Basel Problem/Proof 9

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Theorem

$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.


Proof

Let $f \left({x}\right)$ be the real function defined on $\left({0 \,.\,.\, 2 \pi}\right)$ as:

$f \left({x}\right) = \begin{cases} \left({x - \pi}\right)^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end{cases}$

From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:

$f \left({x}\right) \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos n x} {n^2} + \left({\frac {\left({-1}\right)^n \pi} n + \frac {2 \left({\left({-1}\right)^n - 1}\right)} {\pi n^3} }\right) \sin n x}\right)$


Setting $x = 0$:

\(\displaystyle f \left({0}\right)\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos 0} {n^2} + \left({\frac {\left({-1}\right)^n \pi} n + \frac {2 \left({\left({-1}\right)^n - 1}\right)} {\pi n^3} }\right) \sin 0}\right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({0 - \pi}\right)^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos 0} {n^2} }\right)\) Sine of Zero is Zero
\(\displaystyle \leadsto \ \ \) \(\displaystyle \pi^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) Cosine of Zero is One
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 2 - \frac {\pi^2} 3\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 6\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)

$\blacksquare$


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