Fourier Series/Identity Function over Minus Pi to Pi

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Proof 1

From Odd Power is Odd Function, $x$ is a Odd Function.

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

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

where:

\(\displaystyle b_n\) \(=\) \(\displaystyle \frac 2 \pi \int_0^\pi x \sin n x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \pi \intlimits {\frac {\sin n x} {n^2} - \frac {x \cos n x} n} 0 \pi\) Primitive of $x \sin n x$, Fundamental Theorem of Calculus
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 \pi \intlimits {\frac{x \cos n x} n} 0 \pi\) Sine of Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle -\frac 2 \pi \paren {\frac{\pi \cos n \pi} n - \frac {0 \cos 0} n}\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac {2 \pi \cos n \pi} {\pi n}\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac {2 \paren {-1}^n} n\) Cosine of Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \paren {-1}^{n + 1} } n\)


Substituting for $b_n$ in $(1)$:

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

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:

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


Substituting for $\lambda = \pi$ gives:

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

as required.

$\blacksquare$