Find the sum of the first two nonzero terms of the Fourier s

Solution

this is easy way to find the fourier series sum First find a0: a0 = 1/(2L) Integral [-L to L] f(x) dx Observe that f(x) is odd - the above integral is 0, so a0 = 0. For a1, a2, ..., an, you integrate: an = 1/L Integral [-L to L] f(x) cos(n pi x/L) dx But note that the \"an\" are all zero because f(x) is odd and cos (n pi x/L) is even, so the product is odd. To find b1, b2,...,bn, you just integrate: bn = 1/L Integral [-L to L] f(x) sin(n pi x/L) dx As in the video, note that now f(x) is odd and sin (n pi x/L) is odd, so the product is even. Therefore you can write: bn = 2/L Integral [0 to L] f(x) sin(n pi x/L) dx Recall that the function has 2 segments, from x=0,1, where f(x)=0 and from x=1,2 where f(x) = 1. You can write the integral as the sum of two integrals: bn = 2/L Integral [0 to 1] f(x) sin(n pi x/L) dx + 2/L Integral [1 to L] f(x) sin(n pi x/L) dx Filling in the values f(x) and L: bn = 2/2 Integral [0 to 1] (0) sin(n pi x/2) dx + 2/2 Integral [1 to 2] (1) sin(n pi x/2) dx The first term disappears, so we\'re left with: bn = Integral [1 to 2] sin(n pi x/2) dx Integrating: bn = (-2/n pi)cos(n pi x/2) ] [x = 2 to 1] bn = (-2/n pi)(cos(n pi) - cos(n pi/2))