Fourier Series/Identity Function over Minus Pi to Pi/Proof 1
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Theorem
For $x \in \openint {-\pi} \pi$:
- $\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$
Proof
From Odd Power is Odd Function, $x$ is an odd function.
By Fourier Series for Odd Function over Symmetric Range, we have:
- $\ds x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where:
\(\ds b_n\) | \(=\) | \(\ds \frac 2 \pi \int_0^\pi x \sin n x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \pi \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 \pi\) | Primitive of $x \sin n x$, Fundamental Theorem of Calculus | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 2 \pi \intlimits {\frac{x \cos n x} n} 0 \pi\) | Sine of Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 2 \pi \paren {\frac{\pi \cos n \pi} n - \frac {0 \cos 0} n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {2 \pi \cos n \pi} {\pi n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {2 \paren {-1}^n} n\) | Cosine of Multiple of Pi | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {-1}^{n + 1} } n\) |
Substituting for $b_n$ in $(1)$:
- $\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$
as required.
$\blacksquare$