7.5 Functions of Any Period \(P=2L\)
If \(f\) has period \(P = 2L\), the substitution \(v = \frac {\pi x}{L}\) turns it into a function of \(v\) with period \(2\pi \). Applying Euler’s formulas in \(v\) and changing back to \(x\) gives the Fourier series \[f(x) = a_0 + \sum ^{\infty }_{n=1} \left (a_n \cos \frac {n\pi x}{L} + b_n \sin \frac {n\pi x}{L}\right ),\] with coefficients \[a_0 = \frac {1}{2L}\int ^{L}_{-L}f(x)\,dx ,\qquad a_n = \frac {1}{L}\int ^{L}_{-L} f(x) \cos \frac {n\pi x}{L}\,dx ,\qquad b_n = \frac {1}{L}\int ^{L}_{-L} f(x)\sin \frac {n\pi x}{L}\,dx .\] Theorem 7.4.1 holds as before.
Example 7.5.1. Find the Fourier series of the function of period \(4\) given by \[f(x)= \begin {cases} 0, & -2 < x <-1,\\ k, & -1 < x < 1,\\ 0, & 1 < x < 2 . \end {cases}\]
Solution. Here \(P = 2L = 4\), so \(L = 2\). Then \[a_0 = \frac {1}{4}\int ^2_{-2}f(x)\,dx = \frac {1}{4}\int ^1_{-1} k\, dx = \frac {k}{2},\] \[a_n = \frac {1}{2}\int ^1_{-1}k\cos \frac {n\pi x}{2}\,dx = \frac {k}{2}\cdot \frac {4}{n\pi }\sin \frac {n\pi }{2} = \frac {2k}{n\pi }\sin \frac {n\pi }{2} ,\] so \(a_n = 0\) for even \(n\), \(a_n = \frac {2k}{n\pi }\) for \(n = 1, 5, 9, \dots \), and \(a_n = -\frac {2k}{n\pi }\) for \(n = 3, 7, 11, \dots \). Since \(f\) is even, \(b_n = 0\). Hence \[f(x) = \frac {k}{2} + \frac {2k}{\pi }\left ( \cos \frac {\pi x}{2} - \frac {1}{3}\cos \frac {3\pi x}{2} + \frac {1}{5}\cos \frac {5\pi x}{2} - \cdots \right ) .\]
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