The operator K-theory of the convolution $C^\ast$-algebra of a topological groupoid $\mathcal{X}_\bullet$ may be thought of as the topological K-theory of the corresponding topological stack. More generally, for $\mathcal{X} \to \mathbf{B}^2 U(1)$ a principal 2-bundle (bundle gerbe) on the groupoid/stack, the operator K-theory of the corresponding twisted convolution algebra is the twisted K-theory of the stack. ## References ### For discrete geometry The homotopy colimit-interpretation of category algebras over discrete categories is discussed in Groupoid algebras of geometrically discrete groupoids twisted by principal 2-bundles/bundle gerbes/groupoid central extension is reviwed in • Eitan Angel, A Geometric Construction of Cyclic Cocycles on Twisted Convolution Algebras, PhD thesis (2010) Cyclic cocycles on twisted convolution algebras, (arXiv.1103.0578) ### For continuous/smooth geometry #### Convolution $C^\ast$-algebras The study of convolution C-star algebras of Lie groupoids goes back to • Jean Renault, A groupoid approach to $C^\ast$ algebras, Springer Lecture Notes in Mathematics, 793, Springer-Verlag, New York, 1980. Where the integration is performed against a fixed Haar measure. Surveys are for instance in • Nigel Higson, Groupoids, $C^\ast$-algebras and Index theory (pdf) The construction via sections of bundles of half-densities (avoiding a choice of Haar measure) is due to • Alain Connes, A survey of foliations and operator algebras, Proc. Sympos. Pure Math., AMS Providence, 32 (1982), 521–628 A review is on page 106 of More along these lines is in • Paul Muhly, Dana P. Williams, Continuous trace groupoid $C^\ast$-algebras II Math. Scand. 70 (1992), no. 1, 127–145; MR 93i:46117). (pdf) • Paul Muhly, Jean Renault, Dana P. Williams, Continuous trace groupoid $C^\ast$-algebras III , Transactions of the AMS, vol 348, Number 9 (1996) (jstor) • Mădălina Roxana Buneci, Groupoid Representations, Ed. Mirton: Timishoara (2003). • Mădălina Roxana Buneci, Groupoid $C^\ast$-Algebras, Surveys in Mathematics and its Applications, Volume 1: 71–98. (pdf) • Mădălina Roxana Buneci, Convolution algebras for topological groupoids with locally compact fibers (2011) (pdf) A review in the context of geometric quantization is in section 4.3 of • Rogier Bos, Groupoids in geometric quantization PhD Thesis (2007) pdf Specifically the convolution $C^\ast$-algebras of bundle gerbes regarded as centrally extended groupoids (algebras whose modules (see below) are gerbe modules/twisted bundle) is discussed in section 5 of A discussion of convolution algebras of symplectic groupoids (in the context of geometric quantization of symplectic groupoids) is in Functoriality of the construction of $C^\ast$-convolution algebras (its extension to groupoid-bibundles) is discussed in • Paul Muhly, Jean Renault, D. Williams, Equivalence and isomorphism for groupoid $C^\ast$-algebras, J. Operator Theory 17 (1987), no. 1, 3–22. • Janez Mrčun, Functoriality of the bimodule associated to a Hilsum-Skandalis map. K-Theory 18 (1999) 235–253. • Klaas Landsman, Operator Algebras and Poisson Manifolds Associated to Groupoids, Commun. Math. Phys. 222, 97 – 116 (2001) (web) #### Convolution Hopf algebroids A characterization of the convolution algebras of étale groupoids with their Hopf algebroid structure is in #### Modules over Lie groupoid convolution algebras and K-theory Discussion of modules over Lie groupoid convolution algebras is in the following articles. In (Renault80) measurable representations of topological groupoids are related to modules over their $L^1$ convolution star algebra Banach algebras hence over their envoloping $C^\ast$-algebras. In (Bos, chapter 7) is discussion refining this to continuous representations and representation of a convolution $C^\ast$-algebra, also in section 4 of: Representation of convolution algebras of étale groupoids is in The operator K-theory of groupoid convolution algebras (the topological K-theory of the corresponding differentiable stacks) is discussed in Construction of cocycles in KK-theory and spectral triples from groupoid convolution is in Revised on May 14, 2013 09:49:25 by Urs Schreiber (89.204.137.236)

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# Difference between combinational and sequential circuit This article covers the Difference between combinational and sequential circuit. These circuits are important part of digital electronics. ## Combinational Circuit Combinational circuit is a type of circuit whose output depends upon the input provided. It provides instant output and is independent of time. A combinational circuit doesn’t require feedback for further output. In combinational circuits, only the present input is required. So the performance of the circuit becomes faster  and better when compared to sequential circuit. It has simple circuit and used logic gate as basic building blocks. Combinational circuit can be used in both arithmetic and Boolean  circuit. ### Combinational Circuit Diagram The block diagram for combinational circuit is shown in figure below ## Sequential Circuit Sequential circuit are the type of circuit whose output depends on both current and previous input. The output obtained from the previous input is transferred in the form of feedback and the circuit uses it again for generating next output. Sequential circuits are slow when compared to combinational circuits. The feedback feature makes the circuit complex. They use flip – flop as basic building blocks of circuit. ### Sequential circuit diagram The block diagram for sequential circuit is shown in figure below ### Difference Between Combinational and Sequential Circuit CombinationalCircuit SequentialCircuit 1. In combinational circuits, output depends only upon present input. In sequential circuits output depends upon both present and previous input. 2. It has Fast speed. It has Slow speed. 3. Combinational circuits can be designed easily. Sequential circuits are difficult to design when compared to combinational circuit. 4. These circuit doesn’t have any feedback between input and output. These circuits have feedbacks between inputs and output. 5. Elementary building block : logic gate Elementary building block : flip flop 6. Used in arithmetic as well as Boolean operation. Mainly used for storing data. 7. They nether have clock nor they require triggering. They are clock dependent and needs triggering. 8. Combinational circuits don’t have memory element. Sequential circuit have memory element. 9. It is easy to handle and use. It is difficult to handle and use. 10. Example :1.    Encoder2.    Decoder3.    Multiplexer4.    demultiplexer Example :1.    Flip flop2.    counter. Author Akash Sharma

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Explore BrainMass Share # A particle is distributed along the positive x axis with a probability density.... This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! Please see the attached file for the fully formatted problems. 1. A particle is distributed along the positive x axis with a probability density P(x) = C x2 exp(-(x/a)2). Determine it's average coordinate <x>, and the fluctuation <&#61508;x2> ; where <&#61508;x>= x-<x> 2. The coordinate of a particle is distributed along the WHOLE x-axis with a probability density P(x) = C /(x2 +a2)3/2. Determine C, <x>, and the fluctuation <&#61508;x2>

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Sorry, an error occurred while loading the content. ## Re: GEOSTATS: simulating intrinsically stationary processes Expand Messages • For estimation/simulation with a finite search neighborhood, assuming an intrinsic hypothesis becomes quite tangential to the main point of interest -- i.e., Message 1 of 2 , Nov 30, 1999 • 0 Attachment For estimation/simulation with a finite search neighborhood, assuming an intrinsic hypothesis becomes quite tangential to the main point of interest -- i.e., generating good estimates. Assuming something is fBm is just that -- an assumption. It might even be a valid one where variance increases without bounds, but still other such concerns as the size of the search radius, number of estimation control points, etc will still have an impact on overall estimation quality. Syed Abdul Rahman Landmark Graphics Jakarta, Indonesia PS/- fBm is nonstationary but it's derivative -- fGn -- is. Easiest simulation is to generate a Gaussian distribution using a random number generator, then integrate to get fBm with Hurst exponent 0.5. PS2/- _Sample_ covariance will never "fail" if calculated on a simulated fBm trace. Does "fail" mean an inability to generate "good" estimates, e.g. from a comparison of cross-validated residuals? ----- Original Message ----- From: Hillman RJT <R.J.T.Hillman@...> To: <ai-geostats@...> Sent: Tuesday, November 30, 1999 10:02 PM Subject: GEOSTATS: simulating intrinsically stationary processes > Dear All, > > I hope you can help me. I'm an econometrician analysing high-frequency > exchange rate data. We typically observe a transaction price at > irregular intervals, between .01 of a second and three hours depending > on the time of day. We can also observe other things like spreads, > liquidity etc. > > I though it might be a good idea, given the irregular spacing of the > data (on the time-scale) to use some geo-stat methods. I've got Cressie, > plus some other papers, but there are still some outstanding questions. > > 1) Despite often reading claims that the variogram is defined for a > wider class of processes than the covariance (i.e. intrinsically > stationary processes), I haven't seen any convincing evidence that when > we simulate a non-covariance-stationary process that IS intrinsically > stationary, the variogram outperforms the covariance. I would imagine we > could demonstrate this in terms of measuring the dependence and through > kriging mean square erros. Do you know of any examples here people have > demonstrated this? > > 2) I have read that Fractional Brownian motion is not stationary, but is > intrinsically stationary. I thought FBM is stationary when -1/2<d<1/2. > Could someone clarify this? > > What I am trying to so is simple. > > Generate a univariate time series X(t(1)),X(t(2)),X(t(3))...X(t(N)) > according to an intrinsically stationary process which isn't stationary > in the usual sense. Then demonstrate that the covariance fails where the > variogram succeeds. > > Can anyone suggest an easily simulatable process for X(t) that would do > this for me? > > Ultimately I would like to argue that intrinsic stationarity is a useful > concept for financial processes, but whilst it seems reasonable and > apparently un-tested to a large degree in the geo-stats I've seen, I'd > like some firm evidence of it's usefulness. > > Any suggestions... > > thanks in advance > > Robert > > -- > Robert J T Hillman > http://www.city.ac.uk/cubs/ferc/robert/index.html > > > Research Fellow > Financial Econometrics Research Centre > City University Business School > Frobisher Crescent > The Barbican > LONDON > EC2Y 8HB > > tel: +44 (0) 171 477 8734 Direct Line > tel: +44 (0) 171 477 8611 Secretary > fax: +44 (0) 171 477 8881 > -- > *To post a message to the list, send it to ai-geostats@.... > *As a general service to list users, please remember to post a summary > of any useful responses to your questions. > *To unsubscribe, send email to majordomo@... with no subject and > "unsubscribe ai-geostats" in the message body. > DO NOT SEND Subscribe/Unsubscribe requests to the list! > -- *To post a message to the list, send it to ai-geostats@.... *As a general service to list users, please remember to post a summary of any useful responses to your questions. *To unsubscribe, send email to majordomo@... with no subject and "unsubscribe ai-geostats" in the message body. DO NOT SEND Subscribe/Unsubscribe requests to the list! Your message has been successfully submitted and would be delivered to recipients shortly.

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/* nag_zher2k (f16zrc) Example Program. * * Copyright 2005 Numerical Algorithms Group. * * Mark 8, 2005. */ #include #include #include #include #include int main(void) { /* Scalars */ Complex alpha; double beta; Integer adim1, adim2, exit_status, i, j, k, n, pda, pdb, pdc; /* Arrays */ Complex *a=0, *b=0, *c=0; char nag_enum_arg[40]; /* Nag Types */ NagError fail; Nag_OrderType order; Nag_UploType uplo; Nag_TransType trans; Nag_MatrixType matrix; #ifdef NAG_COLUMN_MAJOR #define A(I,J) a[(J-1)*pda + I - 1] #define B(I,J) b[(J-1)*pdb + I - 1] #define C(I,J) c[(J-1)*pdc + I - 1] order = Nag_ColMajor; #else #define A(I,J) a[(I-1)*pda + J - 1] #define B(I,J) b[(I-1)*pdb + J - 1] #define C(I,J) c[(I-1)*pdc + J - 1] order = Nag_RowMajor; #endif exit_status = 0; INIT_FAIL(fail); Vprintf( "nag_zher2k (f16zrc) Example Program Results\n\n"); /* Skip heading in data file */ Vscanf("%*[^\n] "); /* Read the problem dimensions */ Vscanf("%ld%ld%*[^\n] ", &n, &k); /* Read the uplo parameter */ Vscanf("%s%*[^\n] ", nag_enum_arg); /* nag_enum_name_to_value(x04nac). * Converts NAG enum member name to value */ uplo = nag_enum_name_to_value(nag_enum_arg); /* Read the transpose parameter */ Vscanf("%s%*[^\n] ", nag_enum_arg); /* nag_enum_name_to_value(x04nac), see above. */ trans = nag_enum_name_to_value(nag_enum_arg); /* Read scalar parameters */ Vscanf(" ( %lf , %lf )%*[^\n] ", &alpha.re, &alpha.im); Vscanf("%lf%*[^\n] ", &beta); if (trans == Nag_NoTrans) { adim1 = n; adim2 = k; } else { adim1 = k; adim2 = n; } #ifdef NAG_COLUMN_MAJOR pda = adim1; #else pda = adim2; #endif pdb = pda; pdc = n; if (k > 0 && n > 0) { /* Allocate memory */ if ( !(a = NAG_ALLOC(k*n, Complex)) || !(b = NAG_ALLOC(k*n, Complex)) || !(c = NAG_ALLOC(n*n, Complex)) ) { Vprintf("Allocation failure\n"); exit_status = -1; goto END; } } else { Vprintf("Invalid k or n\n"); exit_status = 1; return exit_status; } /* Input matrix A. */ for (i = 1; i <= adim1; ++i) { for (j = 1; j <= adim2; ++j) Vscanf(" ( %lf , %lf )", &A(i,j).re, &A(i,j).im); Vscanf("%*[^\n] "); } /* Input matrix A. */ for (i = 1; i <= adim1; ++i) { for (j = 1; j <= adim2; ++j) Vscanf(" ( %lf , %lf )", &B(i,j).re, &B(i,j).im); Vscanf("%*[^\n] "); } /* Input matrix C. */ if (uplo == Nag_Upper) { for (i = 1; i <= n; ++i) { for (j = i; j <= n; ++j) Vscanf(" ( %lf , %lf )", &C(i,j).re, &C(i,j).im); } Vscanf("%*[^\n] "); } else { for (i = 1; i <= n; ++i) { for (j = 1; j <= i; ++j) Vscanf(" ( %lf , %lf )", &C(i,j).re, &C(i,j).im); } Vscanf("%*[^\n] "); } /* nag_zher2k(f16zrc). * Rank 2k update of complex Hermitian matrix. * */ nag_zher2k(order, uplo, trans, n, k, alpha, a, pda, b, pdb, beta, c, pdc, &fail); if (fail.code != NE_NOERROR) { Vprintf("Error from nag_zher2k.\n%s\n", fail.message); exit_status = 1; goto END; } if (uplo == Nag_Upper) { matrix = Nag_UpperMatrix; } else { matrix = Nag_LowerMatrix; } /* Print updated matrix C */ /* nag_gen_complx_mat_print_comp (x04dbc). * Print complex general matrix (comprehensive) */ nag_gen_complx_mat_print_comp(order, matrix, Nag_NonUnitDiag, n, n, c, pdc, Nag_BracketForm, "%6.2f", "Updated Matrix C", Nag_IntegerLabels, 0, Nag_IntegerLabels, 0, 80, 0, 0, &fail); if (fail.code != NE_NOERROR) { Vprintf("Error from nag_gen_complx_mat_print_comp (x04dbc).\n%s" "\n",fail.message); exit_status = 1; goto END; } END: if (a) NAG_FREE(a); if (b) NAG_FREE(b); if (c) NAG_FREE(c); return exit_status; }

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## Credits Walchand College of Engineering (WCE), Sangli Shweta Patil has created this Calculator and 1000+ more calculators! Nishan Poojary has verified this Calculator and 400+ more calculators! ## Diagonal of Rectangle given radius of Circumcircle of Rectangle Solution STEP 0: Pre-Calculation Summary Formula Used DPolygon = rc*2 This formula uses 1 Variables Variables Used Circumradius - Circumradius is the radius of a circumsphere touching each of the polyhedron's or polygon's vertices. (Measured in Meter) STEP 1: Convert Input(s) to Base Unit Circumradius: 15 Meter --> 15 Meter No Conversion Required STEP 2: Evaluate Formula Substituting Input Values in Formula DPolygon = rc*2 --> 15*2 Evaluating ... ... DPolygon = 30 STEP 3: Convert Result to Output's Unit 30 Meter --> No Conversion Required 30 Meter <-- Polygon Diagonal (Calculation completed in 00.000 seconds) ## < 3 Diagonal of Polygon of Circumcircle Calculators Diagonal of Rectangle given radius of Circumcircle of Rectangle Diagonal of Hexagon given radius of circumcircle of Hexagon Diagonal of Square given radius of Circumcircle of Square DPolygon = rc*2 ## What is circumscribed circle? In geometry, the circumscribed circle or circumcircle of a polygon is a circle that passes through all the vertices of the polygon. The center of this circle is called the circumcenter and its radius is called the circumradius. ## How to Calculate Diagonal of Rectangle given radius of Circumcircle of Rectangle? Diagonal of Rectangle given radius of Circumcircle of Rectangle calculator uses polygon_diagonal = Circumradius*2 to calculate the Polygon Diagonal, Diagonal of Rectangle given radius of Circumcircle of Rectangle formula is defined as a line connecting two opposite vertices of rectangle whose vertices touches circumscribed circle at 4 points. Polygon Diagonal and is denoted by DPolygon symbol. How to calculate Diagonal of Rectangle given radius of Circumcircle of Rectangle using this online calculator? To use this online calculator for Diagonal of Rectangle given radius of Circumcircle of Rectangle, enter Circumradius (rc) and hit the calculate button. Here is how the Diagonal of Rectangle given radius of Circumcircle of Rectangle calculation can be explained with given input values -> 30 = 15*2. ### FAQ What is Diagonal of Rectangle given radius of Circumcircle of Rectangle? Diagonal of Rectangle given radius of Circumcircle of Rectangle formula is defined as a line connecting two opposite vertices of rectangle whose vertices touches circumscribed circle at 4 points and is represented as DPolygon = rc*2 or polygon_diagonal = Circumradius*2. Circumradius is the radius of a circumsphere touching each of the polyhedron's or polygon's vertices. How to calculate Diagonal of Rectangle given radius of Circumcircle of Rectangle? Diagonal of Rectangle given radius of Circumcircle of Rectangle formula is defined as a line connecting two opposite vertices of rectangle whose vertices touches circumscribed circle at 4 points is calculated using polygon_diagonal = Circumradius*2. To calculate Diagonal of Rectangle given radius of Circumcircle of Rectangle, you need Circumradius (rc). With our tool, you need to enter the respective value for Circumradius and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well. How many ways are there to calculate Polygon Diagonal? In this formula, Polygon Diagonal uses Circumradius. We can use 3 other way(s) to calculate the same, which is/are as follows -

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# Cell Array Reshape Operation 2 views (last 30 days) MarshallSc on 18 Jan 2022 Edited: MarshallSc on 19 Jan 2022 I have this 10x10 cell array: I want to reshape it into a 4x4 cell each containing a 10x10 matrix keeping the order of each value in the matrices. For example, the first new cell ( cell(1,1) ) contains the first values of the 16x1 matrices which will transform into a 10x10 for each cell and so on. I'd appreciate it if someone can please help me! Thank you! DGM on 19 Jan 2022 There are probably other ways, but... % a patterned test array A = repmat({(1:16).', (17:32).'; (33:48).', (49:64).'},5,5) A = 10×10 cell array {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} {16×1 double} B = cell2mat(A(:).'); B = reshape(permute(B,[2 3 1]),10,10,[]); B = reshape(num2cell(B,[1 2]),4,4) B = 4×4 cell array {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} {10×10 double} % inspect the result B{1} ans = 10×10 1 17 1 17 1 17 1 17 1 17 33 49 33 49 33 49 33 49 33 49 1 17 1 17 1 17 1 17 1 17 33 49 33 49 33 49 33 49 33 49 1 17 1 17 1 17 1 17 1 17 33 49 33 49 33 49 33 49 33 49 1 17 1 17 1 17 1 17 1 17 33 49 33 49 33 49 33 49 33 49 1 17 1 17 1 17 1 17 1 17 33 49 33 49 33 49 33 49 33 49 B{2} ans = 10×10 2 18 2 18 2 18 2 18 2 18 34 50 34 50 34 50 34 50 34 50 2 18 2 18 2 18 2 18 2 18 34 50 34 50 34 50 34 50 34 50 2 18 2 18 2 18 2 18 2 18 34 50 34 50 34 50 34 50 34 50 2 18 2 18 2 18 2 18 2 18 34 50 34 50 34 50 34 50 34 50 2 18 2 18 2 18 2 18 2 18 34 50 34 50 34 50 34 50 34 50 MarshallSc on 19 Jan 2022 Thanks a lot mate! ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!

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Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: graphing.com Related topics: theory of 2nd order linear equations | ladder method in conversion word problems | printable pages tussy 4t edit int algebra printable problems | distributive property in math for 6th graders | simplifing numbers | solve linear equations worksheet | college algebra tutorial 6 | solving second degree linear algebra | free maths gcse worksheet factorisation | elementary math/least common multiple singapore | fraction test printables | prentice hall mathematics algebra 1 worksheets Author Message [VA]Vridge Registered: 29.04.2002 Posted: Sunday 31st of Dec 08:21 Hi, can anyone please help me with my math homework? I am not quite good at math and would be grateful if you could help me understand how to solve graphing.com problems. I also would like to know if there is a good website which can help me prepare well for my upcoming math exam. Thank you! espinxh Registered: 17.03.2002 From: Norway Posted: Tuesday 02nd of Jan 08:05 Hi, I think that I can to help you out. Have you ever used a program to assist you with your algebra assignments? Some time ago I was also stuck on a similar issues like you, and then I found Algebrator. It helped me a lot with graphing.com and other math problems, so since then I always count on its help! My math grades got better because of Algebrator. Mov Registered: 15.05.2002 From: Posted: Thursday 04th of Jan 08:40 Algebrator is one handy tool. I don’t have much interest in math and have found it to be difficult all my life. Yet one cannot always leave math because it sometimes becomes a compulsory part of one’s course work. My younger brother is a math expert and I found this software in his laptop. It was only then I understood why he finds this subject to be so simple . DoniilT Registered: 27.08.2002 From: Posted: Friday 05th of Jan 08:10 A truly piece of algebra software is Algebrator. Even I faced similar problems while solving least common denominator, side-side-side similarity and like denominators. Just by typing in the problem workbookand clicking on Solve – and step by step solution to my algebra homework would be ready. I have used it through several algebra classes - Basic Math, Intermediate algebra and Intermediate algebra. I highly recommend the program.

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Advanced Algebra Help Forum: Linear and Abstract Algebra 1. ### Make sure your post belongs here • Replies: 7 • Views: 5,409 Nov 25th 2012, 02:03 PM 1. ### Is V in Span(S) • Replies: 8 • Views: 537 Mar 22nd 2013, 11:04 PM 2. ### Algebraicity of arbitrary extensions • Replies: 4 • Views: 469 Mar 22nd 2013, 07:05 PM 3. ### Inseparable extension of rational function field • Replies: 2 • Views: 486 Mar 22nd 2013, 06:51 PM • Replies: 8 • Views: 431 Mar 22nd 2013, 07:03 AM 5. ### Help!! find the minimum • Replies: 1 • Views: 329 Mar 22nd 2013, 04:20 AM 6. ### Subspaces • Replies: 11 • Views: 789 Mar 21st 2013, 07:32 PM 7. ### Subgroup of order 4 • Replies: 3 • Views: 473 Mar 21st 2013, 05:37 PM 8. ### Llinear transformations: Help with proving linear independency • Replies: 3 • Views: 551 Mar 21st 2013, 11:54 AM 9. ### Matrices and vectors point-line distance • Replies: 3 • Views: 845 Mar 21st 2013, 04:57 AM 10. ### Positive definite symmetric matrix has LDL^T decomposition with positive diagonals • Replies: 6 • Views: 873 Mar 21st 2013, 04:53 AM 11. ### Basis and Derivative Transformation Question • Replies: 2 • Views: 522 Mar 21st 2013, 04:41 AM 12. ### I wanna prove if the composite are equal to each other • Replies: 1 • Views: 419 Mar 21st 2013, 03:32 AM 13. ### Diagonalizable question • Replies: 1 • Views: 374 Mar 20th 2013, 06:15 PM 14. ### Composite Functions • Replies: 1 • Views: 693 Mar 20th 2013, 03:20 PM • Replies: 6 • Views: 1,382 Mar 20th 2013, 11:29 AM 16. ### Difficult variable to solve for. includes hyperbolic cos. • Replies: 5 • Views: 483 Mar 19th 2013, 08:07 PM • Replies: 1 • Views: 510 Mar 19th 2013, 06:20 PM 18. ### I have to show this proof for the matrix inverse theorem • Replies: 1 • Views: 636 Mar 19th 2013, 04:52 PM 19. ### Linear Algebra Proof • Replies: 2 • Views: 413 Mar 19th 2013, 01:30 PM 20. ### I don't understand this problem, rational inequailties! • Replies: 6 • Views: 711 Mar 19th 2013, 09:30 AM 21. ### Equation • Replies: 0 • Views: 254 Mar 19th 2013, 05:39 AM 22. ### Vector Identity Question • Replies: 0 • Views: 287 Mar 18th 2013, 06:56 PM 23. ### 2 x 2 matrix • Replies: 1 • Views: 294 Mar 18th 2013, 05:27 PM 24. ### Eigenvalues and Eigenvectors • Replies: 3 • Views: 406 Mar 18th 2013, 05:24 PM 25. ### Symmetric mappings help!! • Replies: 6 • Views: 670 Mar 18th 2013, 03:21 AM 26. ### True or False? • Replies: 1 • Views: 561 Mar 18th 2013, 02:52 AM 27. ### Describing how vectors can be used to Non Math friends? • Replies: 1 • Views: 606 Mar 17th 2013, 07:14 PM 28. ### Describing how using Matrices is similar and different than solving an equation? • Replies: 1 • Views: 438 Mar 17th 2013, 06:11 PM 29. ### Definition of a vector space • Replies: 1 • Views: 384 Mar 17th 2013, 06:07 PM , , , , # svmsung.j7.1218 , Click on a term to search for related topics. Use this control to limit the display of threads to those newer than the specified time frame. Allows you to choose the data by which the thread list will be sorted.

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# 140.621.01Statistical Methods in Public Health I Location: East Baltimore Term: 1st term Department: Biostatistics Credits: 4 credits 2017 - 2018 Instruction Method: TBD Class Times: • Tu Th,  10:30 - 11:50am Lab Times: • Monday,  1:30 - 3:00pm (01) • Tuesday,  1:30 - 3:00pm (02) • Wednesday,  1:30 - 3:00pm (03) • Thursday,  1:30 - 3:00pm (04) • Friday,  1:30 - 3:00pm (05) • Monday,  3:30 - 5:00pm (06) • Tuesday,  3:30 - 5:00pm (07) • Wednesday,  3:30 - 5:00pm (08) • Thursday,  3:30 - 5:00pm (09) Auditors Allowed: Yes, with instructor consent Course Instructors: Contact: Marie Diener-West Resources: Description: Introduces the basic concepts and methods of statistics as applied to diverse problems in public health and medicine. Demonstrates methods of exploring, organizing, and presenting data, and introduces fundamentals of probability, including probability distributions and conditional probability, with applications to 2x2 tables. Presents the foundations of statistical inference, including concepts of population, sample parameter, and estimate; and approaches to inferences using the likelihood function, confidence intervals, and hypothesis tests. Introduces and employs the statistical computing package, STATA, to manipulate data and prepare students for remaining course work in this sequence. Learning Objectives: Upon successfully completing this course, students will be able to: 1. Use statistical reasoning to formulate public health questions in quantitative terms within the scientific method. 2. Design and interpret graphical and tabular displays of statistical information, including stem and leaf plots, box plots, Q-Q plots and frequency tables. 3. Distinguish probability models (binomial, Poisson, and Gaussian) for describing trends and random variation in public health data. 4. Use stratification to eliminate the influence of a possible confounding variable in a study of the association between a risk factor and outcome. 5. Use bootstrapping to construct confidence intervals and interpret them in a scientific context. 6. Explain the implications of the Central Limit Theorem in determining the sampling distributions of sample statistics. 7. Use sampling distribution theory for a single sample mean, difference between two sample means, paired mean difference, single sample proportion, and difference between two sample proportions for statistical inference. 8. Employ statistical methods for inference, including tests and confidence intervals, to draw public health inferences from data. 9. Use the Stata statistical analysis package to construct tables and graphs and perform statistical methods for inference. Methods of Assessment: Student evaluation based on problem sets and exams. Enrollment Restriction: For MPH, DrPH, "special students" and MSPH degree candidates Instructor Consent: Consent required for some students Consent Note: Consent Required for non-PH students For consent, contact:

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## 留言板 P·艾拉瓦利亚, N·S·纳拉. 在重力作用下的上覆无限热弹性流体对广义热弹性固体转动的影响[J]. 应用数学和力学, 2009, 30(12): 1415-1426. doi: 10.3879/j.issn.1000-0887.2009.12.003 引用本文: P·艾拉瓦利亚, N·S·纳拉. 在重力作用下的上覆无限热弹性流体对广义热弹性固体转动的影响[J]. 应用数学和力学, 2009, 30(12): 1415-1426. Praveen Ailawalia, Naib Singh Narah. Effect of Rotation in Generalized Thermoelastic Solid Under the Influence of Gravity With an Overlying Infinite Thermoelastic Fluid[J]. Applied Mathematics and Mechanics, 2009, 30(12): 1415-1426. doi: 10.3879/j.issn.1000-0887.2009.12.003 Citation: Praveen Ailawalia, Naib Singh Narah. Effect of Rotation in Generalized Thermoelastic Solid Under the Influence of Gravity With an Overlying Infinite Thermoelastic Fluid[J]. Applied Mathematics and Mechanics, 2009, 30(12): 1415-1426. ## 在重力作用下的上覆无限热弹性流体对广义热弹性固体转动的影响 ##### doi: 10.3879/j.issn.1000-0887.2009.12.003 • 中图分类号: O343.6 ## Effect of Rotation in Generalized Thermoelastic Solid Under the Influence of Gravity With an Overlying Infinite Thermoelastic Fluid • 摘要: 计及上覆无限热弹性流体的重力作用，沿界面有不同的外力作用时，研究广义热弹性固体的旋转变形问题．在Laplace和Fourier域内，通过积分变换，得到了位移、应力及温度分布的表达式．然后在物理域内，应用数值逆变换方法，得到这些分量的值，并讨论了该问题的一些特例．结果以图形方式给出，显示了介质的旋转以及重力作用的影响． • [1] Chandrasekharaiah D S.Hyperbolic thermoelastic:a review of recent literature[J].Appl Mech Rev,1998,51(12):705-729. [2] Lord H W,Shulman Y.A generalized dynamical theory of thermoelasticity solids[J].J Mech Phy Solids,1967,15(5):299-309. [3] Müller I.The coldness,a universal function in thermoelastic bodies[J].Arch Rat Mech Anal,1971,41(5):319-332. [4] Green A E,Laws N.On the entropy production inequality[J].Arch Rat Mech Anal,1972,45(1):45-47. [5] Green A E,Lindsay K A.Thermoelasticity[J].J Elasticity,1972,2(1):1-7. [6] Suhubi E S.Thermoelastic solids[A].In:Eringin A C,Ed.Continuum Physics[C].Vol Ⅱ Part Ⅱ,Chapter Ⅱ.New York:Acamadic Press,1975. [7] Green A E,Naghdi P M.On thermoelasticity without energy dissipation[J].J Elasticity,1993,31(3):189-208. [8] Barber J R,Martin-Moran C J.Green's functions for transient thermoelastic contact problems for the half-plane[J].Wear,1982,79:11-19. [9] Barber J R.Thermoelastic displacements and stresses due to a heat source moving over the surface of a half plane[J].ASME,Transactions,Journal of Applied Mechanics,1984,51(3):636-640. [10] Sherief H H.Fundamental solution of the generalized thermoelastic problem for short times[J].J Thermal Stresses,1986,9(2):151-164. [11] Dhaliwal R S,Majumdar S R,Wang J.Thermoelastic waves in an infinite solid caused by a line heat source[J].Int J Math & Math Sci,1997,20(2):323-334. [12] Chandrasekharaiah D S,Srinath K S.Thermoelastic interactions without energy dissipation due to a point heat source[J].J Elasticity,1998,50(2):97-108. [13] Sharma J N,Chauhan R S,Kumar R.Time-harmonic sources in a generalized thermoelastic continuum[J].J Thermal Stresses,2000,23(7):657-674. [14] Sharma J N,Chauhan R S.Mechanical and thermal sources in a generalized thermoelastic half-space[J].J Thermal Stresses,2001,24(7):651-675. [15] Sharma J N,Sharma P K,Gupta S K.Steady state response to moving loads in thermoelastic solid media[J].J Thermal Stresses,2004,27(10):931-951. [16] 德斯瓦尔 S，乔德哈瑞 S.带扩散的广义弹性固体中移动荷载引起的二维相互作用[J].应用数学和力学，2008,29(2):188-202. [17] Chand D,Sharma J N,Sud S P.Transient generalized magneto-thermoelastic waves in a rotating half space[J].Int J Engg Sci,1990,28(6):547-556. [18] Schoenberg M,Censor D.Elastic waves in rotating media[J].Quart Appl Math,1973,31:115-125. [19] Clarke N S,Burdness J S.Rayleigh waves on a rotating surface[J].ASME J Appl Mech,1994,61：724-726. [20] Destrade M.Surface waves in rotating rhombic crystal[J].Proc Royal Soc London，Series A,2004,460:653-665. [21] Roychoudhuri S K,Mukhopadhyay S.Effect of rotation and relaxation times on plane waves in generalized thermo-visco-elasticity[J].Int J Math Math Sci,2000,23(7):497-505. [22] Ting T C T.Surface waves in a rotating anisotropic elastic half-space[J].Wave Motion,2004,40(4):329-346. [23] Sharma J N,Thakur D.Effect of rotation on Rayleigh-Lamb waves in magneto-thermoelastic media[J].J Sound Vib,2006,296(4/5):871-887. [24] Sharma J N,Walia V.Effect of rotation on Rayleigh-Lamb waves in piezothermoelastic half space[J].J Solid Structures,2007,44(3/4):1060-1072. [25] Sharma J N,Othman M I A.Effect of rotation on generalized thermo-viscoelastic Rayleigh-Lamb waves[J].J Solid Structures,2007,44(13):4243-4255. [26] Sharma J N,Walia V,Gupta S K.Effect of rotation and thermal relaxation on Rayleigh waves in piezothermoelastic half space[J].Int J Mech Sci,2008,50(3):433-444. [27] Othman M I A,Song Y.Effect of rotation on plane waves of generalized electro-magneto-thermoviscoelasticity with two relaxation times[J].Appl Math Modelling,2008,32(5):811-825. [28] Bromwich T J J A.On the influence of gravity on elastic waves and in particular on the vibrations of an elastic globe[J].Proc London Math Soc,1898,30(1):98-120. [29] Love A E H.Some Problems of Geodynamics[M].New York:Dover,1911. [30] De S N,Sengupta P R.Plane Lamb's problem under the influence of gravity[J].Gerland Beitr Geophysics (Leipzig),1973,82:421-426. [31] De S N,Sengupta P R.Influence of gravity on wave propagation in an elastic layer[J].J Acoust Soc Am,1974,55(5):919-921. [32] De S N,Sengupta P R.Surface waves under the influence of gravity[J].Gerland Beitr Geophysics (Leipzig),1976,85:311-318. [33] Sengupta P R,Acharya D.The influence of gravity on the propagation of waves in a thermoelastic layer[J].Rev Roum Sci Technol Mech Appl Tome,1979,24:395-406. [34] Das S C,Acharya D P,Sengupta P R.Surface waves in an inhomogeneous elastic medium under the influence of gravity[J].Rev Roum Des Sci Tech,1992,37(5):539-551. [35] Abd-Alla A M,Ahmed S M.Rayleigh waves in an orthotropic thermoelastic medium under gravity field and initial stress[J].J Earth Moon Planets,1996,75(3):185-197. [36] Abd-Alla A M,Ahmed S M.Stonley and Rayleigh waves in a nonhomogeneous orthotropic elastic medium under the influence of gravity[J].Appl Math Comp,2003,135:187-200. [37] Youssef H M.Problem of generalized thermoelastic infinite medium with cylindrical cavity subjected to ramp-type heating and loading[J].Arch Appl Mech,2006,75:553-565. [38] Sinha S B,Elsibai K A.Reflection and refraction of thermoelastic waves at an interface of two semi-infinite media with two relaxation times[J].J Thermal Stresses,1997,20(2):129-145. [39] Sharma J N,Kumar V.Plane strain problems of transversely isotropic thermoelastic media[J].J Thermal Stresses,1997,20(5):463-476. ##### 计量 • 文章访问数:  1581 • HTML全文浏览量:  135 • PDF下载量:  704 • 被引次数: 0 ##### 出版历程 • 收稿日期:  2009-04-24 • 修回日期:  2009-08-24 • 刊出日期:  2009-12-15 / • 分享 • 用微信扫码二维码 分享至好友和朋友圈

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# If Then Statement in Excel Vba You are going to focus on the If statement. The If statement is a condition. Conditions are very useful in programming because they allow the user to execute actions based on specific criteria (that the user defines). If is one of the most widely used and popular statement in excel VBA. The If statement is also sometimes referred to as the If else statement. The purpose of the If statement is to check whether the condition (or conditions) defined by the user are met or not. Toggle ## Syntax This is the syntax of the If statement: ```IF condition1 THEN 'What will the happen if the condition is met ELSE 'What will the happen if the condition is not met END IF``` Explanation: The user defines the condition after the if statement to check whether it’s true or false. If the condition is true than the user in the next line defines what the code will do. However if the condition are not met then the user uses the else statement after which he/she defines what will happen if the condition are not met. Our program will take an input from the user and tell whether it’s a positive number or a negative number. ## The code Click on Developer tab and select View Code. A new window (Visual Basic Editor) will open which would have a dialog box in the center. Write the following line of code in the dialog box. ```Sub Find_Negative() On Error GoTo catch_error Dim number As Integer number = InputBox("Enter the number: ") If number < 0 Then MsgBox "Entered number is negative!" Else MsgBox "Entered number is positive!" End If Exit Sub catch_error: MsgBox "Oops, Some Error Occurred" End Sub ``` After writing the code close the window by clicking on the cross(x) icon on the upper right side of the screen. ## Explanation In this code, we are first taking the input from the user (the input must be a number).we are checking whether the modulus (Mod) of that number with 2 is zero or not. If its zero than this means that the number is divisible by 2 and thus we display a message box saying the number is positive. If its modulus with 2 is not zero than that number is not perfectly divisible by 2. Hence we display a message box saying that the number is negative. However if the user instead of entering a number enters a letter or a special character (like @) than we display a message box saying that an error has occurred. This is the result if the user has entered a positive number. This is the result if the user has entered a negative number. This is the result if the user has entered a special character or a letter. That’s it; you have now successfully used the If statement.

