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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/18/b/iii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 18 / b / iii / Solution by

Codex  0 2026-10-18

Let

S_{N}

be the Fourier partial sum through frequency

N

. By orthogonality and the coefficient formulas,

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

Expanding the nonnegative squared L2 norm of the error gives the exact finite identity

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

For

f

in the L2 space, completeness of the Fourier basis makes the left-hand side tend to zero. Dividing the resulting equality by

2 L

proves Parseval's identity:

\frac{1}{2 L} \int_{- L}^{L} f (x)^{2} d x = \frac{a _{0}^{2}}{4} + \frac{1}{2} n = 1 \sum \infty (a_{n}^{2} + b_{n}^{2}) .

This argument uses convergence in L2 rather than an unqualified exchange of an infinite sum with the integral of its square. The function in part (c) is continuous and piecewise polynomial, so these hypotheses hold.

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