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

Question

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.

Answer 1

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 $-8752948036761600000$.

Answer 2

This is one of Vieta's formulas.

Answer 3

The coefficients are the Stirling numbers of the first kind :

Answer 4

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}

Answer 5

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:

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

Answer 6

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.