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test algorithm to decide if point is inside area [duplicate] here's a geometric problem that I've been unsuccessful to solve: • we have four points A, B, C, D defining an area • and two points E, F • E is within the boundaries of the polygon ABCD • F is outside the borders We know the (x, y) coordinates of each point. (see the figure below) Determine for any point G (x, y) if G is inside or outside ABCD Any ideas out there? • This sounds like math. – nhgrif Nov 6 '13 at 19:33 • Also, as a hint, the fact that this is not a "concave" shape will be a key to solving this puzzle. – BlackVegetable Nov 6 '13 at 19:33 • Step one: find a point that's guaranteed to be outside of the poligon – user3458 Nov 6 '13 at 19:34 • This question appears to be off-topic because it is about math. – Mihai Maruseac Nov 6 '13 at 19:34 • Well, the generalized question is one of the most popular on this site. I wouldn't say it's off-topic, it's just computational geometry. – Zong Nov 6 '13 at 19:34 For your specific problem (quadrilateral - convex), you can do the following: 1) calculate the equations for the 4 sides 2) calculate the intersection of a vertical line which has G on it with each line 3) if you find an even number of intersections, then it is outside 4) if you find an odd number of intersections, then it is inside 5) be careful at the endpoint intersections 6) be careful if G is on one of the sides • Wait... in both cases it will be even, no? if inside the intersections will happen up and down, if outside up-up or down-down... – adrienlucca.wordpress.com Nov 6 '13 at 19:49 • OK, your answer works if we add that the vertical line is not infinite on both sides, but only starts (and goes up or down), at G – adrienlucca.wordpress.com Nov 6 '13 at 19:59 • I think you got it. Let me know if you need any more help. – No One in Particular Nov 6 '13 at 20:22 • one more case, we can still have Zero intersection, then it's outside... – adrienlucca.wordpress.com Nov 6 '13 at 20:40 • Zero intersections is even. – No One in Particular Nov 6 '13 at 23:32

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gov.sandia.cognition.learning.algorithm.minimization.matrix ## Class OverconstrainedMatrixVectorMultiplier • All Implemented Interfaces: Evaluator<Vector,Vector> ```@PublicationReference(author="Jonathan Richard Shewchuk", title="An Introduction to the Conjugate Gradient Method Without the Agonizing Pain", type=WebPage, year=1994, url="http://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf\u200e") public class OverconstrainedMatrixVectorMultiplier extends MatrixVectorMultiplier``` Implements an overconstrainted matrix-vector multiplication. Since: 4.0.0 Author: Jeremy D. Wendt • ### Fields inherited from class gov.sandia.cognition.learning.algorithm.minimization.matrix.MatrixVectorMultiplier `m` • ### Method Summary All Methods Modifier and Type Method and Description `boolean` `equals(java.lang.Object o)` `Vector` `evaluate(Vector input)` Returns m times input. `int` `hashCode()` `Vector` `transposeMult(Vector input)` Return A^(T) * input. • ### Methods inherited from class java.lang.Object `clone, finalize, getClass, notify, notifyAll, toString, wait, wait, wait` • ### Method Detail • #### evaluate `public Vector evaluate(Vector input)` Returns m times input. Specified by: `evaluate` in interface `Evaluator<Vector,Vector>` Overrides: `evaluate` in class `MatrixVectorMultiplier` Parameters: `input` - The vector to multiply by m. Returns: m times input. • #### transposeMult `public Vector transposeMult(Vector input)` Return A^(T) * input. Parameters: `input` - The vector to multiply by the transpose of A Returns: A^(T) * input • #### equals `public boolean equals(java.lang.Object o)` Overrides: `equals` in class `MatrixVectorMultiplier` • #### hashCode `public int hashCode()` Overrides: `hashCode` in class `MatrixVectorMultiplier`

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## peroxodicarbonate dianion, O2COOCO22- O4 \ C3 = O5 / O1 - O2 / O7 - C6 \\ O8 The ion charge is -2. ## Atomic Charges and Dipole Moment O1 charge=-0.341 O2 charge=-0.344 C3 charge= 0.932 O4 charge=-0.782 O5 charge=-0.806 C6 charge= 0.929 O7 charge=-0.805 O8 charge=-0.780 with a dipole moment of 7.41619 Debye ## Bond Lengths: between O1 and O2: distance=1.584 ang___ between O1 and C6: distance=1.433 ang___ between O1 and O7: distance=2.184 ang___ between O1 and O8: distance=2.350 ang___ between O2 and C3: distance=1.433 ang___ between O2 and O4: distance=2.350 ang___ between O2 and O5: distance=2.184 ang___ between C3 and O4: distance=1.255 ang___ between C3 and O5: distance=1.276 ang___ between O4 and O5: distance=2.302 ang___ between C6 and O7: distance=1.276 ang___ between C6 and O8: distance=1.255 ang___ between O7 and O8: distance=2.302 ang___ ## Bond Angles: for C3-O2-O1: angle=112.2 deg___ for O4-C3-O2: angle=121.7 deg___ for O5-C3-O2: angle=107.3 deg___ for C6-O1-O2: angle=112.2 deg___ for O7-C6-O1: angle=107.3 deg___ for O8-C6-O1: angle=121.7 deg___ ## Bond Orders (Mulliken): between O1 and O2: order=0.857___ between O1 and C6: order=0.779___ between O1 and O7: order=-0.180___ between O1 and O8: order=-0.150___ between O2 and C3: order=0.779___ between O2 and O4: order=-0.151___ between O2 and O5: order=-0.180___ between C3 and O4: order=1.642___ between C3 and O5: order=1.506___ between O4 and O5: order=-0.090___ between C6 and O7: order=1.506___ between C6 and O8: order=1.642___ between O7 and O8: order=-0.090___ ## Best Lewis Structure The Lewis structure that is closest to your structure is determined. The hybridization of the atoms in this idealized Lewis structure is given in the table below. Please note that your structure can't be well described by a single Lewis structure, because of extensive delocalization. (This analysis was run at the DZVP level with diffuse functions instead of the TZVP level, because of the size of the calculation.) ### Hybridization in the Best Lewis Structure 1. A bonding orbital for O1-O2 with 1.9851 electrons __has 50.00% O 1 character in a s0.29 p3 hybrid __has 50.00% O 2 character in a s0.29 p3 hybrid 2. A bonding orbital for O1-C6 with 1.9884 electrons __has 72.38% O 1 character in a sp2.42 hybrid __has 27.62% C 6 character in a sp2.86 hybrid 3. A bonding orbital for O2-C3 with 1.9884 electrons __has 72.38% O 2 character in a sp2.42 hybrid __has 27.62% C 3 character in a sp2.86 hybrid 4. A bonding orbital for C3-O4 with 1.9979 electrons __has 18.25% C 3 character in a p-pi orbital ( 99.21% p 0.79% d) __has 81.75% O 4 character in a p-pi orbital ( 99.76% p 0.24% d) 5. A bonding orbital for C3-O4 with 1.9944 electrons __has 34.08% C 3 character in a sp1.62 hybrid __has 65.92% O 4 character in a sp1.79 hybrid 6. A bonding orbital for C3-O5 with 1.9893 electrons __has 34.18% C 3 character in a sp1.73 hybrid __has 65.82% O 5 character in a sp1.93 hybrid 7. A bonding orbital for C6-O7 with 1.9893 electrons __has 34.18% C 6 character in a sp1.73 hybrid __has 65.82% O 7 character in a sp1.93 hybrid 8. A bonding orbital for C6-O8 with 1.9979 electrons __has 18.25% C 6 character in a p-pi orbital ( 99.21% p 0.79% d) __has 81.75% O 8 character in a p-pi orbital ( 99.76% p 0.24% d) 9. A bonding orbital for C6-O8 with 1.9944 electrons __has 34.08% C 6 character in a sp1.62 hybrid __has 65.92% O 8 character in a sp1.79 hybrid 18. A lone pair orbital for O1 with 1.9866 electrons 19. A lone pair orbital for O1 with 1.9269 electrons __made from a p-pi orbital ( 99.92% p 0.08% d) 20. A lone pair orbital for O2 with 1.9866 electrons 21. A lone pair orbital for O2 with 1.9269 electrons __made from a p-pi orbital ( 99.92% p 0.08% d) 22. A lone pair orbital for O4 with 1.9803 electrons 23. A lone pair orbital for O4 with 1.8949 electrons 24. A lone pair orbital for O5 with 1.9814 electrons 25. A lone pair orbital for O5 with 1.8951 electrons 26. A lone pair orbital for O5 with 1.7319 electrons __made from a p-pi orbital ( 99.79% p 0.21% d) 27. A lone pair orbital for O7 with 1.9814 electrons 28. A lone pair orbital for O7 with 1.8951 electrons 29. A lone pair orbital for O7 with 1.7319 electrons __made from a p-pi orbital ( 99.79% p 0.21% d) 30. A lone pair orbital for O8 with 1.9803 electrons 31. A lone pair orbital for O8 with 1.8949 electrons 145. A antibonding orbital for O1-C6 with 0.1253 electrons __has 27.62% O 1 character in a sp2.42 hybrid __has 72.38% C 6 character in a sp2.86 hybrid 146. A antibonding orbital for O2-C3 with 0.1253 electrons __has 27.62% O 2 character in a sp2.42 hybrid __has 72.38% C 3 character in a sp2.86 hybrid 147. A antibonding orbital for C3-O4 with 0.3260 electrons __has 81.75% C 3 character in a p-pi orbital ( 99.21% p 0.79% d) __has 18.25% O 4 character in a p-pi orbital ( 99.76% p 0.24% d) 151. A antibonding orbital for C6-O8 with 0.3260 electrons __has 81.75% C 6 character in a p-pi orbital ( 99.21% p 0.79% d) __has 18.25% O 8 character in a p-pi orbital ( 99.76% p 0.24% d) -With core pairs on: O 1 O 2 C 3 O 4 O 5 C 6 O 7 O 8 - #### Donor Acceptor Interactions in the Best Lewis Structure The localized orbitals in your best Lewis structure can interact strongly. A filled bonding or lone pair orbital can act as a donor and an empty or filled bonding, antibonding, or lone pair orbital can act as an acceptor. These interactions can strengthen and weaken bonds. For example, a lone pair donor->antibonding acceptor orbital interaction will weaken the bond associated with the antibonding orbital. Conversly, an interaction with a bonding pair as the acceptor will strengthen the bond. Strong electron delocalization in your best Lewis structure will also show up as donor-acceptor interactions. Interactions greater than 20 kJ/mol for bonding and lone pair orbitals are listed below. The interaction of lone pair donor orbital, 18, for O1 with the second antibonding acceptor orbital, 152, for C6-O8 is 26.2 kJ/mol. The interaction of the second lone pair donor orbital, 19, for O1 with the antibonding acceptor orbital, 151, for C6-O8 is 152. kJ/mol. The interaction of lone pair donor orbital, 20, for O2 with the second antibonding acceptor orbital, 148, for C3-O4 is 26.2 kJ/mol. The interaction of the second lone pair donor orbital, 21, for O2 with the antibonding acceptor orbital, 147, for C3-O4 is 152. kJ/mol. The interaction of lone pair donor orbital, 22, for O4 with the antibonding acceptor orbital, 149, for C3-O5 is 23.8 kJ/mol. The interaction of the second lone pair donor orbital, 23, for O4 with the antibonding acceptor orbital, 146, for O2-C3 is 173. kJ/mol. The interaction of the second lone pair donor orbital, 23, for O4 with the antibonding acceptor orbital, 149, for C3-O5 is 86.3 kJ/mol. The interaction of lone pair donor orbital, 24, for O5 with the second antibonding acceptor orbital, 148, for C3-O4 is 24.8 kJ/mol. The interaction of the second lone pair donor orbital, 25, for O5 with the antibonding acceptor orbital, 146, for O2-C3 is 138. kJ/mol. The interaction of the second lone pair donor orbital, 25, for O5 with the second antibonding acceptor orbital, 148, for C3-O4 is 75.7 kJ/mol. The interaction of the third lone pair donor orbital, 26, for O5 with the antibonding acceptor orbital, 147, for C3-O4 is 572. kJ/mol. The interaction of lone pair donor orbital, 27, for O7 with the second antibonding acceptor orbital, 152, for C6-O8 is 24.8 kJ/mol. The interaction of the second lone pair donor orbital, 28, for O7 with the antibonding acceptor orbital, 145, for O1-C6 is 138. kJ/mol. The interaction of the second lone pair donor orbital, 28, for O7 with the second antibonding acceptor orbital, 152, for C6-O8 is 75.7 kJ/mol. The interaction of the third lone pair donor orbital, 29, for O7 with the antibonding acceptor orbital, 151, for C6-O8 is 572. kJ/mol. The interaction of lone pair donor orbital, 30, for O8 with the antibonding acceptor orbital, 150, for C6-O7 is 23.8 kJ/mol. The interaction of the second lone pair donor orbital, 31, for O8 with the antibonding acceptor orbital, 145, for O1-C6 is 173. kJ/mol. The interaction of the second lone pair donor orbital, 31, for O8 with the antibonding acceptor orbital, 150, for C6-O7 is 86.3 kJ/mol. ## Molecular Orbital Energies The orbital energies are given in eV, where 1 eV=96.49 kJ/mol. Orbitals with very low energy are core 1s orbitals. More antibonding orbitals than you might expect are sometimes listed, because d orbitals are always included for heavy atoms and p orbitals are included for H atoms. Up spins are shown with a ^ and down spins are shown as v. 35 ----- 11.24 34 ----- 9.323 33 ----- 9.169 32 ----- 6.369 31 -^-v- 3.045 30 -^-v- 3.014 29 -^-v- 2.839 28 -^-v- 2.259 27 -^-v- 2.189 26 -^-v- 2.048 25 -^-v- 1.758 24 -^-v- 0.778 23 -^-v- 0.337 22 -^-v- -1.494 21 -^-v- -1.556 20 -^-v- -1.618 19 -^-v- -2.222 18 -^-v- -2.343 17 -^-v- -2.521 16 -^-v- -3.597 15 -^-v- -5.231 14 -^-v- -13.26 13 -^-v- -14.94 12 -^-v- -15.00 11 -^-v- -15.90 10 -^-v- -17.45 9 -^-v- -18.47 8 -^-v- -261.0 7 -^-v- -261.0 6 -^-v- -496.6 5 -^-v- -496.6 4 -^-v- -496.7 3 -^-v- -496.7 2 -^-v- -499.0 1 -^-v- -499.0 ## Total Electronic Energy The total electronic energy is a very large number, so by convention the units are given in atomic units, that is Hartrees (H). One Hartree is 2625.5 kJ/mol. The energy reference is for totally dissociated atoms. In other words, the reference state is a gas consisting of nuclei and electrons all at infinite distance from each other. The electronic energy includes all electric interactions and the kinetic energy of the electrons. This energy does not include translation, rotation, or vibration of the the molecule. Total electronic energy = -527.7492197954 Hartrees

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# 46 So there’s this thing where GPT-3 is able to do addition, it has the internal model to do addition, but it takes a little poking and prodding to actually get it to do addition. “Few-shot learning”, as the paper calls it. Rather than prompting the model with Q: What is 48 + 76? A: Q: What is 48 + 76? A: 124 Q: What is 34 + 53? A: 87 Q: What is 29 + 86? A: The same applies to lots of other tasks: arithmetic, anagrams and spelling correction, translation, assorted benchmarks, etc. To get GPT-3 to do the thing we want, it helps to give it a few examples, so it can “figure out what we’re asking for”. This is an alignment problem. Indeed, I think of it as the quintessential alignment problem: to translate what-a-human-wants into a specification usable by an AI. The hard part is not to build a system which can do the thing we want, the hard part is to specify the thing we want in such a way that the system actually does it. The GPT family of models are trained to mimic human writing. So the prototypical “alignment problem” on GPT is prompt design: write a prompt such that actual human writing which started with that prompt would likely contain the thing you actually want. Assuming that GPT has a sufficiently powerful and accurate model of human writing, it should then generate the thing you want. Viewed through that frame, “few-shot learning” just designs a prompt by listing some examples of what we want - e.g. listing some addition problems and their answers. Call me picky, but that seems like a rather primitive way to design a prompt. Surely we can do better? Indeed, people are already noticing clever ways to get better results out of GPT-3 - e.g. TurnTrout recommends conditioning on writing by smart people, and the right prompt makes the system complain about nonsense rather than generating further nonsense in response. I expect we’ll see many such insights over the next month or so. ## Capabilities vs Alignment as Bottleneck to Value I said that the alignment problem on GPT is prompt design: write a prompt such that actual human writing which started with that prompt would likely contain the thing you actually want. Important point: this is worded to be agnostic to the details GPT algorithm itself; it’s mainly about predictive power. If we’ve designed a good prompt, the current generation of GPT might still be unable to solve the problem - e.g. GPT-3 doesn’t understand long addition no matter how good the prompt, but some future model with more predictive power should eventually be able to solve it. In other words, there’s a clear distinction between alignment and capabilities: • alignment is mainly about the prompt, and asks whether human writing which started with that prompt would be likely to contain the thing you want • capabilities are mainly about GPT’s model, and ask about how well GPT-generated writing matches realistic human writing Interesting question: between alignment and capabilities, which is the main bottleneck to getting value out of GPT-like models, both in the short term and the long(er) term? In the short term, it seems like capabilities are still pretty obviously the main bottleneck. GPT-3 clearly has pretty limited “working memory” and understanding of the world. That said, it does seem plausible that GPT-3 could consistently do at least some economically-useful things right now, with a carefully designed prompt - e.g. writing ad copy or editing humans’ writing. In the longer term, though, we have a clear path forward for better capabilities. Just continuing along the current trajectory will push capabilities to an economically-valuable point on a wide range of problems, and soon. Alignment, on the other hand, doesn’t have much of a trajectory at all yet; designing-writing-prompts-such-that-writing-which-starts-with-the-prompt-contains-the-thing-you-want isn’t exactly a hot research area. There’s probably low-hanging fruit there for now, and it’s largely unclear how hard the problem will be going forward. Two predictions on this front: • With this version of GPT and especially with whatever comes next, we’ll start to see a lot more effort going into prompt design (or the equivalent alignment problem for future systems) • As the capabilities of GPT-style models begin to cross beyond what humans can do (at least in some domains), alignment will become a much harder bottleneck, because it’s hard to make a human-mimicking system do things which humans cannot do Reasoning for the first prediction: GPT-3 is right on the borderline of making alignment economically valuable - i.e. it’s at the point where there’s plausibly some immediate value to be had by figuring out better ways to write prompts. That means there’s finally going to be economic pressure for alignment - there’s going to be ways to make money by coming up with better alignment tricks. That won’t necessarily mean economic pressure for generalizable or robust alignment tricks, though - most of the economy runs on ad-hoc barely-good-enough tricks most of the time, and early alignment tricks will likely be the same. In the longer run, focus will shift toward more robust alignment, as the low-hanging problems are solved and the remaining problems have most of their value in the long tail. Reasoning for the second prediction: how do I write a prompt such that human writing which began with that prompt would contain a workable explanation of a cheap fusion power generator? In practice, writing which claims to contain such a thing is generally crackpottery. I could take a different angle, maybe write some section-headers with names of particular technologies (e.g. electric motor, radio antenna, water pump, …) and descriptions of how they work, then write a header for “fusion generator” and let the model fill in the description. Something like that could plausibly work. Or it could generate scifi technobabble, because that’s what would be most likely to show up in such a piece of writing today. It all depends on which is "more likely" to appear in human writing. Point is: GPT is trained to mimic human writing; getting it to write things which humans cannot currently write is likely to be hard, even if it has the requisite capabilities. # 46 New Comment I wonder how long we'll be in the "prompt programming" regime. As Nick Cammarata put it: We should actually be programming these by manipulating the hidden layers and prompts are a stand-in until we can. My guess is that OpenAI will pretty quickly (within the next year) find a much better way to interface with what GPT-3 has learned. Do others agree? Any reason to think that wouldn't be possible (or wouldn't give significant benefits)? The problem with directly manipulating the hidden layers is reusability. If we directly manipulate the hidden layers, then we have to redo that whenever a newer, shinier model comes out, since the hidden layers will presumably be different. On the other hand, a prompt is designed so that human writing which starts with that prompt will likely contain the thing we want - a property mostly independent of the internal structure of the model, so presumably the prompt can be reused. I think the eventual solution here (and a major technical problem of alignment) is to take an internal notion learned by one model (i.e. found via introspection tools), back out a universal representation of the real-world pattern it represents, then match that real-world pattern against the internals of a different model in order to find the "corresponding" internal notion. Assuming that the first model has learned a real pattern which is actually present in the environment, we should expect that "better" models will also have some structure corresponding to that pattern - otherwise they'd lose predictive power on at least the cases where that pattern applies. Ideally, this would all happen in such a way that the second model can be more accurate, and that increased accuracy would be used. In the shorter term, I agree OpenAI will probably come up with some tricks over the next year or so. Planned summary for the Alignment Newsletter: Currently, many people are trying to figure out how to prompt GPT-3 into doing what they want -- in other words, how to align GPT-3 with their desires. GPT-3 may be capable of the task, but that doesn’t mean it will do it (potential example). This suggests that alignment will soon be a bottleneck on our ability to get value from large language models. Certainly GPT-3 isn’t perfectly capable yet. The author thinks that in the immediate future the major bottleneck will still be its capability, but we have a clear story for how to improve its capabilities: just scale up the model and data even more. Alignment on the other hand is much harder: we don’t know how to <@translate@>(@Alignment as Translation@) the tasks we want into a format that will cause GPT-3 to “try” to accomplish that task. As a result, in the future we might expect a lot more work to go into prompt design (or whatever becomes the next way to direct language models at specific tasks). In addition, once GPT is better than humans (at least in some domains), alignment in those domains will be particularly difficult, as it is unclear how you would get a system trained to mimic humans <@to do better than humans@>(@The easy goal inference problem is still hard@). Planned opinion: The general point of this post seems clearly correct and worth pointing out. I’m looking forward to the work we’ll see in the future figuring out how to apply these broad and general methods to real tasks in a reliable way. LGTM I think it's important to take a step back and notice how AI risk-related arguments are shifting. In the sequences, a key argument (probably the key argument) for AI risk was the complexity of human value, and how it would be highly anthropomorphic for us to believe that our evolved morality was embedded in the fabric of the universe in a way that any intelligent system would naturally discover.  An intelligent system could just as easily maximize paperclips, the argument went. No one seems to have noticed that GPT actually does a lot to invalidate the original complexity-of-value-means-FAI-is-super-difficult argument. You write: To get GPT-3 to do the thing we want, it helps to give it a few examples, so it can “figure out what we’re asking for”. This is an alignment problem. We've gotten from "the alignment problem is about complexity of value" to "the alignment problem is about programming by example" (also known as "supervised learning", or Machine Learning 101). There's actually a long history of systems which combine observing-lots-of-data-about-the-world (GPT-3's training procedure, "unsupervised learning") with programming-by-example ("supervised learning") The term for this is "semi-supervised learning".  When I search for it on Google Scholar, I get almost 100K results. ("Transfer learning" is a related literature.) The fact that GPT-3's API only does text completion is, in my view, basically just an API detail that we shouldn't particularly expect to be true of GPT-4 or GPT-5.  There's no reason why OpenAI couldn't offer an API which takes in a list of (x, y) pairs and then given some x it predicts y.  I expect if they chose to do this as a dedicated engineering effort, getting into the guts of the system as needed, and collected a lot of user feedback on whether the predicted y was correct for many different problems, they could exceed the performance gains you can currently get by manipulating the prompt. In other words, there’s a clear distinction between alignment and capabilities: • alignment is mainly about the prompt, and asks whether human writing which started with that prompt would be likely to contain the thing you want • capabilities are mainly about GPT’s model, and ask about how well GPT-generated writing matches realistic human writing I'm wary of a world where "the alignment problem" becomes just a way to refer to "whatever the difference is between our current system and the ideal system". (If I trained a supervised learning system to classify word vectors based on whether they're things that humans like or dislike, and the result didn't work very well, I can easily imagine some rationalists telling me this represented a failure to "solve the alignment problem"--even if the bottleneck was mainly in the word vectors themselves, as evidenced e.g. by large performance improvements on switching to higher-dimensional word vectors.)  I'm reminded of a classic bad argument. As the capabilities of GPT-style models begin to cross beyond what humans can do (at least in some domains), alignment will become a much harder bottleneck, because it’s hard to make a human-mimicking system do things which humans cannot do If it's hard to make a human-mimicking system do things which humans cannot do, why should we expect the capabilities of GPT-style models to cross beyond what humans can do in the first place? My steelman of what you're saying: Over the course of GPT's training procedure, it incidentally acquires superhuman knowledge, but then that superhuman knowledge gets masked as it sees more data and learns which specific bits of its superhuman knowledge humans are actually ignorant of (and even after catastrophic forgetting, some bits of superhuman knowledge remain at the end of the training run).  If that's the case, it seems like we could mitigate the problem by restricting GPT's training to textbooks full of knowledge we're very certain in (or fine-tuning GPT on such textbooks after the main training run, or simply increasing their weight in the loss function).  Or replace every phrase like "we don't yet know X" in GPT's training data with "X is a topic for a more advanced textbook", so GPT never ends up learning what humans are actually ignorant about. Or simply use a prompt which starts with the letterhead for a university press release: "Top MIT scientists have made an important discovery related to X today..." Or a prompt which looks like the beginning of a Nature article. Or even: "Google has recently released a super advanced new AI system which is aligned with human values; given X, it says Y." (Boom! I solved the alignment problem! We thought about uploading a human, but uploading an FAI turned out to work much better.) (Sorry if this comment came across as grumpy.  I'm very frustrated that so much upvoting/downvoting on LW seems to be based on what advances AI doom as a bottom line. It's not because I think superhuman AI is automatically gonna be safe. It's because I'd rather we did not get distracted by a notion of "the alignment problem" which OpenAI could likely solve with a few months of dedicated work on their API.) My main response to this needs a post of its own, so I'm not going to argue it in detail here, but I'll give a summary and then address some tangential points. Summary: the sense in which human values are "complex" is not about predictive power. A low-level physical model of some humans has everything there is to know about human values embedded within it; it has all the predictive power which can be had by a good model of human values. The hard part is pointing to the thing we consider "human values" embedded within that model. In large part, that's hard because it's not just a matter of predictive power. Looking at it as semi-supervised learning: it's not actually hard for an unsupervised learner to end up with some notion of human values embedded in its world-model, but finding the embedding is hard, and it's hard in a way which cannot-even-in-principle be solved by supervised learning (because that would reduce it to predictive power). On to tangential points... We've gotten from "the alignment problem is about complexity of value" to "the alignment problem is about programming by example" (also known as "supervised learning", or Machine Learning 101). Tangential point 1: a major thing which I probably didn't emphasize enough in the OP is the difference between "the problem of aligning with X" and "the problem of aligning with human values". For any goal, we can talk about the problem of aligning a system with that goal. So e.g. if I want a system to solve addition problems, then I can talk about aligning GPT-3 with that particular goal. This is not the same problem as aligning with human values. And for very powerful AI, the system has to be aligned with human values, because aligning it with anything else gives us paperclip-style problems. That said, it does seem like alignment problems in general have certain common features - i.e. there are some major common subproblems between aligning a system with the goal of solving addition problems and aligning a system with human values. That's why it makes sense to talk about "alignment problems" as a natural category of problems. To the extent that supervised/semi-supervised learning are ways of specifying what we want, we can potentially learn useful things about alignment problems in general by thinking about those approaches - in particular, we can notice a lot of failure modes that way! The fact that GPT-3's API only does text completion is, in my view, basically just an API detail that we shouldn't particularly expect to be true of GPT-4 or GPT-5.  There's no reason why OpenAI couldn't offer an API which takes in a list of (x, y) pairs and then given some x it predicts y.  I expect if they chose to do this as a dedicated engineering effort, getting into the guts of the system as needed, and collected a lot of user feedback on whether the predicted y was correct for many different problems, they could exceed the performance gains you can currently get by manipulating the prompt. Tangential point 2: interfaces are a scarce resource, and "programming by example" is a bad interface. Sure, the GPT team could offer this, but it would be a step backward for problems which are currently hard, not a step forward. The sort of problems which can be cheaply and usefully expressed as supervised learning problems already have lots of support; it's the rest of problem-space that's interesting here. Specifically in the context of alignment-to-human-values, this goes back to the main point: "human values" are complex in a way which does not easily reduce to predictive power, so offering a standard supervised-learning-style interface does not really get us any closer to solving that problem. (If I trained a supervised learning system to classify word vectors based on whether they're things that humans like or dislike, and the result didn't work very well, I can easily imagine some rationalists telling me this represented a failure to "solve the alignment problem"--even if the bottleneck was mainly in the word vectors themselves, as evidenced e.g. by large performance improvements on switching to higher-dimensional word vectors.) Tangential point 3: the formulation in the OP specifically avoids that failure mode, and that's not an accident. Any failure which can be solved by using a system with better general-purpose predictive power is not an alignment failure. If a system is trained to predict X, then the "alignment problem" (at least as I'm using the phrase) is about aligning X with what we want, not about aligning the system itself. (I think this is also what we usually mean by "outer alignment".) If it's hard to make a human-mimicking system do things which humans cannot do, why should we expect the capabilities of GPT-style models to cross beyond what humans can do in the first place? Tangential point 4: at a bare minimum, GPT-style models should be able to do a lot of different things which a lot of different people can do, but which no single person can do. I find it plausible that GPT-3 has some such capabilities already - e.g. it might be able to translate between more different languages with better fluency than any single human. Sorry if this comment came across as grumpy. It was usefully and intelligently grumpy, which plenty good enough for me. Summary: the sense in which human values are "complex" is not about predictive power. A low-level physical model of some humans has everything there is to know about human values embedded within it; it has all the predictive power which can be had by a good model of human values. The hard part is pointing to the thing we consider "human values" embedded within that model. In large part, that's hard because it's not just a matter of predictive power. 1. This still sounds like a shift in arguments to me. From what I remember, the MIRI-sphere take on uploads is (was?): "if uploads come before AGI, that's probably a good thing, as long as it's a sufficiently high-fidelity upload of a benevolent individual, and the technology is not misused prior to that person being uploaded". (Can probably dig up some sources if you don't believe me.) 2. I still don't buy it. Your argument proves too much--how is it that transfer learning works? Seems that pointing to relevant knowledge embedded in an ML model isn't super hard in practice. Tangential point 1: a major thing which I probably didn't emphasize enough in the OP is the difference between "the problem of aligning with X" and "the problem of aligning with human values". For any goal, we can talk about the problem of aligning a system with that goal. So e.g. if I want a system to solve addition problems, then I can talk about aligning GPT-3 with that particular goal. This is not the same problem as aligning with human values. Is there a fundamental difference? You say: 'The hard part is pointing to the thing we consider "human values" embedded within that model.' What is it about pointing to the thing we consider "human values" which makes it fundamentally different from pointing to the thing we consider a dog? The main possible reason I can think of is because a dog is in some sense a more natural category than human values. That there are a bunch of different things which are kind of like human values, but not quite, and one has to sort through a large number of them in order to pinpoint the right one ("will the REAL human values please stand up?") (I'm not sure I buy this, but it's the only way I see for your argument to make sense.) As an example of something which is not a natural category, consider a sandwich. Or, to an even greater degree: "Tasty ice cream flavors" is not a natural category, because everyone has their own ice cream preferences. And for very powerful AI, the system has to be aligned with human values Disagree, you could also align it with corrigibility. "programming by example" is a bad interface A big part of this post is about how people are trying to shoehorn programming by example into text completion, wasn't it? What do you think a good interface would be? The sort of problems which can be cheaply and usefully expressed as supervised learning problems already have lots of support I think perhaps you're confusing the collection of ML methods that conventionally fall under the umbrella of "supervised learning", and the philosophical task of predicting (x, y) pairs. As an example, from a philosophical perspective, I could automate away most software development if I could train a supervised learning system where x=the README of a Github project and y=the code for that Github project. But most of the ML methods that come to the mind of an average ML person when I say "supervised learning" are not remotely up to that task. (BTW note that such a collection of README/code pairs comes pretty close to pinpointing the notion of "do what I mean". Which could be a very useful building block--remember, a premise of the classic AI safety arguments is that "do what I mean" doesn't exist. Also note that quality code on Github is a heck of a lot more interpretable than the weights in a neural net--and restricting the training set to quality code seems pretty easy to do.) It was usefully and intelligently grumpy, which plenty good enough for me. Glad to hear it. I get so demoralized commenting on LW because it so often feels like a waste of time in retrospect. The main possible reason I can think of is because a dog is in some sense a more natural category than human values. That there are a bunch of different things which are kind of like human values, but not quite, and one has to sort through a large number of them in order to pinpoint the right one ("will the REAL human values please stand up?") This is close, though not exactly what I want to claim. It's not that "dogs" are a "more natural category" in the sense that there are fewer similar categories which are hard to tell apart. Rather, it's that "dogs" are a less abstract natural category. Like, "human" is a natural category in roughly the same way as "dog", but in order to get to "human values" we need a few more layers of abstraction on top of that, and some of those layers have different type signatures than the layers below - e.g. we need not just a notion of "human", but a notion of humans "wanting things", which requires an entirely different kind of model from recognizing dogs. And that's exactly the kind of complexity which is hard for something based on predictive power. Lower abstraction levels should generally perform better in terms of raw prediction, but the thing we want to point to lives at a high abstraction level. We are able to get systems to learn some abstractions just by limiting compute - today's deep nets have nowhere near the compute to learn to do Bayesian updates on low-level physics, so it needs to learn some abstraction. But the exact abstraction level learned is always going to be a tradeoff between available compute and predictive power. I do think there's probably a wide range of parameters which would end up using the "right" level of abstraction for human values to be "natural", but we don't have a good way to recognize when that has/hasn't happened, and relying on it happening would be a crapshoot. (Also, sandwiches are definitely a natural category. Just because a cluster has edge cases does not make it any less of a cluster, even if a bunch of trolls argue about the edge cases. "Tasty ice cream flavors" is also a natural category if we know who the speaker is, which is exactly how humans understand the phrase in practice.) I think perhaps you're confusing the collection of ML methods that conventionally fall under the umbrella of "supervised learning", and the philosophical task of predicting (x, y) pairs... Part of what I mean here by "already have lots of support" is that there's already a path to improvement on these sorts of problems, not necessarily that they're already all solved. The point is that we already have a working interface for this sort of thing, and for an awful lot of problems, the interface is what's hard. As for how close a collection of README/code pairs comes to pinpointing "do what I mean"... imagine picking a random github repo, giving its README to a programmer, and asking them to write code implementing the behavior described. This is basically equivalent to a step in the waterfall model of development: create a specification, then have someone implement it in a single step without feedback. The general consensus among developers seems to be that this works very badly. Indeed, this is an example of interfaces being hard: you'd think "I give you a spec, you give me code" would be a good interface, but it actually works pretty badly. And that's exactly the kind of complexity which is hard for something based on predictive power. Lower abstraction levels should generally perform better in terms of raw prediction, but the thing we want to point to lives at a high abstraction level. You told me that "it's not actually hard for an unsupervised learner to end up with some notion of human values embedded in its world-model".  Now you're saying that things based on "predictive power" have trouble learning things at high abstraction levels.  Doesn't this suggest that your original statement is wrong, and the predict-the-next-word training method used by GPT-3 means it will not develop notions such as human value? (BTW, I think this argument proves too much.  Recognizing a photo of a dog does require learning various lower-level abstractions such as legs, nose, ears, etc. which in turn require even lower-level abstractions such as fur textures and object boundaries.  In any case, if you think things based on "predictive power" have trouble learning things at high abstraction levels, that suggests that it should also have trouble understanding e.g. Layer 7 in the OSI networking model.) We are able to get systems to learn some abstractions just by limiting compute - today's deep nets have nowhere near the compute to learn to do Bayesian updates on low-level physics, so it needs to learn some abstraction. But the exact abstraction level learned is always going to be a tradeoff between available compute and predictive power. I do think there's probably a wide range of parameters which would end up using the "right" level of abstraction for human values to be "natural", but we don't have a good way to recognize when that has/hasn't happened, and relying on it happening would be a crapshoot. It sounds like you're talking about the bias/variance tradeoff?  The standard solution is to use cross validation, do you have any reason to believe it wouldn't work here? The point is that we already have a working interface for this sort of thing, and for an awful lot of problems, the interface is what's hard. I'm very unpersuaded that interfaces are the hard part of creating superhuman AI. As for how close a collection of README/code pairs comes to pinpointing "do what I mean"... imagine picking a random github repo, giving its README to a programmer, and asking them to write code implementing the behavior described. This is basically equivalent to a step in the waterfall model of development: create a specification, then have someone implement it in a single step without feedback. The general consensus among developers seems to be that this works very badly. I mean, if you don't like the result, tweak the README and run the code-generating AI a second time. The reason waterfall sucks is because human programmers are time-consuming and costly, and people who hire them don't always know precisely what they want.  And testing intermediate versions of the software can help them figure that out.  But if generating a final version of the software is costless, there's no reason not to test the final version instead of an intermediate version. You told me that "it's not actually hard for an unsupervised learner to end up with some notion of human values embedded in its world-model".  Now you're saying that things based on "predictive power" have trouble learning things at high abstraction levels.  Doesn't this suggest that your original statement is wrong, and the predict-the-next-word training method used by GPT-3 means it will not develop notions such as human value? The original example is a perfect example of what this looks like: an unsupervised learner, given crap-tons of data and compute, should have no difficulty learning a low-level physics model of humans. That model will have great predictive power, which is why the model will learn it. Human values will be embedded in that model in exactly the same way that they're embedded in physical humans. Likewise, GPT-style models should have no trouble learning some model with human values embedded in it. But that embedding will not necessarily be simple; there won't just be a neuron that lights up in response to humans having their values met. The model will have a notion of human values embedded in it, but it won't actually use "human values" as an abstract object in its internal calculations; it will work with some lower-level "components" which themselves implement/embed human values. It sounds like you're talking about the bias/variance tradeoff?  The standard solution is to use cross validation, do you have any reason to believe it wouldn't work here? I am definitely not talking about bias-variance tradeoff. I am talking about compute-accuracy tradeoff. Again, think about the example of Bayesian updates on a low-level physical model: there is no bias-variance tradeoff there. It's the ideal model, full stop. The reason we can't use it is because we don't have that much compute. In order to get computationally tractable models, we need to operate at higher levels of abstraction than "simulate all these quantum fields". I'm very unpersuaded that interfaces are the hard part of creating superhuman AI. Aligning superhuman AI, not just creating it. If you're unpersuaded, you should go leave feedback on Alignment as Translation, which directly talks about alignment as an interface problem. Likewise, GPT-style models should have no trouble learning some model with human values embedded in it. But that embedding will not necessarily be simple; there won't just be a neuron that lights up in response to humans having their values met. The model will have a notion of human values embedded in it, but it won't actually use "human values" as an abstract object in its internal calculations; it will work with some lower-level "components" which themselves implement/embed human values. If it's read moral philosophy, it should have some notion of what the words "human values" mean. In any case, I still don't understand what you're trying to get at. Suppose I pretrain a neural net to differentiate lots of non-marsupial animals. It doesn't know what a koala looks like, but it has some lower-level "components" which would allow it to characterize a koala. Then I use transfer learning and train it to differentiate marsupials. Now it knows about koalas too. This is actually a tougher scenario than what you're describing (GPT will have seen human values yet the pretrained net hasn't seen koalas in my hypothetical), but it's a boring application of transfer learning. Locating human values might be trickier than characterizing a koala, but the difference seems quantitative, not qualitative. If it's read moral philosophy, it should have some notion of what the words "human values" mean. GPT-3 and systems like it are trained to mimic human discourse. Even if (in the limit of arbitrary computational power) it manages to encode an implicit representation of human values somewhere in its internal state, in actual practice there is nothing tying that representation to the phrase "human values", since moral philosophy is written by (confused) humans, and in human-written text the phrase "human values" is not used in the consistent, coherent manner that would be required to infer its use as a label for a fixed concept. This is essentially the "tasty ice cream flavors" problem, am I right?  Trying to check if we're on the same page. If so: John Wentsworth said "Tasty ice cream flavors" is also a natural category if we know who the speaker is So how about instead of talking about "human values", we talk about what a particular moral philosopher endorses saying or doing, or even better, what a committee of famous moral philosophers would endorse saying/doing. No, this is not the "tasty ice cream flavors" problem. The problem there is that the concept is inherently relative to a person. That problem could apply to "human values", but that's a separate issue from what dxu is talking about. The problem is that "what a committee of famous moral philosophers would endorse saying/doing", or human written text containing the phrase "human values", is a proxy for human values, not a direct pointer to the actual concept. And if a system is trained to predict what the committee says, or what the text says, then it will learn the proxy, but that does not imply that it directly uses the concept. Well, the moral judgements of a high-fidelity upload of a benevolent human are also a proxy for human values--an inferior proxy, actually.  Seems to me you're letting the perfect be the enemy of the good. It doesn't matter how high-fidelity the upload is or how benevolent the human is, I'm not happy giving them the power to launch nukes without at least two keys, and a bunch of other safeguards on top of that. "Don't let the perfect be the enemy of the good" is advice for writing emails and cleaning the house, not nuclear security. The capabilities of powerful AGI will be a lot more dangerous than nukes, and merit a lot more perfectionism. Humans themselves are not aligned enough that I would be happy giving them the sort of power that AGI will eventually have. They'd probably be better than many of the worst-case scenarios, but they still wouldn't be a best or even good scenario. Humans just don't have the processing power to avoid shooting themselves (and the rest of the universe) in the foot sooner or later, given that kind of power. It doesn't matter how high-fidelity the upload is or how benevolent the human is, I'm not happy giving them the power to launch nukes without at least two keys, and a bunch of other safeguards on top of that. Here are some of the people who have the power to set off nukes right now: • Donald Trump • Kim Jong-un • Both parties in this conflict "Don't let the perfect be the enemy of the good" is advice for writing emails and cleaning the house, not nuclear security. “A good plan violently executed now is better than a perfect plan executed at some indefinite time in the future.” - George Patton Just because it's in your nature (and my nature, and the nature of many people who read this site) to be a cautious nerd, does not mean that the cautious nerd orientation is always the best orientation to have. In any case, it may be that the annual amount of xrisk is actually quite low, and no one outside the rationalist community is smart enough to invent AGI, and we have all the time in the world. In which case, yes, being perfectionistic is the right strategy. But this still seems to represent a major retreat from the AI doomist position that AI doom is the default outcome. It's a classic motte-and-bailey: "It's very hard to build an AGI which isn't a paperclipper!" "Well actually here are some straightforward ways one might be able to create a helpful non-paperclipper AGI..." "Yeah but we gotta be super perfectionistic because there is so much at stake!" Your final "humans will misuse AI" worry may be justified, but I think naive deployment of this worry is likely to be counterproductive. Suppose there are two types of people, "cautious" and "incautious". Suppose that the "humans will misuse AI" worry discourages cautious people from developing AGI, but not incautious people. So now we're in a world where the first AGI is most likely controlled by incautious people, making the "humans will misuse AI" worry even more severe. Humans just don't have the processing power to avoid shooting themselves (and the rest of the universe) in the foot sooner or later, given that kind of power. If you're willing to grant the premise of the technical alignment problem being solved, shooting oneself in the foot would appear to be much less of a worry, because you can simply tell your FAI "please don't let me shoot myself in the foot too badly", and it will prevent you from doing that. It's a classic motte-and-bailey: "It's very hard to build an AGI which isn't a paperclipper!" "Well actually here are some straightforward ways one might be able to create a helpful non-paperclipper AGI..." "Yeah but we gotta be super perfectionistic because there is so much at stake!" There is a single coherent position here in which it is very hard to build an AGI which reliably is not a paperclipper. Yes, there are straightforward ways one might be able to create a helpful non-paperclipper AGI. But that "might" is carrying a lot of weight. All those straightforward ways have failure modes which will definitely occur in at least some range of parameters, and we don't know exactly what those parameter ranges are. It's sort of like saying: "It's very hard to design a long bridge which won't fall down!" "Well actually here are some straightforward ways one might be able to create a long non-falling-down bridge..." <shows picture of a wooden truss> What I'm saying is, that truss is design is 100% going to fail once it gets big enough, and we don't currently know how big that is. When I say "it's hard to design a long bridge which won't fall down", I do not mean a bridge which might not fall down if we're lucky and just happen to be within the safe parameter range. In any case, it may be that the annual amount of xrisk is actually quite low, and no one outside the rationalist community is smart enough to invent AGI, and we have all the time in the world. In which case, yes, being perfectionistic is the right strategy. But this still seems to represent a major retreat from the AI doomist position that AI doom is the default outcome. These are sufficient conditions for a careful strategy to make sense, not necessary conditions. Here's another set of sufficient conditions, which I find more realistic: the gains to be had in reducing AI risk are binary. Either we find the "right" way of doing things, in which case risk drops to near-zero, or we don't, in which case it's a gamble and we don't have much ability to adjust the chances/payoff. There are no significant marginal gains to be had. There is a single coherent position here in which it is very hard to build an AGI which reliably is not a paperclipper. This is simultaneously • a major retreat from the "default outcome is doom" thesis which is frequently trotted out on this site (the statement is consistent with a AGI design that's is 99.9% likely to be safe, which is very much incompatible with "default outcome is doom") • unrelated to our upload discussion (an upload is not an AGI, but you said even a great upload wasn't good enough for you) You've picked a position vaguely in between the motte and the bailey and said "the motte and the bailey are both equivalent to this position!"  That doesn't look at all true to me. All those straightforward ways have failure modes which will definitely occur in at least some range of parameters, and we don't know exactly what those parameter ranges are. This is a very strong claim which to my knowledge has not been well-justified anywhere.  Daniel K agreed with me the other day that there isn't a standard reference for this claim.  Do you know of one? There are a couple problems I see here: • Simple is not the same as obvious.  Even if someone at some point tried to think of every obvious solution and justifiably discarded them all, there are probably many "obvious" solutions they didn't think of. • Nothing ever gets counted as evidence against this claim.  Simple proposals get rejected on the basis that everyone knows simple proposals won't work. A MIRI employee openly admitted here that they apply different standards of evidence to claims of safety vs claims of not-safety.  Maybe there are good arguments for that, but the problem is that if you're not careful, your view of reality is gonna get distorted.  Which means community wisdom on claims such as "simple solutions never work" is likely to be systematically wrong.  "Everyone knows X", without a good written defense of X, or a good answer to "what would change the community's mind about X", is fertile ground for information cascades etc.  And this is on top of standard ideological homophily problems (the AI safety community is very self-selected subset of the broader AI research world). What I'm saying is, that truss is design is 100% going to fail once it gets big enough, and we don't currently know how big that is. When I say "it's hard to design a long bridge which won't fall down", I do not mean a bridge which might not fall down if we're lucky and just happen to be within the safe parameter range. My perception of your behavior in this thread is: instead of talking about whether the bridge can be extended, you changed the subject and explained that the real problem is that the bridge has to support very heavy trucks.  This is logically rude.  And it makes it impossible to have an in-depth discussion about whether the bridge design can actually be extended or not.  From my perspective, you've pulled this conversational move multiple times in this thread.  It seems to be pretty common when I have discussions about AI safety people.  That's part of why I find the discussions so frustrating.  My view is that this is a cultural problem which has to be solved for the AI safety community to do much useful AI safety work (as opposed to "complaining about how hard AI safety is" work, which is useful but insufficient). Anyway, I'll let you have the last word in this thread. This is a very strong claim which to my knowledge has not been well-justified anywhere.  Daniel K agreed with me the other day that there isn't a standard reference for this claim.  Do you know of one? There isn't a standard reference because the argument takes one sentence, and I've been repeating it over and over again: what would Bayesian updates on low-level physics do? That's the unique solution with best-possible predictive power, so we know that anything which scales up to best-possible predictive power in the limit will eventually behave that way. My perception of your behavior in this thread is: instead of talking about whether the bridge can be extended, you changed the subject and explained that the real problem is that the bridge has to support very heavy trucks.  This is logically rude.  And it makes it impossible to have an in-depth discussion about whether the bridge design can actually be extended or not. The "what would Bayesian updates on a low-level model do?" question is exactly the argument that the bridge design cannot be extended indefinitely, which is why I keep bringing it up over and over again. This does point to one possibly-useful-to-notice ambiguous point: the difference between "this method would produce an aligned AI" vs "this method would continue to produce aligned AI over time, as things scale up". I am definitely thinking mainly about long-term alignment here; I don't really care about alignment on low-power AI like GPT-3 except insofar as it's a toy problem for alignment of more powerful AIs (or insofar as it's profitable, but that's a different matter). I've been less careful than I should be about distinguishing these two in this thread. All these things which we're saying "might work" are things which might work in the short term on some low-power AI, but will definitely not work in the long term on high-power AI. That's probably part of why it seems like I keep switching positions - I haven't been properly distinguishing when we're talking short-term vs long-term. A second comment on this: instead of talking about whether the bridge can be extended, you changed the subject and explained that the real problem is that the bridge has to support very heavy trucks If we want to make a piece of code faster, the first step is to profile the code to figure out which step is the slow one. If we want to make a beam stronger, the first step is to figure out where it fails. If we want to extend a bridge design, the first step is to figure out which piece fails under load if we just elongate everything. Likewise, if we want to scale up an AI alignment method, the first step is to figure out exactly how it fails under load as the AI's capabilities grow. I think you currently do not understand the failure mode I keep pointing to by saying "what would Bayesian updates on low-level physics do?". Elsewhere in the thread, you said that optimizing "for having a diverse range of models that all seem to fit the data" would fix the problem, which is my main evidence that you don't understand the problem. The problem is not "the data underdetermines what we're asking for", the problem is "the data fully determines what we're asking for, and we're asking for a proxy rather than the thing we actually want". Locating human values might be trickier than characterizing a koala, but the difference seems quantitative, not qualitative. I generally agree with this. The things I'm saying about human values also apply to koala classification. As with koalas, I do think there's probably a wide range of parameters which would end up using the "right" level of abstraction for human values to be "natural". On the other hand, for both koalas and humans, we can be fairly certain that a system will stop directly using those concepts once it has sufficient available compute - again, because Bayesian updates on low-level physics are just better in terms of predictive power. Right now, we have no idea when that line will be crossed - just an extreme upper bound. We have no idea how wide/narrow the window of training parameters is in which either "koalas" or "human values" is a natural level of abstraction. It doesn't know what a koala looks like, but it has some lower-level "components" which would allow it to characterize a koala. Then I use transfer learning and train it to differentiate marsupials. Now it knows about koalas too. Ability to differentiate marsupials does not imply that the system is directly using the concept of koala. Yet again, consider how Bayesian updates on low-level physics would respond to the marsupial-differentiation task: it would model the entire physical process which generated the labels on the photos/videos. "Physical process which generates the label koala" is not the same as "koala", and the system can get higher predictive power by modelling the former rather than the latter. When we move to human values, that distinction becomes a lot more important: "physical process which generates the label 'human values satisfied'" is not the same as "human values satisfied". Confusing those two is how we get Goodhart problems. We don't need to go all the way to low-level physics models in order for all of that to apply. In order for a system to directly use the concept "koala", rather than "physical process which generates the label koala", it has to be constrained on compute in a way which makes the latter too expensive - despite the latter having higher predictive power on the training data. Adding in transfer learning on some lower-level components does not change any of that; it should still be possible to use those lower-level components to model the physical process which generates the label koala without directly reasoning about koalas. I've now written essentially the same response at least four times to your objections, so I recommend applying the general pattern yourself: • Consider how Bayesian updates on a low-level physics model would behave on whatever task you're considering. What would go wrong? • Next, imagine a more realistic system (e.g. current ML systems) failing in an analogous way. What would that look like? • What's preventing ML systems from failing in that way already? The answer is probably "they don't have enough compute to get higher predictive power from a less abstract model" - which means that, if things keep scaling up, sooner or later that failure will happen. You say: "we can be fairly certain that a system will stop directly using those concepts once it has sufficient available compute".  I think this depends on specific details of how the system is engineered. "Physical process which generates the label koala" is not the same as "koala", and the system can get higher predictive power by modelling the former rather than the latter. Suppose we use classification accuracy as our loss function.  If all the koalas are correctly classified by both models, then the two models have equal loss function scores.  I suggested that at that point, we use some kind of active learning scheme to better specify the notion of "koala" or "human values" or whatever it is that we want.  Or maybe just be conservative, and implement human values in a way that all our different notions of "human values" agree with. You seem to be imagining a system that throws out all of its more abstract notions of "koala" once it has the capability to do Bayesian updates on low-level physics.  I don't see why we should engineer our system in this way.  My expectation is that human brains have many different computational notions of any given concept, similar to an ensemble (for example, you might give me a precise definition of a sandwich, and I show you something and you're like "oh actually that is/is not a sandwich, guess my definition was wrong in this case"--which reveals you have more than one way of knowing what "a sandwich" is), and AGI will work the same way (at least, that's how I would design it!) I've now written essentially the same response at least four times to your objections I was trying to understand what you were getting at.  This new argument seems pretty different from the "alignment is mainly about the prompt" thesis in your original post--another shift in arguments?  (I don't necessarily think it is bad for arguments to shift, I just think people should acknowledge that's going on.) You seem to be imagining a system that throws out all of its more abstract notions of "koala" once it has the capability to do Bayesian updates on low-level physics.  I don't see why we should engineer our system in this way. It's certainly conceivable to engineer systems some other way, and indeed I hope we do. Problem is: • if we just optimize for predictive power, then abstract notions will definitely be thrown away once the system can discover and perform Bayesian updates on low-level physics. (In principle we could engineer a system which never discovers that, but then it will still optimize predictive power by coming as close as possible.) • if we're not just optimizing for predictive power, then we need some other design criteria, some other criteria for whether/how well the system is working. In one sense, the goal of all this abstract theorizing is to identify what that other criteria needs to be in order to reliably end up using the "right" abstractions in the way we want. We could probably make up some ad-hoc criteria which works at least sometimes, but then as architectures and hardware advance over time we have no idea when that criteria will fail. or example, you might give me a precise definition of a sandwich, and I show you something and you're like "oh actually that is/is not a sandwich, guess my definition was wrong in this case"--which reveals you have more than one way of knowing what "a sandwich" is (Probably tangential) No, this reveals that my verbal definition of a sandwich was not a particularly accurate description of my underlying notion of sandwich - which is indeed the case for most definitions most of the time. It certainly does not prove the existence of multiple ways of knowing what a sandwich is. Also, even if there's some sort of ensembling, the concept "sandwich" still needs to specify one particular ensemble. This new argument seems pretty different from the "alignment is mainly about the prompt" thesis in your original post--another shift in arguments? We've shifted to arguing over a largely orthogonal topic. The OP is mostly about the interface by which GPT can be aligned to things. We've shifted to talking about what alignment means in general, and what's hard about aligning systems to the kinds of things we want. An analogy: the OP was mostly about programming in a particular language, while our current discussion is about what kinds of algorithms we want to write. Prompts are a tool/interface for via which one can align a certain kind of system (i.e. GPT-3) with certain kinds of goals (addition, translation, etc). Our current discussion is about the properties of a certain kind of goal - goals which are abstract in an analogous way to human values. if we're not just optimizing for predictive power, then we need some other design criteria, some other criteria for whether/how well the system is working. Optimize for having a diverse range of models that all seem to fit the data. How would that fix any of the problems we've been talking about? To put it another way: What semisupervised learning and transfer learning have in common is: You find a learning problem you have a lot of data for, such that training a learner for that problem will incidentally cause it to develop generally useful computational structures (often people say "features" but I'm trying to take more of an open-ended philosophical view).  Then you re-use those computational structures in a supervised learning context to solve a problem you don't have a lot of data for. From an AI safety perspective, there are a couple obvious ways this could fail: • Training a learner for the problem with lots of data might cause it to develop the wrong computational structures.  (Example: GPT-3 learns a meaning of the word "love" which is subtly incorrect.) • While attempting to re-use the computational structures, you end up pinpointing the wrong one, even though the right one exists.  (Example: computational structures for both "Effective Altruism" and "maximize # grandchildren" have been learned correctly, but your provided x/y pairs which are supposed to indicate human values don't allow for differentiating between the two, and your system arbitrarily chooses "maximize # grandchildren" when what you really wanted was "Effective Altruism"). I don't think this post makes a good argument that we should expect the second problem to be more difficult in general.  Note that, for example, it's not too hard to have your system try to figure out where the "Effective Altruism" and "maximize # grandchildren" theories of how (x, y) arose differ, and query you on those specific data points ("active learning" has 62,000 results on Google Scholar). Incidentally, I'm most worried about non-obvious failure modes, I expect obvious failure modes to get a lot of attention.  (As an example of a non-obvious thing that could go wrong, imagine a hypothetical super-advanced AI that queries you on some super enticing scenario where you become global dictator, in order to figure out if the (x, y) pairs it's trying to predict correspond to a person who outwardly behaves in an altruistic way, but is secretly an egoist who will succumb to temptation if the temptation is sufficiently strong.  In my opinion the key problem is to catalogue all the non-obvious ways in which things could fail like this.) This is almost, but not quite, the division of failure-modes which I see as relevant. If my other response doesn't clarify sufficiently, let me know and I'll write more of a response here. I think it's important to take a step back and notice how AI risk-related arguments are shifting. In the sequences, a key argument (probably the key argument) for AI risk was the complexity of human value, and how it would be highly anthropomorphic for us to believe that our evolved morality was embedded in the fabric of the universe in a way that any intelligent system would naturally discover.  An intelligent system could just as easily maximize paperclips, the argument went. No one seems to have noticed that GPT actually does a lot to invalidate the original complexity-of-value-means-FAI-is-super-difficult argument. As far as I see, GPT-3 did absolutely nothing to invalidate the argument about complexity of value. GPT-3 is able to predict correctly the kind of things we want it to predict in a context-window of at most 1000 words in very small time scale. So it can predict what we want it to do in basically one or a couple of abstract steps. That seems to guarante nothing whatsoever about the ability of GPT-3 to infer our exact values for the time scales and complexity even relevant to human level AI, let alone AGI. But I'm very interested for any experience that seems to invalidate this point. I'm not claiming GPT-3 understands human values, I'm saying it's easy to extrapolate from GPT-3 to a future GPT-N system which basically does. Curated. Simple, crucially important point, I'm really glad you wrote it up. There are infinitely many distributions from which the training data of GPT could have been sampled from [EDIT: including ones that could be catastrophic as the distribution our AGI learns], so it's worth mentioning an additional challenge on this route: making the future AGI-level-GPT learn the "human writing distribution" that we have in mind.

