fourier series for f(x)= 1+x+x^2 in (-π,π)

Question

Fourier series

Answer 1

f(x)= 1+x+x^2 in (-π,π)
a0 = (1/π) ∫(x = -π to π) 1+x+x^2 dx
a0=2(3+π^2)/3π

For n > 1:
a_n = (1/π) ∫(x = -π to π) 1+x+x^2 cos(nx) dx
a_n = [{(2(π^2n^2-2)sin (πn)+4πn cos (πn))/n^3}+2π]/π

b_n = (1/π) ∫(x = -π to π) 1+x+x^2 sin(nx) dx
b_n = 2

Therefore,
f(x) ~ (1/2)[(2(3+π^2))/3π] + Σ(n=1 to ∞) [[{(2(π^2n^2-2)sin (πn)+4πn cos (πn))/n^3}+2π]/π cos(nx) + 2 sin(nx)]

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