Fourier Series/Identity Function over Minus Pi to Pi

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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 1

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.

$\blacksquare$

Proof 2

By Fourier Series for Identity Function over Symmetric Range, the function $f: \openint {-\lambda} \lambda \to \R$ defined as:

$\forall x \in \openint {-\lambda} \lambda: \map f x = x$

has a Fourier series:

$\map f x \sim \dfrac {2 \lambda} \pi \ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$

Substituting for $\lambda = \pi$ gives:

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

as required.

$\blacksquare$