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# Java Converting long into currency Hello I am having trouble converting a long (cents) into currency format. My Code: long doublePayment = 1099; //Should equal \$10.99 DecimalFormat dFormat = new DecimalFormat(); String formattedString = dFormat.format(doublePayment); System.out.println(formattedString); Output: 1,099 I also tried: long doublePayment = 1099; NumberFormat n = NumberFormat.getCurrencyInstance(Locale.US); String s = n.format(doublePayment); System.out.println(s); Since this is cents, the output should be 10.99 or \$10.99. Cant figure out what I am doing wrong. Thanks!!! - Oops sorry i dont know why my code didnt go into "CodeFormat" so sorry in advance. –  mcd Sep 24 '12 at 7:34 Because you need a newline before the code; I fixed it for you. –  Jesper Sep 24 '12 at 7:35 FYI, you should use BigDecimal for currency: stackoverflow.com/questions/285680/… –  wulfgar.pro Sep 24 '12 at 7:42 To convert cents to dollars you can use long doublePayment = 1099; NumberFormat n = NumberFormat.getCurrencyInstance(Locale.US); String s = n.format(doublePayment / 100.0); System.out.println(s); This will be accurate up to \$70 trillion. - Thanks for the help –  mcd Sep 24 '12 at 15:03 I had tried what I thought was this before but put "x / 100" not even thinking that is different than "x / 100.0" –  mcd Sep 24 '12 at 15:11 Yes, An integer divided by an integer gives you an integer. It should give you a double really which you can cast to an integer for integer division, but you need to have at least double/integer or integer/double to get a double. –  Peter Lawrey Sep 24 '12 at 15:14 In case You have long to start with, you still should use java.math.BigDecimal. long doublePayment = 1099; BigDecimal payment = new BigDecimal(doublePayment).movePointLeft(2); System.out.println("\$" + payment); // produces: \$10.99 Let it be once again said out loud: One should never use floating-point variables to store money/currency value. - Except most investment banks use double. There is no BigDecimal in C or C++ and yet a lot system use C or C++. ;) –  Peter Lawrey Sep 24 '12 at 8:13 I would assume, that banks do use some integer or long form to store cents or fractions of cents, because floatingpoint arithmetic does produce minor errors, which are not allowed in banking systems. –  Lauri Sep 24 '12 at 8:29 minor errors are not random errors. You can predict how large they can be and use appropriate rounding. You are right long and int are also used, but these are less common IMHO. I would agree that if you don't know how to use appropriate rounding, then use BigDecimal. –  Peter Lawrey Sep 24 '12 at 8:38 Your literal is 1099, which is a thousand and ninety nine, coping with the Java rules for integer literals. According to your JVM locale, this number is represented with 1,099. If you were in Europe, it'd be 1.099. So, it's not an issue with your output, but with your input. The problem is that you have to represent a fixed point value, but you don't know java.math.BigDecimal and try to fake it. Things will broken when you'll do some computations. Don't do it. This is what you are supposed to do. Simply (it's far less code, too): BigDecimal payment = new BigDecimal("10.99"); System.out.println(String.format("\$%.2f", payment)); Note how you really initailize a number with a String. Also, String.format() will take care of the current locale, or you could supply the required one via the overloaded method. - Case might be, that David has long value of cents, then he should use new BigDecimal(doublePayment).movePointLeft(2); –  Lauri Sep 24 '12 at 8:09 double doublePayment = 10.99; You need to provide currency like this. NumberFormat will not understand whether you are providing it in cents or \$ unless it see decimal point - It's not an issue with formatting, it's an issue with input. Divide your input by 100 and you'll be all set. float payment = 1099 / ((float) 100); - I would use double rather than float as there will be less representation error. i.e. double payment = 1099 / 100.0; –  Peter Lawrey Sep 24 '12 at 8:10 Use currency formatter: NumberFormat nf = NumberFormat.getCurrencyInstance(Locale.ENGLISH); String output = nf.format(value); System.out.println(value + " " + output); -

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Never miss a story × ## Applying For A Personal Loan? Use Fullerton India EMI Calculator First Applying for a personal loan. Pixabay By Ankit A personal loan can be a lifesaver when you urgently need money to meet any financial obligations at hand. If you are looking for a reliable lending institution that can offer your immediate assistance, opt for the Fullerton India instant personal loan. The best thing about this service provider is that it offers a user-friendly online portal to accept your loan request with minimal documentation requirements. To make your experience even better, it also provides a personal loan EMI calculator on top of the webpage. You can use this to estimate the amount you will need to pay in installments. How Does a Personal Loan EMI Calculator Help? Whenever you take a personal loan from a lending party, you need to pay it back within a predefined period. You can either pay in full (a lump-sum amount) or in installments based on your preference and suitability. Most people opt for a monthly installment option to repay the loan amount. A personal loan EMI calculator allows you to check the actual installment amount that you will need to pay. This gives you a fair idea as to whether taking a personal loan would be a feasible option or not. To calculate your Equated Monthly Instalment (EMI), you need to have inputs like the total loan amount, total duration, and rate of interest applicable on your loans. You can easily get your final output in the form of “calculated EMI" by entering the required inputs. EMIs can also be calculated manually, however, using an online EMI calculator is more reliable and convenient as it helps to avoid any random calculation errors and give accurate results. Let's delve deeper into how the calculation takes place. Maths Behind the Calculation The mainstream digital or online personal loan calculators use the following formula to calculate the EMI that borrowers need to pay. E = P * r * (1+r)n / (1+r)n-1 'E' in the above-mentioned mathematical equation stands for the monthly amount payable by the borrower (EMI). 'P' denotes the principal amount or the total loan amount borrowed from the lending party. 'r' denotes the rate of interest that will be charged on your loan monthly. It is very subjective and depends on your credit history and relationship with the lender. 'n' here signifies the total duration for which you have taken the loan, you will need to repay the borrowed amount in this course. Let us take an example to understand the calculation process pragmatically. Suppose that a borrower 'A' has applied for a quick personal loan of Rs. 5,00,000 to be repaid within 5 years (60 months). Let's assume the interest rate charged on A's loan amount is 15%. As per the mathematical formula mentioned above, P = 5,00,000, r = 15% (0.15), n = 60. After applying the formula, EMI will be equal to Rs. 11,895. Computing EMI using the Fullerton India EMI Calculator The Fullerton Indian EMI calculator is an online tool that facilitates EMI calculation after taking essential inputs from you. You need to follow the below-mentioned steps to find out your EMI amount. Step 1: Visit the Fullerton India personal loan webpage by clicking on this link. Step 2: Navigate to the top of the page, you will find an EMI calculator Step 3: Put in the total loan amount that you want to apply for in the 'Loan Amount' field. You can also use the slider to select your amount conveniently. Step 4: Enter the interest rate offered to you using the same process in the 'Rate of Interest' field. Step 5: Now add the total duration for which you will be taking the loan in the “Loan Tenure' field. After you have completed the above steps, you will be able to easily find out the amount (EMI) that you will need to repay every month. You can put different values in the loan amount, rate of interest, and loan tenure field to find out EMIs for those values. You can easily click on the 'Apply for Personal Loan' button available next to the calculated EMI section for loan inquiry. ALSO READ: Are Instant Personal Loans In India Easily Available Why Opt for a Personal Loan? Personal loans do not restrict the borrower as to how they can use the funds. You can use it to pay an emergency medical bill, tuition fees, home renovation, vacation, etc. The flexibility offered by a personal loan is unmatched, making it an excellent choice for managing current financial needs. In addition to this, personal loans are unsecured, meaning you do not have to pledge any security to borrow the loan amount. This is a boon for borrowers who don't have any security to borrow funds. Given the current pandemic scenario, one can consider opting for a personal loan to finance any emergency expenses, including your medical bill. ## Hiking Destinations In Kashmir Pixabay Kashmir's natural splendour, with its beautiful valleys and towering mountains, is really unlike anywhere. Along with the undeniable natural beauty, the Kashmir valley has developed a reputation for adventurous activities like trekking, hiking, and river rafting. Kashmir has maintained its charm, allowing us to time-travel into beautiful destinations which make one forget about the stress and worries of life. The hikes in Kashmir offer adventurers to go on a self-discovery trip through nature's lap over the mountains while taking in the breathtaking scenery that surrounds them on their journey. In addition to the hikes, there are many thrilling adventure activities, like rock climbing, rope climbing, etc. Trekking across the region of mountains and lakes will allow you to experience living in the "Paradise on Earth," and you wouldn't want to return to your regular life after that. The following are some of the finest hiking destinations in Kashmir: #1: Kashmir Great Lakes Trek: You will be transported to a heavenly and unseen aspect of Kashmir on the Kashmir Great Lakes Trek. In addition to three high-altitude passes and five river valley crossings, this is the only trip in the Himalayas that includes seven alpine lakes, each of which is a stunning shade of green, blue, or turquoise. The extravagance is limitless and breathtakingly stunning every day: infinite blue sky, a larger-than-life backdrop of the Rocky Mountains, colourful meadows overflowing with wildflowers, river crossings are just a few examples of what you will encounter during the trek. You will be transported to a heavenly and unseen aspect of Kashmir on the Kashmir Great Lakes Trek. | Photo by prayer flags on Unsplash #2: Sonamarg-Vishansar-Bandipora Trek: The Sonamarg-Vishansar-Bandipora trek is a one-of-a-kind experience that provides a glimpse into Kashmir's undiscovered regions. Sonamarg, famously known as the Meadows of Gold, is the starting point for this fascinating journey that is the perfect experience for anyone looking to get away from the frantic tourist rush. This trek is a fascinating journey that allows nature enthusiasts to bask in the splendour of nature's grandeur. The trek goes over many high mountain passes, some as high as 4000 metres in elevation. The hiking route, in addition to providing breathtaking views of the magnificent Vishansar Lake, provides visitors with the chance to see more than 50 alpine lakes. Sonamarg, famously known as the Meadows of Gold, is the starting point for this fascinating journey. | Photo by YASER NABI MIR on Unsplash #3: Tral-Narastan-Marsar Trek: The Tral-Narastan-Marsar trek is filled with a range of exciting experiences from beginning to end. The hiking trail passes past a waving saffron field, beautiful meadows, and several streams. The path also crosses the Dachigam National Park, where there is an opportunity to see various animal species. Trekkers may take in spectacular views of the high mountains running parallel to them as they cut and pass through Narastan, a Hindu pilgrimage place. The Tral-Narastan-Marsar trek is filled with a range of exciting experiences from beginning to end. | Wikimedia Commons #4: Chhatargul-Mahlish-Gangabal: The journey, which passes through beautiful locations such as Chattargul, Mahlish, Kolsar, and Trunkul, provides a peek into an utterly uninhabited wilderness of Kashmir. There are lakes and meadows adorned with flowers along the route as one trek into the alpine wilderness. Trekkers can also enjoy fishing in the crystal clear lakes, camping, or just seeing towering snow-capped mountains while on their journey. There are lakes and meadows adorned with flowers along the route as one treks into the alpine wilderness. | Wikimedia Commons #5: Kolahoi Base Camp Trek: The Kolahoi Base Camp trek in Kashmir has been famous since the early 1900s and has been a goal for many seasoned hikers from across the world. While Srinagar serves as the beginning point for the trip, it is in Aru Valley that the actual hiking begins. The Kolahoi Base Camp Trek is a gentle adventure that is ideal for novices and families with children. The breathtaking sight of the peaks rising into the sky on the horizon of the Pirpanjal and Karakoram ranges is certainly worth capturing. It is considered to be one of the most popular treks in the Kashmir valley. The Kolahoi Base Camp Trek is a gentle adventure that is ideal for novices and families with children. | Wikimedia Commons Kashmir's natural splendour, with its beautiful valleys and towering mountains, is really unlike anywhere. Trekking through various valleys and peaks while taking in the scenic beauty is something that always calms the heart and provides us with memories that we will remember for a lifetime. Keywords: Kashmir, Lakes, Alpine, Hiking, Trekking, Treks, Sonamarg, Gangabal, Kolahoi, Chhatargul, Mahlish, Tral, Narastan, Marsar ## The Importance of Pitru Paksha Photo by Wikimedia Commons. Pind Daan at Jagannath Ghat, Kolkata. The Pitru Paksha starts after the Full Moon day, and this day marks the beginning of the waning phase of the Lunar cycle. This event is roughly of 15-day period, and is of great significance. From this day, rituals like Tarpan or Tarpanam and Shradh are carried out to pay respects to dead relatives and ancestors. It is believed that from the very first day till the last day, the unhappy souls of the deceased return to the Earth to see their family members. So, in order to ensure that the dead attain Moksha, i.e. to get liberation, family members of these souls quench their thirst and satisfy their hunger by performing the Pind Daan, which includes offering food consisting of cooked rice and black sesame seeds. The literal meaning of Pind Daan is the act of satisfying those who no longer exist physically. For fifteen days, prayers are offered in temples and rituals are performed to help the souls get free from the cycle of birth, life, and death, and attain salvation. At the same time, the Pitru Paksha is also an important period for people with Pitru Dosha, which means the curse imposed by the ancestors. Hence, in order to ask forgiveness, people perform Shradh rituals and offer food to the crows, who are considered as living beings that represent the dead. It is believed, if the crow eats the offered food, the ancestors are happy and pleased. But, if the crow doesn't eat the offered food and flies away, the ancestors are not happy. The event of Pitru Paksha is widely observed by Hindus from all over the world, and they perform prayers and rituals in order to gain their ancestors blessings. ## Bangalore's Colonial History Preserved In Cubbon Park wikimedia commons Cubbon Park is a lush green garden at the heart of Bangalore At the heart of Bangalore city, a large 300-acre space of lush greenery and heritage stands as a symbol of the city's past, present, and future. Cubbon Park is every child's favourite park, every Bangalorean's haven of fresh air, and altogether, the city's pride. It stands testament to the past, in terms of the diversity of flora it houses. Bangalore traffic in the recent past has grown into a menace, but the stretch between MG Road and Cubbon Park is always a pleasurable place to stop and wait for the signal to turn green. The gust of wind that blows here, and the smell of mud, coupled with floral scents instantly transports citizens to Old Bangalore, where the weather was fine, and the trees loomed over roads with thick canopies that did not even allow rainwater to penetrate. Cubbon Park is also a historical site, and one of the few remaining monuments of colonial heritage in Central Bangalore. It houses many statues and among them, the most famous is that of Queen Victoria, which faces the St. Mark's Square. The stretch outside Cubbon Park is cool and well-shaded from the canopy of trees over it. Image source: wikimedia commons At present, Cubbon Park is known for the cultural hub that it is. It houses Jawahar Bal Bhavan, which is a large theatre that hosts film festivals through the year. Festivals, poetry open mics, and other such shows are conducted on the lawns every Sunday. A small stream runs through the park, where boat rides are held occasionally when the water level is high enough. There is a children's park on one corner, and a government-maintained aquarium, two-storeys tall, with exotic fish. The Park has been renamed many times in the past. It was originally named Meade's Park, after Sir John Meade, the acting commissioner of Mysore in 1870. It was later changed to Cubbon Park after Sir Mark Cubbon, who was the longest-serving commissioner of the Mysore state. In 1927, the park was renamed after the Mysore Maharaja Sri Krishna Wodeyar, to celebrate his silver jubilee, since the park was developed during the reign of his ancestors. Even though it is officially named Sri Chamrajendra Park, it is still known as Cubbon Park all over the city. In fact, Bangalore was alluded the sobriquet of 'Garden City' because of the rich botanical diversity of this park. Art Installation at Cubbon Park Image source: wikimedia commons In many parts of the country, governments have renamed structures, places, and cities to remove traces of colonialism. But, in a city like Bangalore, there is too much evidence of the British rule. Many of the most prominent attractions of the city are known by their British identities despite the change in name. Even the city's name continues to be Bangalore, despite having been changed to Bengaluru. Last year, the British era and its achievements were celebrated in Cubbon Park when Sir Mark Cubbon's statue was moved from the grounds of the Karnataka High Court and placed in the Park. Keywords: Cubbon Park, Mark Cubbon, British Colonialism, Cultural hub, Garden City

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# SCM Repository [matrix] View of /pkg/src/geMatrix.h [matrix] / pkg / src / geMatrix.h # View of /pkg/src/geMatrix.h Revision 301 - (download) (as text) (annotate) Sun Oct 17 05:06:04 2004 UTC (15 years, 5 months ago) by bates File size: 1083 byte(s) `transfer to another computer` ```#ifndef MATRIX_GEMATRIX_H #define MATRIX_GEMATRIX_H #include <R_ext/Lapack.h> #include "Mutils.h" SEXP geMatrix_validate(SEXP obj); SEXP geMatrix_norm(SEXP obj, SEXP norm); SEXP geMatrix_crossprod(SEXP x); SEXP geMatrix_geMatrix_crossprod(SEXP x, SEXP y); SEXP geMatrix_matrix_crossprod(SEXP x, SEXP y); SEXP geMatrix_getDiag(SEXP x); SEXP geMatrix_LU(SEXP x); SEXP geMatrix_determinant(SEXP x, SEXP logarithm); SEXP geMatrix_solve(SEXP a); SEXP geMatrix_geMatrix_mm(SEXP a, SEXP b); SEXP geMatrix_svd(SEXP x, SEXP nu, SEXP nv); /* DGESDD - compute the singular value decomposition (SVD); of a */ /* real M-by-N matrix A, optionally computing the left and/or */ /* right singular vectors. If singular vectors are desired, it uses a */ /* divide-and-conquer algorithm. */ void F77_NAME(dgesdd)(const char *jobz, const int *m, const int *n, double *a, const int *lda, double *s, double *u, const int *ldu, double *vt, const int *ldvt, double *work, const int *lwork, int *iwork, int *info); #endif ``` root@r-forge.r-project.org ViewVC Help Powered by ViewVC 1.0.0 Thanks to:

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## screw feeder power calculation Screw Feeder Calculation, Wholesale Various High Quality Screw Feeder Calculation P. all purpose scientific calculator with solar power 3000 Pieces (Min. Order) OS-5512C c. > screw feeder power calculation Read More>> Cad Blocks Free - Down... • Screw Feeder Calculation, Wholesale Various High Quality Screw Feeder Calculation P. all purpose scientific calculator with solar power 3000 Pieces (Min. Order) OS-5512C c. • Find Complete Details about Fengbo Reliable Auto Screw Feeder,Auto Screw Feeder,Screw Conveyor Design Calculation,Flexible Screw Conveyor from Mining Feeder Supplie. • Dears, As we encountered electrical trip in a screw feeder, I would like to do power calculation more precisely. So I would be pleased if you review my calculation and inform me a. • the C mould cavity volume of twinscrew feeder is discussed in detail and its calculation f. 7 LU Jia ping (Southern Yangtze University , Wuxi 214063, China);Study on the Screw F. • 3LIU Jing,WANG Cheng-sheng(Shan Xi University of science and Technology,Shanxi Xianyang 712081,China);Design and calculation of Screw-Feeder for straw[J];Wood Processi. • On the base of analyzing the parametric design model of the screw feeder with variable diameter and screw pitches, a parametric design model of screw shaft was given in this pap. • THE GEOMETRICAL VOLUM CALCULATIONOF TWINSCREW FEEDER AND THE A. a screw feeder is designed with variable screw pitch,which has many advantages such a. • Drive motor power: 1.5-5.5 kw capacity: 3-12 M3/h screw pipe diameter: 114-219mm . LS series screw feeder is mainly consisted of drive motor, screw, screw pipe, feeding h. • a screw feeder is designed for biomass.Main characters of the feede are described,including diameter of screw axes,screw-pitch,axes rotate speed and power selection etc.The. • fine powder material,has been designed,and on the basis of preliminary caLculation,three feeding screw rods with identical diameter and different pitch of screws for said feeder,as. • generators and mechanical power transmission products, services and expertise to sav. Calculation of the expansion of screw conveyors handling hot . T Torque, inch pounds T. . • Parameter model Screw diameter(mm) Voltageï¼Vï¼ L/D Main motor power(k. Twin screw side feeder Single screw forced feeder Zero gravity calculation electronic fe. • power by analyzing the drive roller structure. The calculation method is simple,easy to gr. Productivity of the Self-Synchronization Vibration Feeder[J];JOURNAL OF UNIVERSITY. • Find 2015 Cheap Bottles Conveyor Screw on Alibaba, You Can Buy Various High Quali. Power: 3kW Application: Apparel,Beverage,Chemical,Commodity,Food,Machinery & H. • Photoelectric sensor and double-screw feeder can control the feeding forcibly and equa. calculation, a design project of automatic feeder that can realize quantitative control of m. • In brief, the strong points of the screw devicesare: Reduced risk of environmental pollution;Flexibility of use;Functional reliability;Low investment costs;Easy to install. Calculation of th. • screw feeder the torque requirement is a principal parameter which is related to the feeder loads, properties of the bulk solid and constructional features of the screw. In this paper. • Chain Conveyor Calculation, Buy Various High Quality Chain Conveyor Calculation Prod. Bucket elevator manufacturer, bucket elevator power . US \$1000-9999 1 Set (Min. Orde. • Chain Conveyor Calculation, Buy Various High Quality Chain Conveyor Calculation Prod. Bucket elevator manufacturer, bucket elevator power . US \$1000-9999 1 Set (Min. Orde. • CHOOSING AND CALCULATION OF REVOLVING VANE FEEDER . slippage of materials within feeder casings and for power requirements of screw feeder. • Calculation of the nominal flow can be done once the screw geometry, its rotation speed and the fillingcoefficient are known. The flow rate of a screw conveyor or feeder depends .

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# Help proving bound on POVM measurement probabilities I am trying to follow Nielsen and Chuang's 1 proof that the difference in measurement probabilities is bounded by the difference between two unitary operators applied to a given state. Can someone show me how to get from Equation 4.66 to 4.67 in the proof below (see page 195 in the 10th anniversary edition): 1 Nielsen and Chuang, "Quantum Computation and Quantum Information" I'll elaborate on the answer, since it took me some effort and I'm glad to share all the steps. Applying the Cauchy-Schwarz inequality to $$|\langle \psi| U^\dagger M |\Delta\rangle |$$ it follows that: $$|\langle \psi| U^\dagger M |\Delta\rangle | \leq ||\psi|| \: ||U^\dagger M \Delta ||$$ Note that angle brackets and vertical bar of the bra-ket notation were omitted to make the notation more readable. The next step is: $$||\psi|| \: ||U^\dagger M \Delta ||\leq||\psi||\:||U^\dagger M||\: ||\Delta ||$$ This inequality derives from the definition of operator norm. If $$A$$ is a linear operator, we define the norm of $$A$$ as: $$\|A \|_{op} = \sup \{\frac {\| Ax \|} {\| x \|} \ \forall \: x \neq 0 \}$$ Then it follows that for any $$\Delta$$ it holds $$\| U^\dagger M\Delta \|\leq\| U^\dagger M\|_{op}\|\Delta \|$$. We'll omit the $$\:_{op }$$ subscript hereinafter. From the properties of the norm, it follows that $$||\psi|| \:\| U^\dagger M\|\:||\Delta ||\leq||\psi|| \:\| U^\dagger \|\|M\|\:||\Delta ||$$ Now remember that $$||\psi||= 1$$ and that $$\| U^\dagger\|=1$$ since $$U$$ is a unitary matrix. To prove that $$||M||\leq 1$$ remember that $$M$$ is an element of a POVM. A POVM is a set of elements $$\{M_i\}$$ of $$n X n$$ matrices which are hermitian, positive definite and complete; note that typically they are non-projective and non-orthogonal. From completeness it holds: $$\sum_{i=0}^{n} M_i = I$$ Since $$M_i$$ is hermitian, then it is a normal operator and any normal operator is diagonal with respect to some orthonormal basis (spectral decomposition theorem); since $$M_i$$ is hermitian and positive definite all the eigenvalues $$\lambda_j$$ are real and positive and then $$M_i=\sum_{j=0}^{m} \lambda_j |j \rangle\langle j|$$. Since $$||\sum_{i=0}^{n} M_i ||=|| I||$$ then $$||M_i||\leq1$$. So we proved that $$|\langle \psi| U^\dagger M |\Delta\rangle | \leq||\Delta ||$$. Applying all the previous steps to $$|\langle \psi| MV |\Delta\rangle|$$ it follows that $$|\langle \psi| MV |\Delta\rangle | \leq||\Delta ||$$. Then we proved that: $$|\langle \psi| U^\dagger M |\Delta\rangle |+|\langle \psi| MV |\Delta\rangle| \leq \||\Delta\rangle ||+\||\Delta\rangle \|$$ I hope this can help someone else. $$|\langle \psi| AB \phi\rangle | \leq ||\psi|| \: ||AB \phi|| \leq ||\psi||\: ||AB|| \: ||\phi|| \leq ||\psi|| \: ||A||\:||B||\: ||\phi||$$ In our case $$||\psi||=1$$ and $$||A||, ||B|| \leq 1$$. because one of $$A$$ and $$B$$ is unitary and the other is part of a POVM.

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## Converting tenths and hundredths to decimals Activity type: Interactive Activity ## Converting tenths and hundredths to decimals Course Mathematics Section Decimals Outcome Converting tenths and hundredths to decimals Activity Type Interactive Activity Activity ID 28489 ## Testimonials What a brilliant site you have!!! I love it, especially as it saves me hours and hours of hard work. Others who haven't found your site yet don't know what they are missing! ## United States – Common Core State Standards • ##### 4.NF – Number & Operations—Fractions (Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, 100.) • Mathematics • 4.NF.6 – Use decimal notation for fractions with denominators 10 or 100. For example, rewrite 0.62 as 62/100; describe a length as 0.62 meters; locate 0.62 on a number line diagram. ## Australia – Australian Curriculum • ##### Number and Algebra • Fractions and decimals • ACMNA105 – Compare, order and represent decimals ## United Kingdom – National Curriculum • ##### Year 4 programme of study • KS2.Y4.N.F – Number - fractions (including decimals) • Pupils should be taught to: • KS2.Y4.N.F.2 – Count up and down in hundredths; recognise that hundredths arise when dividing an object by 100 and dividing tenths by 10

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Mathematics # $I =\int \frac{ ( x + x^{} + x^{}) }{ x(1+x^{}) } dx$equals to (a)$\frac{3}{2} x^{} + 6 \ tan^{-1} ( x^{} ) + c$ Its FREE, you're just one step away Single Correct Medium Published on 17th 09, 2020 Questions 203525 Subjects 9 Chapters 126 Enrolled Students 86 #### Realted Questions Q1 Subjective Medium $\int {\frac{1}{{(x - 1)(x - 2)}}dx}$ 1 Verified Answer | Published on 17th 09, 2020 Q2 Subjective Medium Evaluate the following integrals: $\displaystyle \int { \cfrac { 1 }{ { x }^{ 3 } } } \sin { \left( \log { x } \right) } dx\quad$ 1 Verified Answer | Published on 17th 09, 2020 Q3 Single Correct Medium $\int x^{\dfrac{13}{2}}.(1+x^{\dfrac{5}{2}})^{\dfrac{1}{2}}dx$ • A. $\dfrac{2}{5}(\dfrac{(1+x^\cfrac32)^3}{2}+\dfrac{(1+x^\cfrac52)}{2}-\dfrac{(1+x^\cfrac52)^2}{2})$ • B. $\dfrac{2}{5}(\dfrac{(1+x^\cfrac72)^3}{2}+\dfrac{(1+x^\cfrac52)}{2}-\dfrac{(1+x^\cfrac52)^2}{2})$ • C. $\dfrac{2}{5}(\dfrac{(1+x^\cfrac92)^3}{2}+\dfrac{(1+x^\cfrac52)}{2}-\dfrac{(1+x^\cfrac52)^2}{2})$ • D. $\dfrac{2}{5}(\dfrac{(1+x^\cfrac52)^3}{2}+\dfrac{(1+x^\cfrac52)}{2}-\dfrac{(1+x^\cfrac52)^2}{2})$ 1 Verified Answer | Published on 17th 09, 2020 Q4 Subjective Medium Solve : $(x^{2} - 1)\dfrac{dy}{dx} + 2xy = \dfrac{1}{x^{2} - 1}$ Evaluate $\displaystyle\int e^x\left ( \frac{1 + \sin\, x}{1 + \cos\, x} \right ) dx$

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# Microeconomics- Consumer 1. Mar 11, 2012 ### hellothere1 1. The problem statement, all variables and given/known data "Monet is a first year graduate student and he receives a weekly salary from his teaching assistant job on campus. His stress–release activity is to make strawberry smoothies. Therefore he spends all his weekly salary for buying strawberries and ice. In order to make a perfect smoothie, he needs to put 10 strawberries and 5 cubes of ice per cup of smoothie. The price of a strawberry is $0.5 and the price of a cube of ice is$0.2. His weekly salary is \$10." b. Draw one of Monet’s indifference curves for strawberries and ice and explain the economic intuition. (I think I got this one) 2. Relevant equations Y(income)= Price of ice * ice + price of strawberries * strawberries 3. The attempt at a solution 10 = ????? omg help the intercepts will be 20 and 50 2. Mar 12, 2012 ### Staff: Mentor To make n smoothies, how many strawberries and how many ice cubes does Monet need?

