Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 15

Use the real Fourier series convention

g = a_{0} /2 + \sum_{n \geq 1} [a_{n} cos (nπ x) + b_{n} sin (nπ x)]

for period two. Since

g

b_{n} = 0

, and

a_{0} = \int_{- 1}^{1} x^{2} d x = 2/3

. Twice applying integration by parts, with

k = nπ

, gives

a_{n} = 2 \int_{0}^{1} x^{2} cos (k x) d x = 2 [\frac{x ^{2} s i n ( k x )}{k} + \frac{2 x c o s ( k x )}{k ^{2}} - \frac{2 s i n ( k x )}{k ^{3}}]_{0}^{1} = \frac{4 ( - 1 ) ^{n}}{π ^{2} n ^{2}} .

Hence

g (x) = \frac{1}{3} + \frac{4}{π ^{2}} n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}} cos (nπ x) .

The periodic extension is continuous at its endpoints, so the series takes the specified endpoint values as well as the interior values.

b

Apply Parseval's identity in the same coefficient convention:

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

The left-hand side is

\int_{- 1}^{1} x^{4} d x = 2/5

. Thus

2/5 = 2/9 + (16/ π^{4}) \sum_{n \geq 1} n^{- 4}

n = 1 \sum \infty \frac{1}{n ^{4}} = \frac{π ^{4}}{90} .

Again the function is even, so all sine coefficients vanish. Its constant term is

a_{0} /2 = sinh 1

. With

k = nπ

\int cosh x cos (k x) d x = \frac{s i n h x c o s ( k x ) + k c o s h x s i n ( k x )}{1 + k ^{2}} .

Thus

a_{n} = 2 (- 1)^{n} sinh 1/ (1 + n^{2} π^{2})

. In the requested factored notation,

A_{0} = 1, A_{n} (x) = \frac{2 ( - 1 ) ^{n} cos ( nπ x )}{1 + n ^{2} π ^{2}} .

This is the period-two specialization of the Fourier series of periodic hyperbolic sine and cosine. The function values match at the endpoints, although its derivative has a jump in the periodic extension.

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