 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/17/a/solution

With the standard real normalization, the Fourier series of a period-

2 L

f (x) \sim \frac{a _{0}}{2} + \sum_{n = 1}^{\infty} [a_{n} cos \frac{nπ x}{L} + b_{n} sin \frac{nπ x}{L}],

where

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

S_{N}

, Fourier orthogonality eliminates all cross terms. The constant term contributes

2 L (a_{0} /2)^{2}

, while every squared sine or cosine integrates to

L

\int_{- L}^{L} S_{N} (x)^{2} d x = L \frac{a _{0}^{2}}{2} + L \sum_{n = 1}^{N} (a_{n}^{2} + b_{n}^{2}) .

L^{2} [- L, L]

makes

S_{N} \to f

in norm. Passing to the limit proves Parseval's identity:

\int_{- L}^{L} f (x)^{2} d x = L [\frac{a _{0}^{2}}{2} + n = 1 \sum \infty (a_{n}^{2} + b_{n}^{2})] .

Square integrability is the needed hypothesis here; periodicity alone is not an integrability or convergence condition.

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