Content deleted Content added
Reverted 3 edits by Contribute.Math: I'm afraid you've missed the point here; in fact, it was better briefer...but this really needs a proper source. There are also lots of problems with the presentation (layout of formulas, tone, etc)...but it's possibly at the point of diminishing returns on adding yet another proof to the article (there are many more out there for sure) (TW) |
It took into account the two previous remarks from the users who modified the page. I hope that this time I met their requirements. |
||
Line 26:
Using the [[Weierstrass factorization theorem]], it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials (which Euler assumed as a [[heuristic]] for expanding an infinite degree [[polynomial]] in terms of its roots, but is in general not always true for general <math>P(x)</math>):<ref>A priori, since the left-hand-side is a [[polynomial]] (of infinite degree) we can write it as a product of its roots as
:<math>\begin{align}
\sin(x) & = x (x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2) \cdots \\
Line 298 ⟶ 299:
and this completes the proof.
== A rigorous proof with [[Wallis' integrals]]==
First of all, <math>\displaystyle \int ^{\frac{\pi }{2}}_{0} \operatorname{Arcsin}( \sin\ t) \ dt\ =\int ^{\frac{\pi }{2}}_{0} t\ dt\
=\frac{\pi ^{2}}{8}
</math>
Now let's use the [[List of mathematical series|power series of Arcsin]] :
<math>\begin{align}
& \int ^{\frac{\pi }{2}}_{0} \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \sin( t)^{2n+1} dt\ & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \int ^{\frac{\pi }{2}}_{0} \sin( t)^{2n+1} dt\\
\\
& & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \frac{\left( n!\ 2^{n}\right)^{2}}{( 2n+1) !}\\
& & = & \ \ \sum ^{+\infty }_{n=0}\frac{1}{( 2n+1)^{2}} \\
& & = & \ \ \frac{\pi ^{2}}{8}
\end{align}
</math>
The sum and the integral signs were interchanged thanks to Beppo Levi's [[monotone convergence theorem]] for Lebesgue integral. Then, [[Wallis' integrals]] enabled us to integrate the powers of sine.
Let's separate even and odd numbers :
<math>\displaystyle \sum ^{+\infty }_{n=0}\frac{1}{n^{2}} =\sum ^{+\infty }_{n=1}\frac{1}{( 2n)^{2}} \ +\sum ^{+\infty }_{n=0}\frac{1}{( 2n+1)^{2}} \
=\frac{1}{4}\sum ^{+\infty }_{n=0}\frac{1}{n^{2}} +\displaystyle \ \frac{\pi ^{2}}{8}
</math>
Finally,
<math>\displaystyle \sum ^{+\infty }_{n=0}\frac{1}{n^{2}}
=\frac{4}{3}\frac{\pi ^{2}}{8} \ =\frac{\pi ^{2}}{6}
</math>
==Other identities==
| |||