mm-41 === -- I hope you guys are a bit friendlier than the folks at rec.math... 24 people.. 12 males, 12 females. If they reproduce and their children reproduce and their children reproduce, after 100 yrs, what would be the population of their group? They start reproducing when they are 18 and its possible to do so until they are 45 but I really only want them to have about 5 kids at a time..so they would stop repro'ing at about 25 I suppose..I said 45 as its more down to earth. There is enough food and very little desease. The folks live to about 90yrs old. The original 24 are 18yrs old. have to avoid cross breeding. this isn't Tassie :) This isn't a test. I'm just curious about the outcome..No hidden agendas or whatever. hope you can help gra --- Checked by AVG anti-virus system (http://www.grisoft.com). === ---- I came up with this while an undergrad in 1974. I showed it to a couple of profs who suggested I look it up in Riordan's book. I didn't find anything like it there. Has anyone seen the equivalent of it? The identity is (apologies for the notation) n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} Here's how I came up with it. I was churning through Fermat's Last Theorem and examining the differences of x^n. The n! on the left is the nth derivative of x^n. The business on the right comes from constructing the nth difference: (x+n)^n-(x+n-1)^n [(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n] {[(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n]}-{[(x+n-1)^n-(x+n-2)^n]-[(x+n-2)^ n-(x+n-3)^n]} : Proving this (if indeed it holds) amounts to proving that the coefficient for [(x+n-k)^n] is {[(-1)^k] times [n choose k]}. Heuristically that seems obvious but I have never come up with a rigorous proof. Can anyone offer one, or a counterexample? === > The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} The right sides have x in them, while the left sides don't... do you want to extract the coefficient of x^n here? J === No, but one may set x=0 The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} The right sides have x in them, while the left sides don't... do you > want to extract the coefficient of x^n here? > === > I came up with this while an undergrad in 1974. I showed it to a couple > of profs who suggested I look it up in Riordan's book. I didn't find > anything like it there. Has anyone seen the equivalent of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by > [(k!)(n-k)!]} Here's how I came up with it. I was churning through Fermat's Last > Theorem and examining the differences of x^n. The n! on the left is the > nth derivative of x^n. The business on the right comes from constructing > the nth difference: (x+n)^n-(x+n-1)^n > [(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n] > {[(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n]}-{[(x+n-1)^n-(x+n-2)^n]-[(x+n-2)^ n-(x+n-3)^n]} > : > : > : Proving this (if indeed it holds) amounts to proving that the coefficient > for [(x+n-k)^n] is {[(-1)^k] times [n choose k]}. Heuristically that > seems obvious but I have never come up with a rigorous proof. Can anyone > offer one, or a counterexample? Just for grins, I entered this into xmaxima, which I have just started playing around with, and I couldn't get it to perform any simplification. However, I tested it for a few values of n, including n = 125, and the identity held everywhere. It returned almost immediately for n = 25, but when I jumped to n = 125, it had to think for the better part of a minute on my 533 MHz AlphaPC. My guess is that you're absolutely right, but I have never seen that identity. What a great find! Have you tried putting this into any other more powerful symbolic math packages like Maple or Mathematica? Perhaps one of them would recognize the identity and then show you the way to a proof. === Robert, My n_C_k is your [n choose k]. Consider the binomial expansion: (1-x)^n = Sum(k=0 to n) (-x)^k n_C_k = Sum(k=0 to n) (-1)^k x^k n_C_k (1) For n>0 and x=1 this gives 0 = Sum(k=0 to n) (-1)^k n_C_k (2) Take the first derivative of (1): -n (1-x)^(n-1) = Sum(k=0 to n) (-1)^k k x^(k-1) n_C_k For n>1 and x=1 this gives 0 = Sum(k=0 to n) (-1)^k k n_C_k (3) Take the second derivative of (1) n (n-1) (1-x)^(n-2) = Sum(k=0 to n) (-1)^k k (k-1) x^(k-2) n_C_k Note that actually the k=0 and k=1 terms on the right are both zero, but I am formally keeping them. For a good proof you will have to justify doing this, so that when you use k (k-1) = k^2 - k and separate the sum into the sum of two sums, you can then use (3) to kill the sum with the k. For n>2 and x=1, and using (3) 0 = Sum(k=0 to n) (-1)^k k^2 n_C_k The nth derivative will give (-1)^n n! = Sum(k=0 to n) (-1)^k k (k-1)... (k-n+1) x^(k-n) n_C_k So when x=1 and using all the previous identities we have (-1)^n n! = Sum(k=0 to n) (-1)^k k^n n_C_k Now your sum your sum = Sum(k=0 to n) (-1)^k (x+n-k)^n n_C_k can be rewritten using the identity n_C_n-k = n_C_k, and letting k -> n-k, as your sum = (-1)^n Sum(k=0 to n) (-1)^k (x+k)^n