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I have recently been driving myself mad solving equations using falling factorials. A typical example would be \(\displaystyle n^5 = A n(n - 1)(n - 2)(n - 3)(n - 4) + B n(n - 1)(n - 2)(n - 3) + C n(n - 1)(n - 2) + D n(n - 1) + E n\) and finding A, B, C, D, and E. We can write this a bit more neatly as \(\displaystyle n^5 = A \dfrac{n!}{(n - 5)!} + B \dfrac{n!}{(n - 4)!} + C \dfrac{n!}{(n - 3)!} + D \dfrac{n!}{(n - 2)!} + E \dfrac{n!}{(n - 1)!}\) which saves some typing space for things like my graphing calculator or W|A. I have found a way to proceed but it gets quickly out of hand with larger n, which is the information I need to use to find a general formula for the coefficients. Now, since I know how to do them and have developed some expertise at it I have been working on how to get a computer to do the job for me. The best I've got is to plug it into Mathematica and have it simplify as much as I can. They I can choose a value for A. Then simplify. Choose a value for B. etc. This works and gets me a solution to the equation fairly easily but I am not quite satisfied with it. Is it possible to either use Linear Algebra (or something) to solve for the coefficients (or perhaps a matrix that can be solved) or is there a way to get Mathematica to do it for me? I don't know how to do either. Thanks! -Dan
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The problem was to try to isolate one of the unknowns and solve for it. But I've seen the closed form expression for both of the Stirling numbers... that's apparently not going to happen here. I've tried to come up with a vector or matrix approach but it doesn't seem to work. No problem. Thanks for the help! -Dan
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I thought about this problem more and I doubted if my approach was useful. Doing the example given made it more concrete. Following the idea to set up a system of equations to get a matrix, we get the following. \(\displaystyle \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 0& 1\\ -10 & 1 & 0 & 0 & 0 & 0 \\ 35 & -6 & 1 & 0 & 0 & 0 \\ -50 & 11 & -3 & 1 & 0 & 0 \\ 24 & -6 & 2 & -1 & 1 & 0 \end{array} \) Row reducing, we get \(\displaystyle \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & 0 & 10 \\ 0 & 0 & 1 & 0 & 0 & 25 \\ 0 & 0 & 0 & 1 & 0 & 15 \\ 0 & 0 & 0 & 0 & 1 & 1 \end{array} \) So A=1, B=10, C=25, D=15, E=1 So as long as you have a program to do RREF for you and populate the initial matrix with signed Stirling numbers of the first kind, you can solve directly for your variables.
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Jomo said: A Physicist who doesn't look at vectors? What's next, a mathematician who doesn't look at numbers (then again, I had a math professor who said that numbers get in the way of real mathematics) Click to expand... I was referring to a vector space with a basis \(\displaystyle \{1 , x^1 , x^2, x^3, \text{ ...} \}\). It might seem silly since I've done Hermite polynomials, Laguerre polynomials, etc. but when I see something like \(\displaystyle \mathbb{R} [x]\) I don't think, hey \(\displaystyle a x^2\) is a vector. I should think it, because it is and I know it. But I don't. -Dan
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I was going to say A(1, -10, 35, -50, 24) + B(0,1,-6,11,-6) + C(0,0,1,-3,2) + D(0,0,0,1,-1) + E(0,0,0,0,1) = (1,0,0,0,0) but MacstersUndead beat me to it (by hours). And I am not a Physicist!
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A Physicist who doesn't look at vectors? What's next, a mathematician who doesn't look at numbers (then again, I had a math professor who said that numbers get in the way of real mathematics)
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MacstersUndead said: I found out that falling factorial coefficients are related to signed Stirling numbers of the first kind, so you can generate those numbers and set up a system of equations which can be represented by a matrix. In your example, we have \(\displaystyle A(n^5 - 10n^4 + 35n^3 - 50n^2 + 24n) + B(n^4 - 6n^3 + 11n^2 - 6n) + C (n^3 - 3n^2 + 2) + D(n^2 - n) + En = n^5\) We can represent this as a linear equation (keeping only the coefficients that are signed Stirling numbers), with a table of signed values similar to Stirling numbers of the first kind - Wikipedia In your example, we have (imagine the vectors as columns). A(1, -10, 35, -50, 24) + B(0,1,-6,11,-6) + C(0,0,1,-3,2) + D(0,0,0,1,-1) + E(0,0,0,0,1) = (1,0,0,0,0) Click to expand... Well, it looks like brute force is out. I had suspected a relationship involving the Stirling numbers but I hadn't had time to investigate very far. Even if I ignored the derivation the explicit formula is going to be too complicated to work with well. The vector idea is a good one! I hadn't thought to look at it like that, though I've seen it used before. (This is what happens to a Physicist!) Thanks for the help! -Dan
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I found out that falling factorial coefficients are related to signed Stirling numbers of the first kind, so you can generate those numbers and set up a system of equations which can be represented by a matrix. In your example, we have \(\displaystyle A(n^5 - 10n^4 + 35n^3 - 50n^2 + 24n) + B(n^4 - 6n^3 + 11n^2 - 6n) + C (n^3 - 3n^2 + 2) + D(n^2 - n) + En = n^5\) We can represent this as a linear equation (keeping only the coefficients that are signed Stirling numbers), with a table of signed values similar to Stirling numbers of the first kind - Wikipedia In your example, we have (imagine the vectors as columns). A(1, -10, 35, -50, 24) + B(0,1,-6,11,-6) + C(0,0,1,-3,2) + D(0,0,0,1,-1) + E(0,0,0,0,1) = (1,0,0,0,0)
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