Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 18 / c

a_{0} = a_{n} = 0

. Here

L = 1

b_{n} = 2 \int_{0}^{1} x (1 - x) sin (nπ x) d x .

Write

k = nπ

. One integration by parts has zero endpoint terms because

x (1 - x)

vanishes at both endpoints:

b_{n} = \frac{2}{k} \int_{0}^{1} (1 - 2 x) cos (k x) d x .

\int_{0}^{1} (1 - 2 x) cos (k x) d x = \frac{2}{k} \int_{0}^{1} sin (k x) d x = \frac{2 ( 1 - c o s k )}{k ^{2}} .

Therefore

b_{n} = 4 (1 - (- 1)^{n}) / (n^{3} π^{3})

, zero for even

n

and

8/ (n^{3} π^{3})

for odd

n

f (x) = \frac{8}{π ^{3}} m = 0 \sum \infty \frac{sin (( 2 m + 1 ) π x )}{( 2 m + 1 ) ^{3}} .

The coefficients are absolutely summable, and the periodic function is continuous even at the joins

x = \pm 1

, so the displayed Fourier series represents it everywhere.

ii

The square is even, and direct integration gives

\int_{- 1}^{1} f (x)^{2} d x = 2 \int_{0}^{1} x^{2} (1 - x)^{2} d x = 2 (\frac{1}{3} - \frac{1}{2} + \frac{1}{5}) = \frac{1}{15} .

\frac{1}{30} = \frac{1}{2} \sum_{m = 0}^{\infty} \frac{64}{π ^{6} ( 2 m + 1 ) ^{6}} .

Hence

m = 0 \sum \infty \frac{1}{( 2 m + 1 ) ^{6}} = \frac{π ^{6}}{960}, A = \frac{1}{960} .

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