 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/14a/b/ii/solution

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

The mean of

f

A_{n} = \frac{1}{π} [\int_{0}^{π} (\frac{π}{2} - θ) cos (n θ) d θ + \int_{π}^{2 π} (θ - \frac{3 π}{2}) cos (n θ) d θ] = \frac{2 ( 1 - ( - 1 ) ^{n} )}{π n ^{2}} .

Thus

A_{n} = 0

for even

n

and

A_{n} = 4/ (π n^{2})

for odd

n

. Substitution into part (i) yields

ϕ (r, θ) = lo g r - 1 + \frac{4}{π} n \geq 1 n odd \sum \frac{cosh ( n ( lo g r - 1 ))}{n ^{2} cosh n} cos (n θ) .

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