 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/17/c/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 17 / c / Solution by

Codex  0 2026-10-18

Take the odd extension

H (x) = x (π - ∣ x ∣)

on

[- π, π]

and then extend with period

2 π

. It agrees with the required arch for

0 \leq x \leq π

and is continuous at its joins. Its Fourier sine series has

b_{n} = \frac{2}{π} \int_{0}^{π} x (π - x) sin n x d x .

For clarity, let

h = x (π - x)

, so

h^{'} = π - 2 x

and

h^{''} = - 2

. Two integrations by parts, with the first endpoint term zero, give

\int_{0}^{π} h sin n x d x = \frac{1}{n} \int_{0}^{π} h^{'} cos n x d x = - \frac{1}{n ^{2}} \int_{0}^{π} h^{''} sin n x d x = \frac{2 [ 1 - ( - 1 ) ^{n} ]}{n ^{3}} .

Consequently

b_{n} = 4 [1 - (- 1)^{n}] / (π n^{3})

: the even coefficients vanish and the odd ones are

8/ (π n^{3})

. This proves

h (x) = \frac{8}{π} n = 0 \sum \infty \frac{sin (( 2 n + 1 ) x )}{( 2 n + 1 ) ^{3}}, 0 \leq x \leq π .

The Fourier sine series of an odd quadratic arch converges to this continuous extension; its coefficients are also absolutely summable. Finally evaluate at

x = π /2

. The left side is

π^{2} /4

and the sine factors are

(- 1)^{n}

, so

n = 0 \sum \infty \frac{( - 1 ) ^{n}}{( 2 n + 1 ) ^{3}} = \frac{π ^{3}}{32} .

The evaluation is at a continuity point, so it does not invoke a jump midpoint.

 Read the full article

 New to topics? Read the docs here!