Fourier Series--Triangle

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

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

The Coefficients are therefore

$$a_0 = {1\over L}\left[{\int_0^{2L/m} {nx\over 2L}\,dx+\int_{2L/m}^{2L} {n\over 1-n}\left({{x\over 2L}-1}\right)\,dx}\right]$$

$$= 1$$

$$a_n = {m\left[{1-m-\cos(2\pi n)+m\cos\left({2n\pi\over m}\right)}\right]\over 2(m-1)n^2\pi^2}$$

$$= {m^2\left[{\cos\left({2n\pi\over m}\right)-1}\right]\over 2(m-1)n^2\pi^2}$$

$$b_n = {m\left[{m\sin\left({2\pi n\over m}\right)-\sin(2\pi n)}\right]\over 2(m-1)n^2\pi^2}$$

$$= {m^2\sin\left({2\pi n\over m}\right)\over 2(m-1)n^2\pi^2}.$$

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

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

$$b_n = 0,$$

giving

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

See also Fourier Series