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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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    It would be kind to provide links to [Basel problem](http://en.wikipedia.org/wiki/Basel_problem#Euler_attacks_the_problem) and [Newton's identities](http://en.wikipedia.org/wiki/Newton%27s_identities)2011-12-23
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    "Newton's identities" look like a red herring in this argument. (Sometimes people try to compress an idea in a mathematical argument with a named result, but such hints can be more puzzling than the direct proof.) I would change the wiki article to read: "if we formally multiply out this product and collect all the $x^2$ terms, we see that the $x^2$ coefficient of ..." Play a few finite products to convince yourself.2011-12-23
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    @Jerry: a few more details, and I think that could be an answer... :)2011-12-23
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    Still not understanding. I done some products, but have nothing to do with the final series2011-12-24

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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 $\frac{\sin z}{z}$.