# Finding the coefficient on the $x$ term of ${\prod_{n = 1}^{20}(x-n)}.$

I am trying to find the coefficient on the $x$ term of $\displaystyle{\prod_{n = 1}^{20}(x-n)}$. The issue is that the binomial theorem can't be applied since our $b$ value is changing from term to term. Is there any simple way to do this problem, perhaps a way to change the expression so that the binomial theorem applies? I've tried to do that, and tried looking for a pattern on similar expressions, but I haven't come up with anything. Any help you might have would be appreciated.

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It will be the 19th elementary symmetric polynomial in your roots. See en.m.wikipedia.org/wiki/Elementary_symmetric_polynomial for symmetric polynomials. – Marc Jun 1 '14 at 17:11
A concept popularly known as sum of roots of a polynomial can be of help. – fermesomme Jun 1 '14 at 17:22

The coefficient is given by $$\sum_{k=1}^{20}\prod_{n\not=k\atop n=1,\dots,20}{(-n)}=-\sum_{k=1}^{20}\frac{20!}{k}=-20! H_{20}$$ where $H_k$ is the $k$-th Harmonic number. Those can be looked up in tables (see A001008 and A002805): $H_{20}=\frac{55835135}{15519504}$ and thus the coefficient is given by $-8752948036761600000$.

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This answer just gets better and better on a moment-by-moment basis! Thanks for adding the value of $H_{20}$! – Robert Lewis Jun 1 '14 at 18:30
Just a question, is there a simple way to calculate the constant of the expanded form? – recursive recursion Jun 1 '14 at 19:47
@Dominik, Also, I'm not sure how you got your first expression. Could you please explain that – recursive recursion Jun 1 '14 at 19:51
You get terms with a single power in $x$ from the product $\prod_{n=1}^{20}(x-n)$ by taking $19$ non-$x$ factors and one $x$. The sum is the sum over the different choices of those non-$x$ and $x$ factors. More formally you could try to prove that $(-1)^{N+1} N! H_{N}$ is the $x$-coefficient of $\prod_{n=1}^N (x-n)$ by induction. – Dominik Jun 1 '14 at 20:14

This is one of Vieta's formulas.

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The coefficient must be less than $-20!$, so this can't be correct? – copper.hat Jun 1 '14 at 17:22
Not to put too fine a point on it, but this answer is incorrect; the coefficient of $x^{19}$ is in fact $-\sum_{n = 1}^{20} n = -(20 \times 21)/2 = -210$; for the coefficient of $x$, see the answer given by Dominik. I do believe it is correct. – Robert Lewis Jun 1 '14 at 17:31
@RobertLewis: Unfortunately no 'Med' time for me this morning... – copper.hat Jun 1 '14 at 17:37
@Robert Lewis. Shoot! You're right! I'll remove everything but the vieta's formula link. – Avi Steiner Jun 1 '14 at 17:40
@Avi Steiner: well done! Glad we got it right! – Robert Lewis Jun 1 '14 at 19:40

One way is to just use Maclaurin's formula: $$f(x) = f(0) + \frac{f'(0)}{1!} x + \ldots$$ In this case: \begin{align} f'(x) &= \sum_{1 \le n \le m} \prod_{\substack{1 \le k \le m\\k \ne n}} (x - k) \\ f'(0) &= (-1)^{m - 1} m! H_m \end{align}

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The question has been well answered by others. However, I would like to point out that this polynomial has a name---Wilkinson's polynomial---and its own Wikipedia article. It was put forward by Wilkinson as an example of an apparently innocuous polynomial with remarkable numerical-analytic properties.

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This polynomial is a lot more interesting than I thought! Thanks for giving me a name to google. – recursive recursion Jun 1 '14 at 21:29

A generalization to other coefficients (and limits different from $20$) is given in the generating functions of Stirling numbers of the first kind, Pochammer symbols:

$$(x)_n:=\prod_{k=0}^{n-1}(x-k)=\sum_{k=0}^n s(n,k)x^k.$$

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The coefficients are the Stirling numbers of the first kind :

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