 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/18/a/solution

For a real

2 π

-periodic square-integrable function, use the Fourier series convention

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

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

The trigonometric orthogonality relations are

\int_{- π}^{π} cos n x cos m x d x = π δ_{nm}

for positive

n, m

, with the same formula for sines; sine-cosine products integrate to zero. The constant has squared integral

2 π

and is orthogonal to every positive harmonic. Thus the squared norm of the

N

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

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

Square integrability is the natural hypothesis for this finite norm identity; additional smoothness can be used when pointwise convergence is required.

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