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So I have a factored polynomial of the form $(z-a)(z-b)(z-c)\ldots(z-n)$ for $n$ an even positive integer. Thus the coefficient of $z^k$ for $0 \le k < n$ will be the sum of all distinct $n-k$ element products taken from the set $\{a,b,\ldots,n\}$ multiplied by $(-1)^k$, I hope that makes sense, please ask if you need more clarification.

I'm trying to put these coefficients into a row vector with the first column containing the constant coefficient (which would be $abc\ldots n$) and the last column containing the coefficient for $z^n$ (which would be 1).

I imagine there is a way to brute force this with a ton of nested loops, but I'm hoping there is a more efficient way. This is being done in Matlab (which I'm not that familiar with) and I know Matlab has a ton of algorithms and functions, so maybe its got something I can use. Can anyone think of a way to do this?

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You could use the DocPolynom class: mathworks.co.uk/help/techdoc/matlab_oop/f3-28024.html –  Jonathan May 6 '12 at 23:30
This comment needs revision. Let the coefficient vector be $[c_1, c_2, \ldots, c_n],$ with $c_1 = 1$ and $c_n = ab\cdots n.$ Then $ c_k = (-1)^k p_k,$ where $p_k$ is a sum of $\binom{n}{k}$ terms; each term is a product of $k$ different elements from $\{a,b, \dots, n \}.$ –  user2468 May 6 '12 at 23:37
Ok thanks J.D. you're right, I edited it so now I mention the alternating sign. –  Blood Pudding May 6 '12 at 23:42
possible duplicate of Algorithm(s) for computing an elementary symmetric polynomial –  J. M. May 7 '12 at 1:39

1 Answer 1

up vote 1 down vote accepted

There is a formula involving the 'elementary symmetric polynomials' for the coefficients of the polynomial. See here. If you have the mupad extension for matlab, use polylib::elemSym.
Mupad used to be free until Matlab bought it, so you have a decent chance to have it installed with your Matlab (what's the point of buying a package like Mupad if you don't give it to people afterwards). So it's different from say the Maple integration in Matlab (that probably requires an additional Maple license).

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Ok I think I do have Mupad, I'm going to see if I can get it to work for me, thanks. –  Blood Pudding May 6 '12 at 23:51

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