Find the Fourier series of f(x) = [π - x/2]^2, -π < x < -2π. Hence deduce that π^2 = 1/1^2 + 1/2^2 + 1/3^2 + ...
Added by Catherine R.
Step 1
Since the function is defined for $-2\pi < x < 0$, we can extend it to a periodic function on the interval $0 < x < 2\pi$ by defining $f(x) = [\pi - (x - 2\pi)/2]^2$ for $0 < x < 2\pi$. Now, the function is periodic with period $2\pi$. The Fourier series of a Show more…
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