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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 4 / a / Solution by

Codex  0 2026-10-25

For a nonzero integer

k

, direct integration gives

\int_{- π}^{π} cos (k x) d x = 0, \int_{- π}^{π} sin (k x) d x = 0,

whereas

\int_{- π}^{π} 1 d x = 2 π

. The product-to-sum formulas

2 cos (m x) cos (n x) 2 sin (m x) sin (n x) 2 sin (m x) cos (n x) = cos ((m - n) x) + cos ((m + n) x), = cos ((m - n) x) - cos ((m + n) x), = sin ((m + n) x) + sin ((m - n) x)

therefore establish Fourier orthogonality: for positive integers

m, n

,

\int_{- π}^{π} cos (m x) cos (n x) d x \int_{- π}^{π} sin (m x) sin (n x) d x \int_{- π}^{π} sin (m x) cos (n x) d x = π δ_{mn}, = π δ_{mn}, = 0.

Here

δ_{mn}

is the Kronecker delta. In particular the constant function is orthogonal to every nonconstant sine and cosine in this system. Integrating the given Fourier series, then multiplying it by

cos (m x)

or

sin (m x)

and integrating, isolates its Fourier coefficients. Using the permitted interchange of the sum and integral gives

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

If

f

is an even function, then

f (x) sin (n x)

is an odd function, while

f (x) cos (n x)

is an even function. Thus

b_{n} = 0, a_{n} = \frac{2}{π} \int_{0}^{π} f (x) cos (n x) d x (n \geq 0) .

The Fourier series consequently reduces to a Fourier cosine series.

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