How do I compute ∫[f(x)]^2 dx from -π to π given the following?
Given:
f(x) = ∑(cosnx)/(2^n), n = 0,1,2,...
Parseval's identity:
(1/(2π)) ∫(-π to π) (f(x))^2 dx = (a_0)^2 + (1/2)∑((a_n)^2 + (b_n)^2), n = 1,2,3,...
Table of values:
∑1/(n^2) = (π^2)/6, n=1,2,3,...
∑1/(n^4) = (π^4)/90, n=1,2,3,...
∑1/(n^6) = (π^6)/945, n=1,2,3,...
∑1/(n^8) = (π^8)/9450, n=1,2,3,...
∑1/(n^10) = (π^10)/93555, n=1,2,3,...
My book gives that ∫(-π to π) [f(x)]^2 dx = 7π/3
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You don't need anything in that table; just the geometric series formula. The sequence of Fourier coefficients of your f(x) is given by
a_n = 1/2^n, n = 0, 1, 2, 3, 4, ...
b_n = 0, n = 0, 1, 2, 3, 4, ...
You could see this by computing the integrals, but I think you are supposed to notice this just by looking at it (f(x) is defined in terms of a Fourier series expansion). Because of this, the right hand side of Parseval's identity is
(1/2^0)^2 + (1/2) the sum from n = 1 to infinity of (1/2^n)^2
= 1 + (1/2) the sum from n = 1 to infinity of 1/2^(2n)
= 1 + (1/2) the sum from n = 1 to infinity of 1/(2^2)^n
= 1 + (1/2) the sum from n = 1 to infinity of 1/4^n
[this sum is a geometric series with first term a = 1/4^1 = 1/4 and common ratio of terms r = 1/4, so by the geometric series formula its sum is a/(1 - r) = (1/4)/(1 - 1/4) = (1/4)/(3/4) = 1/3]
= 1 + (1/2) (1/3)
= 7/6
The left hand side of Parseval's identity is 1/(2pi) * [the integral you want]. So multiplying both sides of Parseval's identity by 2pi, you learn that [the integral you want] = (7/6) * 2pi = 7pi/3.