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

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Special Case of Fourier Series for Square Wave

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

$\forall x \in \R: \map S x = \begin {cases}

1 & : x \in \openint 0 \pi \\ -1 & : x \in \openint {-\pi} 0 \\ \map S {x + 2 \pi} & : x < -\pi \\ \map S {x - 2 \pi} & : x > +\pi \end {cases}$


Then its half-range Fourier sine series can be expressed as:

\(\ds \map S x\) \(\sim\) \(\ds \frac 4 \pi \sum_{r \mathop = 0}^\infty \frac 1 {2 r + 1} {\sin x}\)
\(\ds \) \(=\) \(\ds \frac 4 \pi \paren {\sin x + \dfrac {\sin 3 x} 3 + \dfrac {\sin 5 x} 5 + \dotsb}\)


Proof

From Fourier Series for Square Wave, the real function $\map f x$ defined on the open interval $\openint {-l} l$ as:

$\map f x = \begin {cases} -1 & : x \in \openint {-l} 0 \\ 1 & : x \in \openint 0 l \end {cases}$


has a half-range Fourier sine series which can be expressed as:

\(\ds \map f x\) \(\sim\) \(\ds \frac 4 \pi \sum_{r \mathop = 0}^\infty \frac 1 {2 r + 1} {\sin \frac {\pi x} l}\)
\(\ds \) \(=\) \(\ds \frac 4 \pi \paren {\sin \frac {\pi x} l + \dfrac 1 3 \sin \frac {3 \pi x} l + \dfrac 1 5 \sin \frac {5 \pi x} l + \dotsb}\)

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

$\blacksquare$


Sources