 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/17/b/solution

Over one period, the graph rises linearly from the limiting point

(- π, 0)

(0, π)

, then drops to zero and stays there up to

π

. Repeat this pattern every

2 π

. The given open endpoints do not fix values at the jumps, but changing individual point values does not change any Fourier coefficient.

Only the negative half-interval contributes. The constant coefficient is

a_{0} = \frac{1}{π} \int_{- π}^{0} (x + π) d x = \frac{π}{2} .

For

n \geq 1

a_{n} = \frac{1}{π} [\frac{( x + π ) s i n n x}{n} + \frac{c o s n x}{n ^{2}}]_{- π}^{0} = \frac{1 - ( - 1 ) ^{n}}{π n ^{2}},

and

b_{n} = \frac{1}{π} [- \frac{( x + π ) c o s n x}{n} + \frac{s i n n x}{n ^{2}}]_{- π}^{0} = - \frac{1}{n} .

Consequently the requested Fourier series is

\frac{π}{4} + n = 1 \sum \infty [\frac{1 - ( - 1 ) ^{n}}{π n ^{2}} cos n x - \frac{1}{n} sin n x] .

It equals the given function on each open continuity interval. By the Dirichlet-Jordan convergence theorem, at

x = 2 kπ

it takes the jump midpoint

π /2

, while at

x = (2 k + 1) π

it takes zero because the periodic one-sided limits agree there.

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