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I do not understand this line of Wikipedia's page on the Basel problem:

If we formally multiply out this product and collect all the $x^2$ terms (we are allowed to do so because of Newton's identities), we see that the $x^2$ coefficient of $\sin(x)/x$ is...

Which Newton identity? How?

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    Still not understanding. I done some products, but have nothing to do with the final series2011-12-24

1 Answers 1

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As Jerry says, it appears someone was trying to be pedantic.

What is meant is that for a polynomial $f(z)=z^n+a_{n-1}z^{n-1}+\dots+a_0$ with roots $\alpha_1, \dots, \alpha_n$, we have:

$a_{n-1} = -\sum_{1 \leq i \leq n}\alpha_i$ $a_{n-2} = \sum_{1 \leq i_i < i_2 \leq n} \alpha_{i_1}\alpha_{i_2}$ $\dots$ $a_0 = (-1)^n\alpha_1 \dots \alpha_n.$

You can see this by expanding $(z-\alpha_1)\dots (z-\alpha_n)$ and comparing with the coefficients of $f(z)$.

To solve the Basel problem, Euler did the same thing with the product expansion of $\frac{\sin z}{z}$, as if it were an infinite degree polynomial, and comparing with the Taylor expansion of $\frac{\sin z}{z}$.