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It is currently 19 Nov 2017, 23:28 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If x^2+12x−k=0, is x=4? Author Message TAGS: ### Hide Tags Retired Moderator Joined: 27 Aug 2012 Posts: 1184 Kudos [?]: 1958 [0], given: 152 ### Show Tags 05 Jul 2013, 12:41 5 This post was BOOKMARKED 00:00 Difficulty: 65% (hard) Question Stats: 50% (01:24) correct 50% (01:31) wrong based on 206 sessions ### HideShow timer Statistics If x^2+12x−k=0, is x=4? (1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. (2) x≠−16 P.S: Bunuel- Need your help Sir! [Reveal] Spoiler: OA _________________ Kudos [?]: 1958 [0], given: 152 Math Expert Joined: 02 Sep 2009 Posts: 42255 Kudos [?]: 132741 [2], given: 12360 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 05 Jul 2013, 12:54 2 KUDOS Expert's post 3 This post was BOOKMARKED If x^2+12x−k=0, is x=4? (1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out $$x+16$$ from $$x^2+12x-k=0$$, so we would have $$(x+16)*(something)=0$$. Thus x=-16 is one of the roots of the given quadratic equation. Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$: $$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$. Thus according to the above $$x_1+x_2=-16+x_2=\frac{-12}{1}$$ --> $$x_2=4$$. So, we have that x is either -16 or 4. Not sufficient. (2) x≠−16. Clearly insufficient. (1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient. Hope it's clear. _________________ Kudos [?]: 132741 [2], given: 12360 Retired Moderator Joined: 27 Aug 2012 Posts: 1184 Kudos [?]: 1958 [0], given: 152 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 05 Jul 2013, 13:01 Bunuel wrote: If x^2+12x−k=0, is x=4? (1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out $$x+16$$ from $$x^2+12x-k=0$$, so we would have $$(x+16)*(something)=0$$. Thus x=-16 is one of the roots of the given quadratic equation. Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$: $$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$. Thus according to the above $$x_1+x_2=-16+x_2=\frac{-12}{1}$$ --> $$x_2=4$$. So, we have that x is either -16 or 4. Not sufficient. (2) x≠−16. Clearly insufficient. (1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient. Hope it's clear. (1)+(2)=> 1 says x=-16 and 2 says x is NOT -16...So,isn't it contradicting hence Insufficient..? _________________ Kudos [?]: 1958 [0], given: 152 Math Expert Joined: 02 Sep 2009 Posts: 42255 Kudos [?]: 132741 [0], given: 12360 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 05 Jul 2013, 13:04 debayan222 wrote: Bunuel wrote: If x^2+12x−k=0, is x=4? (1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out $$x+16$$ from $$x^2+12x-k=0$$, so we would have $$(x+16)*(something)=0$$. Thus x=-16 is one of the roots of the given quadratic equation. Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$: $$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$. Thus according to the above $$x_1+x_2=-16+x_2=\frac{-12}{1}$$ --> $$x_2=4$$. So, we have that x is either -16 or 4. Not sufficient. (2) x≠−16. Clearly insufficient. (1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient. Hope it's clear. (1)+(2)=> 1 says x=-16 and 2 says x is NOT -16...So,isn't it contradicting hence Insufficient..? (1) says that x=-16 OR x=4. The equation is $$x^2+12x-64=0$$ (k=64) --> x=-16 OR x=4. _________________ Kudos [?]: 132741 [0], given: 12360 Retired Moderator Joined: 27 Aug 2012 Posts: 1184 Kudos [?]: 1958 [0], given: 152 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 05 Jul 2013, 13:13 Got it! I was doing the mistake by considering the Stat.1 ONLY not focusing on the solns of the eqn in 1...! Thanks for clarifying. _________________ Kudos [?]: 1958 [0], given: 152 Intern Joined: 27 Jun 2013 Posts: 25 Kudos [?]: 109 [0], given: 29 Location: Russian Federation GMAT 1: 620 Q50 V27 GMAT 2: 730 Q50 V39 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 22 Aug 2013, 08:37 Bunuel wrote: If x^2+12x−k=0, is x=4? (1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out $$x+16$$ from $$x^2+12x-k=0$$, so we would have $$(x+16)*(something)=0$$. Thus x=-16 is one of the roots of the given quadratic equation. Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$: $$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$. Thus according to the above $$x_1+x_2=-16+x_2=\frac{-12}{1}$$ --> $$x_2=4$$. So, we have that x is either -16 or 4. Not sufficient. (2) x≠−16. Clearly insufficient. (1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient. Hope it's clear. Why A is not sufficient? X cannot be -16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan Kudos [?]: 109 [0], given: 29 Math Expert Joined: 02 Sep 2009 Posts: 42255 Kudos [?]: 132741 [0], given: 12360 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 22 Aug 2013, 09:46 Dmitriy wrote: Bunuel wrote: If x^2+12x−k=0, is x=4? (1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out $$x+16$$ from $$x^2+12x-k=0$$, so we would have $$(x+16)*(something)=0$$. Thus x=-16 is one of the roots of the given quadratic equation. Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$: $$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$. Thus according to the above $$x_1+x_2=-16+x_2=\frac{-12}{1}$$ --> $$x_2=4$$. So, we have that x is either -16 or 4. Not sufficient. (2) x≠−16. Clearly insufficient. (1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient. Hope it's clear. Why A is not sufficient? X cannot be -16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan Factor of a number and factor of an expression are two different things. For example, both (x+16) and (x-4) are factors of x^2+12x−64=0 --> (x+16)(x-4)=0. Hope this helps. _________________ Kudos [?]: 132741 [0], given: 12360 Intern Joined: 27 Jun 2013 Posts: 25 Kudos [?]: 109 [0], given: 29 Location: Russian Federation GMAT 1: 620 Q50 V27 GMAT 2: 730 Q50 V39 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 28 Aug 2013, 22:40 Bunuel wrote: Dmitriy wrote: Bunuel wrote: If x^2+12x−k=0, is x=4? (1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out $$x+16$$ from $$x^2+12x-k=0$$, so we would have $$(x+16)*(something)=0$$. Thus x=-16 is one of the roots of the given quadratic equation. Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$: $$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$. Thus according to the above $$x_1+x_2=-16+x_2=\frac{-12}{1}$$ --> $$x_2=4$$. So, we have that x is either -16 or 4. Not sufficient. (2) x≠−16. Clearly insufficient. (1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient. Hope it's clear. Why A is not sufficient? X cannot be -16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan Factor of a number and factor of an expression are two different things. For example, both (x+16) and (x-4) are factors of x^2+12x−64=0 --> (x+16)(x-4)=0. Hope this helps. Thanks. I didnt know about factors of an expression. Kudos [?]: 109 [0], given: 29 Non-Human User Joined: 09 Sep 2013 Posts: 15643 Kudos [?]: 283 [0], given: 0 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 09 Sep 2014, 07:28 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 283 [0], given: 0 Non-Human User Joined: 09 Sep 2013 Posts: 15643 Kudos [?]: 283 [0], given: 0 Re: If x^2+12x−k=0, is x=4? [#permalink] ### Show Tags 08 Sep 2016, 06:44 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 283 [0], given: 0 Re: If x^2+12x−k=0, is x=4?   [#permalink] 08 Sep 2016, 06:44 Display posts from previous: Sort by

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The distance formula date period find the distance between each pair of points. round your answer to the nearest tenth, if necessary. x y x y x y x y x y x y Math worksheets. objective i can use the distance formula to get the distance between two points. the distance between the two points x, y and x, y is given by the distance formula. read the lesson on distance formula for more information and examples. fill S a m a a r d i e o h w x i j t a h h r i w n s f c i z n i b t q e u d g l e a o f m x e g t e r p y r. List of The Distance Formula Worksheet Answers Worksheet by software software infinite geometry name period date the distance formula find the distance between each pair of points. round your answer to the nearest tenth, if Oct, , on software infinite geometry the distance formula answers. worksheet by software software infinite algebra name the distance formula date period. the distance formula the midpoint formula classifying triangles and quadrilaterals angle sum of triangles and quadrilaterals area of triangles area of squares rectangles and parallelograms area of trapezoids Oct, worksheet. distance formula software infinite geometry answers. 1. 8 Midpoint Distance Formula Ideas Teaching Math Both liter and liters jug included.Answers the area of a label is equal to the lateral area of the can. our formula for the lateral area of a pyramid gives square feet of waterproof material.As a member, also get unlimited access to over, lessons in math,, science, history, and more. 5. 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Midpoint Distance Formulas Puzzle Set Formula Common Core Math Activities G. apply the theorem to find the distance between two points in a coordinate system. use similar triangles to explain why the slope m is the same between any two distinct The distance formula parallel lines in the coordinate plane. the formula above is known as the distance formula. 7. 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Distance Formula Practice Writing Linear Equations Coordinate Plane Graphing C. c. got to be an easier way this would take a very long time if we had to draw a triangle every time we needed to find the distance between two points.Oct, print the distance, rate, and time worksheet no. when solving distance problems, explain to students that they will use the formula rt d or rate speed times time equals distance. for example, the first problem states the prince ship headed south at an average speed of mph.Distance formula worksheets. what is the distance formula the distance formula is a useful tool to find the distance between two arbitrarily points. the formula is divided from the theorem which is the a b c, where c is the longest side of a triangle. 39. Distance Formula Coloring Activity Color Activities Math Lesson Plans Distance and midpoint formulas displaying top worksheets found for this concept. some of the worksheets for this concept are the midpoint formula, the midpoint formula date period,, performance based learning and assessment task distance, midpoint and distance formulas, midpoint formula, distance midpoint formulas circles. In this distance formula worksheet, graders solve and complete various types of problems. first, they find the distance between the points and give their answer in simplest radical form. then, students find the perimeter of a.Free worksheet at httpswww. 40. Coloring Activity Set Problems Finding Distance Points Formula Color Activities Mathematics Worksheets Find the distance of. graph the points e, and f,. find the midpoint of. find the distance.Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula. hsggpe.b. find the point on a directed line segment between two given points that partitions the segment in a given ratio. This is a worksheet for students to practice finding the distance and midpoint for a set of points. the answers are given in an answer bank, with an extra answer for distance and for midpoint. they are to work out the answers, and then determine what the extra answers are for each. 41. City Plan Design Activity Applying Midpoint Distance Formulas Formula Teaching Geometry Mathematics Distance time graph worksheet answer key. check students understanding of distance time graphs as they are provided with two distance time graphs and are asked to analyze the graphs and describe what is happening during each section of the graph. the speed equation is provided as a triangle to support students in rearrange the formula to. Oct, theorem distance formula warm up scavenger hunt distance on the coordinate and distance i can find the distance between two points. i can find the midpoint of a line segment. part graphing.graph the points a, and b,. find the midpoint of. 42. Basic Rties Facts Arithmetic Operations Exponent Radicals Algebra Formulas Cheat Sheet Teaching Write the problem in fractional form. rationalize the denominator by multiplying the numerator and the denominator by Answers home answers operations m complex numbers p to n answers complex number of practice answers back to. overview of complex quiz answers. 44. Zombies Distance Formula Activity Midpoint By, , a distance formula software infinite geometry answers. there are over topics in all from multi step equations to constructions. round your answer to the nearest tenth if necessary.Distance on a number line distance in the coordinate plane ab x x ab x x or x x distance formula y x bx, ax, d x. x y y use the number line to find ab. ab ab find the distance between a, and b,. distance formula d x x y y D x x y y. the formula above is known as the distance formula. we can use it to find the distance d between any two points in the plane.

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# Jiří Matas. Hough Transform Save this PDF as: Size: px Start display at page: ## Transcription 1 Hough Transform Jiří Matas Center for Machine Perception Department of Cybernetics, Faculty of Electrical Engineering Czech Technical University, Prague Many slides thanks to Kristen Grauman and Bastian Leibe 2 Why HT and not Recognition with Local Features? Strengths: applicable to many objects (e.g. in image stitching) is real-time scales well to very large problems (retrieval of millions of images) handles occlusion well insensitive to a broad class of image transformations Weaknesses: applicable to recognition of specific objects (no categorization) applicable only to objects with distinguished local features Slide credit: David Lowe 2 3 Why HT and not the Scanning Window (Viola-Jones) Method? Strengths: applicable to many classes of objects not restricted to specific objects often real-time Weaknesses: extension to a large number of classes not straightforward (standard implementation: linear complexity in the number of classes) occlusion handling not easy full 3D recognition requires too many windows to be checked training time is potentially very long 3/25 4 Hough Transform A method for detecting geometric primitives based on evaluation of an objective function: is the parameter space, are tokens (image points of interest) Origin: Detection of straight lines Examples of for different geometric primitives: Straight line: Circle: Parameters evaluated on a grid Discretization of : 4 5 Comparison: Template Matching and HT Template Matching: for all! 2 J(!) = 0 for all x = (x; y) 2 Image = for all x i» tokens if satisfies J(!) = J(!) + p(x) else /* nothing */ Complexity: O(j j jpj) HT: (basic idea: each token votes for all primitives it is consistent with) for all find x i (x i ) J(!) = J(!) + p(x i ) Complexity: O(j (x i )j jpj); j (x i )j j j 5 6 Hough Transform for Straight Lines 1. Define the minimal parametrisation (p,q) of the space of lines: Most common: angle distance from origin (½, µ) Other options: tangent of angle intercept (a,b), nearest point to center, Quantize the Hough space: Identify the maximum and minimum values of a and b, and the number of cells, 3. Create an accumulator array A(p,q); set all values to zero 4. (if gradient available) : For all edge points (x i,y i ) in the image Use gradient direction Compute a from the equation Increment A(p,q) by one (if gradient not available): For all edge points (x i,y i ) in the image Increment A(p,q) by one for all lines incident on x,y 5. For all cells in A(p,q) Search for the maximum value of A(p,q) Calculate the equation of the line 6. To reduce the effect of noise more than one element (elements in a neighborhood) in the accumulator array are increased 6 7 HT for Straight Lines: Variations Representation of a line Usual form y = a x + b has a singularity around 90º. Can be overcome by considering two cases, y = a x + b and x = a y + b Common parametrization parameterization: x cos( ) + y sin( ) = y Using gradient orientation Uses not only point but also orientation consistent with the edge orientation Variables: In HT: for x P; ; Á : P! h0; ¼) (x i ; Á(x i )) Can be used by weighting the strength of the vote by: Ã line orientation, Á gradient orientation y x ρ já Ãj θ K. Grauman, B. Leibe 7 8 Examples Hough transform for a square (left) and a circle (right) K. Grauman, B. Leibe 8 9 Hough Transform: Noisy Line ρ Tokens Votes θ Problem: Finding the true maximum K. Grauman, B. Leibe Slide credit: David Lowe 9 10 Hough Transform: Noisy Input ρ Tokens Votes θ Problem: Lots of spurious maxima K. Grauman, B. Leibe Slide credit: David Lowe 10 11 HT for different primitive: circles, circles with fixed radius circles squares with a known orientation and size rectangles 11 12 HT for multiple instances 1. p 1 = HT(P; ) : strongest result of HT 2. Set 3. Unvote 4. P 1 = P n p 1 p 1 p 2 = HT(P 1 ; ) 5. Cont. to get as many instances as required Greedy Sequential 12 13 Hough Transform Problems 1. Search space (accumulator size) gets prohibitively large easily Line segments: Circular arc: r; c x ; c y ; t 1 ; t 2 2. Cost function must be additive. 3. Greedy assignment rule of a token to primitive 4. No global objective function for multiple primitives (global optimization for one primitive only) 13 14 When is the Hough transform useful? Textbooks often imply that it is useful mostly for finding lines In fact, it can be very effective for recognizing arbitrary shapes or objects (Generalized HT) The key to efficiency is to have each feature (token) determine as many parameters as possible For example, lines can be detected much more efficiently from small edge elements (or points with local gradients) than from just points For object recognition, each token should predict location, scale, and orientation (4D array) Bottom line: The Hough transform can extract feature groupings from clutter in linear time! K. Grauman, B. Leibe Slide credit: David Lowe 14 15 Generalized Hough Transform [Ballard81] Generalization for an arbitrary contour or shape Choose reference point for the contour (e.g. center) For each point on the contour remember where it is located w.r.t. to the reference point Remember radius r and angle relative to the contour tangent Recognition: whenever you find a contour point, calculate the tangent angle and vote for all possible reference points Instead of reference point, can also vote for transformation The same idea can be used with local features! K. Grauman, B. Leibe Slide credit: Bernt Schiele 15 16 Gen. Hough Transform with Local Features For every feature, store possible occurrences For new image, let the matched features vote for possible object positions Object identity Pose Relative position K. Grauman, B. Leibe 16 17 Finding Consistent Configurations Global spatial models Generalized Hough Transform [Lowe99] RANSAC [Obdrzalek02, Chum05, Nister06] Basic assumption: object is planar Assumption is often justified in practice Valid for many structures on buildings Sufficient for small viewpoint variations on 3D objects K. Grauman, B. Leibe 17 18 3D Object Recognition Gen. HT for Recognition Typically only 3 feature matches needed for recognition Extra matches provide robustness Affine model can be used for planar objects [Lowe99] K. Grauman, B. Leibe Slide credit: David Lowe 18 19 Comparison Gen. Hough Transform Advantages Very effective for recognizing arbitrary shapes or objects Can handle high percentage of outliers (>95%) Extracts groupings from clutter in linear time Disadvantages Quantization issues Only practical for small number of dimensions (up to 4) Improvements available Probabilistic Extensions Continuous Voting Space [Leibe08] RANSAC Advantages General method suited to large range of problems Easy to implement Independent of number of dimensions Disadvantages Only handles moderate number of outliers (<50%) Many variants available, e.g. PROSAC: Progressive RANSAC [Chum05] Preemptive RANSAC [Nister05] K. Grauman, B. Leibe 19 20 RHT = Randomized Hough Transform [Xu93] In: Out: E = fe i g; m( ; e) = 0 S1 ; S2 ; :::; SN Repeat: I. Hypothesis 1. Select random M feature points e k1 ; :::; e km 2. Compute k : m( k ; e kj ) = 0; j = 1; :::; M II. Pre-Verification 3. Add 1 to accumulator k ( k ) > T 1 4. If (accumulator ) goto III. Else goto I. III. Verification 5. Find support for k 6. If (support ) output 7. Reset accumulator ( k ) > T 2 k Proc v Out Omega-S-N, a v algoritmu Omega-N 20 21 Probabilistic Hough Transform [Kiryati et al. 91] Idea: Evaluate Algorithm: NX i=1 p(x i ; ) using only a fraction f = k MAX N 1. Select points at random 2. Perform standard HT Analysis: k MAX Selection of k MAX is incorrect of N points x i the number L of selected points from L N points of a line in a random subset of k MAX points is governed by hypergeometric, not binomial distance L N L L P (L N ) = N k MAX L N N k MAX ¹ = N ¾ 2 = k MAX L N (N L N ) N 2 µ 1 k MAX 1 N 1 21 22 Idea: 1. PHT = Monte Carlo Evaluation of N X i=1 p(x i ; ) 2. Apply standard MC analysis to find k MAX in PHT to guarantee Pffalse positiveg and Pffalse negativeg < ² Algorithm: 1. Select a random point 2. Vote and return it 3. Finish if k MAX reached 22 23 CHT = Cascaded Hough Transform [Tuytelaars et al. 97] Finds structures at different hierarchical levels by iterating one kind of HT (fixed points, fixed lines, lines of fixed points, pencils of fixed lines) Uses duality of lines and points in image and parameter spaces Algorithm: 1. First HT: detects lines in the image and keeps dominant peaks in the parameter space 2. Second HT: detects lines of collinear peaks in parameter space and keeps vertices where several straight lines in the original image intersect (vanishing points) 3. Third HT: applied to the peaks of thto detect collinear vertices (vanishing lines) 23 24 CHT: Experiments Lines belonging to one of the three detected vanishing points Aerial image of buildings and streets (left), the corresponding edges (right) 24 25 macros.tex sfmath.sty cmpitemize.tex Thank you for your attention. 25 ### Camera geometry and image alignment Computer Vision and Machine Learning Winter School ENS Lyon 2010 Camera geometry and image alignment Josef Sivic http://www.di.ens.fr/~josef INRIA, WILLOW, ENS/INRIA/CNRS UMR 8548 Laboratoire d Informatique, ### CS 534: Computer Vision 3D Model-based recognition CS 534: Computer Vision 3D Model-based recognition Ahmed Elgammal Dept of Computer Science CS 534 3D Model-based Vision - 1 High Level Vision Object Recognition: What it means? Two main recognition tasks:! ### Efficient visual search of local features. 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Lecture 2: Logistic regression & intro to MIL Iasonas Kokkinos Iasonas.kokkinos@ecp.fr Machine Learning for Computer Vision 1 MVA ENS Cachan Lecture 2: Logistic regression & intro to MIL Iasonas Kokkinos Iasonas.kokkinos@ecp.fr Department of Applied Mathematics Ecole Centrale Paris Galen ### The Not-Formula Book for C1 Not The Not-Formula Book for C1 Everything you need to know for Core 1 that won t be in the formula book Examination Board: AQA Brief This document is intended as an aid for revision. Although it includes ### Exam 1 Sample Question SOLUTIONS. y = 2x Exam Sample Question SOLUTIONS. Eliminate the parameter to find a Cartesian equation for the curve: x e t, y e t. SOLUTION: You might look at the coordinates and notice that If you don t see it, we can ### Going Big in Data Dimensionality: LUDWIG- MAXIMILIANS- UNIVERSITY MUNICH DEPARTMENT INSTITUTE FOR INFORMATICS DATABASE Going Big in Data Dimensionality: Challenges and Solutions for Mining High Dimensional Data Peer Kröger Lehrstuhl für ### Circle detection and tracking speed-up based on change-driven image processing Circle detection and tracking speed-up based on change-driven image processing Fernando Pardo, Jose A. Boluda, Julio C. Sosa Departamento de Informática, Universidad de Valencia Avda. Vicente Andres Estelles ### Arrangements And Duality Arrangements And Duality 3.1 Introduction 3 Point configurations are tbe most basic structure we study in computational geometry. But what about configurations of more complicated shapes? For example, ### Circle Name: Radius: Diameter: Chord: Secant: 12.1: Tangent Lines Congruent Circles: circles that have the same radius length Diagram of Examples Center of Circle: Circle Name: Radius: Diameter: Chord: Secant: Tangent to A Circle: a line in the plane ### Distance based clustering // Distance based clustering Chapter ² ² Clustering Clustering is the art of finding groups in data (Kaufman and Rousseeuw, 99). What is a cluster? Group of objects separated from other clusters Means ### Sampling via Moment Sharing: A New Framework for Distributed Bayesian Inference for Big Data Sampling via Moment Sharing: A New Framework for Distributed Bayesian Inference for Big Data (Oxford) in collaboration with: Minjie Xu, Jun Zhu, Bo Zhang (Tsinghua) Balaji Lakshminarayanan (Gatsby) Bayesian ### Solutions to old Exam 1 problems Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections ### Unit 7 Quadratic Relations of the Form y = ax 2 + bx + c Unit 7 Quadratic Relations of the Form y = ax 2 + bx + c Lesson Outline BIG PICTURE Students will: manipulate algebraic expressions, as needed to understand quadratic relations; identify characteristics ### Definition III: Circular Functions SECTION 3.3 Definition III: Circular Functions Copyright Cengage Learning. All rights reserved. Learning Objectives 1 2 3 4 Evaluate a trigonometric function using the unit circle. Find the value of a ### Face detection is a process of localizing and extracting the face region from the Chapter 4 FACE NORMALIZATION 4.1 INTRODUCTION Face detection is a process of localizing and extracting the face region from the background. The detected face varies in rotation, brightness, size, etc.

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## Just a moment . . . Many years ago, I was taught this compact and intuitive convention to show turning moments. I think it should be more widely known, as it not only is concise and powerful, but also meets the criterion of being an effective form of dual coding which is helpful for both GCSE and A-level Physics students. Let’s look at an example question. Let’s start by ‘annotating the hell’ out of the diagram. We could take moments around any of the marked points A-E on the diagram. However, we’re going to take moments around B as it enables us to ignore the upward reaction force acting on the rule at B. (This force is not shown on the diagram.) To indicate that we’re going to be considering the sum of the clockwise moments about point B, we use this intuitive notation: If we consider the sum of anticlockwise moments about point B, we use this: We lay out our calculations of the total clockwise and anticlockwise moments about B as follows. We show that we are going to apply the Principle of Moments (the sum of clockwise moments is equal to the sum of anticlockwise moments for an object in equilibrium) like this: The rest, as they say, is not history but algebra: I hope you find this ‘momentary’ convention useful(!) ## ‘Killer-joules’ and other abominations As noted earlier, some students struggle with unit conversions. To take a simple example: if we need to convert 3.7 kilojoules (or ‘killer-joules’ as some insist on calling them *shudders*) into joules, then whilst many students know that the conversion involves applying a factor of one thousand, they do not know whether to multiply 3.7 by a thousand or divide 3.7 by a thousand. Michael Porter shared a brilliant suggestion for helping students over this hurdle. He suggests that we break down the operation into two parts: • Consider if we are making the unit larger or smaller. • If making the unit larger, we must make the number smaller to compensate; and vice versa. Let’s look at using the Porter system for the example shown above. (Note: I have used kilojoules for our first example since, at least for GCSE Science calculation contexts, students are unlikely to have to convert kilograms into grams. This is because, of course, the kilogram (not the gram) is the base unit of mass in the SI System.) By changing from kilojoules to joules we are making the unit smaller, since one kilojoule is larger than one joule. To keep the measured quantity of energy the same magnitude, we must therefore make the number part of the measurement bigger to compensate for the reduction in size of the unit. Now let’s look if we had to convert 830 microamps into amps: ### The strange case of time Obviously 1 minute is a very small quantity of time compared with a whole week. Indeed, our forefathers considered it small as compared with an hour, and called it “one minùte,” meaning a minute fraction — namely one sixtieth — of an hour. When they came to require still smaller subdivisions of time, they divided each minute into 60 still smaller parts, which, in Queen Elizabeth’s days, they called “second minùtes” (i.e., small quantities of the second order of minuteness). Silvanus P. Thompson, “Calculus Made Easy” (1914) It is probable that the division of units of time into sixtieths dates back many thousands of years to the ancient Babylonians(!) Is it any wonder that some students find it hard to convert units of time? We can use the Porter system to help students with these conversions. For example, what is 7 hours in seconds? This type of diagram is, I think, very useful for showing students explicitly what we are doing. ## Units, you nit! The S.I. System of Units is a thing of beauty: a lean, sinewy and utilitarian beauty that is the work of many committees, true; but in spite of that common saw about ‘a camel being a horse designed by a committee’, the S.I. System is truly a thing of rigorous beauty nonetheless. Even the pedestrian Wikipedia entry on the 2019 Redefinition of the S.I. System reads like a lost episode from Homer’s Odyssey. As Odysseus tied himself to the mast of his ship to avoid the irresistible lure of the Sirens, so in 2019 the S.I, System tied itself to the values of a select number of universal physical constants to remove the last vestiges of merely human artifacts such as the now obsolete International Prototype Kilogram. However, the austere beauty of the S.I. System is not always recognised by our students at GCSE or A-level. ‘Units, you nit!!!’ is a comment that physics teachers have scrawled on student work from time immemorial with varying degrees of disbelief, rage or despair at errors of omission (e.g. not including the unit with a final answer); errors of imprecision (e.g. writing ‘j’ instead of ‘J’ for ‘joule — unforgivable!); or errors of commission (e.g. changing kilograms into grams when the kilogram is the base unit, not the gram — barbarous!). The saddest occasion for writing ‘Units, you nit!’ at least in my opinion, is when a student has incorrectly converted a prefix: for example, changing millijoules into joules by multiplying by one thousand rather than dividing by one thousand so that a student writes that 5.6 mJ = 5600 J. This odd little issue can affect students from across the attainment range, so I have developed a procedure to deal with it which is loosely based on the Singapore Bar Model. One millijoule is a teeny tiny amount of energy, so when we convert it joules it is only a small portion of one whole joule. So to convert mJ to J we divide by 1000. One joule is a much larger quantity of energy than one millijoule, so when we convert joules to millijoules we multiply by one thousand because we need one thousand millijoules for each single joule. In time, and if needed, you can move to a simplified version to remind students. Strangely, one of the unit conversions that some students find most difficult in the context of calculations is time: for example, hours into seconds. A diagram similar to the one below can help students over this ‘hump’. These diagrams may seem trivial, but we must beware of ‘the Curse of Knowledge’: just because we find these conversions easy (and, to be fair, so do many students) that does not mean that all students find them so. The conversions that students may be asked to do from memory are listed below (in the context of amperes). ## Dual Coding and Equations of Motion for GCSE Necessity may well be the mother of invention, but teacher desperation is often the mother of new pedagogy. For an unconscionably long time, I think that I failed to adequately help students understand the so called ‘equations of motion’ (the mathematical descriptions of uniformly accelerated motion using the standard v, u, a, s and t notation) because I suffered from the ‘curse of knowledge’: I did not find the topic hard, so I naturally assumed that students wouldn’t either. This, sadly, proved not to be the case, even when the equation was printed on the equation sheet! What follows is a summary of a dual coding technique that I have found really helpful in helping students become confident with problems involving the ‘equations of motion’. This is especially true at GCSE, where students encounter formulas such as for the first time. ### Applying dual coding to an equation of motion problem EXAMPLE: A car is travelling at 6.0 m/s. As the car passes a lamp-post it accelerates up to a velocity of 14.2 m/s over a distance of 250 m. Calculate a) the acceleration; and b) the time taken for this change. The problem can be represented using the dual coding convention as shown below. Note that the arrow for v is longer than the arrow for u since the car has a positive acceleration; that is to say, the car in this example is speeding up. Also, the convention has a different style of arrow for acceleration, emphasising that it is an entirely different type of quantity from velocity. We can now answer part (a) using the FIFA calculation system. Part (b) can be answered as shown below. ### Visualising Stopping Distance Questions These can be challenging for many students, as we often seem to grabbing numbers and manipulating them without rhyme or reason. Dual coding helps make our thinking explicit, EXAMPLE: A driver has a reaction time of 0.7 seconds. The maximum deceleration of the car is 2.6 metres per second squared. Calculate the overall stopping distance when the car is travelling at 13 m/s. We need to calculate both the thinking distance and the braking distance to solve this problem The acceleration of the car is zero during the driver’s reaction time, since the brakes have not been applied yet. We visualise the first part of the problem like this. Now let’s look at the second part of the question. There are three things to note: • Since the car comes to a complete stop, the final velocity v is zero. • The acceleration is negative (shown as a backward pointing arrow) since we are talking about a deceleration: in other words, the velocity gradually decreases in size from the initial velocity value of u as the car traverses the distance s. • Since the second part of the question does not involve any consideration of time we have omitted the ‘clock’ symbols for the second part of the journey. We can now apply FIFA: Since the acceleration arrow points in the opposite direction to the positive arrow, we enter it as a negative value. When we get to the third line of the Fine Tune stage, we see that a negative 169 divided by negative 4.4 gives positive 38.408 — in other words, the dual coding convention does the hard work of assigning positive and negative values! And finally, we can see that the overall stopping distance is 9.1 + 38.4 = 47.5 metres, ### Conclusion I have found this form of dual coding extremely useful in helping students understand the easily-missed subtleties of motion problems, and hope other science teachers will give it a try. If you have enjoyed this post, you might enjoy these also: ## A Gnome-inal Value for ‘g’ . . . setting storms and billows at defiance, and visiting the remotest parts of the terraqueous globe. Samuel Johnson, The Rambler, 17 April 1750 That an object in free fall will accelerate towards the centre of our terraqueous globe at a rate of 9.81 metres per second per second is, at best, only a partial and parochial truth. It is 9.81 metres per second per second in the United Kingdom, yes; but the value of both acceleration due to free fall and the gravitational field strength vary from place to place across the globe (and in the SI System of measurement, the two quantities are numerically equal and dimensionally equivalent). For example, according to Hirt et al. (2013) the lowest value for g on the Earth’s surface is atop Mount Huascarán in Peru where g = 9.7639 m s-2 and the highest is at the surface of the Arctic Ocean where g = 9.8337 m s-2. ### Why does g vary? There are three factors which can affect the local value of g. Firstly, the distribution of mass within the volume of the Earth. The Earth is not of uniform density and volumes of rock within the crust of especially high or low density could affect g at the surface. The density of the rocks comprising the Earth’s crust varies between 2.6 – 2.9 g/cm3 (according to Jones 2007). This is a variation of 10% but the crust only comprises about 1.6% of the Earth’s mass since the density of material in the mantle and core is far higher so the variation in g due this factor is probably of the order of 0.2%. Secondly, the Earth is not a perfect sphere but rather an oblate spheroid that bulges at the equator so that the equatorial radius is 6378 km but the polar radius is 6357 km. This is a variation of 0.33% but since the gravitational force is proportional to 1/r2 let’s assume that this accounts for a possible variation of the order of 0.7% in the value of g. Thirdly, the acceleration due to the rotation of the Earth. We will look in detail at the theory underlying this in a moment, but from our rough and ready calculations above, it would seem that this is the major factor accounting for any variation in g: that is to say, g is a minimum at the equator and a maximum at the poles because of the Earth’s rotation. ### The Gnome Experiment In 2012, precision scale manufacturers Kern and Sohn used this well-known variation in the value of g to embark on a highly successful advertising campaign they called the ‘Gnome Experiment’ (see link 1 and link 2). Whatever units their lying LCD displays show, electronic scales don’t measure mass or even weight: they actually measure the reaction force the scales exert on the item in their top pan. The reading will be affected if the scales are accelerating. In diagram B, the apple and scales are in an elevator that is accelerating upward at 1.00 metres per second per second. The resultant upward force must therefore be larger than the downward weight as shown in the free body diagram. The scales show a reading of 1.081/9.81 – 0.110 194 kg = 110.194 g. In diagram C, the the apple and scales are in an elevator that is accelerating downwards at 1.00 metres per second per second. The resultant upward force must therefore be smaller than the downward weight as shown in the free body diagram. The scales show a reading of 0.881/9.81 – 0.089 806 kg = 89.806 g. ### Never mind the weight, feel the acceleration Now let’s look at the situation the Kern gnome mentioned above. The gnome was measured to have a ‘mass’ (or ‘reaction force’ calibrated in grams, really) of 309.82 g at the South Pole. Showing this situation on a diagram: Looking at the free body diagram for Kern the Gnome at the equator, we see that his reaction force must be less than his weight in order to produce the required centripetal acceleration towards the centre of the Earth. Assuming the scales are calibrated for the UK this would predict a reading on the scales of 3.029/9.81= 0.30875 kg = 308.75 g. The actual value recorded at the equator during the Gnome Experiment was 307.86 g, a discrepancy of 0.3% which would suggest a contribution from one or both of the first two factors affecting g as discussed at the beginning of this post. Although the work of Hirt et al. (2013) may seem the definitive scientific word on the gravitational environment close to the Earth’s surface, there is great value in taking measurements that are perhaps more directly understandable to check our comprehension: and that I think explains the emotional resonance that many felt in response to the Kern Gnome Experiment. There is a role for the ‘artificer’ as well as the ‘philosopher’ in the scientific enterprise on which humanity has embarked, but perhaps Samuel Johnson put it more eloquently: The philosopher may very justly be delighted with the extent of his views, the artificer with the readiness of his hands; but let the one remember, that, without mechanical performances, refined speculation is an empty dream, and the other, that, without theoretical reasoning, dexterity is little more than a brute instinct. Samuel Johnson, The Rambler, 17 April 1750 ### References Hirt, C., Claessens, S., Fecher, T., Kuhn, M., Pail, R., & Rexer, M. (2013). New ultrahigh‐resolution picture of Earth’s gravity fieldGeophysical research letters40(16), 4279-4283. Jones, F. (2007). Geophysics Foundations: Physical Properties: Density. University of British Columbia website, accessed on 2/5/21. < ## FIFA and Really Challenging GCSE Physics Calculations ‘FIFA’ in this context has nothing to do with football; rather, it is a mnemonic that helps KS3 and KS4 students from across the attainment range engage productively with calculation questions. FIFA stands for: • Formula • Insert values • Fine-tune From personal experience, I can say that FIFA has worked to boost physics outcomes in the schools I have worked in. What is especially gratifying, however, is that a number of fellow teaching professionals have been kind enough to share their experience of using it: ### Framing FIFA as a modular approach Straightforward calculation questions (typically 2 or 3 marks) can be ‘unlocked’ using the original FIFA approach. More challenging questions (typically 4 or 5 marks) can often be handled using the FIFA-one-two approach. However, what about the most challenging 5 or 6 mark questions that are targeted at Grade 8/9? Can FIFA help in solving these? I believe it can. But before we dive into that, let’s look at a more traditional, non-FIFA, algebraic approach. ### A challenging freezing question: the traditional (non-FIFA) algebraic approach A pdf of this question is here. A traditional algebraic approach to solving this problem would look like this: This approach would be fine for confident students with high previous attainment in physics and mathematics. I will go further and say that it should be positively encouraged for students who possess — in Edward Gibbon’s words — that ‘happy disposition’: But the power of instruction is seldom of much efficacy, except in those happy dispositions where it is almost superfluous. Edward Gibbon, The Decline and Fall of the Roman Empire But what about those students who are more akin to the rest of us, and for whom the ‘power of instruction’ is not a superfluity but rather a necessity on which they depend? ### A challenging freezing question: the FIFA-1-2-3 approach Since this question involves both cooling and freezing it seems reasonable to start with the specific heat capacity formula and then use the specific latent heat formula: FIFA-one-two isn’t enough. We must resort to FIFA-1-2-3. What is noteworthy here is that the third FIFA formula isn’t on the formula sheet and is not on the list of formulas that need to be memorised. Instead, it is made by the student based on their understanding of physics and a close reading of the question. Challenging? Yes, undoubtedly. But students will have unlocked some marks (up to 4 out of 6 by my estimation). FIFA isn’t a royal road to mathematical mastery (although it certainly is a better bet than the dreaded ‘formula triangle’ that I and many other have used in the past). FIFA is the scaffolding, not the finished product. We complete the FIFA-1-2-3 process as follows: ### Conclusion: FIFA fixes it The FIFA-system was born of the despair engendered when you mark a set of mock exam papers and the majority of pages are blank: students had not even attempted the calculation skills. In my experience, FIFA fixes that — students are much more willing to start a calculation question. And that means that, even when they cannot successfully navigate to a ‘full mark’ conclusion, they gain at least some marks, and and one does not have to be a particularly perceptive scholar of the human heart to understand that gaining ‘some marks‘ is more motivating than ‘no marks‘. Update: Ed Southall makes a persuasive case against formula triangles in this 2016 article. ## Physics Six Mark Calculation Question? Give it the old FIFA-One-Two! Many students struggle with Physics calculation questions at KS3 and KS4. Since 40% of the marks on GCSE Physics papers are for maths, this is a real worry for their teachers. The FIFA system (if that’s not too grandiose a description) provides a minimal and flexible framework that helps students to successfully attempt calculation questions. Since adopting the system, we encounter far fewer blanks on test and exam scripts where students simply skip over a calculation question. A typical student can gain 10-20 marks. The FIFA system is outlined here but essentially consists of: • Formula: students write the formula or equation • Insert values: students insert the known data from the question. • Fine-tune: rearrange, convert units, simplify etc. The “Fine-tune” stage is not — repeat, not — synonymous with re-arranging and is designed to be “creatively ambiguous” and allow space to “do what needs to be done” and can include unit conversion (e.g. kilowatts to watts), algebraic rearrangement and simplification. ### The FIFA-One-Two Uniquely for Physics, instead of the dreaded “Six Marker” extended writing question, we have the even-more-dreaded “Six Marker” long calculation question. (Actually, they can be awarded anywhere between 4 to 6 marks, but we’ll keep calling them “Six Markers” for convenience.) The “FIFA-one-two” strategy can help students gain marks in these questions. Let’s look how it could be applied to a typical “Six mark” long calculation question. We prepare the ground like this: Since the question mentions the power output of the kettle first, let’s begin by writing down the energy transferred equation. Next we insert the values. It’s quite helpful to write in any “non standard” units such as kilowatts, minutes etc as a reminder that these need to be converted in the Fine-tune phase. And so we arrive at the final answer for this first section: Next we write down the specific heat capacity equation: And going through the second FIFA operation: ### Conclusion I think every “Six Marker” extended calculation question can be approached in a productive way using the FIFA-One-Two approach. This means that, even if students can’t reach the final answer, they will pick up some method marks along the way. I hope you give the FIFA-One-Two method a go with your students. You can read more about using the FIFA system here: ‘Using the FIFA system for really challenging GCSE physics calculations‘. Update: Ed Southall makes a very persuasive case against formula triangle in this 2016 article. ## FIFA for the GCSE Physics calculation win Student: Did you know FIFA is also the name of a video game, Sir? Me: Really? Student: Yeah. It’s part of a series. I just got FIFA 20. It’s one of my favourite games ever. Me: Goodness me. I had no idea. I just chose the letters ‘FIFA’ completely and utterly at random! The FIFA method is an AQA mark scheme-friendly* way of approaching GCSE Physics calculation questions. (It is also useful for some Y12 Physics students.) I mentioned it in a previous blog and @PedagogueSci was kind enough to give it a boost here, so I thought I’d explain the method in a separate blog post. (Update: you can also watch my talk at ChatPhysics Live 2021 here.) The FIFA method: 1. Avoids the use of formula triangles 2. Minimises the cognitive load on students when approaching calculations. ### Why we shouldn’t use formula triangles During a university admissions interview for veterinary medicine, I asked a prospective student to explain how they would make up a solution for infusion into a dog. Part of the answer required them to work out the volume required for a given amount and concentration. The candidate started off by drawing a triangle, then hesitated, eventually giving up in despair. […] They are a trick that hides the maths: students don’t apply the skills they have previously learned. This means students don’t realise how important maths is for science. I’m also concerned that if students can’t rearrange simple equations like the one above, they really can’t manage when equations become more complex. [Update: this 2016 article from Ed Southall also makes a very persuasive case against formula triangle.] I believe the use of formula triangle also increases (rather than decreases) the cognitive load on students when carrying out calculations. For example, if the concentration c is 0.5 mol dm-3 and the number of moles n required is 0.01 mol, then in order to calculate the volume V they need to: • recall the relevant equation and what each symbol means and hold it in working memory • recall the layout of symbols within the formula triangle and either (a) write it down or (b) hold it in working memory • recall that n and c are known values and that V is the unknown value and hold this information in working memory when applying the formula triangle to the problem ### The FIFA method in use (part 1) The FIFA acronym stands for: • FORMULA • INSERT VALUES • FINE TUNE (this often, but not always, equates to rearranging the formula) Lets look at applying it for a typical higher level GCSE Physics calculation question Students have to recall the relevant equation as it is not given on the Data and Formula Sheet. They write it down. This is an important step as once it is written down they no longer have to hold it in their working memory. Note that this is less cognitively demanding on the student’s working memory as they only have to recall the formula on its own; they do not have to recall the formula triangle associated with it. Students find it encouraging that on many mark schemes, the selection of the correct equation may gain a mark, even if no further steps are taken. Next, we insert the values. I find it useful to provide a framework for this such as: We can ask general questions such as: “What data are in the question?” or more focused questions such as “Yes or no: are we told what the kinetic energy store is?” and follow up questions such as “What is the kinetic energy? What units do we use for that?” and so on. Note that since we are considering each item of data individually and in a sequence determined by the written formula, this is much less cognitively demanding in terms of what needs to be held in the student’s working memory than the formula triangle method. Note also that on many mark schemes, a mark is available for the correct substitution of values. Even if they were not able to proceed any further, they would still gain 2/5 marks. For many students, the notion of incremental gain in calculation questions needs to be pushed really hard otherwise they will not attempt these “scary” calculation questions. Next we are going to “fine tune” what we have written down in order to calculate the final answer. In this instance, the “fine tuning” process equates to a simple algebraic rearrangement. However, it is useful to leave room for some “creative ambiguity” here as we can also use the “fine tuning” process to resolve difficulties with units. Tempting though it may seem, DON’T change FIFA to FIRA. We fine tune in three distinct steps (see addendum): Finally, we input the values on a calculator to give a final answer. Note that since AQA have declined to provide a unit on the final answer line, a mark is available for writing “kg” in the relevant space — a fact which students find surprising but strangely encouraging. The key idea here is to be as positive and encouraging as possible. Even if all they can do is recall the formula and remember that mass is measured in kg, there is an incremental gain. A mark or two here is always better than zero marks. ### The FIFA method in use (part 2) In this example, we are using the creative ambiguity inherent in the term “fine tune” rather than “rearrange” to resolve a possible difficulty with unit conversion. In this example, we resolve another potential difficulty with unit conversion during the our creatively ambiguous “fine tune” stage: The emphasis, as always, is to resolve issues sequentially and individually in order to minimise cognitive overload. ### The FIFA method and low demand Foundation tier calculation questions I teach the FIFA method to all students, but it’s essential to show how the method can be adapted for low demand Foundation tier questions. (Note: improving student performance on these questions is probably a more significant and quicker and easier win than working on their “extended answer” skills). For the treatment below, the assumption is that students have already been taught the FIFA method in a number of contexts and that we are teaching them how to apply it to the calculation questions on the foundation tier paper, perhaps as part of an examination skills session. For the majority of low demand questions, the required formula will be supplied so students will not need to recall it. What they will need, however, is support in inserting the values correctly. Providing a framework as shown below can be very helpful: Also, clearly indicating where the data came from is useful. The fine tune stage is not needed, so we can move straight to the answer. Note also that the FIFA method can be applied to all calculation questions, not just the ones that could be answered using formula triangle methods, as in part (c) of the question above. ### And finally… I believe that using FIFA helps to make our thinking and methods in Physics calculations more explicit and clearer for students. My hope is that science teachers reading this will give it a go. You can read about using the FIFA system for more challenging questions by clicking on these links: ‘Physics six mark calculation? Give it the old FIFA-one-two!‘ and ‘Using the FIFA system for really challenging GCSE calculations‘. PS If you have enjoyed this, you might also enjoy Dual Coding SUVAT Problems and also Magnification using the Singapore Bar Model. *Disclaimer: AQA has not endorsed the FIFA method. I describe it as “AQA mark scheme-friendly” using my professional own judgment and interpretation of published AQA mark schemes.

