 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/5a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

f

has zero mean and its cosine coefficients vanish:

a_{0} = 0, a_{n} = 0.

b_{n} = \frac{1}{π} (\int_{0}^{π} sin (n θ) d θ - \int_{π}^{2 π} sin (n θ) d θ),

b_{n} = ⎩ ⎨ ⎧ \frac{4}{πn}, 0, n odd, n even .

Away from the corners,

F^{'} = f

. Differentiating the Fourier series shows that

n B_{n} = a_{n}

and

- n A_{n} = b_{n}

. Also,

A_{0} = \frac{1}{π} \int_{0}^{2 π} F (θ) d θ = π .

Consequently

B_{n} = 0, A_{n} = ⎩ ⎨ ⎧ - \frac{4}{π n ^{2}}, 0, n odd, n even,

and

F (θ) = \frac{π}{2} - \frac{4}{π} \sum_{r = 0}^{\infty} \frac{c o s (( 2 r + 1 ) θ )}{( 2 r + 1 ) ^{2}} .

Evaluating the series for

f

θ = π /2

r = 0 \sum \infty \frac{( - 1 ) ^{r}}{2 r + 1} = \frac{π}{4} .

Evaluating the continuous Fourier series for

F

θ = 0

gives

0 = \frac{π}{2} - \frac{4}{π} \sum_{r = 0}^{\infty} \frac{1}{( 2 r + 1 ) ^{2}},

and hence

r = 0 \sum \infty \frac{1}{( 2 r + 1 ) ^{2}} = \frac{π ^{2}}{8} .

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