Series Expansion for Pi over Root 2/Proof 1
Jump to navigation
Jump to search
Theorem
- $\ds \frac \pi {\sqrt 2} = \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }$
Proof
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
- $\map f x = \begin{cases} \sin \dfrac x 2 & : 0 \le x < \pi \\ -\sin \dfrac x 2 & : \pi < x \le 2 \pi \end{cases}$
From Half-Range Fourier Sine Series: $\sin \dfrac x 2$ over $\closedint 0 \pi$, $-\sin \dfrac x 2$ over $\closedint 0 {2 \pi}$, we have:
- $\map f x \sim \ds \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin n x} {4 n^2 - 1}$
Setting $x = \dfrac {\pi} 2$, we have:
\(\ds \sin \dfrac \pi 4\) | \(=\) | \(\ds \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\sqrt 2} 2\) | \(=\) | \(\ds \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\) | Sine of $\dfrac \pi 4$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac \pi {\sqrt 2}\) | \(=\) | \(\ds 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\) | rearranging |
When $n$ is even, $\dfrac {n \pi} 2$ is an integer multiple of $\pi$.
Hence, in this case, from Sine of Multiple of Pi:
- $\sin \dfrac {n \pi} 2 = 0$
When $n$ is odd it can be expressed as $n = 2 r - 1$ for $r \ge 1$.
Hence we have:
\(\ds \frac \pi {\sqrt 2}\) | \(=\) | \(\ds 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {\paren {2 n + 1} \paren {2 n - 1} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{\paren {2 r - 1} - 1} \frac {\paren {2 r - 1} \sin \dfrac {\paren {2 r - 1} \pi} 2 } {\paren {2 \paren {2 r - 1} + 1} \paren {2 \paren {2 r - 1} - 1} }\) | from above: terms in even $n$ vanish | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r} \frac {\paren {2 r - 1} \map \sin {\paren {r - 1} + \frac 1 2} \pi} {\paren {4 r - 1} \paren {4 r - 3} }\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 \sum_{r \mathop = 1}^\infty \frac {\paren {2 r - 1} \map \sin {\paren {r - 1} + \frac 1 2} \pi} {\paren {4 r - 1} \paren {4 r - 3} }\) | $\paren {-1}^{2 r} = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 \sum_{r \mathop = 1}^\infty \frac {\paren {2 r - 1} \paren {-1}^{r - 1} } {\paren {4 r - 1} \paren {4 r - 3} }\) | Sine of Half-Integer Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {2 \paren {r - \frac 1 2} } {16 r^2 - 16 r + 3}\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }\) | simplifying |
$\blacksquare$
Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $2$.