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

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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)],

where the Fourier coefficients are

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) .

For a square-integrable periodic function, Parseval's identity is

\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}) .

The Fourier coefficients require integrability, and the stated Parseval identity requires square integrability. The expansion agrees with

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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