Editing Basel problem (section)

===Generalizations of Euler's method using elementary symmetric polynomials===

Using formulae obtained from [[elementary symmetric polynomial]]s,<ref>Cf., the formulae for generalized Stirling numbers proved in: {{citation|last1=Schmidt|first1=M. D.|title=Combinatorial Identities for Generalized Stirling Numbers Expanding f-Factorial Functions and the f-Harmonic Numbers|journal=J. Integer Seq.|date=2018|volume=21|issue=Article 18.2.7|url=https://cs.uwaterloo.ca/journals/JIS/VOL21/Schmidt/schmidt18.html}}</ref> this same approach can be used to enumerate formulae for the even-indexed [[zeta constants|even zeta constants]] which have the following known formula expanded by the [[Bernoulli numbers]]:
<math display=block>\zeta(2n) = \frac{(-1)^{n-1} (2\pi)^{2n}}{2 \cdot (2n)!} B_{2n}. </math>

For example, let the partial product for <math>\sin(x)</math> expanded as above be defined by <math>\frac{S_n(x)}{x} := \prod\limits_{k=1}^n \left(1 - \frac{x^2}{k^2 \cdot \pi^2}\right)</math>. Then using known [[Newton's identities#Expressing elementary symmetric polynomials in terms of power sums|formulas for elementary symmetric polynomial]]s (a.k.a., Newton's formulas expanded in terms of [[power sum]] identities), we can see (for example) that
<math display=block>
\begin{align}
\left[x^4\right] \frac{S_n(x)}{x} & = \frac{1}{2\pi^4}\left(\left(H_n^{(2)}\right)^2 - H_n^{(4)}\right) \qquad \xrightarrow{n \rightarrow \infty} \qquad \frac{1}{2\pi^4}\left(\zeta(2)^2-\zeta(4)\right) \\[4pt]
& \qquad \implies \zeta(4) = \frac{\pi^4}{90} = -2\pi^4 \cdot [x^4] \frac{\sin(x)}{x} +\frac{\pi^4}{36} \\[8pt]
\left[x^6\right] \frac{S_n(x)}{x} & = -\frac{1}{6\pi^6}\left(\left(H_n^{(2)}\right)^3 - 2H_n^{(2)} H_n^{(4)} + 2H_n^{(6)}\right) \qquad \xrightarrow{n \rightarrow \infty} \qquad \frac{1}{6\pi^6}\left(\zeta(2)^3-3\zeta(2)\zeta(4) + 2\zeta(6)\right) \\[4pt]
& \qquad \implies \zeta(6) = \frac{\pi^6}{945} = -3 \cdot \pi^6 [x^6] \frac{\sin(x)}{x} - \frac{2}{3} \frac{\pi^2}{6} \frac{\pi^4}{90} + \frac{\pi^6}{216},
\end{align}
</math>

and so on for subsequent coefficients of <math>[x^{2k}] \frac{S_n(x)}{x}</math>. There are [[Newton's identities#Expressing power sums in terms of elementary symmetric polynomials|other forms of Newton's identities]] expressing the (finite) power sums <math>H_n^{(2k)}</math> in terms of the [[elementary symmetric polynomial]]s, <math>e_i \equiv e_i\left(-\frac{\pi^2}{1^2}, -\frac{\pi^2}{2^2}, -\frac{\pi^2}{3^2}, -\frac{\pi^2}{4^2}, \ldots\right), </math> but we can go a more direct route to expressing non-recursive formulas for <math>\zeta(2k)</math> using the method of [[elementary symmetric polynomial]]s. Namely, we have a recurrence relation between the elementary symmetric polynomials and the [[Power sum symmetric polynomial|power sum polynomials]] given as on [[Newton's identities#Comparing coefficients in series|this page]] by
<math display=block>(-1)^{k}k e_k(x_1,\ldots,x_n) = \sum_{j=1}^k (-1)^{k-j-1} p_j(x_1,\ldots,x_n)e_{k-j}(x_1,\ldots,x_n),</math>

which in our situation equates to the limiting recurrence relation (or [[generating function]] convolution, or [[Cauchy product|product]]) expanded as
<math display=block> \frac{\pi^{2k}}{2}\cdot \frac{(2k) \cdot (-1)^k}{(2k+1)!} = -[x^{2k}] \frac{\sin(\pi x)}{\pi x} \times \sum_{i \geq 1} \zeta(2i) x^i. </math>

Then by differentiation and rearrangement of the terms in the previous equation, we obtain that
<math display=block>\zeta(2k) = [x^{2k}]\frac{1}{2}\left(1-\pi x\cot(\pi x)\right). </math>