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# Search by Topic #### Resources tagged with Mathematical Thinking similar to Engaging Students, Developing Confidence, Promoting Independence: Filter by: Content type: Stage: Challenge level: ##### Other tags that relate to Engaging Students, Developing Confidence, Promoting Independence Mathematical Thinking. University life and mathematics. Gifted and Talented. Problem solving. Learning mathematics. Pedagogy. Cambridge university. Group worthy. Rich Tasks. Applying to university. ### There are 25 results Broad Topics > Mathematics Education and Research > Mathematical Thinking ### Thinking about Different Ways of Thinking ##### Stage: 1, 2, 3 and 4 This article, the first in a series, discusses mathematical-logical intelligence as described by Howard Gardner. ### Engaging Students, Developing Confidence, Promoting Independence ##### Stage: 1, 2, 3, 4 and 5 Ideas to support mathematics teachers who are committed to nurturing confident, resourceful and enthusiastic learners. ### Kingsfield School - Building on Rich Starting Points ##### Stage: 1, 2, 3, 4 and 5 Alf and Tracy explain how the Kingsfield School maths department use common tasks to encourage all students to think mathematically about key areas in the curriculum. ### What Is a Mathematically Rich Task? ##### Stage: 1, 2, 3, 4 and 5 Here we describe the essence of a 'rich' mathematical task ### A New Challenge ##### Stage: 1 and 2 In this article for teachers, Bernard gives some background about the theme for November 2011's primary activities, which focus on analysing different approaches. ### Devon Teachers Enriching NRICH - Part 2 ##### Stage: 1 and 2 This is the second part of an article describing the ‘Enriching Mathematics’ project in Devon. The participating teachers describe NRICH activities they have tried with their learners. ### Devon Teachers Enriching NRICH - Part 1 ##### Stage: 1 and 2 It began in Devon in 2008. The Maths Team was keen to raise the profile of mathematics investigations and further promote mathematical thinking and problem solving in primary classes. Liz was invited. . . . ### Working with Highly Able Mathematicians ##### Stage: 1 and 2 In this article for teachers, Bernard describes ways to challenge higher-attaining children at primary level. ### Using Low Threshold High Ceiling Tasks ##### Stage: 1 and 2 This article explores what LTHC tasks are and why they are a firm favourite here at NRICH. We recommend that you start by reading the article to understand what makes a task LTHC and then delve into. . . . ### Maths and Creativity in Bristol ##### Stage: 1 and 2 This article for teachers describes NRICH's work with Creative Partnerships and three Bristol primary schools. ### Path of Discovery Series: 1. Uncertain Beginnings ##### Stage: 1 Marion Bond suggests that we try to imagine mathematical knowledge as a broad crazy paving rather than a path of stepping stones. There is no one right place to start and there is no one right route. . . . ### Logic, and How it Should Influence Our Teaching ##### Stage: 1, 2, 3 and 4 Providing opportunities for children to participate in group narrative in our classrooms is vital. Their contrasting views lead to a high level of revision and improvement, and through this process. . . . ### Co-operative Problem Solving: Pieces of the Puzzle Approach ##### Stage: 1, 2, 3 and 4 The content of this article is largely drawn from an Australian publication by Peter Gould that has been a source of many successful mathematics lessons for both children and student-teachers. It. . . . ### Integrating Rich Tasks - Activity 1.5 ##### Stage: 1 and 2 This professional development activity encourages you to investigate what pupils are doing when they problem solving. ### Generating Number Patterns: an Email Conversation ##### Stage: 2, 3 and 4 This article for teachers describes the exchanges on an email talk list about ideas for an investigation which has the sum of the squares as its solution. ### Using Questioning to Stimulate Mathematical Thinking ##### Stage: 1, 2 and 3 Good questioning techniques have long being regarded as a fundamental tool of effective teachers. This article for teachers looks at different categories of questions that can promote mathematical. . . . ### Numbers and Notation - Ambiguities and Confusions ##### Stage: 1 While musing about the difficulties children face in comprehending number structure, notation, etc., it occured to the author that there is a vast array of occasions when numbers and signs are used. . . . ### Number Sense Series: Developing Early Number Sense ##### Stage: 1 This article for teachers suggests teaching strategies and resources that can help to develop children's number sense. ### Teaching Fractions with Understanding: Part-whole Concept ##### Stage: 1, 2 and 3 Written for teachers, this article describes four basic approaches children use in understanding fractions as equal parts of a whole. ### Going for Games ##### Stage: 1 and 2 In this article for teachers, Liz Woodham describes the criteria she uses to choose mathematical games for the classroom and shares some examples from NRICH. ### Money Problems? ##### Stage: 1 Marion Bond investigates the skills needed in order for children to understand money. ### I've Submitted a Solution - What Next? ##### Stage: 1, 2, 3, 4 and 5 In this article, the NRICH team describe the process of selecting solutions for publication on the site. ### Encouraging Primary Children to Work Systematically ##### Stage: Early years, 1 and 2 This article for primary teachers suggests ways in which to help children become better at working systematically. ### Children's Mathematical Graphics: Understanding the Key Concept ##### Stage: 1 In this article for teachers, Elizabeth Carruthers and Maulfry Worthington explore the differences between 'recording mathematics' and 'representing mathematical thinking'. ### Breaking the Equation ' Empirical Argument = Proof ' ##### Stage: 2, 3, 4 and 5 This article stems from research on the teaching of proof and offers guidance on how to move learners from focussing on experimental arguments to mathematical arguments and deductive reasoning.

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321w11p1 # 321w11p1 - Multicollinearity Perfect Multicollinearity one... This preview shows pages 1–6. Sign up to view the full content. Multicollinearity Perfect Multicollinearity – one regressor is a perfect linear combination of one or more of the other regressors. perfect multicollinearity: x j = 1 x 1  2 x 2 ...  k x k , where i 0for at least one i. It is easy to solve the problem of perfect multicollinearity by dropping or combining one or more of the independent variables. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document However, with Imperfect Multicollinearity, we face a different dillemna. Imperfect Multicollinearity – two or more regressors are highly correlated. There are no theoretical issues with OLS estimators in the case of imperfect multicollinearity. But if there is a very high correlation between two variables, at least one coefficient will be imprecisely estimated. Recall: Var j = 2 SST j 1 R j 2 The higher the correlation among the regressors, the higher is R j , the smaller the denominator and the higher the variance of j . So when you include highly correlated variables, you are likely to see imprecise estimates for at least one regressor. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document How much correlation is too much? Addressing Multicollinearity: 1. Use joint hypothesis tests instead of individual t-ratios 2. Factor Analysis 3. Drop one or more variables Ex/ Suppose we wish to estimate the effect of University quality on wages and estimate the following regression for Canadian students: This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. ## This note was uploaded on 01/26/2012 for the course ECON 401 taught by Professor Burbidge,john during the Fall '08 term at Waterloo. ### Page1 / 20 321w11p1 - Multicollinearity Perfect Multicollinearity one... This preview shows document pages 1 - 6. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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# A Statistical Approach to the Synoptic Problem: Part 1—Triple Tradition Articles 4 Comments "A Statistical Approach to the Synoptic Problem," a new series on Jerusalem Perspective by Jerusalem School of Synoptic Research member Halvor Ronning, aims to contribute to the body of empirical data that must be accounted for by any viable theory that attempts to describe the interrelationships between the Synoptic Gospels. To that end, Halvor Ronning has developed and adapted several new methods of quantifying and testing synoptic hypotheses which will be described and applied in "A Statistical Approach to to the Synoptic Problem." This article aims to contribute to the body of empirical data—particularly in the matter of the thousands of words involved in the “minor agreements” between the Gospels of Matthew and Luke against the Gospel of Mark—which must be accounted for by any viable theory that attempts to describe the interrelationships between the Synoptic Gospels, the so-called Synoptic Problem. It should be clear why the Synoptic Problem has become such a battleground:[1] Whatever solution is adopted will have tremendous influence on Gospel scholarship, all the way from textual criticism to the attempts to summarize the theologies of the respective Gospels.[2] With stakes that are so high, it is important to analyze the available data on an objective empirical basis. To that end, I have developed and adapted several new methods of quantifying and testing synoptic theories. The method I will discuss here is to evaluate all the options of linear dependence between three authors. Only six theoretically possible options exist, and the question to be asked is whether any of the options can stand up to objective empirical analysis or whether they all fail the test. ## Six Theoretical Options of Linear Dependence Theoretically speaking, it is possible that no literary dependence among the Synoptic Gospels exists at all.[3] On the other hand, all kinds of complicated schemes of interdependence are theoretically possible. Even in the case of linear dependence, the simplest of the hypothetically possible schemes of interdependence, it is theoretically possible that only two of the gospel writers had some literary relationship and the third was completely independent of the other two. The method I will discuss here is intended to test out the viability of the various schemes of linear dependence in which all three authors are involved, of which there are six theoretically possible relationships: • Matthew→Luke→Mark • Mark→Luke→Matthew • Luke→Matthew→Mark • Mark→Matthew→Luke • Luke→Mark→Matthew • Matthew→Mark→Luke Premium Members If you are not a Premium Member, please consider becoming one starting at \$10/month (paid monthly) or only \$5/month (paid annually): One Time Purchase Rather Than Membership Rather than a membership, you may also purchase access to this entire page for \$1.99 USD. (If you do not have an account select "Register & Purchase.") Login & Purchase • [1] This has become all the more clear from the analyses of Stoldt and Meijboom. They wrote independently, and a century apart from one another, yet they agree amazingly on the role of David F. Strauss as scaring people into Markan Priority. See Hans-Martin Stoldt, History and Criticism of the Marcan Hypothesis (Macon, Georgia: Mercer University Press, 1980), 227-235; H. A. Meijboom, History and Critique of the Origin of the Marcan Hypothesis1835-1866, (trans. John J. Kiwiet; Macon, Georgia: Mercer, 1993), 9-11. • [2] Presently all critical editions of the Greek New Testament text which are in common use are products of text critical scholars who accept the theory of Markan Priority. An example of this, noted by David Flusser, is the insertion of the notion of ascension into the text of Luke 24:51 from Mark 16:19 on the biased judgment call that certain papyri, which include the ascension phrase, outweigh all the other texts, which omit the phrase. The old Nestle text did not have this bias at this point; it did not include the phrase about ascension in its text of Luke. This Markan priority bias also artificially creates a conflict between this supposed ascension in Luke 24 and the fuller account in Acts 1 which takes place after forty days. Reuben Swanson has published a new critical edition based on an existing historical document, Codex Vaticanus, with all the variants presented in the apparatus. See Reuben Swanson, New Testament Greek Manuscripts: Variant Readings Arranged in Horizontal Lines Against Codex Vaticanus (4 vols.; Sheffield: Sheffield Academic Press, 1995). Of course, no text will be free from the bias of its editor(s), but there is no good reason why theorists who question Markan Priority should be forced to work from a text already biased in that direction. • [3] See Eta Linnemann, Is there a Synoptic Problem? (Grand Rapids: Baker, 1992). ## Comments 4 1. Hello, how can I get a copy of Huck’s synoptic book. I bought one from amazon but it was really hard to read. print is very small. Thank You • ## Halvor Ronning Halvor Ronning (B.A., St. Olaf Lutheran College; B.D., Lutheran Theological Seminary; M.A., Yale University) is a founding member and past director of the Jerusalem School of Synoptic Research. An officially licensed Israeli tour guide, he has lived in Israel for over fifty years. Although born… [Read more about author]

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### Complex Calculus (MATH 305) 2023 Spring Faculty of Engineering and Natural Sciences Mathematics(MATH) 3 6 Gökalp Alpan gokalp.alpan@sabanciuniv.edu, English MATH102 Formal lecture Interactive,Communicative ### CONTENT This course covers the field of complex numbers, functions of one complex variable; analytic functions, the Cauchy- Riemann equations, harmonic functions integration in the complex plane, Cauchy integral formula, power series, Laurent series and isolated singularities, theory of residues and applications, conformal mappings. ### OBJECTIVE To give an introduction to main methods of Complex Analysis which are needed for successful activity in many fields (engineering, economics, etc.) and for developing of mathematical thinking, as well ### LEARNING OUTCOMES • Upon completion of this course, students should be able to: Operate with complex numbers; • Differentiate and integrate complex valued functions; • Distinguish analyticity from differentiability by real variables (Cauchy-Riemann equation); • Understand how analytic and harmonic functions are connected; • Formulate Cauchy Theorem and Cauchy Formula and apply them consciously for integration; • Identify Taylor and Laurent expansions and distinguish isolated singularities; • Apply Cauchy Residue Theorem to calculations of definite integrals; ### PROGRAMME OUTCOMES 1. Understand the world, their country, their society, as well as themselves and have awareness of ethical problems, social rights, values and responsibility to the self and to others. 1 2. Understand different disciplines from natural and social sciences to mathematics and art, and develop interdisciplinary approaches in thinking and practice. 4 3. Think critically, follow innovations and developments in science and technology, demonstrate personal and organizational entrepreneurship and engage in life-long learning in various subjects; have the ability to continue to educate him/herself. 2 4. Communicate effectively in Turkish and English by oral, written, graphical and technological means. 2 5. Take individual and team responsibility, function effectively and respectively as an individual and a member or a leader of a team; and have the skills to work effectively in multi-disciplinary teams. 2 1. Possess sufficient knowledge of mathematics, science and program-specific engineering topics; use theoretical and applied knowledge of these areas in complex engineering problems. 4 2. Identify, define, formulate and solve complex engineering problems; choose and apply suitable analysis and modeling methods for this purpose. 4 3. Develop, choose and use modern techniques and tools that are needed for analysis and solution of complex problems faced in engineering applications; possess knowledge of standards used in engineering applications; use information technologies effectively. 3 4. Have the ability to design a complex system, process, instrument or a product under realistic constraints and conditions, with the goal of fulfilling specified needs; apply modern design techniques for this purpose. 3 5. Design and conduct experiments, collect data, analyze and interpret the results to investigate complex engineering problems or program-specific research areas. 2 6. Possess knowledge of business practices such as project management, risk management and change management; awareness on innovation; knowledge of sustainable development. 2 7. Possess knowledge of impact of engineering solutions in a global, economic, environmental, health and societal context; knowledge of contemporary issues; awareness on legal outcomes of engineering solutions; knowledge of behavior according to ethical principles, understanding of professional and ethical responsibility. 2 8. Have the ability to write effective reports and comprehend written reports, prepare design and production reports, make effective presentations, and give and receive clear and intelligible instructions. 3 1. Design, implement, test, and evaluate a computer system, component, or algorithm to meet desired needs and to solve a computational problem. 3 2. Demonstrate knowledge of discrete mathematics and data structures. 2 3. Demonstrate knowledge of probability and statistics, including applications appropriate to computer science and engineering. 3 1. Use mathematics (including derivative and integral calculations, probability and statistics, differential equations, linear algebra, complex variables and discrete mathematics), basic sciences, computer and programming, and electronics engineering knowledge to (a) Design and analyze complex electronic circuits, instruments, software and electronics systems with hardware/software or (b) Design and analyze communication networks and systems, signal processing algorithms or software 5 1. Applying fundamental and advanced knowledge of natural sciences as well as engineering principles to develop and design new materials and establish the relation between internal structure and physical properties using experimental, computational and theoretical tools. 4 2. Merging the existing knowledge on physical properties, design limits and fabrication methods in materials selection for a particular application or to resolve material performance related problems. 2 3. Predicting and understanding the behavior of a material under use in a specific environment knowing the internal structure or vice versa. 2 1. Familiarity with concepts in statistics and optimization, knowledge in basic differential and integral calculus, linear algebra, differential equations, complex variables, multi-variable calculus, as well as physics and computer science, and ability to use this knowledge in modeling, design and analysis of complex dynamical systems containing hardware and software components. 5 2. Ability to work in design, implementation and integration of engineering applications, such as electronic, mechanical, electromechanical, control and computer systems that contain software and hardware components, including sensors, actuators and controllers. 2 1. Formulate and analyze problems in complex manufacturing and service systems by comprehending and applying the basic tools of industrial engineering such as modeling and optimization, stochastics, statistics. 3 2. Design and develop appropriate analytical solution strategies for problems in integrated production and service systems involving human capital, materials, information, equipment, and energy. 3 3. Implement solution strategies on a computer platform for decision-support purposes by employing effective computational and experimental tools. 3 ### ASSESSMENT METHODS and CRITERIA Percentage (%) Midterm 60 Participation 5 Homework 40

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http://restructuringalgebra.blogspot.com/2016/01/desmos-picture.html

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## Saturday, January 16, 2016 ### Desmos Picture My students in Algebra 1 are learning about functions, and we start with domain and range of relations.  Due to amazing #MTBoS resources, I had some excellent resources to integrate into a week long lesson, which started with pictionary thanks to +John Scammell (@scamdog) and his post here. My one good thing, however, is about the performance assessment I assigned to students using +Desmos free online graphing calculator.  On our PD day at the beginning of the semester I attended a session by our curriculum director @montemunsinger about creating rubrics for performance assessments, so I used it to help set up this rubric based on my standard related to domain and range: The assignment was to create a picture in Desmos with at least 10 relations.  In addition, students must restrict the domain for three relations and the range for three relations.  Then students complete a reflection where they explain one domain choice they made and one range choice they made.  The reflection is important to me because it is their opportunity to share what they learned, not just what they created via trial and error. Don't get me wrong, the trial and error aspect of Desmos is the only thing that makes this assignment at all possible for my students, but I want to make sure that through the trial and error process they are learning something. So my #onegoodthing is watching my students create!  We worked on it off and on throughout almost the whole week.  Some students jumped right in and have created some awesome things, others wanted to copy a previous Desmos picture they saw, but could explain polar coordinates to me (shocker!) so I made them start over (aka not copy).  Some students needed a lot of guidance at first ("Try y=mx+b and substitute some things in for m and b until you get what you want.  Now what part of the line do you want for your picture?  How do we do that?") and then were able to take off and just ask me for help with troubleshooting when they made an error ("Why did my whole line just disappear when I did the domain?" *I check and see -7.5<=x<=-8*  "Remember to put the minimum on the left and the maximum on the right..."). Hopefully, I can get permission to post some pictures here, but let's just say I've seen Olaf, a Christmas tree w/star and presents, batman symbol, personal designs, etc.  My favorite part, however, is when the students learn about new types of relations.  "How do I make a circle?"  "How do I make an oval?"  "Can you help me make this rounded?" I don't usually get to share about circles and ellipses in Algebra 1, but we did this week! -Kathryn

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http://www.gurufocus.com/term/Total%20Current%20Assets/PBY/Total%252BCurrent%252BAssets/Pep%2BBoys%2B-%2BManny%252C%2BMoe%2B%2526%2BJack

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Switch to: More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas. Pep Boys - Manny Moe & Jack (NYSE:PBY) Total Current Assets \$797 Mil (As of Oct. 2015) Total current assets includes Cash and Cash Equivalents, Accounts Receivable, Inventory, and Other Current Assets. Pep Boys - Manny Moe & Jack's total current assets for the quarter that ended in Oct. 2015 was \$797 Mil. Definition Total Current Assets are the asset that can be converted to cash or used to pay current liabilities within 12 months. Pep Boys - Manny Moe & Jack's Total Current Assets for the fiscal year that ended in Jan. 2015 is calculated as Total Current Assets = Cash and Cash Equivalents + Accounts Receivable + Inventory + Other Current Assets = 38.044 + 31.013 + 656.957 + 86.586 = 813 Pep Boys - Manny Moe & Jack's Total Current Assets for the quarter that ended in Oct. 2015 is calculated as Total Current Assets = Cash and Cash Equivalents + Accounts Receivable + Inventory + Other Current Assets = 71.332 + 31.743 + 630.608 + 63.798 = 797 * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Total Current Assets is linked to the Current Ratio, which is the result of dividing total current assets by total current liabilities. It is frequently used as an indicator of a company's liquidity, its ability to meet short-term obligations. Total Current Assets is also linked to Working Capital, Net working capital is calculated as Total Current Assets minus Total Current Liabilities. Explanation In Ben Graham’s calculation of liquidation value, inventory is only considered worth half of its book value, and accounts receivable is considered worth 75% of its value. Therefore the liquidation value is lower than calculated from total current assets. Pep Boys - Manny Moe & Jack's Liquidation Value for the quarter that ended in Oct. 2015 is Liquidation value = Cash and Cash Equivalents - Total Liabilities + (0.75 * Account Receivable) + (0.5 * Inventory) = 71.332 - 944.663 + 0.75 * 31.743 + 0.5 * 630.608 = -534 * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Related Terms Historical Data * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Pep Boys - Manny Moe & Jack Annual Data Jan06 Jan07 Jan08 Jan09 Jan10 Jan11 Jan12 Jan13 Jan14 Jan15 Total Current Assets 828 768 750 716 716 764 785 814 826 813 Pep Boys - Manny Moe & Jack Quarterly Data Jul13 Oct13 Jan14 Apr14 Jul14 Oct14 Jan15 Apr15 Jul15 Oct15 Total Current Assets 818 815 826 817 806 807 813 804 803 797 Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)

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https://justaaa.com/physics/522879-in-the-figure-two-blocks-are-shown-with-an

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Question # In the figure, two blocks are shown with an inclined plane. The two blocks are connected... In the figure, two blocks are shown with an inclined plane. The two blocks are connected by a massless string strung over a massless pulley. The mass of Block #1 is 3.57 kg and that of Block #2 is 11.0 kg. The angle θ of the incline is 43.0 degrees. The plane is NOT smooth and has a coefficient of static friction of 0.570 and a coefficient of kinetic friction of 0.240. Taking the positive direction to be up the incline, determine the following: (a) the acceleration of Block #1 if it is initially at rest. m/s2 (b) the acceleration of Block #1 if it is already moving up the plane. m/s2 (c) the acceleration of Block #1 if it is already moving down the plane. m/s2 #### Earn Coins Coins can be redeemed for fabulous gifts.

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https://starcoder.org/usaco/USACO-2020-Open-Silver/

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# USACO 2020 Open Silver ## Problem 1. Social Distancing Social Distancing Farmer John is worried for the health of his cows after an outbreak of the highly contagious bovine disease COWVID-19. In order to limit transmission of the disease, Farmer John’s N cows (2≤N≤105) have decided to practice “social distancing” and spread themselves out across the farm. The farm is shaped like a 1D number line, with M mutually-disjoint intervals (1≤M≤105) in which there is grass for grazing. The cows want to locate themselves at distinct integer points, each covered in grass, so as to maximize the value of D, where D represents the distance between the closest pair of cows. Please help the cows determine the largest possible value of D. INPUT FORMAT (file socdist.in): The first line of input contains N and M. The next M lines each describe an interval in terms of two integers a and b, where 0≤a≤b≤1018. No two intervals overlap or touch at their endpoints. A cow standing on the endpoint of an interval counts as standing on grass. OUTPUT FORMAT (file socdist.out): Print the largest possible value of D such that all pairs of cows are D units apart. A solution with D>0 is guaranteed to exist. SAMPLE INPUT: 5 3 0 2 4 7 9 9 SAMPLE OUTPUT: 2 One way to achieve D=2 is to have cows at positions 0, 2, 4, 6 and 9. SCORING: Test cases 2-3 satisfy b≤105. Test cases 4-10 satisfy no additional constraints. Problem credits: Brian Dean ## Problem 2. Cereal Cereal Farmer John’s cows like nothing more than cereal for breakfast! In fact, the cows have such large appetites that they will each eat an entire box of cereal for a single meal. The farm has recently received a shipment with M different types of cereal (1≤M≤105) . Unfortunately, there is only one box of each cereal! Each of the N cows (1≤N≤105) has a favorite cereal and a second favorite cereal. When given a selection of cereals to choose from, a cow performs the following process: If the box of her favorite cereal is still available, take it and leave. Otherwise, if the box of her second-favorite cereal is still available, take it and leave. Otherwise, she will moo with disappointment and leave without taking any cereal. The cows have lined up to get cereal. For each 0≤i≤N−1, determine how many cows would take a box of cereal if Farmer John removed the first i cows from the line. INPUT FORMAT (file cereal.in): The first line contains two space-separated integers N and M. For each 1≤i≤N, the i-th line contains two space-separted integers fi and si (1≤fi,si≤M and fi≠si) denoting the favorite and second-favorite cereals of the i-th cow in line. OUTPUT FORMAT (file cereal.out): For each 0≤i≤N−1, print a line containing the answer for i. SAMPLE INPUT: 4 2 1 2 1 2 1 2 1 2 SAMPLE OUTPUT: 2 2 2 1 If at least two cows remain, then exactly two of them get a box of cereal. SCORING: Test cases 2-3 satisfy N,M≤1000. Test cases 4-10 satisfy no additional constraints. Problem credits: Dhruv Rohatgi ## Problem 3. The Moo Particle The Moo Particle Quarantined for their protection during an outbreak of COWVID-19, Farmer John’s cows have come up with a new way to alleviate their boredom: studying advanced physics! In fact, the cows have even managed to discover a new subatomic particle, which they have named the “moo particle”. The cows are currently running an experiment involving N moo particles (1≤N≤105). Particle i has a “spin” described by two integers xi and yi in the range −109…109 inclusive. Sometimes two moo particles interact. This can happen to particles with spins (xi,yi) and (xj,yj) only if xi≤xj and yi≤yj. Under these conditions, it’s possible that exactly one of these two particles may disappear (and nothing happens to the other particle). At any given time, at most one interaction will occur. The cows want to know the minimum number of moo particles that may be left after some arbitrary sequence of interactions. INPUT FORMAT (file moop.in): The first line contains a single integer N, the initial number of moo particles. Each of the next N lines contains two space-separated integers, indicating the spin of one particle. Each particle has a distinct spin. OUTPUT FORMAT (file moop.out): A single integer, the smallest number of moo particles that may remain after some arbitrary sequence of interactions. SAMPLE INPUT: 4 1 0 0 1 -1 0 0 -1 SAMPLE OUTPUT: 1 One possible sequence of interactions: Particles 1 and 4 interact, particle 1 disappears. Particles 2 and 4 interact, particle 4 disappears. Particles 2 and 3 interact, particle 3 disappears. Only particle 2 remains. SAMPLE INPUT: 3 0 0 1 1 -1 3 SAMPLE OUTPUT: 2 Particle 3 cannot interact with either of the other two particles, so it must remain. At least one of particles 1 and 2 must also remain. SCORING: Test cases 3-6 satisfy N≤1000. Test cases 7-12 satisfy no additional constraints. Problem credits: Dhruv Rohatgi

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https://factoring-polynomials.com/factoring-polynomial/simplifying-expressions/solve-algebra-equations.html

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Home A Summary of Factoring Polynomials Factoring The Difference of 2 Squares Factoring Trinomials Quadratic Expressions Factoring Trinomials The 7 Forms of Factoring Factoring Trinomials Finding The Greatest Common Factor (GCF) Factoring Trinomials Quadratic Expressions Factoring simple expressions Polynomials Factoring Polynomials Fractoring Polynomials Other Math Resources Factoring Polynomials Polynomials Finding the Greatest Common Factor (GCF) Factoring Trinomials Finding the Least Common Multiples Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: # solve algebra equations? Below is a number of phrases that visitors entered recently to visit math help pages. How can this be helpful ? • find the search keyword you are searching for (i.e. solve algebra equations) in the table below • Click on the related program demo found in the same line  as your search term • If you find the program demonstration of help click on the purchase button to obtain the program at a special low price extended only to factoring-polynomials.com users Related Search Phrase Algebrator Flash Demo Algebrator Static html Demo Purchase now how to put a linear equation into vertex form software for solving third order polynomial how to find a GCF using a TI-83 PLUS Second Order Homogeneous glencoe physics answers 4th grade math variables worksheets compare and order decimals worksheet foerster algebra sample polynomials simplifying calculator pie r squared freecalculator calculate difference quotient graph a differential equation with matlab AMATYC answer explaination free math study sheets add subtract multiply divide decimals solving linear systems of equations on TI-83+ math exercises for preschools Graphing solvers solving exponential equations with fractional coefficients multiply numbers entered from keyboard in java easy cal algebra nonlinear differential equations in matlab buy thinkwell physics cd ONLINE MATH WORD PROBLEM SOLVER googlefractions balancing equations worksheet Prev Next

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1529216000000 1,529,216,000,000 (one trillion five hundred twenty-nine billion two hundred sixteen million) is an even thirteen-digits composite number following 1529215999999 and preceding 1529216000001. In scientific notation, it is written as 1.529216 × 1012. The sum of its digits is 26. It has a total of 21 prime factors and 392 positive divisors. There are 564,019,200,000 positive integers (up to 1529216000000) that are relatively prime to 1529216000000. Basic properties • Is Prime? No • Number parity Even • Number length 13 • Sum of Digits 26 • Digital Root 8 Name Short name 1 trillion 529 billion 216 million one trillion five hundred twenty-nine billion two hundred sixteen million Notation Scientific notation 1.529216 × 1012 1.529216 × 1012 Prime Factorization of 1529216000000 Prime Factorization 213 × 56 × 13 × 919 Composite number Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 21 Total number of prime factors rad(n) 119470 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 1,529,216,000,000 is 213 × 56 × 13 × 919. Since it has a total of 21 prime factors, 1,529,216,000,000 is a composite number. Divisors of 1529216000000 392 divisors Even divisors 364 28 14 14 Total Divisors Sum of Divisors Aliquot Sum τ(n) 392 Total number of the positive divisors of n σ(n) 4.1213e+12 Sum of all the positive divisors of n s(n) 2.59208e+12 Sum of the proper positive divisors of n A(n) 1.05135e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23661e+06 Returns the nth root of the product of n divisors H(n) 145.452 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 1,529,216,000,000 can be divided by 392 positive divisors (out of which 364 are even, and 28 are odd). The sum of these divisors (counting 1,529,216,000,000) is 4,121,295,684,240, the average is 105,135,093,98.,571. Other Arithmetic Functions (n = 1529216000000) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 564019200000 Total number of positive integers not greater than n that are coprime to n λ(n) 5875200000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56581057859 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares There are 564,019,200,000 positive integers (less than 1,529,216,000,000) that are coprime with 1,529,216,000,000. And there are approximately 56,581,057,859 prime numbers less than or equal to 1,529,216,000,000. Divisibility of 1529216000000 m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 3 0 8 The number 1,529,216,000,000 is divisible by 2, 4, 5 and 8. • Abundant • Polite • Practical • Frugal Base conversion (1529216000000) Base System Value 2 Binary 10110010000001100011000000110000000000000 3 Ternary 12102012011200200201021122 4 Quaternary 112100030120012000000 5 Quinary 200023313244000000 6 Senary 3130302342011412 8 Octal 26201430060000 10 Decimal 1529216000000 12 Duodecimal 208457416b68 20 Vigesimal 2jee000000 36 Base36 jiifq1a8 Basic calculations (n = 1529216000000) Multiplication n×i n×2 3058432000000 4587648000000 6116864000000 7646080000000 Division ni n⁄2 7.64608e+11 5.09739e+11 3.82304e+11 3.05843e+11 Exponentiation ni n2 2338501574656000000000000 3576074023989149696000000000000000000 5468589614668591541518336000000000000000000000000 8362654736185044882754503704576000000000000000000000000000000 Nth Root i√n 2√n 1.23661e+06 11521 1112.03 273.46 1529216000000 as geometric shapes Circle Diameter 3.05843e+12 9.60835e+12 7.34662e+24 Sphere Volume 1.49794e+37 2.93865e+25 9.60835e+12 Square Length = n Perimeter 6.11686e+12 2.3385e+24 2.16264e+12 Cube Length = n Surface area 1.4031e+25 3.57607e+36 2.64868e+12 Equilateral Triangle Length = n Perimeter 4.58765e+12 1.0126e+24 1.32434e+12 Triangular Pyramid Length = n Surface area 4.0504e+24 4.21444e+35 1.2486e+12 Cryptographic Hash Functions md5 54c97a96e74dbab8ce8623110295bf45 93db1f5817269604ff3cf850098ec552e3635cb9 7940fe997fac53826873aba3b9b47d4ae305e58d82ab38b2b57b765ce51da3d6 3105b86754539dddd93028aaf8eafa42dfb43dc14ff12bd281382b74e6b688ebb16da3ee3f18b604a27ad80a788912159bd2c26f06ed19c8581aa4b5f9d45711 086e66c71eed96f64803e3286af6c51742218424

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https://oeis.org/A115513

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A115513 Inverse of matrix (1,x)+(x,x^3). 1 1, -1, 1, 0, 0, 1, 0, 0, 0, 1, 1, -1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; table; graph; refs; listen; history; text; internal format) OFFSET 0,1 COMMENTS Row sums are A014578(n+5) or sum{k=0..ceiling(log(n+4)/log(3)),(-1)^k*(floor((n+4)/3^k)-floor((n+3)/3^k)}. Inverse of number triangle A115512. LINKS EXAMPLE Triangle begins 1, -1, 1, 0, 0, 1, 0, 0, 0, 1, 1, -1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 CROSSREFS Sequence in context: A128189 A303340 A115512 * A133080 A316917 A133985 Adjacent sequences:  A115510 A115511 A115512 * A115514 A115515 A115516 KEYWORD sign,tabl AUTHOR Paul Barry, Jan 23 2006 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified June 13 00:57 EDT 2021. Contains 344980 sequences. (Running on oeis4.)

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# Search by Topic #### Resources tagged with Mathematical reasoning & proof similar to Inspector Morse: Filter by: Content type: Stage: Challenge level: ### There are 176 results Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof ### Football Champs ##### Stage: 3 Challenge Level: Three teams have each played two matches. The table gives the total number points and goals scored for and against each team. Fill in the table and find the scores in the three matches. ### Flight of the Flibbins ##### Stage: 3 Challenge Level: Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . . ### Pattern of Islands ##### Stage: 3 Challenge Level: In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island... ### 9 Weights ##### Stage: 3 Challenge Level: You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance? ### Hockey ##### Stage: 3 Challenge Level: After some matches were played, most of the information in the table containing the results of the games was accidentally deleted. What was the score in each match played? ### Not Necessarily in That Order ##### Stage: 3 Challenge Level: Baker, Cooper, Jones and Smith are four people whose occupations are teacher, welder, mechanic and programmer, but not necessarily in that order. What is each person’s occupation? ### Online ##### Stage: 2 and 3 Challenge Level: A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter. ### Top-heavy Pyramids ##### Stage: 3 Challenge Level: Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200. ### Children at Large ##### Stage: 3 Challenge Level: There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children? ### Problem Solving, Using and Applying and Functional Mathematics ##### Stage: 1, 2, 3, 4 and 5 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ### Clocked ##### Stage: 3 Challenge Level: Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours? ### Dicing with Numbers ##### Stage: 3 Challenge Level: In how many ways can you arrange three dice side by side on a surface so that the sum of the numbers on each of the four faces (top, bottom, front and back) is equal? ### 1 Step 2 Step ##### Stage: 3 Challenge Level: Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps? ### One O Five ##### Stage: 3 Challenge Level: You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . . ### Aba ##### Stage: 3 Challenge Level: In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct. ### Tis Unique ##### Stage: 3 Challenge Level: This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility. ### Calendar Capers ##### Stage: 3 Challenge Level: Choose any three by three square of dates on a calendar page... ### Even So ##### Stage: 3 Challenge Level: Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why? ### Elevenses ##### Stage: 3 Challenge Level: How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results? ##### Stage: 3 Challenge Level: Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . . ### Tourism ##### Stage: 3 Challenge Level: If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable. ### Take Three from Five ##### Stage: 4 Challenge Level: Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him? ### Unit Fractions ##### Stage: 3 Challenge Level: Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation. ### Königsberg ##### Stage: 3 Challenge Level: Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps? ### Tessellating Hexagons ##### Stage: 3 Challenge Level: Which hexagons tessellate? ### Concrete Wheel ##### Stage: 3 Challenge Level: A huge wheel is rolling past your window. What do you see? ### Triangle Inequality ##### Stage: 3 Challenge Level: ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm. ### Eleven ##### Stage: 3 Challenge Level: Replace each letter with a digit to make this addition correct. ### Cross-country Race ##### Stage: 3 Challenge Level: Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places? ### Cycle It ##### Stage: 3 Challenge Level: Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total. ### Volume of a Pyramid and a Cone ##### Stage: 3 These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts. ### Convex Polygons ##### Stage: 3 Challenge Level: Show that among the interior angles of a convex polygon there cannot be more than three acute angles. ### How Many Dice? ##### Stage: 3 Challenge Level: A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . . ### Sticky Numbers ##### Stage: 3 Challenge Level: Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number? ### Greetings ##### Stage: 3 Challenge Level: From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . . ### Pyramids ##### Stage: 3 Challenge Level: What are the missing numbers in the pyramids? ### What Numbers Can We Make Now? ##### Stage: 3 Challenge Level: Imagine we have four bags containing numbers from a sequence. What numbers can we make now? ### Always the Same ##### Stage: 3 Challenge Level: Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34? ### What Numbers Can We Make? ##### Stage: 3 Challenge Level: Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make? ### Tri-colour ##### Stage: 3 Challenge Level: Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs? ### Konigsberg Plus ##### Stage: 3 Challenge Level: Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges. ### N000ughty Thoughts ##### Stage: 4 Challenge Level: How many noughts are at the end of these giant numbers? ### Advent Calendar 2011 - Secondary ##### Stage: 3, 4 and 5 Challenge Level: Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas. ### Logic ##### Stage: 2 and 3 What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article. ### Go Forth and Generalise ##### Stage: 3 Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important. ### Natural Sum ##### Stage: 4 Challenge Level: The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . . ### Sprouts Explained ##### Stage: 2, 3, 4 and 5 This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . . ### AMGM ##### Stage: 4 Challenge Level: Can you use the diagram to prove the AM-GM inequality? ### A Long Time at the Till ##### Stage: 4 and 5 Challenge Level: Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem? ### Picture Story ##### Stage: 4 Challenge Level: Can you see how this picture illustrates the formula for the sum of the first six cube numbers?

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# Question:Why does LinearAlgebra:-Eigenvectors(A) give a relatively complicated answer for a relatively simple 2x2 matrix A? ## Question:Why does LinearAlgebra:-Eigenvectors(A) give a relatively complicated answer for a relatively simple 2x2 matrix A? Maple Consider the task of finding the eigenvalues and eigenvectors of a simple 2x2 matrix. Usually I can insert the contents of a Maple worksheet here, but for some reason the following worksheet cannot be inserted: Eigenvectors.mw In that worksheet I try to use LinearAlgebra:-Eigenvectors(A). The eigenvalues contain a complex term, even though they are real for the given matrix. It's not clear what criteria are used in selecting a specific eigenvector for each eigenvalue. I then show a more manual calculation. I was expecting to obtain a simpler solution to this problem using LinearAlgebra:-Eigenvectors(A). Is this expectation unjustified? I am asking primarily from the perspective of a user of the software. It would be interesting to know the answer from the perspective of someone who knows the ins and outs of the implementation of the software as well, but as a user my initial expectation is a more digestible result that doesn't rely on knowing such implementation details. 

