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Fourier Series--Triangle

\begin{figure}\begin{center}\BoxedEPSF{FourierSeriesTriangle.epsf scaled 670}\end{center}\end{figure}

\begin{figure}\begin{center}\BoxedEPSF{FourierSeriesTriangle3.epsf scaled 670}\end{center}\end{figure}

Let a string of length $2L$ have a $y$-displacement of unity when it is pinned an $x$-distance which is ($1/m$)th of the way along the string. The displacement as a function of $x$ is then

{mx\over 2L} & $0\leq x\leq {2L\over m}$\cr
...\left({{x\over 2L}-1}\right)& ${2L\over m} \leq x\leq 2L$.\cr}

The Coefficients are therefore
$\displaystyle a_0$ $\textstyle =$ $\displaystyle {1\over L}\left[{\int_0^{2L/m} {nx\over 2L}\,dx+\int_{2L/m}^{2L} {n\over 1-n}\left({{x\over 2L}-1}\right)\,dx}\right]$  
  $\textstyle =$ $\displaystyle 1$  
$\displaystyle a_n$ $\textstyle =$ $\displaystyle {m\left[{1-m-\cos(2\pi n)+m\cos\left({2n\pi\over m}\right)}\right]\over 2(m-1)n^2\pi^2}$  
  $\textstyle =$ $\displaystyle {m^2\left[{\cos\left({2n\pi\over m}\right)-1}\right]\over 2(m-1)n^2\pi^2}$  
$\displaystyle b_n$ $\textstyle =$ $\displaystyle {m\left[{m\sin\left({2\pi n\over m}\right)-\sin(2\pi n)}\right]\over 2(m-1)n^2\pi^2}$  
  $\textstyle =$ $\displaystyle {m^2\sin\left({2\pi n\over m}\right)\over 2(m-1)n^2\pi^2}.$  

The Fourier series is therefore

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

If $m=2$, then $a_n$ and $b_n$ simplify to

$\displaystyle a_n$ $\textstyle =$ $\displaystyle -{4\over n^2\pi^2}\sin^2({\textstyle{1\over 2}}n\pi)=-{4\over n^2...
...{ll} 0 & \mbox{$n=0$, 2, \dots}\\  1 & \mbox{$n=1$, 3, \dots}\end{array}\right.$  
$\displaystyle b_n$ $\textstyle =$ $\displaystyle 0,$  


f_2(x)={\textstyle{1\over 2}}-{4\over\pi^2}\sum_{n=1,3,5,\dots}^\infty {1\over n^2}\cos\left({n\pi x\over L}\right).

See also Fourier Series

© 1996-9 Eric W. Weisstein