n_C_k (4) Use the binomial expansion as (x+k)^n = Sum(j=0 to n) k^j x^(n-j) n_C_j and put that into (4) and rearrange to get your sum = (-1)^n Sum(j=0 to n) x^(n-j) n_C_j Sum(k=0 to n) (-1)^k k^j n_C_k In that sum on k my identities show that the sum is zero for j = 0, 1, 2, ..., (n-1) and is (-1)^n n! when j=n, so your sum = (-1)^n x^0 n_C_n (-1)^n n! = n! as you surmised. Bob > I came up with this while an undergrad in 1974. I showed it to a couple of > profs who suggested I look it up in Riordan's book. I didn't find anything > like it there. Has anyone seen the equivalent of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} Here's how I came up with it. I was churning through Fermat's Last Theorem > and examining the differences of x^n. > The n! on the left is the nth derivative of x^n. > The business on the right comes from constructing the nth difference: > (x+n)^n-(x+n-1)^n > [(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n] {[(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n]}-{[(x+n-1)^n-(x+n-2)^n]-[(x+n-2)^ n - > (x+n-3)^n]} > : > : > : Proving this (if indeed it holds) amounts to proving that the coefficient for > [(x+n-k)^n] is {[(-1)^k] times [n choose k]}. Heuristically that seems > obvious but I have never come up with a rigorous proof. Can anyone offer one, or a counterexample? > === 2, ..., (n-1) and is (-1)^n n! when j=n, so I know there is a great combinatorial argument for this, but it escapes me at the moment. I had just come around to basically this same proof when I noticed yours. Nicely done. Basically, sum for i = 0 to n of (-1)^(n-i) * C(n,i) * i^j is 0 when j is less than n, and nonzero for j >= n -- where I'm using C(n,i) to mean n choose i or (n!/(i!(n-i)!)). Now that I think about it, I think that it's something like j * (j-1) * (j-2) * .... (j-n+1), which is 0 when j is less than n, and greater than zero when j is greater than or equal to n. Now the trick is to remember the combinatorial argument. So is this a Bob/Robert discussion or what? === I came up with this while an undergrad in 1974. I showed it to a > couple of profs who suggested I look it up in Riordan's book. I > didn't find anything like it there. Has anyone seen the equivalent > of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} You can simplify things by replacing (x+n) by x -- since the result should not depend on x this should not change the answer. In this form it is Sum[k = 0 to n]((-1)^k * C(n,k) * (x-k)^n), which you can find as equation (19) of: http://mathworld.wolfram.com/BinomialSums.html (Note that they ascribe the method of solution to exactly the sort of process you described.) Geoff. - - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. - - === Messrs Kimble, Delaney, & Bailey, I pore over Mr. Delaney's equations, but I will do so. Mr. Bailey is indeed a wonder to come up with a web site with the particular equation; I do not think I will play poker with him anytime soon. Ruiz 1996! Flattering. I was unaware of mathworld.wolfram.com, xmaxima, and Maple. Looking forward to exploring those. I am getting back into math following a 25 year career in programming (mainly APL). Robert Pullman > I came up with this while an undergrad in 1974. I showed it to a > couple of profs who suggested I look it up in Riordan's book. I > didn't find anything like it there. Has anyone seen the equivalent > of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} You can simplify things by replacing (x+n) by x -- since the result > should not depend on x this should not change the answer. In this form > it is Sum[k = 0 to n]((-1)^k * C(n,k) * (x-k)^n), which you can find as > equation (19) of: http://mathworld.wolfram.com/BinomialSums.html (Note that they ascribe the method of solution to exactly the sort of > process you described.) Geoff. ---- --- > Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- > ftww@maths.usyd.edu.au | Gameplayer by vocation. > ---- --- > === I need an algoritm to solve the following problem by programming in Visual Basic. We play a competition with this basic assumptions: - In that competition will play N teams (e.g. 14). - There are S sessions ( e.g. 6). - A session counts R rounds. After each round the teams get another opponent. - A team will meet as many as other teams as there are rounds. this are mostly 6 teams. - When the number of teams for a session is odd, during each round there is one team that don't play and thus don't meet another team. - Which teams will meet each other depends on the place of the teams in the play-scheme. For each number of teams that will participate there is a scheme. - It is possible that one or more teams don't participate in a session. The number of absent teams is rarely 4 or more. - We count the frequency of the meetings between the teams. - Before the start of a session we want the present teams place in the scheme so that the frequenty of the meetings between the teams will optimize. For further explanation an example: After 2 sessions the matrix of the