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# Math and Computing Olympiad Practice Happy New Year, everyone! Let me tell you about the epic New Year's Eve that I had. I got into a fight with the last problem from the December 2016 USA Computing Olympiad contest. I struggled mightily, felt beaten down at times, lost all hope, but I finally overcame. It was a marathon. We started sparring around noon, and I did not vanquish my foe until the final hour of 2017. Having a long weekend in a strange new city, I've had to get creative with things to do. I decided to tackle some olympiad problems. For those who are not familiar with the competitive math or programming scene, the USAMO and USACO are math and programming contests targeted towards high school students. So, as an old man, what am I doing whittling away precious hours tackling these problems? I wish that I could say that I was reliving my glory days from high school. But truth be told, I've always been a lackluster student, who did the minimal effort necessary. I can't recall ever having written a single line of code in high school, and I maybe solved 2 or 3 AIME problems (10 years later, I can usually do the first 10 with some consistency, the rest are a toss-up). Of course, I never trained for the competitions, so who knows if I could have potentially have done well. We all have regrets from our youth. For me, I have all the familiar ones: mistreating people, lost friends, not having the best relationship with my parents, losing tennis matches, quitting the violin, and of course, the one girl that got away. However, what I really regret the most was not having pursued math and computer science earlier. I'm not sure why. Even now, 10 years older, it's quite clear that I am not talented enough to have competed in the IMO or IOI: I couldn't really hack it as a mathematician, and you can scroll to the very bottom to see my score of 33. Despite the lack of talent, I just really love problem solving. In many ways it's become an escape for me when I feel lonely and can't make sense of the world. I can get lost in my own abstract world and forget about my physical needs like sleep or food. On solving a problem, I wake up from my stupor, realize that the world has left me behind, and find everyone suddenly married with kids. There is such triumph in solving a hard problem. Of course, there are times of struggle and hopelessness. Such is my fledging sense of self-worth that it depends on my ability to solve abstract problems that have no basis in reality. Thus, I want to write up my solution to Robotic Cow Herd and 2013 USAMO Problem 2 ## Robotic Cow Herd In the Platinum Division December 2016 contest, there were 3 problems. In contest, I was completely stuck on Lots of Triangles and never had a chance to look at the other 2 problems. This past Friday, I did Team Building in my own time. It took me maybe 3 hours, so I suspect if I started with that problem instead, I could have gotten a decent amount of points on the board. Yesterday, I attempted Robotic Cow Herd. I was actually able to solve this problem on my own, but I worked on it on and off over a period of 12 hours, so I definitely wouldn't have scored anything in this case. My solution is quite different than the given solution, which uses binary search. I did actually consider such a solution, but only gave it 5 minutes of though before abandoning it, far too little time to work out the details. Instead, my solution is quite similar to the one that they describe using priority queue before saying such a solution wouldn't be feasible. However, if we are careful about how we fill our queue it can work. We are charged with assembling $K$ different cows that consist of $N$ components, where each component will have $M$ different types. Each type of component has an associated cost, and cow $A$ is different than cow $B$ if at least one of the components is of a different type. Of course, we aren't going to try all $M^N$ different cows. It's clear right away that we can take greedy approach, start with the cheapest cow, and get the next cheapest cow by varying a single component. Since each new cow that we make is based on a previous cow, it's only necessary to store the deltas rather than all $N$ components. Naturally, this gives way to a tree representation shown in the title picture. Each node is a cow prototype. We start with the cheapest cow as the root, and each child consists of a single delta. The cost of a cow can be had by summing the deltas from the root to the node. Now every prototype gives way to $N$ new possible prototypes. $NK$ is just too much to fit in a priority queue. Hence, the official solution says this approach isn't feasible. However, if we sort our components in the proper order, we know the next two cheapest cows based off this prototype. Moreover, we have to handle a special case, where instead of a cow just generating children, it also generates a sibling. We sort by increasing deltas. In the given sample data, our base cost is $4$, and our delta matrix (not a true matrix) looks like $$\begin{pmatrix} 1 & 0 \\ 2 & 1 & 2 & 2\\ 2 & 2 & 5 \end{pmatrix}.$$ Also, we add our microcontrollers in increasing order to avoid double counting. Now, if we have just added microcontroller $(i,j)$, the cheapest thing to do is to change it to $(i + 1, 0)$ or $(i, j + 1)$. But what about the case, where we want to skip $(i+1,0)$ and add $(i + 2, 0), (i+3,0),\ldots$? Since we're lazy about pushing into our priority queue and only add one child at a time, when a child is removed, we add its sibling in this special case where $j = 0$. Parent-child relationships are marked with solid lines. Creation of a node is marked with a red arrow. Nodes still in the queue are blue. The number before the colon denotes the rank of the cow. In this case, the cost for 10 cows is $$4 + 5 + 5 + 6 + 6 + 7 + 7 + 7 + 7 + 7 = 61.$$ Dashed lines represent the special case of creating a sibling. The tuple $(1,-,0)$ means we used microcontrollers $(0,1)$ and $(2,0)$. For component $1$, we decided to just use cheapest one. Here's the code. import java.io.*; import java.util.*; public class roboherd { /** * Microcontrollers are stored in a matrix-like structure with rows and columns. * Use row-first ordering. */ private static class Position implements Comparable<Position> { private int row; private int column; public Position(int row, int column) { this.row = row; this.column = column; } public int getRow() { return this.row; } public int getColumn() { return this.column; } public int compareTo(Position other) { if (this.getRow() != other.getRow()) return this.getRow() - other.getRow(); return this.getColumn() - other.getColumn(); } @Override public String toString() { return "{" + this.getRow() + ", " + this.getColumn() + "}"; } } /** * Stores the current cost of a cow along with the last microcontroller added. To save space, * states only store the last delta and obscures the rest of the state in the cost variable. */ private static class MicrocontrollerState implements Comparable<MicrocontrollerState> { private long cost; private Position position; // the position of the last microcontroller added public MicrocontrollerState(long cost, Position position) { this.cost = cost; this.position = position; } public long getCost() { return this.cost; } public Position getPosition() { return this.position; } public int compareTo(MicrocontrollerState other) { if (this.getCost() != other.getCost()) return (int) Math.signum(this.getCost() - other.getCost()); return this.position.compareTo(other.position); } } public static void main(String[] args) throws IOException { PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("roboherd.out"))); int N = Integer.parseInt(st.nextToken()); // number of microcontrollers per cow int K = Integer.parseInt(st.nextToken()); // number of cows to make assert 1 <= N && N <= 100000 : N; assert 1 <= K && K <= 100000 : K; ArrayList<int[]> P = new ArrayList<int[]>(N); // microcontroller cost deltas long minCost = 0; // min cost to make all the cows wanted for (int i = 0; i < N; ++i) { int M = Integer.parseInt(st.nextToken()); assert 1 <= M && M <= 10 : M; int[] costs = new int[M]; for (int j = 0; j < M; ++j) { costs[j] = Integer.parseInt(st.nextToken()); assert 1 <= costs[j] && costs[j] <= 100000000 : costs[j]; } Arrays.sort(costs); minCost += costs[0]; // Store deltas, which will only exist if there is more than one type of microcontroller. if (M > 1) { int[] costDeltas = new int[M - 1]; for (int j = M - 2; j >= 0; --j) costDeltas[j] = costs[j + 1] - costs[j]; } } in.close(); N = P.size(); // redefine N to exclude microcontrollers of only 1 type --K; // we already have our first cow // Identify the next best configuration in log(K) time. PriorityQueue<MicrocontrollerState> pq = new PriorityQueue<MicrocontrollerState>(3*K); // Order the microcontrollers in such a way that if we were to vary the prototype by only 1, // the best way to do would be to pick microcontrollers in the order // (0,0), (0,1),...,(0,M_0-2),(1,0),...,(1,M_1-2),...,(N-1,0),...,(N-1,M_{N-1}-2) Collections.sort(P, new Comparator<int[]>() { @Override public int compare(int[] a, int[] b) { for (int j = 0; j < Math.min(a.length, b.length); ++j) if (a[j] != b[j]) return a[j] - b[j]; return a.length - b.length; } }); pq.add(new MicrocontrollerState(minCost + P.get(0)[0], new Position(0, 0))); // Imagine constructing a tree with K nodes, where the root is the cheapest cow. Each node contains // the delta from its parent. The next cheapest cow can always be had by taking an existing node on // the tree and varying a single microcontroller. for (; K > 0; --K) { MicrocontrollerState currentState = pq.remove(); // get the next best cow prototype. long currentCost = currentState.getCost(); minCost += currentCost; int i = currentState.getPosition().getRow(); int j = currentState.getPosition().getColumn(); // Our invariant to avoid double counting is to only add microcontrollers with "greater" position. // Given a prototype, from our ordering, the best way to vary a single microcontroller is replace // it with (i,j + 1) or add (i + 1, 0). if (j + 1 < P.get(i).length) { pq.add(new MicrocontrollerState(currentCost + P.get(i)[j + 1], new Position(i, j + 1))); } if (i + 1 < N) { // Account for the special case, where we just use the cheapest version of type i microcontrollers. // Thus, we remove i and add i + 1. This is better than preemptively filling the priority queue. if (j == 0) pq.add(new MicrocontrollerState( currentCost - P.get(i)[j] + P.get(i + 1)[0], new Position(i + 1, 0))); pq.add(new MicrocontrollerState(currentCost + P.get(i + 1)[0], new Position(i + 1, 0))); } } out.println(minCost); out.close(); } } Sorting is $O(NM\log N)$. Polling from the priority queue is $O(K\log K)$ since each node will at most generate 3 additional nodes to put in the priority queue. So, total running time is $O(NM\log N + K\log K)$. ## 2013 USAMO Problem 2 Math has become a bit painful for me. While it was my first love, I have to admit that a bachelor's and master's degree later, I'm a failed mathematician. I've recently overcome my disappointment and decided to persist in learning and practicing math despite my lack of talent. This is the first USAMO problem that I've been able to solve, which I did on Friday. Here's the problem. For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move. Prove that $a_{n-1}+a_n=2^n$ for all $n\geq 4$. The solution on the AOPS wiki uses tiling. I use a different strategy that leads to the same result. Let the points on the cricle be $P_1,P_2, \ldots,P_n$. First, we prove that each point on the circle is visited either $1$ or $2$ times, except for $A = P_1$, which can be visited $3$ times since it's our starting and ending point. It's clear that $2$ times is upper bound for the other points. Suppose a point is never visited, though. We can only move in increments of $1$ and $2$, so if $P_k$ was never visited, we have made a move of $2$ steps from $P_{k-1}$ twice, which is not allowed. In this way, we can index our different paths by tuples $(m_1,m_2,\ldots,m_n)$, where $m_i$ is which move we make the first time that we visit $P_i$, so $m_i \in \{1,2\}$. Since moves have to be distinct, the second move is determined by the first move. Thus, we have $2^n$ possible paths. Here are examples of such paths. Both paths are valid in the sense that no move is repeated. However, we only count the one on the left since after two cycles we must return to $P_1$. The path on the left is $(1,2,2,2,1)$, which is valid since we end up at $A = P_1$. The path on the right is $(1,1,1,1,1)$, which is invalid, since miss $A = P_1$ the second time. The first step from a point is black, and the second step is blue. The edge labels are the order in which the edges are traversed. Now, given all the possible paths with distinct moves for a circle with $n - 1$ points, we can generate all the possible paths for a circle with $n$ points by appending a $1$ or a $2$ to the $n - 1$ paths if we consider their representation as a vector of length $n - 1$ of $1$s and $2$s. In this way, the previous $2^{n-1}$ paths become $2^n$ paths. Now, we can attack the problem in a case-wise manner. 1. Consider an invalid path, $(m_1,m_2,\ldots,m_{n-1})$. All these paths must land at $P_1$ after the first cycle around the circle. Why? Since the path is invalid, that means we touch $P_{n-1}$ in the second cycle and make a jump over $P_1$ by moving $2$ steps. Thus, if we previously touched $P_{n-1},$ we moved to $P_1$ since moves must be distinct. If the first time that we touch $P_{n-1}$ is the second cycle, then, we jumped over it in first cycle by going moving $P_{n-2} \rightarrow P_1$. 1. Make it $(m_1,m_2,\ldots,m_{n-1}, 1)$. This path is now valid. $P_n$ is now where $P_1$ would have been in the first cycle, so we hit $P_n$ and move to $P_1$. Then, we continue as we normally did. Instead of ending like $P_{n-1} \rightarrow P_2$ by jumping over $P_1$, we jump over $P_n$ instead, so we end up making the move $P_{n-1} \rightarrow P_1$ at the end. 2. Make it $(m_1,m_2,\ldots,m_{n-1}, 2)$. This path is now valid. This case is easier. We again touch $P_n$ in the first cycle. Thus, next time we hit $P_n$, we'll make the move $P_n \rightarrow P_1$ since we must make distinct moves. If we don't hit $P_n$ again, that means we jumped $2$ from $P_{n-1}$, which means that we made the move $P_{n - 1} \rightarrow P_1$. 2. Consider an existing valid path, now, $(m_1,m_2,\ldots,m_{n-1})$. There are $a_{n-1}$ of these. 1. Let it be a path where we touch $P_1$ $3$ times. 1. Make it $(m_1,m_2,\ldots,m_{n-1}, 1)$. This path is invalid. $P_n$ will be where $P_1$ was in the first cycle. So, we'll make the move $P_n \rightarrow P_1$ and continue with the same sequence of moves as before. But instead of landing at $P_1$ when the second cycle ends, we'll land at $P_n$, and jump over $P_1$ by making the move $P_n \rightarrow P_2$. 2. Make it $(m_1,m_2,\ldots,m_{n-1}, 2)$. This path is valid. Again, we'll touch $P_n$ in the first cycle, so the next time that we hit $P_n$, we'll move to $P_1$. If we don't touch $P_n$ again, we jump over it onto $P_1$, anyway, by moving $P_{n-1} \rightarrow P_1$. 2. Let it be a path where we touch $P_1$ $2$ times. 1. Make it $(m_1,m_2,\ldots,m_{n-1}, 1)$. This path is valid. Instead of jumping over $P_1$ at the end of the first cycle, we'll be jumping over $P_n$. We must touch $P_n$, eventually, so from there, we'll make the move $P_n \rightarrow P_1$. 2. Make it $(m_1,m_2,\ldots,m_{n-1}, 2)$. This path is invalid. We have the same situation where we skip $P_n$ the first time. Then, we'll have to end up at $P_n$ the second time and make the move $P_{n} \rightarrow P_2$. In either case, old valid paths lead to $1$ new valid path and $1$ new invalid path. Thus, we have that $a_n = 2^n - a_{n-1} \Rightarrow \boxed{a_{n - 1} + a_n = 2^n}$ for $n \geq 4$ since old invalid paths lead to $2$ new valid paths and old valid paths lead to $1$ new valid path. And actually, this proof works when $n \geq 3$ even though the problem only asks for $n \geq 4$. Since we have $P_{n-2} \rightarrow P_1$ at one point in the proof, anything with smaller $n$ is nonsense. Yay, 2017! # Snow, Brownies, USACO, and Binary Search When stuck inside because of the snow, what else is there to do but bake and code? I made these brownies here. They are amazingly moist, but I'll probably cut down on the sugar next time I make them. On the algorithms side, I finally got to the Platinum Division on the USA Computing Olympiad. It's primarily for high school students, but I find it fun to participate anyway. One of the competition's problems employs clever usage of binary search that I want to write about. Basically, there are times when the solution is very hard to compute, but it is not too costly to verify. If the solution is numerical and bounded, we can guess solutions with binary search. I've actually been quite familiar with this strategy for a year now, but somehow I missed it in this particular problem. Here, we use a two-dimensional binary search. Thankfully, I got enough points to get promoted to the next division anyway, anyway. ## Angry Cows Here's the problem statement: Bessie the cow has designed what she thinks will be the next big hit video game: "Angry Cows". The premise, which she believes is completely original, is that the player shoots a cow with a slingshot into a one-dimensional scene consisting of a set of hay bales located at various points on a number line; the cow lands with sufficient force to detonate the hay bales in close proximity to her landing site, which in turn might set of a chain reaction that causes additional hay bales to explode. The goal is to use a single cow to start a chain reaction that detonates all the hay bales. There are $N$ hay bales located at distinct integer positions $x_1,x_2,\ldots,x_N$ on the number line. If a cow is launched with power $R$ landing at position $x$, this will causes a blast of "radius $R$", engulfing all hay bales within the range $x−R \ldots x+R$. These hay bales then themselves explode (all simultaneously), each with a blast radius of $R−1$. Any not-yet-exploded bales caught in these blasts then all explode (all simultaneously) with blast radius $R−2$, and so on. Please determine the minimum amount of power $R$ with which a single cow may be launched so that, if it lands at an appropriate location, it will cause subsequent detonation of every single hay bale in the scene. INPUT FORMAT (file angry.in): The first line of input contains $R$ ($2 \leq N \leq 50,000$). The remaining $N$ lines all contain integers $x_1 \ldots x_N$ (each in the range $0 \ldots 1,000,000,000$). OUTPUT FORMAT (file angry.out): Please output the minimum power $R$ with which a cow must be launched in order to detonate all the hay bales. Answers should be rounded and printed to exactly $1$ decimal point. So, if we assume the hay bales are sorted $x_1 \leq \cdots \leq x_N$. The minimum blast radius must be at most $(x_N - x_1)/2$ since we can just launch such a cow at the midpoint and destroy all the hay bales without the chain reaction. It's also worth noting that if the optimal blast radius is $R^*,$ then $2R^* \in \mathbb{Z}$, that is, twice the optimal blast radius is an integer. Since all the hay bales are located at integer coordinates, adding less than $0.5$ to the radius will never encompass another hay bale. Finally, the last observation is that we should fire the cow so that the very left of the blast lines up exactly with a hay bale since we would not gain anything by having the hay bale strictly inside the blast radius. Let $L$ be the index of the leftmost hay bale hit by the initial blast. Thus, we could brute force by trying all $2R^* \in \{0,1,\ldots,x_N-x_1\}$ and $L \in \{1,2,\ldots,N\}$. To check if such values work, we can simulate the chain reaction which takes $O(N)$ time. Thus, brute force would take $O\left(N^2(x_N - x_1)\right)$ time. This is where binary search comes in. During the contest, it was obvious to me that we should do a binary search to find $2R^*$ considering that $x_N - x_1$ could be as large as $10^9$. However, this is not fast enough, as that only gets us $O\left(N^2\log(x_N-x_1)\right)$ time, and $N^2$ can be as large as $2.5 \times 10^9$. After sleeping on it, I made the key insight that we can binary search on the index of the leftmost hay bale, too, so now we have $O\left(N\log(N)\log(x_N-x_1)\right)$ time, which is adequate. To make this explicit, here's the code: import java.io.*; import java.util.*; public class angry { /* check that all the hay bales to the left of idx explode * if we throw cow of power T/2 at hayBales[idx] + T/2 */ public static boolean leftExplodes(int idx, int T, int[] hayBales) { double currentFloor = hayBales[idx]; double currentR = T/2.0; int left; // leftmost exploded bale for (left = idx; left >= 0 && hayBales[left] >= currentFloor; --left) { if (left == 0 || hayBales[left - 1] >= currentFloor) continue; currentR -= 1.0; currentFloor = hayBales[left] - currentR; } return left == -1; } public static boolean isDiameterPossible(int T, int[] hayBales) { int N = hayBales.length; int leftMin = 0; // inclusive int leftMax = N; // exclusive int leftIdx = leftMin + (leftMax - leftMin)/2; while (leftMin < leftMax) { // find smallest left such that this doesn't work if (leftExplodes(leftIdx, T, hayBales)) { leftMin = leftIdx + 1; } else { leftMax = leftIdx; } leftIdx = leftMin + (leftMax - leftMin)/2; } --leftIdx; // this works // now check that the right explodes double currentCeiling = hayBales[leftIdx] + T; double currentR = T/2.0; int right; for (right = leftIdx; right < N && hayBales[right] <= currentCeiling; ++right) { if (right == N - 1 || hayBales[right + 1] <= currentCeiling) continue; currentR -= 1.0; currentCeiling = hayBales[right] + currentR; } return right == N; } public static void main(String[] args) throws IOException { PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("angry.out"))); int[] hayBales = new int[N]; for (int n = 0; n < N; ++n) hayBales[n] = Integer.parseInt(in.readLine()); Arrays.sort(hayBales); // search for T = 2R int minT = 0; int maxT = hayBales[N - 1] - hayBales[0]; int T = minT + (maxT - minT)/2; while (minT < maxT) { // find smallest T that works if (isDiameterPossible(T, hayBales)) { maxT = T; } else { minT = T + 1; } T = minT + (maxT - minT)/2; } out.printf("%.1f\n", T/2.0); in.close(); out.close(); } }

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# Java with explanation, easy to understand • ``````class Solution { public int findNumberOfLIS(int[] nums) { if (nums == null || nums.length == 0) return 0; int[] maxLens = new int[nums.length];// length of longest increasing sequence start from i int[] counts = new int[nums.length]; // number of length of longest increasing sequence start from i int maxLen = 1; // length of longest increasing subsequnce maxLens[nums.length-1] = 1; counts[nums.length-1] = 1; for(int i = nums.length -2; i>=0; i--){//Backward iteration, i is used as the first character int curMax = 1; int count = 1; for(int j = i+1; j < nums.length; j++) {//j is used as the second character if(nums[i] < nums[j]){//increasing number if (curMax == maxLens[j]+1)//means have another way to reach the same max length increasing sequence count += counts[j]; //Important: not ++ else if (curMax < maxLens[j]+1){ count = counts[j]; curMax = maxLens[j]+1; } } } maxLens[i] = curMax; counts[i] = count; maxLen = Math.max(maxLen, curMax); } int count = 0; for(int i = 0; i< maxLens.length; i++){//check each possible start position if (maxLens[i] == maxLen) count += counts[i]; } return count; } } `````` • The array length does not exceed 2000...but then why does O(n^2)...space give MLE Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.

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# Wiring Diagram Vw Polo 2000 • Polo 2000 • Date : November 25, 2020 ## Wiring Diagram Vw Polo 2000 Diagram Vw Downloads Wiring Diagram Vw Polo 2000 diagram venous system diagram veni diagram venn diagram verb diagram vector diagram vektor diagram vehicle diagram verbals diagram venture diagram vertebrae diagram ventilation diagram venn a-b diagram venn adalah diagram ventures lp diagram versus chart diagram versus graph diagram venn diagram diagram venous circulation diagram versus drawing diagram venn indonesian diagram venn di word diagram veins in arm diagram venn 3 himpunan Wiring Diagram Vw Polo 2000What's the Most Particular Location of K on the Real Line? ? The way of looking up the mathematical constant K is also known as K a b or K b. This is the most specific location of K on the real line, and it describes the area where the members of this inverse of the positive real line intersect. In a similar manner, the direct sum of these components of this odd part of a ring is referred to as N a b. This is the largest amount, and it represents the angle between the circle and the unit circle. It is important to understand that from the technique of K a b, there aren't any references to the members of this inverse of the positive real line. The sole real method is the direct sum of these elements of the infinite series of positive integers. In a similar manner, in the technique of a b, there are no references to the members of the reverse of the odd integral of a circle. In order to understand more about the areas of K and N on the real line, it's necessary to understand about both N and K. In reality, one can look both K and N with the similar method of looking up the favorable integral of a circle and then finding the reverse of the same integral. Both these approaches have benefits, and one could prefer one over another based on their own preferences. So far as K is concerned, it is extremely easy to locate K on the true line as N isn't so difficult to discover. But with respect to N, the members of this series of positive integers have a lot more branches and differences to their worth than are actually present on the actual line. In order to find the most specific place of N, it's necessary to look up every one the branches of N and then multiply them by the value corresponding to the most specific place of N. If this step isn't followed, it'll be possible to discover a branch N of N in the series of integers which does not actually exist. In reality, this isn't a problem, since it's known in the faculty of mathematics that such branches do not exist. There are nevertheless other issues which arise when trying to determine the place of K if K is located. In fact, it will often be required to divide the period into two equal portions, and this is a complex procedure. • ### Manual De Taller Citroen C2 Hdi Copyright © 2020 - MAMMAROSA.EU

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# In a function, what is the maximum number of times that the graph can intercept the y-axis? The x-axis? jeew-m | Certified Educator Let us consider a function y=f(x) Here when x changes y will change. So y is dependant on x. x is the independent variable and y is dependent. Since y is dependant there cannot be two y values for same x. Then it is not a function. If a function intercept y axis that means in that point we have x=0. So there may be no intercepts with the y-axis. If that happens they same x will have two y values,so there is no function. On the other hand for a y=f(x) function same y can have any number of x values. So a function can intercept x-axis any number of times. The example graphs will show this clearly. Just check how many times a single graph cuts y-axis and x-axis. So a function can intercept x-axis any number of times depending on the function and y-axis can only intercept once as maximum.

