Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 318 / 2 / a

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Solution

Substituting the Fourier coefficient s into the Fourier partial sum and summing the finite geometric series gives

s_{n} (f, x) = \frac{1}{2 π} \int_{T} f (t) \sum_{k = - n}^{n} e^{ik (x - t)} d t = \frac{1}{π} \int_{T} D_{n} (x - t) f (t) d t,

where

D_{n} (u) = \frac{sin (( n + \frac{1}{2} ) u )}{2 sin ( u /2 )} .

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