 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/18/a/solution

With the real Fourier series convention,

f (θ) \sim \frac{a _{0}}{2} + n = 1 \sum \infty [a_{n} cos (n θ) + b_{n} sin (n θ)],

a_{n} = \frac{1}{π} \int_{- π}^{π} f (t) cos (n t) d t (n \geq 0), b_{n} = \frac{1}{π} \int_{- π}^{π} f (t) sin (n t) d t (n \geq 1) .

\frac{1}{π} \int_{- π}^{π} ∣ f (t) ∣^{2} d t = \frac{∣ a _{0} ∣ ^{2}}{2} + n = 1 \sum \infty (∣ a_{n} ∣^{2} + ∣ b_{n} ∣^{2}) .

f

in the L2 space sense; for piecewise smooth data it converges pointwise to the average of the two one-sided limits. Without such qualifications an arbitrary periodic function need not have a pointwise-convergent Fourier series.

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