Fourier Series--Right Triangle

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

$$a_0 = {1\over L} \int_0^{2L} {x\over 2L}\,dx = {1\over 2L^2} [{\textstyle{1\over 2}}x^2]_0^L={1\over 4L^2}(2L)^2=1$$

$$a_n = {1\over L} \int_0^{2L} {x\over 2L} \cos\left({n\pi x\over L}\right)\,dx$$

$$= {[2n\pi\cos(n\pi)-\sin(n\pi)]\sin(n\pi)\over n^2\pi^2} = 0$$

$$b_n = {1\over L} \int_0^{2L} {x\over 2L} \sin\left({n\pi x\over L}\right)\,dx$$

$$= {-2n\pi\cos(2n\pi)+\sin(2n\pi)\over 2n^2\pi^2} = -{1\over n\pi}.$$

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