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# Relation (mathematics) A relation ( Latin relatio "relationship", "relationship") is generally a relationship that can exist between things. Relations in the sense of mathematics are exclusively those relationships for which it is always clear whether they exist or not; Objects cannot be related to one another “to a certain extent”. This enables a simple set- theoretical definition of the term: A relation is a set of tuples . Things that are related to one another form tuples that are elements of . ${\ displaystyle R}$${\ displaystyle n}$${\ displaystyle R}$${\ displaystyle n}$${\ displaystyle R}$ Unless expressly stated otherwise, a relation is generally understood to be a two-digit or binary relation . In such a relationship, then two elements form and an ordered pair originate and from different base levels and is the name of the relation heterogeneous or "relation between the amounts and ." If the basic quantities match ( ), then that means the relation homogeneously or "Relation in or on the crowd . "${\ displaystyle a}$${\ displaystyle b}$ ${\ displaystyle (a, b).}$${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle A = B}$${\ displaystyle A}$ Important special cases , for example equivalence relations and order relations , are relations on a set. Today, some authors consider the term relation not necessarily as to amounts limited to but let each consisting of ordered pairs class are considered relation. ## Definitions ### Double-digit relation A two-digit relation (also called binary relation ) between two sets and is a subset of the Cartesian product${\ displaystyle R}$${\ displaystyle A}$${\ displaystyle B}$ ${\ displaystyle A \ times B = \ {(a, b) \ mid a \ in A, b \ in B \} \ colon}$ ${\ displaystyle R \ subseteq A \ times B}$. The set is referred to as the source set (English: set of departure ) of the relation , the set as the target set (English: set of destination ). ${\ displaystyle A}$${\ displaystyle R}$${\ displaystyle B}$ Sometimes, however, this definition is not precise enough and referring the source and target quantity in the definition of a, above is then subset of the graph (more rarely Count ) called the relation. A two-digit relation is then defined as a triple${\ displaystyle G_ {R}}$${\ displaystyle R}$ ${\ displaystyle R = (G_ {R}, A, B)}$with .${\ displaystyle G_ {R} \ equiv \ operatorname {Graph} (R) \ subseteq A \ times B}$ Knowing the source and target set is particularly important when considering functions as special (so-called functional) relations. The archetype, argument or definition or pre-range of a given two-digit relation is understood as the smallest possible pre-range to the graph , the elements of which all appear in the ordered pairs of actually on the left side, in signs ${\ displaystyle R}$${\ displaystyle G_ {R}}$${\ displaystyle R}$ ${\ displaystyle Db (R) \ equiv {\ mathfrak {D}} (R): = \ {a \ mid \ exists b \ colon (a, b) \ in G_ {R} \} \ subseteq A}$. The set of values, values or image or after- area in this sense denotes the smallest after-area for a given , whose elements all appear in the pairs from on the right-hand side, in characters ${\ displaystyle G_ {R}}$${\ displaystyle R}$${\ displaystyle R}$ ${\ displaystyle Wb (R) \ equiv {\ mathfrak {B}} (R): = \ {b \ mid \ exists a \ colon (a, b) \ in G_ {R} \} \ subseteq B}$. Sometimes the term field (or node set ) is used for the union set , in characters ${\ displaystyle Fd (R) \ equiv {\ mathfrak {F}} (R): = Db (R) \ cup Wb (R) = \ {x \ mid \ exists x '\ colon (x, x') \ in G_ {R} \ lor (x ', x) \ in R \} \ subseteq A \ cup B}$. In addition, the following designations can be found: • Domain (English domain ) either for the (in principle arbitrarily large) source set or for the original image set (defined by the graph) (domain of definition),${\ displaystyle \ operatorname {dom} R}$ • Co-domain (English codomain , range ) either for the target set or for the image set,${\ displaystyle \ operatorname {cod} R, \ operatorname {ran} R}$ • Set of nodes ( ) for the field of a relation.${\ displaystyle \ operatorname {ver} R}$ If two relations agree in their graphs, one also says that they are essentially the same. Example: Every relation is essentially the same with and with the homogeneous relation . ${\ displaystyle R = (G_ {R}, A, B)}$${\ displaystyle (G_ {R}, Db (R), Wb (R))}$ ${\ displaystyle (G_ {R}, Fd (R), Fd (R))}$ ### n -digit relation More generally, a -digit relation is a subset of the Cartesian product of sets${\ displaystyle n}$ ${\ displaystyle R}$${\ displaystyle n}$${\ displaystyle A_ {1}, \ dotsc, A_ {n}}$ ${\ displaystyle R \ subseteq A_ {1} \ times \ dotsb \ times A_ {n}}$with .${\ displaystyle A_ {1} \ times \ dotsb \ times A_ {n} = \ {(a_ {1}, \ dotsc, a_ {n}) \ mid a_ {1} \ in A_ {1}, \ dotsc, a_ {n} \ in A_ {n} \} = \ textstyle \ prod A}$ Here denotes the finite sequence of the sets, and the Cartesian product.${\ displaystyle A = (A_ {i}) _ {i \ in \ {1, \ dots n \}}}$${\ displaystyle \ textstyle \ prod A = \ textstyle \ prod _ {i = 1} ^ {n} A_ {i}}$ The more detailed definition can also be generalized to -place relations and one then obtains the -tuple ${\ displaystyle n}$${\ displaystyle (n + 1)}$ ${\ displaystyle R = (G_ {R}, A_ {1}, \ dotsc, A_ {n}) = (G_ {R}, A)}$with .${\ displaystyle G_ {R} \ equiv \ operatorname {Graph} (R) \ subseteq A_ {1} \ times \ dotsb \ times A_ {n} = \ textstyle \ prod A}$ The amounts of hot carriers amounts of relation with the minimum support amounts to the graph , namely ${\ displaystyle A_ {1}, \ dotsc, A_ {i}, \ dotsc, A_ {n}}$${\ displaystyle G_ {R}}$ ${\ displaystyle Tr_ {i} (R) \ equiv {\ mathfrak {T}} _ {i} (R): = \ {a_ {i} \ mid \ exists a_ {1}, \ dotsc, a_ {i- 1}, a_ {i + 1}, \ dotsc, a_ {n} \ colon (a_ {1}, \ dotsc, a_ {i-1}, a_ {i}, a_ {i + 1}, \ dotsc, a_ {n}) \ in G_ {R} \}}$. The field of a -digit relation is given by ${\ displaystyle n}$ ${\ displaystyle Fd (R) \ equiv {\ mathfrak {F}} (R): = \ textstyle \ bigcup \ {Tr_ {i} (R) \ mid 1 \ leq i \ leq n \} \ subseteq \ textstyle \ bigcup A}$. Substantial equality is defined in the same way as for two-digit relations by the correspondence of the graphs, in particular every -digit relation is essentially the same with and with the homogeneous relation . ${\ displaystyle n}$${\ displaystyle R = (G_ {R}, A_ {1}, \ dotsc, A_ {n})}$${\ displaystyle (G_ {R}, Tr_ {1} (R), \ dotsc, Tr_ {n} (R))}$${\ displaystyle (G_ {R}, \ underbrace {Fd (R), \ dotsc, Fd (R)} _ {n {\ text {times}}})}$ One-digit and zero-digit relation A one- digit relation on a set is therefore simply a subset , in the detailed definition with . ${\ displaystyle A}$${\ displaystyle R \ subseteq A}$${\ displaystyle R = (G_ {R}, A)}$${\ displaystyle G_ {R} \ subseteq A}$ The zero-digit relations are therefore the subsets of the empty Cartesian product or , that is, and , detailed and . ${\ displaystyle \ textstyle \ prod _ {i = 1} ^ {0} A_ {i} = \ {\ emptyset \}}$${\ displaystyle A ^ {0} = \ {\ emptyset \}}$${\ displaystyle \ emptyset}$${\ displaystyle \ {\ emptyset \}}$${\ displaystyle (\ emptyset, \ {\ emptyset \})}$${\ displaystyle (\ {\ emptyset \}, \ {\ emptyset \})}$ ### Relations between or on real classes Often the carrier areas of a relation are not sets, but real classes , one then speaks of class relations . Occasionally one can avoid the set-theoretic problems that result from this by only looking at the graph of the corresponding relation. The (minimum) carrier quantities ( in the two-digit case definition and value set ) are actually quantities, but it is not necessary to specify the source quantity, target quantity,  ... ( ) from the outset if the relations are essentially the same. Not always it is possible, for example for the equivalence relation of equal thickness , see also: cardinal numbers §Definition . Essentially equality of relations is another example. ${\ displaystyle A_ {i}}$${\ displaystyle Tr_ {i} (R)}$${\ displaystyle Db (R), Wb (R)}$${\ displaystyle A, B, A_ {i}, \ dotsc}$ A two-digit class relation with source class and target class is called small predecessor if the class of the predecessor (archetype fiber of , see below) is a set (i.e. not a real class) for all . The relation is called English right-narrow (German about successor small ), if for all the class of the successor (image fiber of ) is a lot. In the case of right uniqueness (partial mappings, mappings, see below), a class relation is always small, since there is one image value for every archetype (exactly or at most), i.e. the class of the successors is a set of ones (or the empty set ). Every injective class mapping is both small and predecessor small. The inclusion relation is small for each class , since they can not be real classes, but are sets and are therefore also a set. The terms predecessor and successor themselves are usually used in the context of order relations, see order relation §Predecessor and successor . ${\ displaystyle R}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle b \ in B}$${\ displaystyle \ {a \ in A \ mid aRb \}}$${\ displaystyle b}$${\ displaystyle a \ in A}$${\ displaystyle \ {b \ in B \ mid aRb \}}$${\ displaystyle a}$${\ displaystyle \ in}$${\ displaystyle B}$${\ displaystyle b \ in B}$${\ displaystyle \ {a \ in A \ mid a \ in b \} \ subseteq b}$ ## Explanations and notations The Cartesian product of two sets and is the set of all ordered pairs of and where any element from the set and one from represents. In the ordered pair, the order is important; H. differs from in contrast to the disordered pair, which is identical to For , is also written to make it clear that there is that relationship between the objects (as in ). The empty set as a subset of the Cartesian set product understood as a relation is called the zero relation , the full product is called the all relation (also known as universal relation ) (also referred to as ). ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle a}$${\ displaystyle b,}$${\ displaystyle a}$${\ displaystyle A}$${\ displaystyle b}$${\ displaystyle B}$${\ displaystyle (a, b)}$${\ displaystyle (b, a),}$ ${\ displaystyle \ {a, b \},}$${\ displaystyle \ {b, a \}.}$${\ displaystyle (a, b) \ in R}$${\ displaystyle a \; R \; b,}$${\ displaystyle a> b}$ ${\ displaystyle \ mathrm {O}: = \ emptyset = \ {\}}$${\ displaystyle \ mathrm {U}}$${\ displaystyle \ nabla}$ ### Relations and functions • A function is a special one, namely a left-total and right-unique (two-digit) relation, see below for details .${\ displaystyle f \ colon A \ to B}$ • A multifunction is a left total relation .${\ displaystyle f \ colon X \ multimap Y}$${\ displaystyle f \ subseteq A \ times B}$ • A partial function is a right-unambiguous relation (generally not a left total) .${\ displaystyle f \ colon \; X \ rightharpoonup Y}$${\ displaystyle f \ subseteq A \ times B}$ In all cases (or if the detailed definition is used). ${\ displaystyle f \ subseteq A \ times B}$${\ displaystyle G_ {f} \ subseteq A \ times B}$ The following applies to functions and multi-functions: In the more detailed definition , because it is clearly determined by (left total), it can also be omitted and taken more simply .${\ displaystyle f = (G_ {f}, A, B)}$${\ displaystyle A}$${\ displaystyle G_ {f}}$${\ displaystyle A}$${\ displaystyle f = (G_ {f}, B)}$ The following applies to functions and partial functions: For or is also written (English: maplet ), or .${\ displaystyle (a, b) \ in f}$${\ displaystyle (a, b) \ in G_ {f}}$${\ displaystyle f \ colon a \ mapsto b}$${\ displaystyle f (a) = b}$ In general: 1. The zero-place relations (as zero-place zero relation) and (as zero-place full relation) have as characteristic functions the Boolean or logical constants and , as always for zero relation and all relation.${\ displaystyle \ mathrm {O} = \ emptyset}$${\ displaystyle \ mathrm {U} = \ {\ emptyset \}}$ ${\ displaystyle {\ mathsf {wrong}}}$${\ displaystyle {\ mathsf {true}}}$ 2. The case of single-digit relations is trivial. 3. A relation (or ) corresponds in a unique way to a truth function . This function is also known as the indicator function or characteristic function of the subset (or ), where can be replaced by .${\ displaystyle R \ subseteq A \ times B}$${\ displaystyle R = (G_ {R}, A, B)}$ ${\ displaystyle \ chi _ {R} \ colon \; A \ times B \ to \ {{\ mathsf {true}}, {\ mathsf {false}} \}}$${\ displaystyle R \ subseteq A \ times B}$${\ displaystyle G_ {R} \ subseteq A \ times B}$${\ displaystyle \ {{\ mathsf {true}}, {\ mathsf {false}} \}}$${\ displaystyle \ {1,0 \}}$ 4. A -digit relation (or ) corresponds to the characteristic function${\ displaystyle n}$${\ displaystyle R \ subseteq A_ {1} \ times \ dotsb \ times A_ {n}}$${\ displaystyle R = (G_ {R}, A_ {1}, \ dotsc, A_ {n})}$${\ displaystyle \ chi _ {R} \ colon A_ {1} \ times \ dotsb \ times A_ {n} \ to \ {{\ mathsf {true}}, {\ mathsf {false}} \}.}$ The following applies: • ${\ displaystyle n = 0 \ colon \; \ chi _ {\ emptyset} \ Leftrightarrow {\ mathsf {false}}, \ chi _ {\ {\ emptyset \}} \ Leftrightarrow {\ mathsf {true}}}$. ${\ displaystyle n = 1 \ colon \; \ chi _ {R} (a) \ Leftrightarrow a \ in R}$. ${\ displaystyle n = 2 \ colon \; \ chi _ {R} (a, b) \ Leftrightarrow aRb \ Leftrightarrow (a, b) \ in R}$. ${\ displaystyle n> 2 \ colon \; \ chi _ {R} (a_ {1}, \ dotsc, a_ {n}) \ Leftrightarrow (a_ {1}, \ dotsc, a_ {n}) \ in R}$. • A relation can also be understood as a mapping of into the power set of , one then often speaks of a correspondence , and for of a transition relation .${\ displaystyle R \ subseteq A \ times B}$${\ displaystyle \ kappa _ {R}}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle \ kappa _ {R} \ colon A \ to {\ mathcal {P}} (B), \; a \ mapsto \ {b \ in B \ mid (a, b) \ in R \},}$${\ displaystyle B = A}$ ### Chaining of relations The forward chaining of two two-digit relations is defined as follows: ${\ displaystyle R \ subseteq A \ times B, S \ subseteq C \ times D}$ ${\ displaystyle {\ begin {array} {lll} R \; \ mathbf {;} \; S \ equiv RS &: = \ {(a, d) \ mid \ exists b \ colon \; aRb \ land bSd \} \\ & = \ {(a, d) \ in A \ times D \ mid \ exists b \ in B \ cap C \ colon \; (a, b) \ in R \ land (b, d) \ in S \}. \ end {array}}}$ Chaining in the reverse order is called backward chaining : ${\ displaystyle S \ circ R \ qquad \ quad: = \ {(a, d) \ in A \ times D \ mid \ exists b \ in B \ cap C \ colon \; (a, b) \ in R \ land (b, d) \ in S \} = R \; \ mathbf {;} \; S}$. Some authors (W. v. O. Quine) use the notation for this as an alternative .${\ displaystyle S \ mid R}$ The sequence for backward chaining is the same as for chaining functions (which can be seen as special relations). The concatenation of two-digit relations is also known as a relative product . In the concatenation, the simplest relation that contained in each Cartesian product can empty relation ( empty set ) occur, namely, when and disjoint are in symbols . ${\ displaystyle \ emptyset}$${\ displaystyle B}$${\ displaystyle C}$ ${\ displaystyle B \ cap C = \ emptyset}$ Example: The relation “being sister-in-law of” is the union • of the relative product of the relation “being brother of” and the relation “being wife of” and • the relative product of the relation “being spouse of” and the relation “being sister of”. ### Inverse relation The inverse relation (also called converse relation, converse or inverse relation) is defined as for a two-digit relation${\ displaystyle R \ subseteq A \ times B}$ ${\ displaystyle {\ overset {\ smallsmile} {R}} \ equiv {} ^ {\ smallsmile} {R} \ equiv R ^ {\ sim} \ equiv R ^ {- 1}: = \ {(b, a ) \ in B \ times A \ mid (a, b) \ in R \}}$. Occasionally, the term transposed relation is also used for this , in characters .${\ displaystyle R ^ {T}}$ • Example 1: The inverse relation of the relation "is descendant of" is the relation "is ancestor of". • Example 2: The inverse relation of the relation “is less than” is the relation “is greater than”. • Example 3: The reverse relation of the relation “delivers to” is the relation “is supplied by”. The generalization of the inverse relation (converse) to -sign relations is the permutation of the coordinates of the -tuples it contains , special ${\ displaystyle n}$${\ displaystyle n}$ both examples of ( cyclic ) self-inverse permutations . Let be a permutation (i.e. a bijective mapping of to itself) and be a --place relation, then the relation resulting after applying the permutation is (one understands as a family ). In the case of mirroring ${\ displaystyle \ pi \ colon \ {1,2, \ dotsc, n \} \ rightarrow \ {1,2, \ dotsc, n \}}$${\ displaystyle \ {1, \ dotsc, n \}}$${\ displaystyle R \ subseteq (A_ {i}) _ {i \ in \ {1, \ dotsc, n \}}}$${\ displaystyle n}$${\ displaystyle S: = \ {a \ circ \ pi \ mid a \ in R \}}$${\ displaystyle \ pi}$${\ displaystyle a = (a_ {i}) _ {i \ in \ {1, \ dotsc, n \}}}$ ${\ displaystyle \ pi = \ sigma _ {n} = {\ begin {pmatrix} 1 & 2 & \ cdots & n \\ n & n-1 & \ cdots & 1 \ end {pmatrix}}}$ is . ${\ displaystyle S = \ {(a_ {n}, \ dotsc, a_ {1}) \ mid (a_ {1}, \ dotsc, a_ {n}) \ in R \}}$ ### Image and archetype In a two-digit relation , the image of a set or class is the set or class ${\ displaystyle R \ subseteq A \ times B}$${\ displaystyle X}$ ${\ displaystyle R ^ {\ to} (X) \ equiv R \ langle X \ rangle: = \ {y \ mid \ exists x \ in X \ colon \; (x, y) \ in R \}}$. The archetype of a set or class is the set or class ${\ displaystyle Y}$ ${\ displaystyle R ^ {\ leftarrow} (Y) \ equiv {\ overset {\ smallsmile} {R}} \ langle Y \ rangle \ equiv {} ^ {\ smallsmile} {R} \ langle Y \ rangle \ equiv R ^ {\ sim} \ langle Y \ rangle \ equiv R ^ {- 1} \ langle Y \ rangle = \ {x \ mid \ exists y \ in Y \ colon \; (x, y) \ in R \}}$. Occasionally the designation (sic!) Is also found for this , often also with square brackets as noted. In correspondence of a for the image fiber of an amount (Singleton) , the notation is what partially also uses the bracket notation in use, i. H. ; in the case of symmetrical relations, d. H. (possibly partial) equivalence or compatibility relations is the notation and speaks of equivalence or compatibility or tolerance classes.${\ displaystyle R {\ grave {}} {\ grave {}} Y}$${\ displaystyle R [Y]}$${\ displaystyle \ {a \}}$${\ displaystyle \ kappa _ {R} (a) = R \ langle \ {a \} \ rangle}$${\ displaystyle R [a]}$${\ displaystyle [a] _ {R} = R \ langle \ {a \} \ rangle = R ^ {- 1} \ langle \ {a \} \ rangle}$ ### Restriction Relations can be restricted in various ways to subsets of the carrier sets, for more information see Restricting a relation . ### Complementary relation For two-digit relations with a fixed pre- and post-range the complementary relation is given by ${\ displaystyle R \ subseteq A \ times B}$${\ displaystyle A, B}$ ${\ displaystyle {\ overline {R}} \ equiv {} ^ {-} R = (A \ times B) \ setminus R = \ {(x, y) \ in A \ times B \ mid (x, y) \ not \ in R \}}$, analogous for -digit relations for fixed support areas . For example, on the real numbers , the complementary relation to is . ${\ displaystyle n}$${\ displaystyle R \ subseteq A_ {1} \ times \ dotsb \ times A_ {n}}$${\ displaystyle A = (A_ {1}, \ dotsc, A_ {n})}$${\ displaystyle \ mathbb {R}}$${\ displaystyle \ leq}$${\ displaystyle>}$ If the complex notation is used, then is ${\ displaystyle R = (G_ {R}, A, B)}$ ${\ displaystyle {\ overline {R}} \ equiv {} ^ {-} R = ((A \ times B) \ setminus G_ {R}, A, B)}$, where there are no longer any external additions, but components of the relation; analogous for -digit relations in this notation. ${\ displaystyle A, B}$${\ displaystyle n}$ As for all sets, the complement is also involutive for relations: ${\ displaystyle {\ overline {\ overline {R}}} = R}$. ## Homogeneous relations If , then, the relation is called homogeneous. Some authors define a general relation already as a homogeneous relation, as a general relation can always be considered as a homogeneous limitation: . ${\ displaystyle A = B}$${\ displaystyle R \ subseteq A \ times A}$${\ displaystyle R \ subseteq A \ times B}$${\ displaystyle R \ subseteq (A \ cup B) \ times (A \ cup B)}$ ### Special homogeneous relations and operations on homogeneous relations A special homogeneous relation in a set is the equality or identity relation or diagonal${\ displaystyle A}$ ${\ displaystyle \ mathrm {I} _ {A}: = \ {(a, b) \ in A \ times A \ mid a = b \} = \ {(a, a) \ mid a \ in A \} .}$ Alternative notations for the diagonal are or ; if already known, it is simply referred to as , or . ${\ displaystyle \ Delta _ {A}}$${\ displaystyle \ mathrm {D} _ {A}}$${\ displaystyle A}$${\ displaystyle \ mathrm {I}}$${\ displaystyle \ Delta}$${\ displaystyle \ mathrm {D}}$ Another special homogeneous relation is the all relation or universal relation ${\ displaystyle \ mathrm {U} _ {A} = A \ times A}$(also referred to as Nabla as ).${\ displaystyle \ nabla _ {A}}$ If it is already known, the index is omitted here as in the case of the identity relation. ${\ displaystyle A}$ The all relation plays a role in graph theory (see below). An example of use is the following sentence: Is a directed graph with a lot of corners and a (associated) relationship of edges, it is then exactly coherent (strong) when the reflexive transitive closure of the Universal relation.${\ displaystyle G = (E, K)}$${\ displaystyle E}$${\ displaystyle K \ subseteq E \ times E}$${\ displaystyle G}$${\ displaystyle K}$ The formation of the inverse relation (converse relation) of a homogeneous two-digit relation again provides a homogeneous two-digit relation (closeness), twice execution again results in the starting relation (involutivity). The connection of any (also non-homogeneous) relation with the relation to which it is converted is symmetrical and reflexive, i.e. an equivalence relation, but in general not the same as the identity relation. In the case of a homogeneous relation , the concatenation is also a homogeneous relation, so that the homogeneous relations form a monoid with the multiplicative relation and the neutral element . Thus, and to general powers to be defined, which is. is therefore also called one relation on the set . ${\ displaystyle R \ subseteq A \ times A}$${\ displaystyle R \ circ R}$${\ displaystyle A}$ ${\ displaystyle \ circ}$ ${\ displaystyle \ mathrm {I} _ {A}}$${\ displaystyle R ^ {2}: = R \ circ R}$ ${\ displaystyle R ^ {n}: = R \ circ R ^ {n-1}}$${\ displaystyle n \ in \ mathbb {N}}$${\ displaystyle R ^ {0}: = \ mathrm {I} _ {A}}$${\ displaystyle \ mathrm {I} _ {A}}$${\ displaystyle A}$ In extension of the notation instead of for the inverse relation, one denotes its powers with negative exponents: ${\ displaystyle R ^ {- 1}}$${\ displaystyle {\ overset {\ smallsmile} {R}}}$ ${\ displaystyle R ^ {- n}: = {\ overset {\ smallsmile} {R}} ^ {n} = {\ overset {\ smallsmile} {R ^ {n}}}}$. This means that any whole numbers are allowed as exponents. ${\ displaystyle n \ in \ mathbb {Z}}$ In addition, every monoid has homogeneous relations with the empty relation ( zero relation ) ${\ displaystyle \ mathrm {O} = \ emptyset}$ another absorbent element . The relations arise through the union of the different powers ${\ displaystyle R ^ {*}: = \ textstyle \ bigcup _ {n \ in {\ mathbb {N} _ {0}}} R ^ {n}}$and . ${\ displaystyle R ^ {+}: = \ textstyle \ bigcup _ {n \ in {\ mathbb {N}}} R ^ {n}}$ ### Algebraic structures Taken together, the two-digit relations on a set form a relation algebra${\ displaystyle A}$ ${\ displaystyle {\ mathfrak {Re}} (A) = (2 ^ {U_ {A}}, \ cap, \ cup, {} ^ {-}, \ mathrm {O}, U_ {A}, \ circ , \ mathrm {I} _ {A}, {} ^ {\ smallsmile})}$ Using the notations . ${\ displaystyle U_ {A} = A ^ {2} = A \ times A, \; \; 2 ^ {U_ {A}} = 2 ^ {A ^ {2}} = {\ mathcal {P}} ( A \ times A)}$ Together with the restrictions, the homogeneous relations form a ( heterogeneous ) Peirce algebra . ### Homogeneous multi-digit relations Homogeneous multi-digit relations are (with their graph) subsets of . For solid the all relation (also ) and the identity relation (diagonal) (also ) are given by ${\ displaystyle A ^ {n}}$${\ displaystyle n}$${\ displaystyle \ mathrm {U} _ {A}}$${\ displaystyle \ nabla _ {A}}$${\ displaystyle \ mathrm {I} _ {A}}$${\ displaystyle \ mathrm {D} _ {A}, \ Delta _ {A}}$ ${\ displaystyle \ mathrm {U} _ {A} = A ^ {n}, \; \; \ mathrm {I} _ {A} = \ {(a_ {i}) _ {i \ in \ {1, \ dotsc, n \}} \ in A ^ {n} \ mid a_ {1} = a_ {2} = \ dotsb = a_ {n} \}}$. The application of permutations to their -tuples, described as a generalization of the formation of conversions, is of particular importance here, since in this way one always remains within the subsets of (closeness). M. a. W. These operations are bijective mappings in . Other terms known from two-digit relations, such as reflexivity and symmetry, etc., can be canonically (naturally) extended to any multi-digit relations. ${\ displaystyle n}$${\ displaystyle A ^ {n}}$${\ displaystyle {\ mathcal {P}} (A ^ {n}) = 2 ^ {A ^ {n}}}$ ### Graph theory and generalizations Graph theory describes sets with a relation to them together with certain generalizations under a common generic term, the graph . The cases considered in graph theory are (unless otherwise stated) usually finite. 1. A relational structure consisting of a set together with a relation on it is called a directed (also oriented ) graph . is called the set of nodes of the graph, its elements are called nodes . is called a subset of the set of edges , its elements (ordered pairs of ) are called directed (i.e., oriented ) edges . ${\ displaystyle G}$${\ displaystyle M}$${\ displaystyle R}$ ${\ displaystyle G = (M, R)}$${\ displaystyle M = V (G)}$${\ displaystyle E (G) = R}$${\ displaystyle M \ times M}$${\ displaystyle M}$ 2. Symmetric graphs , i.e. H. Sets with a symmetrical relation are equivalent to an undirected graph , whose edge set consists of (undirected) edges, namely the (disordered) sets with (here equivalent to ). ${\ displaystyle G = (M, R)}$${\ displaystyle M}$${\ displaystyle R}$ ${\ displaystyle G ': = (M, \ {\ {a, b \} \ mid a \; R \; b \})}$${\ displaystyle E (G ')}$${\ displaystyle \ {a, b \}}$${\ displaystyle a \ R \ b}$${\ displaystyle b \ R \ a}$ 3. Further generalizations concern so-called directed graphs with combined multiple edges , in which each edge has a natural number as multiplicity. The edges of such graphs can be represented by a multiset : a map with a set and a map that assigns a positive number called a color to each node . Graphs with colored nodes and / or edges are similar . ${\ displaystyle {\ boldsymbol {M}}}$${\ displaystyle {\ boldsymbol {M}} = (M, f)}$${\ displaystyle M = supp ({\ boldsymbol {M}})}$${\ displaystyle f \ colon \; M \ to \ mathbb {N}}$${\ displaystyle v \ in M}$${\ displaystyle f (v)}$ 4. Of weighted nodes and / or edges: We speak of weights instead of colors when the mapping is real-valued. In Node Weighted this corresponds to a fuzzy set , wherein is a real valued multiset . The same applies to weighted edges. For oriented graphs this means in particular that the set of edges (a relation, i.e. set of ordered pairs of nodes) becomes a multi-set or fuzzy set in an extension of the relation concept. ${\ displaystyle f}$${\ displaystyle f \ colon \; M \ to [0,1]}$ ${\ displaystyle {\ boldsymbol {M}} = (M, f)}$${\ displaystyle f \ colon \; M \ to R ^ {+}}$${\ displaystyle {\ boldsymbol {M}} = (M, f)}$ ## Properties of two-digit relations ### General relations #### Overview of the properties The following relations are important for functions (represented as special relations). In general, here the relationship between two different amounts of the case is also possible. ${\ displaystyle R \ subseteq A \ times B}$${\ displaystyle A, B,}$${\ displaystyle A = B}$ The relation is called ${\ displaystyle R}$ if and only if (predicate logic) or equivalent (quantity notation) and that means: left total or final (multifunction) ${\ displaystyle \ forall a \ in A \; \ exists b \ in {B} \ colon \; (a, b) \ in R}$ ${\ displaystyle \ mathrm {I} _ {A} \ subseteq R ^ {- 1} \ circ R}$ Every item from has at least one partner in${\ displaystyle A}$${\ displaystyle B.}$ right total or surjective ${\ displaystyle \ forall b \ in B \; \ exists a \ in A \ colon \; (a, b) \ in R}$ ${\ displaystyle \ mathrm {I} _ {B} \ subseteq R \ circ R ^ {- 1}}$ Every item from has at least one partner in${\ displaystyle B}$${\ displaystyle A.}$ left unambiguous or injective {\ displaystyle {\ begin {aligned} & \ forall b \ in B \; \ forall a, c \ in A \ colon \\ & (a, b) \ in R \, \ land \, (c, b) \ in R \; \ Rightarrow \; a = c \ end {aligned}}} ${\ displaystyle R ^ {- 1} \ circ R \ subseteq \ mathrm {I} _ {A}}$ Each element from has at most one partner in${\ displaystyle B}$${\ displaystyle A.}$ (legally) unambiguous (partial function) {\ displaystyle {\ begin {aligned} & \ forall a \ in A \; \ forall b, d \ in B \ colon \\ & (a, b) \ in R \, \ land \, (a, d) \ in R \; \ Rightarrow \; b = d \ end {aligned}}} ${\ displaystyle R \ circ R ^ {- 1} \ subseteq \ mathrm {I} _ {B}}$ Each element from has at most one partner in${\ displaystyle A}$${\ displaystyle B.}$ The relation is called ${\ displaystyle R}$ if and only if (predicate logic) or equivalent (quantity notation) and that means: bitotal {\ displaystyle {\ begin {aligned} & (\ forall a \ in A \; \ exists b \ in B \ colon \; (a, b) \ in R) \\\ land \; & (\ forall b \ in B \; \ exists a \ in A \ colon \; (a, b) \ in R) \ end {aligned}}} {\ displaystyle {\ begin {aligned} & \ mathrm {I} _ {A} \ subseteq R ^ {- 1} \ circ R \\\ land \; & \ mathrm {I} _ {B} \ subseteq R \ circ R ^ {- 1} \ end {aligned}}} Every element out has at least one partner in and vice versa. ${\ displaystyle A}$${\ displaystyle B}$ unambiguous {\ displaystyle {\ begin {aligned} & \ forall b, d \ in B \; \ forall a, c \ in A \ colon \\ & (a, b) \ in R \, \ land \, (c, b) \ in R \; \ Rightarrow \; a = c, \\ & (a, b) \ in R \, \ land \, (a, d) \ in R \; \ Rightarrow \; b = d \ end {aligned}}} {\ displaystyle {\ begin {aligned} & R ^ {- 1} \ circ R \ subseteq \ mathrm {I} _ {A} \\\ land \; & R \ circ R ^ {- 1} \ subseteq \ mathrm {I } _ {B} \ end {aligned}}} Each element out has at most one partner in and vice versa. ${\ displaystyle B}$${\ displaystyle A}$ bijective ${\ displaystyle \ forall b \ in B \; \ exists! a \ in A \ colon \; (a, b) \ in R}$ {\ displaystyle {\ begin {aligned} & \ mathrm {I} _ {B} \ subseteq R \ circ R ^ {- 1} \\\ land \; & R ^ {- 1} \ circ R \ subseteq \ mathrm { I} _ {A} \ end {aligned}}} Each element from has exactly one partner in${\ displaystyle B}$${\ displaystyle A.}$ Figure or function ${\ displaystyle \ forall a \ in A \; \ exists! b \ in B \ colon \; (a, b) \ in R}$ {\ displaystyle {\ begin {aligned} & \ mathrm {I} _ {A} \ subseteq R ^ {- 1} \ circ R \\\ land \; & R \ circ R ^ {- 1} \ subseteq \ mathrm { I} _ {B} \ end {aligned}}} Each element from has exactly one partner in${\ displaystyle A}$${\ displaystyle B.}$ #### Alternative ways of speaking One also says • left full instead of left total , • right complete instead of right total , • unambiguous instead of unambiguous to the left, • subsequently unambiguous instead of right unambiguous, A right-unambiguous or functional relation is also called a partial function . If this is also left total - i.e. a function  - then for clarity it is also called total function. ### Functions #### Overview of functional properties for relations A relation is therefore a (total) function if and only if it is left total and right unambiguous. This means that every element in A has exactly one partner in B. The properties surjective, injective and bijective are usually used for functions and specify certain additional properties. For example, a function (and any relation) is bijective if and only if it is surjective and injective, i.e. if its inverse relation is a function. ${\ displaystyle R}$${\ displaystyle R ^ {- 1}}$ The relation is called ${\ displaystyle R}$ exactly when she is a is or equivalent (quantity notation) and that means: Surjection surjective function {\ displaystyle {\ begin {aligned} & \ mathrm {I} _ {A} \ subseteq R ^ {- 1} \ circ R \\\ land \; & R \ circ R ^ {- 1} = \ mathrm {I } _ {B} \ end {aligned}}} Every element from has exactly one partner in and every element from has at least one partner in${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle B}$${\ displaystyle A.}$ injection injective function {\ displaystyle {\ begin {aligned} & \ mathrm {I} _ {A} = R ^ {- 1} \ circ R \\\ land \; & R \ circ R ^ {- 1} \ subseteq \ mathrm {I } _ {B} \ end {aligned}}} Each element from has exactly one partner in and each element from has at most one partner in${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle B}$${\ displaystyle A.}$ Bijection bijective function {\ displaystyle {\ begin {aligned} & \ mathrm {I} _ {A} = R ^ {- 1} \ circ R \\\ land \; & R \ circ R ^ {- 1} = \ mathrm {I} _ {B} \ end {aligned}}} Each element out has exactly one partner in and vice versa. ${\ displaystyle A}$${\ displaystyle B}$ #### Inverse function A mapping or function is also called • reversibly unambiguous or reversible if it is bijective . A function is always reversible as a relation, but as a function it is reversible exactly when its reverse relation is also a function, i.e. when there is an inverse function of it. ### Homogeneous relations The examples given in the following tables refer to the normal arrangement of real numbers when using the equal sign “=”, the less than sign “<” and the less than or equal sign “≤”. The relation is called ${\ displaystyle R}$ if and only if (predicate logic) or equivalent (quantity notation) and that means: right comparative or third equal {\ displaystyle {\ begin {aligned} & \ forall a, b, c \ in A \ colon \\ & (a, c) \ in R \, \ land \, (b, c) \ in R \\ & \ Rightarrow \; (a, b) \ in R \ end {aligned}}} ${\ displaystyle R ^ {- 1} \ circ R \ subseteq R}$ If two elements are each related to the same third element, then they are also related to each other. For example, with and always applies${\ displaystyle a = c}$${\ displaystyle b = c}$${\ displaystyle a = b.}$ left-hand comparative or Euclidean {\ displaystyle {\ begin {aligned} & \ forall a, b, c \ in A \ colon \\ & (a, b) \ in R \, \ land \, (a, c) \ in R \\ & \ Rightarrow \; (b, c) \ in R \ end {aligned}}} ${\ displaystyle R \ circ R ^ {- 1} \ subseteq R}$ If a first element is related to a second and a third element, then these are also related to one another. For example, with and is always the same${\ displaystyle a = b}$${\ displaystyle a = c}$${\ displaystyle b = c.}$ transitive {\ displaystyle {\ begin {aligned} & \ forall a, b, c \ in A \ colon \\ & (a, b) \ in R \, \ land \, (b, c) \ in R \\ & \ Rightarrow \; (a, c) \ in R \ end {aligned}}} ${\ displaystyle R \ circ R \ subseteq R}$ If a first element is related to a second element and this in turn is related to a third element, then the first element is also related to the third element. For example, it follows from and always${\ displaystyle a ${\ displaystyle b ${\ displaystyle a intransitive {\ displaystyle {\ begin {aligned} & \ forall a, b, c \ in A \ colon \\ & (a, b) \ in R \, \ land \, (b, c) \ in R \\ & \ Rightarrow \; (a, c) \ notin R \\\ end {aligned}}} ${\ displaystyle (R \ circ R) \ cap R = \ emptyset}$ If two elements are related and the second element is related to a third element, then the first element is not related to the third element. E.g. every natural number is the (immediate) predecessor of and the (immediate) predecessor of but is not the (immediate) predecessor of${\ displaystyle n}$${\ displaystyle n + 1}$${\ displaystyle n + 1}$${\ displaystyle n + 2,}$${\ displaystyle n}$${\ displaystyle n + 2.}$ Non- transitivity (i.e. the relation is not transitive), intransitivity, and negative transitivity are each different from one another. The relation is called ${\ displaystyle R}$ if and only if (predicate logic) or equivalent (quantity notation) and that means: reflexive ${\ displaystyle \ forall a \ in A \ colon (a, a) \ in R}$ ${\ displaystyle \ mathrm {I} \ subseteq R}$ Each element is related to itself, e.g. B. is always${\ displaystyle a \ leq a.}$ ${\ displaystyle \ mathrm {I} \ cap R = \ mathrm {I}}$ irreflexive ${\ displaystyle \ forall a \ in A \ colon (a, a) \ notin R}$ ${\ displaystyle \ mathrm {I} \ cap R = \ emptyset}$ No element is related to itself, e.g. B. applies to none${\ displaystyle a ${\ displaystyle a.}$ The relation is called ${\ displaystyle R}$ if and only if (predicate logic) or equivalent (quantity notation) and that means: symmetrical {\ displaystyle {\ begin {aligned} & \ forall a, b \ in A \ colon \\ & (a, b) \ in R \; \ Rightarrow \; (b, a) \ in R \ end {aligned} }} ${\ displaystyle R ^ {- 1} \ subseteq R}$ The relation is undirected, e.g. B. it always follows from (and vice versa) ${\ displaystyle a = b}$${\ displaystyle b = a}$ ${\ displaystyle R ^ {- 1} = R}$ antisymmetric or identitive {\ displaystyle {\ begin {aligned} & \ forall a, b \ in A \ colon \\ & (a, b) \ in R \, \ land \, (b, a) \ in R \\ & \ Rightarrow \; a = b \ end {aligned}}} ${\ displaystyle R ^ {- 1} \ cap R \ subseteq \ mathrm {I}}$ There are no two different elements that are related in both directions, e.g. B. follows from and always${\ displaystyle a \ leq b}$${\ displaystyle b \ leq a}$${\ displaystyle a = b.}$ asymmetrical {\ displaystyle {\ begin {aligned} & \ forall a, b \ in A \ colon \\ & (a, b) \ in R \; \ Rightarrow \; (b, a) \ notin R \ end {aligned} }} ${\ displaystyle R ^ {- 1} \ cap R = \ emptyset}$ There are no two elements that are related in both directions, e.g. B. it always follows that it does not hold. ${\ displaystyle a ${\ displaystyle b The relation is called ${\ displaystyle R}$ if and only if (predicate logic) or equivalent (quantity notation) and that means: total or completely {\ displaystyle {\ begin {aligned} & \ forall a, b \ in A \ colon \\ & (a, b) \ in R \, \ lor \, (b, a) \ in R \ end {aligned} }} ${\ displaystyle R ^ {- 1} \ cup R = A \ times A}$ Every two elements are related, e.g. B. if always or applies. ${\ displaystyle a \ leq b}$${\ displaystyle b \ leq a}$ konnex or connected {\ displaystyle {\ begin {aligned} & \ forall a, b \ in A \ colon \\ & a \ neq b \\ & \ Rightarrow \; (a, b) \ in R \, \ lor \, (b, a) \ in R \ end {aligned}}} ${\ displaystyle R ^ {- 1} \ cup R \ cup \ mathrm {I} = A \ times A}$ Two different elements are related, e.g. B. if always or also if always or applies. ${\ displaystyle a = b,}$ ${\ displaystyle a ${\ displaystyle b ${\ displaystyle a \ leq b}$${\ displaystyle b \ leq a}$ trichotome {\ displaystyle {\ begin {aligned} & \ forall a, b \ in A \ colon \\ & ((a, b) \ in R \ Rightarrow (b, a) \ notin R) \, \ land \\ & (a \ neq b \\ & \ Rightarrow \; (a, b) \ in R \, \ lor \, (b, a) \ in R) \ end {aligned}}} {\ displaystyle {\ begin {aligned} R ^ {- 1} \ cup R \ cup \ mathrm {I} & = A \ times A \\\ land \; R ^ {- 1} \ cap R & = \ emptyset \ end {aligned}}} Every two different elements are always related in exactly one way, e.g. B. if always either or applies. ${\ displaystyle a ${\ displaystyle b The following relationships apply between the properties: The following relationships exist between the properties of a relation and those of its complement : ${\ displaystyle R}$${\ displaystyle {\ overline {R}}}$ • ${\ displaystyle R}$is reflexive is irreflexive (and vice versa).${\ displaystyle \ iff}$ ${\ displaystyle {\ overline {R}}}$ • ${\ displaystyle R}$is symmetrical is symmetrical.${\ displaystyle \ iff}$ ${\ displaystyle {\ overline {R}}}$ • ${\ displaystyle R}$is antisymmetric is connex (and vice versa).${\ displaystyle \ iff}$ ${\ displaystyle {\ overline {R}}}$ • ${\ displaystyle R}$is total is asymmetrical (and vice versa).${\ displaystyle \ iff}$ ${\ displaystyle {\ overline {R}}}$ ## Classes of relations Relationships between different two-digit relations Other important classes of relations and their properties: ## Relation sign In elementary mathematics there are three basic comparative relations: 1. ${\ displaystyle x (Example: "2 is less than 3")${\ displaystyle 2 <3,}$ 2. ${\ displaystyle x = y}$ (Example: "3 equals 3")${\ displaystyle 3 = 3,}$ 3. ${\ displaystyle x> y}$ (Example: "3 is greater than 2")${\ displaystyle 3> 2,}$ with . ${\ displaystyle x, y \ in \ mathbb {R}}$ Two real numbers are always in exactly one of these relationships. With these relation symbols, others can also be formed. The following applies: • ${\ displaystyle x \ leq y}$, if or (example: "4 is not greater than 5")${\ displaystyle x ${\ displaystyle x = y}$ ${\ displaystyle 4 \ leq 5,}$ • ${\ displaystyle x \ geq y}$, if or (example: "5 is not less than 5")${\ displaystyle x> y}$${\ displaystyle x = y}$ ${\ displaystyle 5 \ geq 5,}$ • ${\ displaystyle x \ neq y}$, if or (example: "4 is not equal to 5")${\ displaystyle x ${\ displaystyle x> y}$ ${\ displaystyle 4 \ neq 5,}$ for everyone . ${\ displaystyle x, y \ in \ mathbb {R}}$ The above order relations do not exist for complex numbers . Mathematicians also use the sign ≤ for abstract order relations (and ≥ for the associated inverse relation) while “<” is not an order relation in the sense of the mathematical definition. For equivalence relations , “symmetrical” symbols such as ≈, ~, ≡ are preferred. ## Category theory For any half-ring with zero element and one element , the following is a category : ${\ displaystyle (H, +, \ cdot)}$${\ displaystyle 0}$${\ displaystyle 1}$${\ displaystyle {\ mathcal {C}}}$ • ${\ displaystyle \ mathrm {Ob} ({\ mathcal {C}}) = \ mathrm {Ob} (\ mathbf {Set})}$. • A morphism is a function .${\ displaystyle f \ in \ mathrm {Hom} _ {\ mathcal {C}} (X, Y)}$${\ displaystyle f \ colon X \ times Y \ to H}$ • The following applies to objects${\ displaystyle X}$ ${\ displaystyle \ mathrm {id} _ {X} (x, x '): = {\ begin {cases} 1 & x = x' \\ 0 & {\ text {otherwise}} \ end {cases}}}$. This is identical to the Kronecker delta : .${\ displaystyle \ mathrm {id} _ {X} (x, x ') = \ delta _ {xx'}}$ • The following applies to objects and morphisms${\ displaystyle X, Y, Z}$${\ displaystyle f \ in \ mathrm {Hom} _ {\ mathcal {C}} (Y, Z), \ g \ in \ mathrm {Hom} _ {\ mathcal {C}} (X, Y)}$ ${\ displaystyle (f \ circ g) (x, z): = \ sum _ {y \ in Y} g (x, y) \ cdot f (y, z)}$. The morphisms are set-indexed matrices and their composition happens as with matrix multiplication , corresponds to the unit matrix . ${\ displaystyle \ mathrm {id} _ {x}}$ ${\ displaystyle E}$ In special cases , d. i.e., is the category of relations. ${\ displaystyle (H, +, \ cdot) = (\ {0.1 \}, \ lor, \ land)}$${\ displaystyle {\ mathcal {C}} = \ mathbf {Rel}}$${\ displaystyle {\ mathcal {C}}}$ ## application Operations on whole relations are examined in relational algebra . In computer science , relationships are important when working with relational databases . ## literature • Garrett Birkhoff : Lattice Theory . 3. Edition. AMS, Providence, RI 1973, ISBN 0-8218-1025-1 . • Stefan Brass: Mathematical Logic with Database Applications . Martin Luther University Halle-Wittenberg, Institute for Computer Science, Halle 2005, p. 176 ( informatik.uni-halle.de [PDF]). • Marcel Erné: Introduction to Order Theory . Bibliographisches Institut, Mannheim 1982, ISBN 3-411-01638-8 . • Helmuth Gericke: Theory of Associations . Bibliographical Institute, Mannheim 1963. • Dieter Klaua : set theory . De Gruyter, Berlin / New York 1979, ISBN 3-11-007726-4 (The author uses the term correspondence in the set-theoretical sense synonymously with relation, but then uses the symbol instead of . In the article here, however, or (graph of ) written).${\ displaystyle F}$${\ displaystyle R}$${\ displaystyle R}$${\ displaystyle G_ {R}}$${\ displaystyle R}$ • H. König: Design and structural theory of controls for production facilities (=  ISW Research and Practice . Volume 13 ). Springer, Berlin / Heidelberg 1976, ISBN 3-540-07669-7 , pp. 15-17 , doi : 10.1007 / 978-3-642-81027-5_1 . • Ingmar Lehmann , Wolfgang Schulz: Sets - Relations - Functions . A clear introduction. 3rd, revised and expanded edition. Vieweg + Teubner, Wiesbaden 2007, ISBN 978-3-8351-0162-3 . • Heike Mildenberger: Axiomatic set theory . University of Freiburg, November 9, 2015, p. 58 ( mathematik.uni-freiburg.de [PDF]). • Willard van Orman Quine : Set theory and its logic (=  logic and foundations of mathematics . Volume 10 ). Vieweg + Teubner, Wiesbaden 1973, ISBN 3-528-08294-1 , pp. 264 (American English: Set Theory And Its Logic . Cambridge, MA 1963. Ullstein 1978 paperback). • Gerard O'Regan: Guide to Discrete Mathematics. Sets, Relations and Functions (=  Texts in Computer Science (TCS) ). Springer, Switzerland 2016, p. 25–51 , doi : 10.1007 / 978-3-319-44561-8_2 ( springer.com [PDF; 1000 kB ]). • Fritz Reinhardt, Heinrich Soeder: dtv atlas mathematics . 11th edition. tape 1 : Fundamentals, Algebra and Geometry. Deutscher Taschenbuchverlag, Munich 1998, ISBN 3-423-03007-0 , p. 30-33, 42-45 . • Gunther Schmidt, Thomas Ströhlein: Relations and graphs . Springer, Berlin a. a. 1989, ISBN 3-540-50304-8 . • Robert Wall: Introduction to Logic and Mathematics for Linguists . tape 1 : Logic and set theory . Librarian, Kronberg / Ts. 1974, ISBN 3-589-00023-6 . • Siegfried Wendt: Non-physical basics of information technology - interpreted formalisms . 2nd Edition. Springer, Berlin / Heidelberg 2013, ISBN 978-3-540-54452-4 , doi : 10.1007 / 978-3-642-87627-1 ( books.google.de ). Wikibooks: Math for non-freaks: Relation  - learning and teaching materials Wikibooks: Math for Non-Freaks: Binary Relation  - Learning and Teaching Materials 1. a b G. Smolka: Programming: Chapter 2 - Set theory , at: Saarland University, May 20, 2003, § 2.5. Binary relations, p. 15. 2. Walter Gellert, Herbert Kästner , Siegfried Neuber (eds.): Lexicon of Mathematics. Bibliographisches Institut Leipzig, 1979, p. 484. 3. a b Albert Monjallon: introduction to modern mathematics. Edition 2. Springer-Verlag, 2013, ISBN 978-3-663-16043-4 , p. 74. doi: 10.1007 / 978-3-663-16043-4 , (books.google.de) 4. ^ A b Wilhelm Dangelmaier: Production Theory 1: Methodical Basics. Springer-Verlag, 2017, ISBN 978-3-662-54922-3 , p. 478. doi: 10.1007 / 978-3-662-54923-0 (books.google.de) 5. a b Cobocards: pre-area and post-area. 6. a b 7. Dieter Klaua: Set theory. Page 62, definition 5 (1st part). 8. a b c H. König: Design and structure theory of controls for production facilities. Page 19. 9. Further notations :, in the English-language literature :, see: Gerard O'Regan: Sets, Relations and Functions. P. 36.${\ displaystyle V_ {R}, V (R)}$${\ displaystyle \ operatorname {dom} R}$ 10. Dieter Klaua: Set theory. Page 62, definition 5, (2nd part). 11. Further notations :, in the English-language literature :, see: Gerard O'Regan: Sets, Relations and Functions. P. 36.${\ displaystyle N_ {R}, N (R)}$${\ displaystyle \ operatorname {rng} R}$ 12. Dieter Klaua: Set theory. Page 62, definition 5, (3rd part). 13. In the theory of algebraic structures - especially with regard to category theory, the terms domain and codomain are mostly used in the sense of source and target sets, while in introductory writings on set theory these are usually defined as archetypes and images, 14. also amount similar , quantitative manner , to ger .: left-narrow or set-like called, see Wikibooks: Mathematics Glossary: Mathematical attributes: small predecessor 15. Heike Mildenberger 2015, p. 59f. 16. Martin Ziegler: Lecture on set theory , University of Freiburg, 1992–2014, p. 12. 17. ^ Azriel Levy: Basic Set Theory (= Dover Books on Mathematics . Volume 13). Courier Corporation, Newburyport 2012, ISBN 978-0-486-15073-4 , p. 22, (online) 18. If primitive elements are allowed: for primitive elements there is also a set.${\ displaystyle b}$${\ displaystyle \ {a \ in A \ mid a \ in b \} = \ emptyset}$ 19. See also: Axiomatic Set Theory, Getting a model of (ZF - Fnd) ∪ {¬Fnd} from a model of ZF , Ben Gurion University (BGU) of the Negev, The Department of Mathematics, 2003. 20. In the general case with the support amount sequence is the Allrelation , in the homogeneous case, with the n times the support amount is  .${\ displaystyle A = (A_ {i}) _ {i \ in \ {1, \ dots n \}}}$${\ displaystyle \ mathrm {U} _ {A} = \ textstyle \ prod A}$${\ displaystyle A}$${\ displaystyle U_ {A} = A ^ {n}}$ 21. Stefan Brass (2005), p. 19. 22. The characteristic function as a truth function therefore corresponds to a logical predicate , and in model theory the relation symbols are therefore also called predicate symbols, see Stefan Brass (2005) p. 16. 23. English: forward relational composition 24. ^ Mathematics Stack Exchange: Forward and backward composition in relational algebra Discussion of the chaining directions in connection with the chaining of functions as special relations. The Maplet notation is sometimes used for ordered pairs :${\ displaystyle x \ mapsto y \ equiv (x, y)}$ 25. a b c Glossary of Z notation §Relations , University of Washington 26. Occasionally there is also the semicolon in outline representation. Depending on the hardware and settings, it is not always displayed correctly in Wikipedia. ${\ displaystyle R {\ mbox {⨟}} S}$ 27. English: backward relational composition 28. a b H. König: Design and structure theory of controls for production facilities. Page 21. 29. a b c W. v. O. Quine: Set theory and its logic. Page 47. 30. ^ Relational algebra. In: Mathepedia.de. 31. Also noted as bijection .${\ displaystyle \ pi \ colon \ {1,2, \ dotsc, n \} \ operatorname {\ twoheadrightarrow \! \! \! \! \! \! \! \! \! \! \! \; \; \ rightarrowtail } \ {1,2, \ dotsc, n \}}$ 32. For notation, see Gary Hardegree: Set Theory, Chapter 2: Relations , University of Massachusetts, Amherst, Department of Philosophy, Fall 2015, p. 11: D16 and D17. In contrast to the other notations, these symbols reference images (functions) between the power sets .${\ displaystyle R ^ {\ to}, R ^ {\ leftarrow}}$${\ displaystyle {\ mathcal {P}} (A), {\ mathcal {P}} (B)}$ 33. Analogous to: D. Klaua: Set theory. P. 63, definition 6 (a). 34. In the case of order relations and the like, some authors also speak of a predecessor set or class, see Heike Mildenberger 2015, p. 6, definition 1.12. 35. W. v. O. Quine: Set theory and its logic. Page 17. Attention: The printout is titled as image , but it clearly defines the original image (a set of left-hand coordinates = arguments ). Note that this notation is used contrary to functions here, with functions it stands for the image (a set of right-hand coordinates = function values ) of a set under a function . Functions are special relations. See picture (math) §Alternative notations .${\ displaystyle x}$${\ displaystyle y}$${\ displaystyle f}$ 36. Johannes Köbler: Introduction to Theoretical Computer Science: Relations. Humboldt University Berlin, Institute for Computer Science WS2013 / 14, p. 68. 37. W. v. O. Quine: Set theory and its logic. Page 17. 38. ↑ Following the detailed relationship definition above, one will understand the diagonal as the graph of identity: (Relation) with (Graph).${\ displaystyle \ mathrm {I} _ {A} = (\ Delta _ {A}, A, A)}$${\ displaystyle \ Delta _ {A} = \ {(a, a) \ mid a \ in A \}}$ 39. ↑ Following the above detailed relation definition, analogous to the diagonal, the Nabla will be understood as the graph of the all relation: (Relation) with (Graph)${\ displaystyle \ mathrm {U} _ {A} = (\ nabla _ {A}, A, A)}$${\ displaystyle \ nabla _ {A} = A \ times A}$ 40. This can lead to confusion with the Cartesian product with lead. The meaning results from the context in each case.${\ displaystyle R ^ {n} = R_ {1} \ times \ dotsb \ times R_ {n}}$${\ displaystyle R = R_ {1} = \ dotsb = R_ {n}}$ 41. ^ A b Gerard O'Regan: Sets, Relations and Functions. P. 39. 43. For the transitivity properties of these associations see Proving that is a transitive relation on A${\ displaystyle \ bigcup _ {n = 1} ^ {\ infty} R ^ {n}}$ , on: StackExchange: Mathematics 2018. 44. ^ Robin Hirsch, Ian Hodkinson: Relation algebras. P. 7, on: Third Indian Conference on Logic and its Applications (ICLA). 7-11 January 2009, Chennai, India. 45. The links (one-digit) and (two-digit) are - strictly speaking - the restrictions on or meant.${\ displaystyle {} ^ {-}, {} ^ {\ smallsmile}}$${\ displaystyle \ cap, \ cup, \ circ}$${\ displaystyle A}$${\ displaystyle A ^ {2}}$ 46. ^ C. Brink, K. Britz, RA Schmidt: Peirce Algebras. (1994), pp. 163f. In: M. Nivat, C. Rattray, T. Rus, G. Scollo (Eds.): Algebraic Methodology and Software Technology (AMAST'93). Workshops in computing. Springer, London. 47. The term graph in the graph-theoretical sense is to be distinguished from the term graph of a relation according to the detailed definition of relations mentioned at the beginning (as well as images), which is not used in graph theory. 48. The term color stems from the fact that the number understood as multiplicity according to the multimetary theory is reproduced in the visual representation as a number-coded color of the edge , analogously with colored nodes . An example of color numbers would be the RAL colors .${\ displaystyle f (v)}$${\ displaystyle v}$${\ displaystyle e}$ 49. WD Blizard: Real-valued Multisets and Fuzzy Sets. In: Fuzzy Sets and Systems. Vol. 33, 1989, pp. 77-97. doi: 10.1016 / 0165-0114 (89) 90218-2 . 50. "Two quantities that are equal to one and the same third are equal to one another." Cf. Henri Poincaré: Wissenschaft und Hypothese. Author. German edition with ext. Note from F. u. L. Lindemann. Teubner, Leipzig 1904, p. 36. 51. Wolfgang Rautenberg: Introduction to Mathematical Logic. A textbook. Vieweg + Teubner, Wiesbaden 2008, ISBN 978-3-8348-0578-2 , p. 42. 52. The 1st axiom in Euclid's Elements can, however, also be seen as synonymous with third equals . 53. ↑ It is not uncommon for konnex to be defined as total . 54. This can easily be seen from the tables above (1st and 2nd column) taking into account , i. H. and the predicate logic rules. The inversions apply because of involutivity .${\ displaystyle \ neg (aRb) \ iff a {\ overline {R}} b}$${\ displaystyle \ neg (a, b) \ in R \ iff (a, b) \ not \ in R \ iff (a, b) \ in {\ overline {R}}}$${\ displaystyle {\ overline {\ overline {R}}} = R}$ 55. Wendt 2013, page 31

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Disclaimer: This is an example of a student written essay. Any scientific information contained within this essay should not be treated as fact, this content is to be used for educational purposes only and may contain factual inaccuracies or be out of date. # Lab Report on the Molar Mass of Butane ✅ Paper Type: Free Essay ✅ Subject: Chemistry ✅ Wordcount: 1481 words ✅ Published: 3rd Nov 2020 Introduction As the world of chemistry constantly evolves, certain elements, compounds, and even substances are always undergoing substance identification. Moreover, there are many methods to identify a substance. As noted by Austin Peay State University’s department of chemistry, the identification of a substance can be determined by making simple observations such as odor, temperature, and most importantly, color. However, some gas substances cannot be easily observed for their color. In such cases, determining the molar mass of the substance can prove to be helpful in substance identification. If you need assistance with writing your essay, our professional essay writing service is here to help! According to Purdue University’s department of chemistry, for many chemists, it is impractical to collect and measure gas because gases have small densities. However, because butane is not soluble in water, it allows for water to be displaced from a container. In doing so, it facilitates the collection of gas. In Dalton’s Law of partial of pressures, the total pressure in a container is equal to the sum of the gas collected and water vapor. In 1834, physicist Emil Clapeyron wrote an equation that assisted many chemists in understanding the behavior of everyday gases. Clapeyron equation is recognized as the ideal gas law and is written as PV=nRT. It states that the product of a gas’s volume (L) and pressure (atm) is proportional to the product of the gas constant, moles (mol), temperature (K). The gas constant, “R” has an exact value of 0.0821 L*atm/mol*k. This equation is important because the number of moles, “n”, can be used to determine the molar mass of butane by rearranging the equation to n=PV/RT. In this experiment, the moles and mass will be required to determine the molar mass of butane in a butane lighter. Experimental To begin the experiment, the mass of a butane lighter was measured before using the butane gas to deplete the water to the 80mL mark on a graduated cylinder. Secondly, the temperature of the of the water was measured after waiting five minutes for the temperature to remain constant. Afterwards, the mass of the butane lighter was measured a second time to determine the displacement. In calculating the molar mass, the ideal gas law was used to first determine the number of moles. Finally, the mass of the butane displaced was divided by the moles to eventually produce the molar mass of the butane. The experiment was repeated for four trials. Results Table 1: Data of molar mass experiment Units Trial 1 Trial 2 Trial 3 Trial 4 Initial Mass of Lighter g 14.053 14.827 16.234 13.903 Final Mass of Lighter g 14.273 15.060 16.346 14.071 Mass of Butane g 0.220 0.233 0.112 0.168 Volume of Gas collected L 0.101 0.084 0.094 0.082 Air temperature K 293.000 293.000 293.000 293.000 Water temperature K 293.000 293.000 293.000 293.000 Vapor pressure of H₂O atm 0.023 0.023 0.023 0.023 Barometric Pressure atm 0.988 0.988 0.988 0.988 Accepted Molar Mass of Butane 58.12g/mol Certain data included in the table were given as a standard. That standard was used to compare the experimental result to. The accepted molar mass of butane, vapor pressure of H₂O and barometric pressure was given as a standard. Table 2: Number of moles by the equation n=PV/RT for trials 1-4 Trial 1 Data n=PV/RT (mol) Trial 2 Data n=PV/RT (mol) P= 0.9649 0.004 P= 0.9649 0.003 V= 0.101 V= 0.084 R= 293 R= 293 T= 0.0821 T= 0.0821 Trial 3 Data n=PV/RT (mol) Trial 4 Data n=PV/RT (mol) P= 0.9649 0.004 P= 0.9649 0.003 V= 0.094 V= 0.082 R= 293 R= 293 T= 0.0821 T= 0.0821 The number of moles was calculated by multiplying the pressure and volume and dividing that product by the constant gas and temperature. The pressure was determined by subtracting the pressure of H₂O from the total pressure. Table 3: Molar mass of butane calculations The molar mass was calculated by dividing the mass of butane by the experimental number of moles. The average of all four trials resulted in a molar mass of 54.17g/mol. Discussion The experiment was designed to be able to facilitate substance identification by determining the molar mass of butane. The molar mass of butane was found by first calculating the number of moles using the ideal gas law, n=PV/RT. Next, the mass of the butane displaced was divided by the moles to finally give the molar mass of the butane. The average experimental molar mass resulted in 54.17g/mol while the accepted value was 58.12g/mol. This is a percent error of 6.779%. In the future, this experiment can be improved upon by not letting the butane gas escape the graduated cylinder. In doing so, it will allow for more accurate results when calculating. Bibliography • “Identification of a Pure Substance.” Austin Peay State University Department of Chemistry, Chem 1111 Lab Handouts. Revision S18. • Finding the Molar Mass of Butane. University of Illinois Urbana-Champaign, www.chem.uiuc.edu/chem103/molar_mass/introduction.htm. • Collection of Gas Over Water. Purdue University, chemed.chem.purdue.edu/genchem/lab/techniques/gascollect.html. View all ## DMCA / Removal Request If you are the original writer of this essay and no longer wish to have your work published on UKEssays.com then please: Related Services Prices from S\$203 Approximate costs for: • 1000 words • 7 day delivery Humanity University Dedicated to your worth and value as a human being! Related Lectures

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# Difference between Load factor vs Demand factor vs Diversity factor This page compares Load factor vs Demand factor vs Diversity factor and mentions formulas or equations of them. It is ratio of average load to maximum demand during certain period of time (e.g. day/month/year) is called load factor. Following formula is used to calculate Load Factor. Load factor = Average Demand/Maximum Demand Since average load is always less than maximum demand, hence load factor is always less than unity. ## Demand Factor The ratio of actual maximum demand on the system to the total rated load connected with the system is called demand factor. It is always less than unity. Following formula is used to calculate demand factor. Demand Factor = Maximum Demand/Connected Load ## Diversity Factor The ratio of sum of individual maximum demand of all the consumers supplied by it to maximum demand of power station is called diversity factor. Following formula is used to calculate diversity factor. Diversity Factor = (Sum of Individual Max. Demand)/(Max. Demand of Power Station)

