Fourier Series/Sawtooth Wave/Special Cases/Half Interval Pi

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Special Case of Fourier Series for Sawtooth Wave

Let $\map S x$ be the sawtooth wave defined on the real numbers $\R$ as:

$\forall x \in \R: \map S x = \begin {cases} x & : x \in \openint {-\pi} \pi \\ \map S {x + 2 \pi} & : x < -\pi \\ \map S {x - 2 \pi} & : x > +\pi \end {cases}$


Then its Fourier series can be expressed as:

\(\ds \map S x\) \(\sim\) \(\ds 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x\)
\(\ds \) \(=\) \(\ds 2 \paren {\sin x - \frac {\sin 2 x} 2 + \frac {\sin 3 x} 3 + \dotsb}\)


Proof

From Fourier Series for Sawtooth Wave, the sawtooth wave defined on the real numbers $\R$ as:

$\forall x \in \R: \map S x = \begin {cases} x & : x \in \openint {-l} l \\ \map S {x + 2 l} & : x < -1 \\ \map S {x - 2 l} & : x > +1 \end {cases}$


has a Fourier series which can be expressed as:

\(\ds \map S x\) \(\sim\) \(\ds \frac {2 l} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \dfrac {n \pi x} l\)
\(\ds \) \(=\) \(\ds \frac {2 l} \pi \paren {\sin \dfrac {\pi x} l - \frac 1 2 \sin \dfrac {2 \pi x} l + \frac 1 3 \sin \dfrac {3 \pi x} l + \dotsb}\)

The result follows by setting $l = \pi$.

$\blacksquare$


Sources

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