Engineering Board Exam Reviewer · Fourier Series · hard

The Fourier series of f(x)=x2 on (-π,π) is π2/3+4 n=1∞(-1)n/n2 nx. Evaluate it at x=π to find n=1∞ 1n2.

1. 3.29
2. 1.645
3. 0.822
4. 1.234

Answer

B. 1.645

Solution

At x=π, nπ=(-1)n, so π2=π2/3+4 (-1)n(-1)n/n2=π2/3+4 1n2. Hence 1/n2=1/4·2π2/3=π2/6 1.645.

More Fourier Series questions

1. The half-range sine series of f(x)=x on (0,2) is bn π x/2. Find b2.
2. The square wave f(x)=-1 on (-π,0) and f(x)=1 on (0,π) has a jump at x=0. To what value does its Fourier series converge at x=0?
3. Find a1 in the Fourier series of f(x)=|x| on (-π,π).
4. Find the constant term a0/2 of the Fourier series of f(x)=x2 on (-π,π).
5. The square wave f(x)=-1 on (-π,0) and f(x)=1 on (0,π) has Fourier coefficients bn. Find b3.
6. Find b3 in the Fourier series of f(x)=x on (-π,π).
7. Which Fourier coefficients vanish for the odd function f(x)=x on (-π,π)?

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