Fourier Series/Identity Function over Minus Pi to Pi/Proof 1

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For $x \in \openint {-\pi} \pi$:

$\ds x = 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$


From Odd Power is Odd Function, $x$ is an odd function.

By Fourier Series for Odd Function over Symmetric Range, we have:

$\ds x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$


\(\ds b_n\) \(=\) \(\ds \frac 2 \pi \int_0^\pi x \sin n x \rd x\)
\(\ds \) \(=\) \(\ds \frac 2 \pi \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 \pi\) 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.