Fourier Series/Identity Function over Symmetric Range
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Theorem
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $\map f x: \openint {-\lambda} \lambda \to \R$ be the identity function on the open real interval $\openint {-\lambda} \lambda$:
- $\forall x \in \openint {-\lambda} \lambda: \map f x = x$
The Fourier series of $f$ over $\openint {-\lambda} \lambda$ can be given as:
- $\map f x \sim \dfrac {2 \lambda} \pi \ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$
Proof
From Identity Function is Odd Function, $\map f x$ is a odd function.
By Fourier Series for Odd Function over Symmetric Range, we have:
- $\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where:
\(\ds b_n\) | \(=\) | \(\ds \frac 2 \lambda \int_0^\pi \map f x \sin \frac {n \pi x} \lambda \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 \lambda \int_0^\pi x \sin \frac {n \pi x} \lambda \rd x\) | Definition of $\map f x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \lambda} {n \pi} \paren {-1}^{n + 1}\) | Half-Range Fourier Sine Series for Identity Function |
Substituting for $b_n$ in $(1)$:
- $\ds x = \dfrac {2 \lambda} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$
as required.
$\blacksquare$