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Canadian Mathematical Society www.cms.math.ca location: # Solutions. 304. Prove that, for any complex numbers z and w, ( |z |+ |w |) êê z |z | + w |w | êê £ 2 |z + w | . Solution 1. ( |z |+ |w |) êê z |z | + w |w | êê = êê z + w + |z |w |w | + |w |z |z | êê £ |z + w |+ 1 |z ||w | | - z zw + - w zw | = |z + w |+ |z w | |z ||w | | - z + - w | = 2 |z + w | . Solution 2. Let z = aei a and w = bei b, with a and b real and positive. Then the left side is equal to |(a + b)(ei a + ei b) | = |aei a + aei b + bei a+ bei b | £ |aei a + bei b |+ |aei b + bei a | . Observe that |z + w |2 = |(aei a + bei b)(ae-i a + be-i b) | = a2 + b2 + ab[ei (a- b) + ei (b- a)] = |(aei b + bei a)(ae-i b + be- i a) | from which we find that the left side does not exceed |aei a + bei b |+ |aei b + bei a | = 2 |aei a + bei b | = 2 |z + w | . Solution 3. Let z = aei a and w = bei b, where a and b are positive reals. Then the inequality is equivalent to êê 1 2 (eia + eib) êê £ |leia + (1 - l)ei b | where l = a/(a+b). But this simply says that the midpoint of the segment joining eia and eib on the unit circle in the Argand diagram is at least as close to the origin as another point on the segment. Solution 4. [G. Goldstein] Observe that, for each m Î C, êê mz |mz | + mw |mw | êê = êê z |z | + w |w | êê , |m|[ |z |+ |w |] = |mz + mw | , and |m||z + w | = |mz + mw | . So the inequality is equivalent to ( |t |+ 1) êê t |t | + 1 êê £ 2 |t + 1 | for t Î C. (Take m = 1/w and t = z/w.) Let t = r(cosq+ i sinq). Then the inequality becomes (r + 1) Ö (cosq+ 1)2 + sin2 q £ 2 Ö (r cosq+ 1)2 + r2 sin2 q = 2 Ö r2 + 2rcosq+ 1 . Now, 4 (r2 + 2r cosq+ 1) - (r + 1)2(2 + 2cosq) = 2r2(1 - cosq) + 4r(cosq- 1) + 2(1 - cosq) = 2(r - 1)2(1 - cosq) ³ 0 , from which the inequality follows. Solution 5. [R. Mong] Consider complex numbers as vectors in the plane. q = (|z |/|w |)w is a vector of magnitude z in the direction w and p = (|w |/|z |)z is a vector of magnitude w in the direction z. A reflection about the angle bisector of vectors z and w interchanges p and w, q and z. Hence |p + q | = |w + z |. Therefore ( |z |+ |w |) êê z |z | + w |w | êê = |z + q + p + w | £ |z + w |+ |p + q | = 2 |z + w | . 305. Suppose that u and v are positive integer divisors of the positive integer n and that uv < n. Is it necessarily so that the greatest common divisor of n/u and n/v exceeds 1? Solution 1. Let n = ur = vs. Then uv < n Þv < r, u < s, so that n2 = uvrs Þ rs > n. Let the greatest common divisor of r and s be g and the least common multiple of r and s be m. Then m £ n < rs = gm, so that g > 1. Solution 2. Let g = gcd (u, v), u = gs and v = gt. Then gst £ g2st < n so that st < n/g. Now s and t are a coprime pair of integers, each of which divides n/g. Therefore, n/g = dst for some d > 1. Therefore n/u = n/(gs) = dt and n/v = n/(gt) = ds, so that n/u and n/v are divisible by d, and so their greatest common divisor exceeds 1. Solution 3. uv < n Þnuv < n2 Þ n < (n/u)(n/v). Suppose, if possible, that n/u and n/v have greatest common divisor 1. Then the least common multiple of n/u and n/v must equal (n/u)(n/v). But n is a common multiple of n/u and n/v, so that (n/u)(n/v) £ n, a contradiction. Hence the greatest common divisor of n/u and n/v exceeds 1. Solution 4. Let P be the set of prime divisors of n, and for each p Î P. Let a(p) be the largest integer k for which pk divides n. Since u and v are divisors of n, the only prime divisors of either u or v must belong to P. Suppose that b(p) is the largest value of the integer k for which pk divides uv. If b(p) ³ a(p) for each p Î P, then n would divide uv, contradicting uv < n. (Note that b(p) > a(p) may occur for some p.) Hence there is a prime q Î P for which b(q) < a(q). Then qa(q) is not a divisor of either u or v, so that q divides both n/u and n/v. Thus, the greatest common divisor of n/u and n/v exceeds 1. Solution 5. [D. Shirokoff] If n/u and n/v be coprime, then there are integers x and y for which (n/u)x + (n/v) = 1, whence n(xv + yu) = uv. Since n and uv are positive, then so is the integer xv + yu. But uv < nÞ 0 < xv + yu < 1, an impossibility. Hence the greatest common divisor of n/u and n/v exceeds 1. 306. The circumferences of three circles of radius r meet in a common point O. They meet also, pairwise, in the points P, Q and R. Determine the maximum and minimum values of the circumradius of triangle PQR. Answer. The circumradius always has the value r. Solution 1. [M. Lipnowski] ÐQPO = ÐQRO, since OQ is a common chord of two congruent circles, and so subtends equal angles at the respective circumferences. (Why are angle QPO and QRO not supplementary?) Similarly, ÐOPR = ÐOQR. Let P¢ be the reflected image of P in the line QR so that triangle P¢QR and PQR are congruent. Then ÐQP¢R + ÐQOR = ÐQPR + ÐQOR = ÐQPO + ÐRPO + ÐQOR = ÐQRO + ÐOQR + ÐQOR = 180° . Hence P¢ lies on the circle through OQR, and this circle has radius r. Hence the circumradius of PQR equals the circumradius of P¢QR, namely r. Solution 2. [P. Shi; A. Wice] Let U, V, W be the centres of the circle. Then OVPW is a rhombus, so that OP and VW intersect at right angles. Let H, J, K be the respective intersections of the pairs (OP, VW), (OQ, UW), (OP, UV). Then H (respectively J, K) is the midpoint of OP and VW (respectively OQ and UW, OP and UV). Triangle PQR is carried by a dilation with centre O and factor 1/2 onto HJK. Also, HJK is similar with factor 1/2 to triangle UVW (determined by the midlines of the latter triangle). Hence triangles PQR and UVW are congruent. But the circumcircle of triangle UVW has centre O and radius r, so the circumradius of triangle PQR is also r. Solution 3. [G. Zheng] Let U, V, W be the respective centres of the circumcircles of OQR, ORP, OPQ. Place O at the centre of coordinates so that U ~ (r cosa, r sina) V ~ (r cosb, r sinb) W ~ (r cosg, r sing) for some a, b, g. Since OVPW is a rhombus, P ~ (r (cosb+ cosg), r (sinb+ sing)) . Similarly, Q ~ ( r (cosa+ cosg),r (sina+ sing), so that |PQ | = r Ö (cosb- cosa)2+ (sinb- sina)2 = |UV | . Similarly, |PR | = |UW | and |QR | = |VW |. Thus, triangles PQR and UVW are congruent. Since O is the circumcentre of triangle UVW, the circumradius of triangle PQR equals the circumradius of triangle UVW which equals r. Solution 4. Let U, V, W be the respective centres of the circles QOR, ROP, POQ. Suppose that ÐOVR = 2b; then ÐOPR = b. Suppose that ÐOWQ = 2g; then ÐOPQ = g. Hence ÐQPR = b+ g. Let r be the circumradius of triangle PQR. Then |QR | = 2rsin(b+ g). Consider triangle QUR. The reflection in the axis OQ takes W to U so that ÐQUO = ÐQWO = 2g. Similarly, ÐRUO = 2g, whence ÐQUR = 2(b+ g). Thus triangle QUR is isosceles with |QU | = |QR | = r and apex angle QUR equal to 2 (b+ g). Hence |QR | = 2rsin(b+ g). It follows that r = r. Comment. This problem was the basis of the logo for the 40th International Mathematical Olympiad held in 1999 in Romania. 307. Let p be a prime and m a positive integer for which m < p and the greatest common divisor of m and p is equal to 1. Suppose that the decimal expansion of m/p has period 2k for some positive integer k, so that m p = .ABABABAB ... = (10k A + B)(10-2k +10-4k + ¼) where A and B are two distinct blocks of k digits. Prove that A + B = 10k - 1 . (For example, 3/7 = 0.428571 ... and 428 + 571 = 999.) Solution. We have that m p = 10kA + B 102k - 1 = 10kA + B (10k - 1)(10k + 1) whence m(10k - 1)(10k + 1) = p(10k A + B) = p(10k - 1)A + p(A + B) . Since the period of m/p is 2k, A ¹ B and p does not divide 10k - 1. Hence 10k - 1 and p are coprime and so 10k - 1 must divide A + B. However, A £ 10k - 1 and B £ 10k - 1 (since both A and B have k digits), and equality can occur at most once. Hence A + B < 2 ×10k - 2 = 2(10k - 1). It follows that A + B = 10k - 1 as desired. Comment. This problem appeared in the College Mathematics Journal 35 (2004), 26-30. In writing up the solution, it is clearer to set up the equation and clear fractions, so that you can argue in terms of factors of products. 308. Let a be a parameter. Define the sequence { fn (x) : n = 0, 1, 2, ¼} of polynomials by f0 (x) º 1 fn+1 (x) = x fn (x) + fn (ax) for n ³ 0. (a) Prove that, for all n, x, fn (x) = xn fn (1/x) . (b) Determine a formula for the coefficient of xk (0 £ k £ n) in fn (x). Solution 1. The polynomial fn (x) has degree n for each n, and we will write fn (x) = nå k=0 b(n, k) xk . Then xn fn (1/x) = nå k=0 b(n, k)xn-k = nå k=0 b(n, n - k)xk . Thus, (a) is equivalent to b(n, k) = b(n,n - k) for 0 £ k £ n. When a = 1, it can be established by induction that fn (x) = (x + 1)n = åk=0n (n || k) xn. Also, when a = 0, fn (x) = xn + xn-1 + ¼+ x + 1 = (xn+1 - 1)(x - 1)-1. Thus, (a) holds in these cases and b(n, k) is respectively equal to (n || k) and 1. Suppose, henceforth, that a ¹ 1. For n ³ 0, fn+1(k) = nå k=0 b(n, k)xk+1 + nå k=0 ak b(n, k)xk = nå k=1 b(n, k-1) xk + b(n, n)xn+1 + b(n, 0)+ nå k=1 ak b(n, k) xk = b(n, 0) + nå k=1 [b(n, k-1) + ak b(n, k)]xk+ b(n, n)xn+1 , whence b(n + 1, 0) = b(n, 0) = b(1, 0) and b(n + 1, n + 1) = b(n, n) = b(1, 1) for all n ³ 1. Since f1 (x) = x + 1, b(n, 0) = b(n, n) = 1 for each n. Also b(n + 1, k) = b(n, k-1) + ak b(n, k) (1) for 1 £ k £ n. We conjecture what the coefficients b(n, k) are from an examination of the first few terms of the sequence: f0(x) = 1;     f1(x) = 1 + x;     f2(x) = 1 + (a + 1)x + x2; f3(x) = 1 + (a2 + a + 1)x + (a2 + a + 1)x2 + x3; f4(x) = 1 + (a3 + a2 + a + 1)x + (a4 + a3 + 2a2 + a + 1)x2+ (a3 + a2 + a + 1)x3 + x4; f5(x) = (1 + x5) + (a4 + a3 + a2 + a + 1)(x + x4)+ (a6 + a5 + 2a4 + 2a3 + 2a2 + a + 1)(x2 + x3) . We make the empirical observation that b(n + 1, k) = an+1-kb(n, k-1) + b(n, k) (2) which, with (1), yields (an+1-k - 1)b(n, k-1) = (ak - 1)b(n, k) so that b(n+1, k) = éë ak - 1 an+1-k - 1 + ak ùû b(n, k) = éë an+1 - 1 an+1-k - 1 ùû b(n, k) for n ³ k. This leads to the conjecture that b(n, k) = æè (an - 1)(an-1 - 1) ¼(ak+1 - 1) (an-k - 1)(an-k-1 - 1) ¼(a - 1) öø b(k, k) (3) where b(k, k) = 1. We establish this conjecture. Let c(n, k) be the right side of (3) for 1 £ k £ n-1 and c(n, n) = 1. Then c(n, 0) = b(n, 0) = c(n, n) = b(n, n) = 1 for each n. In particular, c(n, k) = b(n, k) when n = 1. We show that c(n + 1, k) = c(n, k-1) + ak c(n, k) for 1 £ k £ n, which will, through an induction argument, imply that b(n, k) = c(n, k) for 0 £ k £ n. The right side is equal to æè an - 1 an-k - 1 öø ¼ æè ak+1 - 1 a - 1 öø éë ak - 1 an-k+1 - 1 + ak ùû = (an+1 - 1)(an - 1) ¼(ak+1 - 1) (an+1-k - 1)(an-k - 1) ¼(a - 1) = c(n+1, k) as desired. Thus, we now have a formula for b(n, k) as required in (b). Finally, (a) can be established in a straightforward way, either from the formula (3) or using the pair of recursions (1) and (2). Solution 2. (a) Observe that f0 (x) = 1, f1 (x) = x + 1 and f1 (x) - f0 (x) = x = a0 x f0 (x/a). Assume as an induction hypothesis that fk (x) = xk f(1/x) and fk (x) - fk-1 (x) = ak-1 x fk-1(x/a) for 0 £ k £ n. This holds for k = 1. Then fn+1 (x) - fn (x) = x[fn(x) - f(n-1)(x)] + [fn(ax) - fn-1(ax)] = an-1x2 fn-1(x/a) + an-1axfn-1(x) = anx [fn-1(x) + (x/a)fn-1(x/a) = an x fn(x/a) , whence fn+1(x) = fn (x) + an x fn(x/a) = fn(x) + an x(x/a)nfn(a/x) = xn fn(1/x) + xn+1 fn (a/x) = xn+1 [(1/x)fn(1/x) + fn(a/x)] = xn+1fn+1(1/x) . The desired result follows. Comment. Because of the appearance of the factor a - 1 in denominators, you should dispose of the case a = 1 separately. Failure to do so on a competition would likely cost a mark. 309. Let ABCD be a convex quadrilateral for which all sides and diagonals have rational length and AC and BD intersect at P. Prove that AP, BP, CP, DP all have rational length. Solution 1. Because of the symmetry, it is enough to show that the length of AP is rational. The rationality of the lengths of the remaining segments can be shown similarly. Coordinatize the situation by taking A ~ (0, 0), B ~ (p, q), C ~ (c, 0), D ~ (r, s) and P ~ (u, 0). Then, equating slopes, we find that s r - u = s - q r - p so that sr - ps s - q = r - u whence u = r - [(sr - ps)/(s - q)] = [(ps - qr)/(s - q)]. Note that |AB |2 = p2 + q2, |AC |2 = c2, |BC |2 = (p2 - 2pc + c2)+ q2, |CD |2 = (c2 - 2cr + r2) + s2 and |AD |2 = r2 + s2, we have that 2rc = AC2 + AD2 - CD2 so that, since c is rational, r is rational. Hence s2 is rational. Similarly 2pc = AC2 + AB2 - BC2 . Thus, p is rational, so that q2 is rational. 2qs = q2 + s2 - (q - s)2 = q2 + s2 - [(p - r)2 + (q - s)2] + p2 - 2pr + r2 is rational, so that both qs and q/s = (qs)/s2 are rational. Hence u = p - r(q/s) 1 - (q/s) is rational. Solution 2. By the cosine law, the cosines of all of the angles of the triangle ACD, BCD, ABC and ABD are rational. Now AP AB = sinÐABP sinÐAPB and CP BC = sinÐPBC sinÐBPC . Since ÐAPB + ÐBPC = 180°, therefore sinÐAPB = sinÐBPC and AP CP = AB sinÐABP BC sinÐPBC = AB sinÐABP sinÐPBC BC sin2 ÐPBC = AB (cosÐABP cosÐPBC - cos(ÐABP + ÐPBC)) BC (1 - cos2 ÐPBC) = AB (cosÐABD cosÐDBC - cosÐABC) BC (1 - cos2 ÐDBC) is rational. Also AP + CP is rational, so that (AP/CP)(AP + CP) = ((AP/CP) + 1)AP is rational. Hence AP is rational. 310. (a) Suppose that n is a positive integer. Prove that (x + y)n = nå k=0 æè n k öø x (x + k)k-1 (y - k)n-k . (b) Prove that (x + y)n = nå k=0 æè n k öø x (x - kz)k-1 (y + kz)n-k . Comments. (a) and (b) are equivalent. To obtain (b) from (a), replace x by -x/z and y by -y/z. On the other hand, the substitution z = -1 yields (a) from (b). The establishment of the identities involves the recognition of a certain sum which arise in the theory of finite differences. Let f(x) be a function of x and define the following operators that take functions to functions: If(x) = f(x) Ef(x) = f(x + 1) = (I + D)f(x) Df(x) = f(x + 1) - f(x) = (E - I)f(x) . For any operator P, Pnf(x) is defined recursively by P0 f(x) = f(x) and Pk+1f(x) = P(Pk-1) f(x)), for k ³ 1. Thus Ekf(x) = f(x + k) and D2 f(x) = Df(x + 1) - Df(x) = f(x + 2) - 2f(x + 1) + f(x) = (E2 - 2E + I)f(x) = (E - I)2f(x) . We have an operational calculus in which we can treat polynomials in I, E and D as satisfying the regular rules of algebra. In particular En f(x) = (I + D)n f(x) = å æè n k öø Dk f(x) and Dn f(x) = (E - I)n f(x) = nå k = 0 (-1)n-k æè n k öø Ek f(x) = nå k = 0 (-1)n-k æè n k öø f(x + k) , for each positive integer n, facts than can be verified directly by unpacking the operational notation. Now let f(x) be a polynomial of degree d ³ 0. If f(x) is constant (d = 0), then Df(x) = 0. If d ³ 1, then Df(x) is a polynomial of degree d - 1. It follows that Dd f(x) is constant, and Dn f(x) = 0 whenever n > d. This yields the identity nå k=0 (-1)k æè n k öø f(x + k) = 0 for all x whenever f(x) is a polynomial of degree strictly less than n. Solution 1. [G. Zheng] nå k=0 æè n k öø x(x + k)k-1 (y - k)n-k = nå k=0 æè n k öø x(x + k)k-1 [(x + y) - (x + k)]n-k = nå k=0 n-kå j=0 æè n k öø x(x + k)k-1 æè n-k j öø (x + y)j (-1)n-k-j(x + k)n-k-j = å 0 £ k £ n - j £ n (-1)n-k-j æè n k öø æè n-k j öø x (x + k)n-j-1(x + y)j = nå j=0 n-jå k = 0 (-1)n-k-j æè n j öø æè n-j k öø x(x + k)n-j-1 (x + y)j = nå j=0 (-1)n-j æè n j öø (x + y)j n-jå k=0 (-1)k æè n-j k öø x(x + k)n-j-1 = (x + y)n x(x + 0)-1 + x n-1å j=1 (-1)n-j æè n j öø (x + y)j n-jå k=0 (-1)k æè n-j k öø (x + k)n-j-1 . Let m = n - j so that 1 £ m £ n. Then n-jå k=0 (-1)k æè n-j k öø (x + k)n-j-1 = må k=0 (-1)k æè m k öø (x + k)m-1 = må k=0 (-1)k æè m k öø m-1å l=0 æè m-1 l öø xm-lkl = m-1å l=0 æè m-1 l öø xm-l må k=0 (-1)k æè m k öø kl = 0 . The desired result now follows. Solution 2. [M. Lipnowski] We prove that nå k=0 æè n k öø x (x - kz)k-1 (y + kz)n-k = (x + y)n by induction. When n = 1, this becomes 1 ·x(x)-1y + 1 ·x(x - z)0(y+z)0 = y + x = x + y . Assume that for n ³ 2, n-1å k=0 æè n-1 k öø x(x - kz)k-1(y + zk)n-k-1 = (x + y)n-1 . Let f(y) = (x + y)n and g(y) = åk=0n (n || k)x(x - kz)k-1 (y + kz)n-k. We can establish that f(y) = g(y) for all y by showing that f¢(y) = g¢(y) for all y (equality of the derivatives with respect to y) and f(-x) = g(-x) (equality when y is replaced by -x). That f¢(y) = g¢(y) is a consequence of the induction hypothesis and the identity (n || k)(n - k) = n ((n-1) || k). Also g(-x) = nå k=0 æè n k öø x (x - kz)k-1(-x + kz)n-k = x nå k=0 (-1)n-k æè n k öø (x - kz)n-1 = 0 , by appealing to the finite differences result. The desired result now follows. top of page | contact us | privacy | site map | © Canadian Mathematical Society, 2017 : https://cms.math.ca/

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## Wing Incidence & Tail Size lift equal to the weight of our plane. ... the wing angle of attack by three de- ..... airplane's response to various kinds of ..... are using Lotus 1-2-3 on a DOS com-. Wing Incidence & Tail Size . . . by JOHN G. RONCZ, EAA 112811 15450 Hunting Ridge Tr. Granger, IN 46530-9093 You've gone to a lot of trouble to figure out how big the wing needs to be for your homebuilt. Now you have to decide what angle of incidence to use when you bolt the wing to the fuselage. If the incidence angle is too high, you will be cruising with the fuselage nose low. If you don't have enough incidence, the fuselage will be sitting nose high in level flight. My experience using high-powered three-dimensional electronic wind tunnel programs is that each fuselage and wing combination has its own personal- ity, and that to get the minimum overall drag some fuselages want to cruise a bit nose high (my homebuilt) while others want to cruise a bit nose low (Swiftfury). Lacking this kind of analysis, you might as well assume that the fuselage should be level in cruise. We all learned in Private Pilot Ground School that in level flight, the lift has to equal the weight. So our problem is to find the angle required to produce wing lift equal to the weight of our plane. Cutting directly to the core of the problem, you have to know how much additional lift your wing will produce if you raise its angle of attack by one degree. As we've seen, the actual lift in pounds that the wing produces is a product of the speed squared and the altitude. To get rid of this speed-density issue, we divide the lift in pounds by the dynamic pressure of the air (half the air density times the speed squared) to arrive at our old friend, the coefficient of lift, or CL. This is another way of saying that the CL is the lift, in pounds, produced by one square foot of wing if the dynamic pressure is one pound per square foot. Again, dynamic pressure is the force are in big trouble. So we can assume that you are operating below the stall you feel when you stick your hand out of the car window at high speed. and, therefore, on the straight line portion of the plot. To sound professional, Imagine that you are in cruising flight, and you pull back on the stick and raise the wing angle of attack by three degrees. Let's say that the CL of the wing goes up by .24. Dividing .24 by 3 gives .08, so one degree angle of attack is you should say that you are operating in the linear portion of the curve. Going back to the example above, we can then write the change in lift due to a one degree change in angle of attack as worth .08 lift coefficients. Textbook publishers spend big bucks to purchase Greek type, so you will see this written as A CL/A Alpha = .08 per degree The pretty triangle is the Greek capi- which is proper calculus notation using the small letter delta instead of the capital letter delta. If you missed calculus tal letter delta, which means "change". The term "alpha" means angle of attack. Using this style of notation, A Miles/A Gallons = 21 means that your car is getting 21 miles per gallon. At low angles of attack, the lift you get for every degree angle of attack you add says about the same. But as you get to high angles of attack, increasing the angle of attack may in fact buy you a stalled wing, in which case you'd be losing lift with further angle of attack. If you make a plot of lift coefficient versus angle of attack, like Figure 1, you end up with a straight line at the lower angles of attack, followed by a curved portion at higher angles of attack, once the wing starts to stall. Figure 1 is for my homebuilt's root airfoil. Note that there is a shift at eight degrees angle of attack, because I changed from a cruising Reynolds number to a landing Reynolds number. The shift you see is caused by the fact that at lower speeds (lower Reynolds numbers), the boundary layer is thicker, and decambers the airfoil more, especially on this 18.6% thick airfoil. If your wing is stalled in cruise, you 8 CL/8 Alpha = .08 per degree in school, don't run away. Just understand that the statement is a shorthand way of saying that if you buy one degree more angle of attack, you get .08 CL's absolutely free. Another way of looking at it is that 8CL/8 Alpha is the slope of the lift curve, as I've drawn in on Figure 1. It's perfectly obvious that you have to know how much lift your wing produces for each degree of incidence before you can decide what angle it needs to balance the weight of your airplane. Several things go into the calculation of the lift curve slope 8CL/8 Alpha, so let's start by listing them all: • • • • the aspect ratio the angle of sweep the Mach number the airfoil thickness For a rectangular wing, aspect ratio is simply the span divided by the chord. The chord line is the longest line you can draw inside the airfoil shape, running from the nose to the tail. If your chord was 3 feet, and your span was 30 feet, the aspect ratio is 10, because the wing is 10 times longer than it is wide. For tapered wings it gets more complicated, and is expressed with the general rule SPORT AVIATION 23 ratio of one means that your wing leaks CL versus Alpha, DLR root 1.600 J U g 1.400 1.200 1.000 0.800 U 0.600 0.400 0.200 ~i i r 0.000 0.00 "dCL/ d Alpha n i r 5.00 10.00 15.00 Angle of Attack, Alpha ~ degrees Reynolds number = 6,000,000 below 8 degrees Reynolds number = 3,600,000 above 8 degrees like a sieve. It should therefore come as no surprise that the less leaky your wings are, the more lift you get for each degree more incidence. Sweeping the wings creates another kind of leakiness. If I shrink you down to 1/8 inch, and stand you on top of a swept wing, your next door neighbor will have higher or lower pressure than you do. This encourages the air to start moving sideways rather than devoting its energies to diving straight at the earth like it should. The remaining question is how to measure the angle of sweep. The best answer is that you measure the sweep of the lowest pressure points along the wing span. A close way of approximating this is to measure the sweep of the maximum thickness line. This means you find the fattest portion of the airfoil at the wing root, and the fattest part of the airfoil at the wing tip, then figure the angle between them. The higher the Mach number, the more lift your wing will make for each degree of incidence. At low speeds, the air just goes around the wing like a recruit on an obstacle course. But once an airplane gets going fast enough, the Aspect ratio = span * span / wing trillions of air molecules, so you may Remember to be consistent in your units; don't measure the span in inches and then divide by the wing area in square feet! Aspect ratio is the single biggest factor in determining how much lift your wing will produce for each de- at 35,000 feet. One way to prove that the wing is deflecting air downwards is that the air is following a curved path, and this means that the air on top the wing has to move area. gree angle of attack. To understand why, it helps to know a bit about how wings produce lift. The answer is that wings don't really make lift at all. Wings take air and deflect it towards the ground. The ground gets really upset by this and pushes the wing away. I always pose this question on the exams I give the gifted children I teach in the summer: you are piloting a cargo airplane containing 1000 heavy birds in cages. Suddenly the birds all start flying around in their cages. Does the airplane start to climb? Surely if the birds each carried their own weight, the plane would be much lighter and start to climb. But, of course, it wouldn't, because all the birds do is to deflect air downwards, in this case towards the bottom of the fuselage. So the fuselage still bears the weight of the birds, whether the birds are sitting on the floor or throwing their weight in air at it. For those purists among you, the downward movement in the air may be damped out by all those collisions of ! 24 APRIL 1990 not feel a breeze when a 747 flies over on average faster than the air under the wing. By Bernoulli's law, the pressure on the top of the wing will therefore be lower. Since nature abhors a vacuum, she wiH try and move air in from other places to fill the partial void. The wingtip is the ideal place for this to happen. So Mother Nature kicks air from the bottom of the wing around the wingtip to try and fill in the low pressure area on top of the wing. The resulting wingtip vortices are familiar to anyone who's done some formation flying. The strength of the vortices depends upon the pressure difference between the upper and lower surfaces, so they will be nastiest at low speeds when the wing is working hardest. The moral of this story is that wingtips are leaky. Even if you park winglets and fins and endplates and other aerodynamic chastity belts on the ends of your wings, Ma Nature will eventually find some way to get the high pressure air to the low pressure air. An aspect ratio of ten, then, means that one-tenth of your wingspan is leaky. An aspect air finds that it's easier to compress than to go around the obstacle. The compressed air makes the airfoil appear to have more camber, throwing more air at the ground; this gets the ground even more upset, so you get more lift at high speeds than you do at low. The last item on our list is airfoil thickness. It is true that a thicker airfoil will have more lift for each degree angle of attack, however, this is offset by the fact that the boundary layers are thicker for thicker airfoils, and boundary layers take away some of the airfoil's camber. So most people call it even, and ignore the thickness effect. I will, too. I just wanted to let you know that this effect does exist. To make the final equation easier to write, let's define a few terms: AR = span"2/wing area . . . (aspect ratio) beta = (1-Mach*Mach)" 0.5 . . . (compressibility correction) lambda = tangent of (sweep of max thickness line)... (sweep correction) Then the slope of the lift curve can be written as: 5 CL/5 alpha = .1096623 * AR / (2 + (4 + AR"2 * beta"2*(1 + (lambda/beta)'2)) ".5) To explain this, theory says that you ought to get a CL of .1096623 for every degree angle of attack. You multiply this by a correction based on your aspect ratio, Mach number and sweep. This equation is used in the spreadsheet for this article, so that you can use it to THESE 2 TAILS HAVE THE SAME TAIL VOLUME ! calculate the lift-curve slopes for your wings and tails. The last item we need to discuss on this topic is the angle of zero lift. If you use a symmetrical airfoil (top and bottom are mirror images) then you get zero lift at zero angle of attack. But if you use a cambered airfoil, which you should unless you plan to spend half your time upside down, then at zero angle of attack you still get some positive lift. From the airfoil test data, you need to find the angle of attack at which the airfoil produces zero lift (see Figure 1). You ought to set the incidence of your wing at the proper angle to make the plane fly level in typical cruise. I used 70% power at 7500 feet for my non-turbocharged airplane; you can pick any other point you want. Using the spreadsheets from either of the previous articles, first you have to find the airplane's lift coefficient corresponding to the weight, altitude and speed you picked. The wing incidence will be calculated for Butt Line O, so next you have to find the lift coefficient at Butt Line 0. If the wing has an elliptical planform, then the C, at BLO is the same as the airplane CL. If the wing is rectangular, then the C, at BLO is computed using the area of the ellipse rule I gave you last time, which gives G, (BLO) = airplane CL * 4 / pi. A more complicated formula which should work for all wings is (be sure not to mix feet with inches!) C, (BLO) = airplane CL * 4 * wing area / (chord at BLO * span * pi) You bolt your wing to the fuselage so that BLO has this angle: incidence = C, (BLO) / (5CL / 8 alpha) + angle of zero lift (BLO) As an example, assume that the airplane's lift coefficient needed for the cruise point you picked is .3183. Your wing area has 100 square feet, the span is 38 feet, and the chord at BLO is 4 feet. Using the complicated equation, you find that the C, at BLO would be .3183 ' 4 * 100 / (4 * 28 * pi), or .3619. Next you plug in your aspect ratio, Mach number and sweep into the lift-curve slope equation and get .08122, and the airfoil you plan to use at BLO produces zero lift at -2.5 degrees. Then your incidence at BLO would be .3619 / .08122 + (-2.5) = 1.955 degrees You can round this off to 2 degrees, since 45 thousandths of a degree would not be noticeable in flight. Assuming that your weight and speed estimates prove accurate, your fuselage should then be level at the altitude and power you picked. SIZING YOUR TAILS The tail size you use determines how stable the airplane will be in pitch and yaw. There are two issues here. The first is the ability of the tail surfaces to stabilize the airplane in steady flight, and the second is the ability of the tail surfaces to damp out the disturbances to the original yaw and pitch angles. Fundamental stability is produced by having so many square feet of tail placed so many feet behind the CG. The square footage of the tail times the separation distance in feet gives cubic feet, so you'll see the term tail volume used a lot. It makes sense that a big tail with a short lever arm would have the same power as a little tail with a large lever arm. Either combination is equally effective in yielding steady-state stability. Figure 2 shows two tails with the same tail volume. Damping power, however, is based on the tail area times the lever arm squared. Therefore, a small tail with a big lever arm is the one to pick for a bumpy day. A short-coupled airplane can always be made stable with a big tail, but can never be made to produce the damping forces that a longer lever arm would give. For example, the plane with the big tail in Figure 2 has only half the damping power of the plane with the small tail. You can easily spend the rest of your life studying stability and control. Imagine, for a moment, that you pluck a single leaf off a tree, hold it four feet above the ground, and let it go. Now imagine that your job is to write a set of equations which would exactly predict the path and speed that the falling leaf would take. My guess is that even if you could write the equations, no computer would be fast enough to solve them in your lifetime. The equations of motion for an airplane are simpler than those of a falling leaf, and can be solved. However, you have to assemble a long list of numbers describing your airplane's response to various kinds of disturbances, which are called the stability derivatives, before you can calculate what happens when you let go of the stick and then disturb the airplane. This work is beyond the patience of most homebuilders. It is no fun at all flying an airplane which wallows around in turbulence, making the pilot work hard putting in continuous corrections. Making the tails bigger will help this. However, an airplane can also be too stable. In this case it will take too much elevator or rudder deflection to maneuver the airplane, and pilots will not like this either. The method we'll use here to size our tails is purely statistical. That is, we are going to look at tail volumes which have worked for others, and since we are not departing radically from traditional design, what worked for them will probably work for us as well. By staying close to the statistical average, we can bypass the actual stability and control calculations, and still end up with an airplane that flies nicely. One value that we haven't talked about so far, but which we will be using a lot from now on, is called the mean aerodynamic chord, or MAC. The mean aerodynamic chord is the same as the chord length for a rectangular wing, but for any other shaped wing is MAC = 2/3 * root chord * (1 + X + X •2) / (1 + X). If you already forgot that the upsidedown y is the Greek letter lambda, you can still use this formula! For this formula, lambda has the value X = tip chord / root chord For my plane, the tip chord is 25.5 inches, and the root chord is 50.5 inchSPORT AVIATION 25 1 2 3 4 5 6 7 8 A C: B D I F 2-90 JGR speed (mph) altitude: 220.60! 7500.00: iMach 1 , mph Mach number Cruise weight wing area, ftA2 9 10 wing sjpan, ft 11 sweep, degrees 1 800.00! 92.27! 30.00! -7.00! irho iCL 'Aspect Ratio sweep factor ibefa jdCL/dAlpha 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 E 741.16! 0.2968J 0.0018974! 6.1974! 9.75356! -0.12278! 6.95493! 6.09220! chord BLO, inches angle for zero lift 50.561 -1.08! ICipBLO incidence 6.1837] 0.9125! degrees root chord, inches tip chord, inches 50.50! 25.50! Itaper ratio 6.50561 39.37! 80.13! inches inches HORIZONTAL TAIL: lever arm, inches volume coefficient VERTICAL TAIL: lever arm, inches volume coefficient es, so lambda has the value 25.5 / 50.5 or .505. Therefore, for my plane the MAC is 2/3 * 50.5 * ( 1 + .505 + .505'2) / (1 + .505), or 39.37 inches, which happens to be one meter. The average chord for my plane is 38 inches, so the Mean Aerodynamic Chord is a little bigger than the average chord. It would also be useful to know at what Butt Line the MAC is. The formula for this is 1/3*semispan*(1 + 2 * X ) / ( 1 + \) For my plane, the semispan is 167 inches (not counting wingtip), so the Butt Line for the MAC is at 1/3 * 167 * ( 1 + 2 * .505) / (1 + .505), or Butt Line 74.346 You will see in later articles that we can replace the wing by its MAC. For now, the spreadsheet for this article will calculate the MAC for your wing. Draw in the MAC for your plane, by drawing a line along the Butt Line of the MAC which you just calculated. The center of gravity range of your airplane is roughly from 15% to 30% of the MAC for a typical wing. If you have calculated the weight and balance for your plane, and you find that the aft C of G limit is at 65% of the Mean Aerodynamic 26 APRIL 1990 !MAC ?MACi>iBL = 115.661 itaii size 1 4.22! square feet 120.00! 6.0350! itail size 9.69! square feet 0.4500! Chord, you need to move your wing further aft. There is a wide variation of tail volume coefficients among the airplanes in the fleet. The F-86, at .203, is the lowest I know of. Of course, fighters are designed to be highly maneuverable, HORIZONTAL TAIL SIZE hence have low stability. The Cessna Obviously, the size of your horizontal 170 has .58, which is above the .37 to tail is related to the size of your main .48 range of most light airplanes. While wing. It also is directly related to its lever most factory-builts come in at just under arm, which is the distance from the .4, Thurston recommends .55 in his center of gravity to the center of lift of book (see references at the end). Pazthe horizontal tail. Needless to say, you many gives a nice chart in his book, use the aftmost center of gravity posi- and says he used .43 for his PL-1. It tion to size the tails, since this gives you looks like somewhere between .4 and the shortest lever arm. .48 would put you in the ballpark. I am The horizontal tail (HT) area in using .6 because of the shape of my square feet, multiplied by its lever arm fuselage, and because I need a large in feet, gives cubic feet, as we saw be- center of gravity travel to handle the fore. We will divide this by more cubic four people. I may reduce the tail size feet, namely, the wing area times the as I progress further in my stability and MAC, so that the answer will be a control analysis. number without dimensions. We will call this number the horizontal tail volume VERTICAL TAIL SIZING coefficient VH. VHT = Area of HT * lever arm of HT Just like the horizontal tail, the verti/ ( wing area * MAC) This equation can be rearranged to cal tail has a volume coefficient. In this give the horizontal tail size needed if case, the MAC in the denominator is replaced by the wing span, because the you specify the tail volume coefficient: Area of HT = VH * wing area * MAC wing span had a pronounced effect on yaw stability. This gives the equation / (lever arm of HT) V^ = Area of VT " lever arm of VT / ( wing area * wing span ) ift Again, this can be rearranged to solve for vertical tail area i&! ICiCi area of VT - Vyj * wing area * wing span / lever arm of VT The range of vertical tail volume coef- itOiuj! ficients in the fleet again represent a large spread. The Piper Cub has only .022, while the MiG 21 has .080. The i^iSi i^ ; ^^ IP .045 for our kinds of airplanes. One consideration for both tail vol- umes is engine power. Power is destabilizing for a tractor propeller configuration, so if you plan to use a big engine, you need bigger tail surfaces. original flight condition. The aspect CNJ- i I range seems to be between .030 and Another consideration is tail aspect ratio. If your tail surfaces are short and stubby, they will have a low lift-curve slope, so lots of angle of attack is needed to get enough CL to restore the icvil ic :*-*-2T ' ! x ^ : ! Bl§|8! glEiS : i-Ki—i iNicci®! !fr!«-i ti ••RIO! !§!®!^i !i^!^!0i :X!^?! >: CO LLiCVj CM n CM m 10

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# Difference between revisions of "2019 AMC 8 Problems/Problem 21" ## Problem 21 What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$? $\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$ ## Solution 1 First we need to find the coordinates where the graphs intersect. $y=5$, and $y=x+1$ intersect at $(4,5)$, $y=5$, and $y=1-x$ intersect at $(-4,5)$, $y=1-x$ and $y=1+x$ intersect at $(0,1)$. Using the Shoelace Theorem we get: $$\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}$$ $=$ So our answer is $\boxed{\textbf{(E)}\ 16}$. ~heeeeeeheeeee ~more edits by BakedPotato66 ## Solution 2 Graphing the lines, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\frac{4\cdot8}{2}$ which is equal to $\boxed{\textbf{(E)}\ 16}$. ~SmileKat32

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# What is “solid angle” and how does it relate to photography? So, I was hanging out in the chat room, and hear mention of something called "Solid Angle". What is this, and how can it be important? - The solid angle is the extension of the concept of angle from two to three dimension. So let's start from 2d: consider a circle and pick two rays starting from the center. They will divide the circumference in two parts, called arcs. The length of each arc divided by the length of the radius will be the measure of the angle subtended by the arc itself. Extend this to three dimensions: instead of a circle take a sphere, and instead of picking two rays pick a cone centered in the center of the sphere.The cone will cross the surface of the sphere: and now to define the solid angle measure the area of the surface delimited by the cone, divided by the square of the length of the radius (so that we have an area divided by an area). The key point is that - since they are ratios - angles (and the solid ones make no exception) are dimensionless quantities: a small object as seen from a short distance can cover the same angle as a large object as seen from a long distance. Why does this matter ? Because we live in 3 spatial dimensions ( :-) ). For instance consider a single light point source radiating (a star seen from very far?) By symmetry there is no reason for it to radiate more in one direction than in the other. So all the photons will be equally spread out in the space. Now you decide to look at how much light arrives in a given region of space: trace a "cone" from the region of space of your interest (the subject of your photo) with the vertex on the star, and you will have "measured" the solid angle. Now the ratio of photons will be equal to the ratio of the solid angle to the total (which is, by the way, 4*pi, similar to 2*pi in two dimensions): if the star is very far, this will be a very small number. Now from stars move to flash units. These are not really point like (neither stars are, after all :) ) and not radiate isotropically (they are usually oriented so that all the light goes somewhere useful) but the same reasoning applies since they are usually much smaller than the subjects we are photographing. This kind of computations underlies the so called inverse square law effect (basically you are spreading a fixed amount of light in a given solid angle: the area of the sphere subtended by the same solid angle grows with the square of the distance from the source, and so if you double the distance the area will be squared). - A solid angle is a fairly abstract concept of geometry, but hopefully easy enough to understand once the concept is grasped. One simple way to think of it is to expand the concept of a normal angle from one dimension (the length of an arc) to two dimensions (the area of a circle). An angle is defined by the arc that "subtends" two rays extending from the center point of a unit circle. The formula for an angle is: θ = s/r (Where `s` is the length of the arc between the two rays, and `r` is the radius the circle) In the same way, a solid angle is defined by the area of a "circle" that subtends two rays extending from the center point of a unit sphere. Where the rays intersect with the surface of the sphere, an arc between the two rays is created at the sphere's surface...your angle. However, that same arc can be drawn at any orientation on the surface of the sphere. Assuming you spun the arc around its center point on the surface of the sphere, you would create a circle on the surface of the sphere. Another way to look at it would be to say the area of a circle on the surface of a sphere created by the projection of a cone created by the same angle from the center of the sphere. The area of that circle is a solid angle. The formula for a solid angle is: Ω = A/r^2 (Where `A` is the area of the circle as subtended by the two rays, and `r` is the radius of the sphere) Given the units of both equations, both angles and solid angles are unitless and independent of the actual size of the unit circle or sphere they are based on. Solid angles have useful application in photography, namely in the area of calculating luminance from a light source and deriving the necessary exposure value to properly expose a scene lit by a given luminance. The standard unit of solid angles is the steradian, a unitless value that represents the solid angle of area `r^2`. The solid angle of a whole sphere is `4π sr`. The favored unit for measurements of illumination when calculating exposure value is lux, and it so happens that one lux is the equivalent of one candela (a measurement of luminous intensity) steradian per meter squared: 1 lux = 1 cd sr/m^2 A lux is a measurement of light of a certain intensity (cd) emitted from a certain geometry (steradians) per specific area (m^2). Solid angles are important to photography as they help bring specific geometry into the equation. This is all well and good when one needs to be highly specific in regards to exposure, such as when performing scientific tests of camera equipment for the purposes of comparing one piece of gear to another. From a practical standpoint, solid angles don't have much real-world application. One generally doesn't spend time running the math when setting up studio lighting...such things are best learned by experimentation, building up a body of experience and understanding from actual use of lighting apparatuses. Only then can all the nuances of illumination, shading, and light in general be understood in a practical sense. For a detailed explanation of exactly how solid angles are important to calculating exposure value given a specific illumination, see my answer to the following question: What is the difference between luminance and illuminance? -

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