 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/2/c/solution

Write the boundary Fourier series as

f = A_{0} + \sum_{n \geq 1} (a_{n} cos n ϕ + b_{n} sin n ϕ)

. Since

R = 2

a_{n} = 2^{n} A_{n}

and

b_{n} = 2^{n} B_{n}

. The mean is

A_{0} = \frac{1}{2 π} (π + \int_{π}^{2 π} cos^{2} ϕ d ϕ) = \frac{3}{4} .

For

n \geq 1

, the integral of

cos n ϕ

on the first semicircle is zero. On the second semicircle, use

cos^{2} ϕ = (1 + cos 2 ϕ) /2

a_{2} = 1/4

and every other

a_{n} = 0

For the sine coefficients, every even

n

gives zero on both semicircles. For odd

n

π b_{n} = \int_{0}^{π} sin n ϕ d ϕ + \int_{π}^{2 π} \frac{1 + cos 2 ϕ}{2} sin n ϕ d ϕ = \frac{2}{n} - \frac{1}{n} - \frac{1}{2} (\frac{1}{n + 2} + \frac{1}{n - 2}) = \frac{4}{n ( 4 - n ^{2} )} .

Thus the requested radial-expansion coefficients are

A_{0} = \frac{3}{4}, A_{n} = {1/16, 0, n = 2, n \neq = 2, B_{n} = ⎩ ⎨ ⎧ \frac{4}{π 2 ^{n} n ( 4 - n ^{2} )}, 0, n odd, n even .

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