Series Expansion for Pi over 8 Root 2
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Theorem
- $\ds \frac \pi {8 \sqrt 2} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {2 n - 1} {\paren {4 n - 1} \paren {4 n - 3} }$
Proof
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:
- $\map f x = \begin {cases}
\cos x & : -\pi < x < 0 \\ -\cos x & : 0 < x < \pi \end {cases}$
From Fourier Series: $\cos x$ over $\openint {-\pi} 0$, $-\cos x$ over $\openint 0 \pi$, we have:
- $\ds \map f x \sim -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$
Setting $x = \dfrac \pi 4$, we have:
\(\ds -\cos \dfrac \pi 4\) | \(=\) | \(\ds -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r \dfrac \pi 4} {4 r^2 - 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cos \dfrac \pi 4\) | \(=\) | \(\ds \frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\sqrt 2} 2\) | \(=\) | \(\ds \frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\) | Cosine of $\dfrac \pi 4$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\pi} {8 \sqrt 2}\) | \(=\) | \(\ds \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\) | rearranging |
When $r$ is even, $\dfrac {r \pi} 2$ is an integer multiple of $\pi$.
Hence, in this case, from Sine of Multiple of Pi:
- $\sin \dfrac {r \pi} 2 = 0$
When $r$ is odd it can be expressed as $r = 2 n - 1$ for $n \ge 1$.
Hence we have:
\(\ds \frac {\pi} {8 \sqrt 2}\) | \(=\) | \(\ds \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {\paren {2 r + 1} \paren {2 r - 1} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \sin \dfrac {\paren {2 n - 1} \pi} 2} {\paren {2 \paren {2 n - 1} + 1} \paren {2 \paren {2 n - 1} - 1} }\) | from above: terms in even $r$ vanish | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \sin \paren {\paren {n - 1} + \frac 1 2} \pi} {\paren {4 n - 1} \paren {4 n - 3} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \paren {-1}^{n - 1} } {\paren {4 n - 1} \paren {4 n - 3} }\) | Sine of Half-Integer Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {2 n - 1} {\paren {4 n - 1} \paren {4 n - 3} }\) | rearranging |
$\blacksquare$
Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $3$.