Basel Problem/Proof 9
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Theorem
- $\ds \map \zeta 2 = \sum_{n \mathop = 1}^\infty {\frac 1 {n^2} } = \frac {\pi^2} 6$
where $\zeta$ denotes the Riemann zeta function.
Proof
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
- $\map f x = \begin {cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$
From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:
- $\map f x \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$
Setting $x = 0$:
\(\ds \map f 0\) | \(=\) | \(\ds \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin 0}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {0 - \pi}^2\) | \(=\) | \(\ds \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} }\) | Sine of Zero is Zero | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \pi^2\) | \(=\) | \(\ds \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) | Cosine of Zero is One | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\pi^2} 2 - \frac {\pi^2} 3\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\pi^2} 6\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) |
$\blacksquare$
Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Chapter One: $\S 2$. Fourier Series: Example $1$