Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 2

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- Solution 2

Solution

On the basic interval the graph is the pair of straight segments joining

(- π, 0)

,

(0, 1)

and

(π, 0)

. Its periodic extension is an even triangular wave.

Figure 1.
The even triangular wave and its periodic extension
. Original plot marking the basic interval and the repeated linear pieces. Details.

The Fourier coefficients satisfy

b_{n} = 0

by evenness, and

a_{0} = \frac{2}{π} \int_{0}^{π} (1 - x / π) d x = 1, a_{n} = \frac{2}{π} \int_{0}^{π} (1 - x / π) cos (n x) d x = \frac{2 [ 1 - ( - 1 ) ^{n} ]}{π ^{2} n ^{2}} .

For the second formula, integration by parts gives

\int_{0}^{π} x cos (n x) d x = [(- 1)^{n} - 1] / n^{2}

. Only odd indices survive, giving this rescaled Fourier cosine series of a triangular wave:

f (x) = \frac{1}{2} + \frac{4}{π ^{2}} m = 0 \sum \infty \frac{cos [( 2 m + 1 ) x ]}{( 2 m + 1 ) ^{2}} .

The pointwise convergence of a piecewise smooth Fourier series applies throughout the interval, including the continuous joined endpoints. The coefficients also imply uniform convergence by the Weierstrass M-test. At

x = 0

, the value is one, so

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

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