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Fourier Series--Right Triangle

\begin{figure}\begin{center}\BoxedEPSF{FourierSeriesRight.epsf scaled 680}\end{center}\end{figure}

Consider a string of length $2L$ plucked at the right end, then

$\displaystyle a_0$ $\textstyle =$ $\displaystyle {1\over L} \int_0^{2L} {x\over 2L}\,dx = {1\over 2L^2} [{\textstyle{1\over 2}}x^2]_0^L={1\over 4L^2}(2L)^2=1$  
$\displaystyle a_n$ $\textstyle =$ $\displaystyle {1\over L} \int_0^{2L} {x\over 2L} \cos\left({n\pi x\over L}\right)\,dx$  
  $\textstyle =$ $\displaystyle {[2n\pi\cos(n\pi)-\sin(n\pi)]\sin(n\pi)\over n^2\pi^2} = 0$  
$\displaystyle b_n$ $\textstyle =$ $\displaystyle {1\over L} \int_0^{2L} {x\over 2L} \sin\left({n\pi x\over L}\right)\,dx$  
  $\textstyle =$ $\displaystyle {-2n\pi\cos(2n\pi)+\sin(2n\pi)\over 2n^2\pi^2} = -{1\over n\pi}.$  

The Fourier series is therefore

f(x)={\textstyle{1\over 2}}-{1\over\pi}\sum_{n=1}^\infty{1\over n}\sin\left({n\pi x\over L}\right).

See also Fourier Series

© 1996-9 Eric W. Weisstein