7.5 Functions of Any Period \(P=2L\)
If a function \(f(x)\) of period \(P = 2L\) has a Fourier series, we claim that this series is \[f(x) = a_0 + \sum ^{\infty }_{n=1} \Big [a_n \cos \frac {n\pi }{L}x + b_n \sin \frac {n\pi }{L}x\Big ]\]
with Fourier coefficients \[a_0 = \frac {1}{2L}\int ^{L}_{-L}f(x)dx\hspace {0.2cm}, \hspace {0.5cm} a_n = \frac {1}{L}\int ^{L}_{-L} f(x) \cos \frac {n\pi }{L} x dx\hspace {0.3cm}, \hspace {0.5cm} b_n = \frac {1}{L}\int ^{L}_{-L} f(x)\sin \frac {n\pi }{L} x dx\]
Find the Fourier series of \(f(x)= \begin {cases} 0, & \text {if}\hspace {0.2cm} -2 < x <-1\\ k, & \text {if}\hspace {0.2cm} -1 < x < 1\\ 0, & \text {if}\hspace {0.2cm} 1 < x < 2\\ \end {cases} \)
Solution.
Here \(P = 2L = 4 \implies L = 2\)
\[a_0 = \frac {1}{2(2)} \int ^2_{-2}f(x)dx = \frac {1}{4}\int ^1_{-1} k dx = \frac {k}{2}\]
\begin {align*} a_n & = \frac {1}{2}\int ^2_{-2}f(x)\cos \frac {n\pi }{2}xdx\\\\ & = \frac {-k}{2}\int ^1_{-1}\cos \frac {n\pi }{2}xdx = \frac {2k}{n\pi }\sin \Big ( \frac {n\pi }{2}\Big )\\\\ \implies \hspace {0.5cm} a_n & = \begin {cases} 0, & \text {if} \hspace {0.3cm} n \hspace {0.3cm} \text {is even}\\\\ \dfrac {2k}{n\pi }, & \text {if}\hspace {0.3cm} n = 1,5, 9, 13,\cdots \cdots \\\\ \dfrac {-2k}{n\pi } & \text {if}\hspace {0.3cm} n = 3,7, 11,15,\cdots \cdots \\ \end {cases}\\ \end {align*}
\[b_n = \frac {k}{2}\int ^1_{-1} \sin \frac {n\pi x}{2}\,dx = 0 \hspace {0.3cm}, n= 1,2,3,\cdots \cdots \cdots \]
Hence, \[f(x) = \frac {k}{2} + \frac {2k}{\pi }\Big ( \cos \frac {\pi }{2}x - \frac {1}{3}\cos \frac {3\pi }{2}x + \frac {1}{5}\cos \frac {5\pi }{2}x + \cdots \cdots \cdots + \cdots \Big )\]
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