meetings between the teams A to L of a competition with 12 teams may be: A B C D E F G H I J K L A - 2 0 2 1 2 2 0 1 0 1 1 B 2 1 0 1 2 0 1 0 2 1 1 1 C 0 0 0 1 1 1 0 0 1 1 0 1 (This team has played only one session) D 2 1 1 1 1 1 0 0 1 2 1 1 E 1 2 1 1 0 2 1 0 0 1 1 2 F 2 0 1 1 2 1 1 0 1 2 1 0 G 2 1 0 0 1 1 1 0 1 2 2 1 H 0 0 0 0 0 0 0 - 0 0 0 0 (This team didn't play the first 2 sessions) I 1 2 1 1 0 1 1 0 0 1 2 2 J 0 1 1 2 1 2 2 0 1 0 1 1 K 1 1 0 1 1 1 2 0 2 1 1 1 L 1 1 1 1 2 0 1 0 2 1 1 1 When a team cannot play a round, because the number of teams for a session is odd, we see in the matrix a meeting with the team itself. See for example the 1 in the element (B, B). Suppose that all 12 teams will be present during session 3. The play-scheme for 12 teams is: Team number meet the teams 1 2, 4, 5, 7, 11, 12 2 1, 3, 5, 7, 9, 11 3 2, 4, 6, 8, 10, 12 4 1, 3, 5, 7, 9,10 5 1, 2, 4, 6, 8, 12 6 3, 5, 7, 9, 10, 11 7 1, 2, 4, 6, 8, 12 8 3, 5, 7, 9, 10, 11 9 2, 4, 6, 8, 10, 12 10 3, 4, 6, 8, 9, 11 11 1, 2, 6, 8, 10, 12 12 1, 3, 5, 7, 9, 11 Which teamnumbers must we attribute to the teams A to L to optimize the meetings between the teams? An attribution is for example: A gets number 3/ B to 5/ C to 1/ D to 12/ .... etc. Has anyone a solution for this optimize-problem? May be based on experience with this kind of competitions. I am very interested. Jan van der Meulen === > I need an algoritm to solve the following problem by programming in Visual > Basic. > Similar school timetabling problems are now solved using genetic algorithms/simulated annealing http://www-2.cs.cmu.edu/Groups/AI/html/faqs/ai/genetic/part3/faq-doc-1.html gtoomey Free ASX end of day data www.float.com.au/data === I am trying to find out whether the function below has ben recognoized, described, its properies listed, etc. The function has one parameter (nu) and two independent variables -- z and tau. I denote it by k_nu(z,tau). Its integral representation is k_nu(z,tau) = int_{0}^{tau} t^{nu-1} exp(-t-z^2/(4t)} dt { integral between t=0 and t=tau of t to the power of (nu-1) times the exponent of [-t - z^2/(4t)] }. This is, of course, completely analogous to the modified Bessel function, K_nu (z), whose integral representation is the above integral with tau -> infinity. Thus, if tau It has been many years since I solved such problems so I'm pretty > rusty. A vessel in the shape of an inverted cone with its axis vertical is > full of water. It is being emptied through a hole at the vertex at a > rate proportional to the square root of the dept at any instant. > Initially the water is 10 meters deep and falls to 9 meters in the > first minute. Find out how long it takes for the vessel to empty. Could you please show all the steps to the solution. > you this problem? === Could anyone help, any functions (except solver) and formular in excel could find the a , b & c thanks a lot! 1.258333 a + 1.516667 b + 1.688889 c = 21200 1.775a + b + c = 21200 a + 2.55b + c = 21200 a + b + 3.066667c = 21200 === >Could anyone help, any functions (except solver) and formular in excel could >find the a , b & c thanks a lot! 1.258333 a + 1.516667 b + 1.688889 c = 21200 >1.775a + b + c = 21200 >a + 2.55b + c = 21200 >a + b + 3.066667c = 21200 Can't be done without more specification on your part. You have 3 variables, a, b and c, but 4 euqations, so your system of equations is overspecified. If any 3 of these are linearly independent, you could use them to determine a, b and c. For instance, put the coefficients for a, b and c from the 2nd through 4th equations into a square matrix, i.e., a 3-by-3 range of cells like A1:C3. 1.775000 1.000000 1.000000 1.000000 2.550000 1.000000 1.000000 1.000000 3.066667 and put 21200 in a 3-row, 1-column range like D1:D3. Then enter the following array formula in a 3-row, 1-column range like E1:E3. =MMULT(MINVERSE(A1:C3),D1:D3) -- 1. Don't attach files to postings in this newsgroup. 2. Snip unnecessary text from quoted text. Indiscriminate quoting is wasteful. 3. Excel 97 & later provides 65,536 rows & 256 columns per worksheet. There are no add-ins or patches that increase them. Need more? Use something else. === Harlan, Harlan Grove find the a , b & c thanks a lot! 1.258333 a + 1.516667 b + 1.688889 c = 21200 >1.775a + b + c = 21200 >a + 2.55b + c = 21200 >a + b + 3.066667c = 21200 Can't be done without more specification on your part. You have 3 variables, a, > b and c, but 4 euqations, so your system of equations is overspecified. If any 3 > of these are linearly independent, you could use them to determine a, b and c. > For instance, put the coefficients for a, b and c from the 2nd through 4th > equations into a square matrix, i.e., a 3-by-3 range of cells like A1:C3. 1.775000 1.000000 1.000000 > 1.000000 2.550000 1.000000 > 1.000000 1.000000 3.066667 and put 21200 in a 3-row, 1-column range like D1:D3. Then enter the following > array formula in a 3-row, 1-column range like E1:E3. =MMULT(MINVERSE(A1:C3),D1:D3) -- > 1. Don't attach files to postings in this newsgroup.