 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/15/solution

Use the real Fourier series convention

f = a_{0} /2 + \sum_{n \geq 1} (a_{n} cos n x + b_{n} sin n x)

b_{n} = 0

. For noninteger real

μ

a_{0} = \frac{2}{π} \int_{0}^{π} cos (μx) d x = \frac{2 s i n π μ}{π μ} .

For

n \geq 1

, the product-to-sum identity gives

a_{n} = \frac{1}{π} (\frac{sin π ( μ - n )}{μ - n} + \frac{sin π ( μ + n )}{μ + n}) = \frac{2 μ ( - 1 ) ^{n} sin π μ}{π ( μ ^{2} - n ^{2} )} .

f (x) = \frac{sin π μ}{π μ} + \frac{2 μ sin π μ}{π} n = 1 \sum \infty \frac{( - 1 ) ^{n} cos n x}{μ ^{2} - n ^{2}} .

O (n^{- 2})

, giving absolute convergence and uniform convergence. The periodic function is continuous at its joins because the two endpoint cosine values coincide, so the Dirichlet-Jordan convergence theorem gives this value at every point.

Putting

μ = 1/3

gives

f (x) = \frac{3 3}{2 π} - \frac{3 3}{π} n = 1 \sum \infty \frac{( - 1 ) ^{n} cos n x}{9 n ^{2} - 1} .

Evaluate at

x = 0

, where

f (0) = 1

, and multiply by

π / 3

\frac{π}{3} = \frac{3}{2} - 3 \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{9 n ^{2} - 1} = \frac{3}{2} - \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{3 n ^{2} - 1/3} .

Comparing with the required form gives

A = 3, B = 3

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