Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 2

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- Solution 2

Solution

The square wave alternates between a low interval of length five and a high interval of length five; its integral average is three. In the real Fourier series convention

f (x) \sim \frac{a _{0}}{2} + \sum_{n = 1}^{\infty} [a_{n} cos \frac{nπ x}{5} + b_{n} sin \frac{nπ x}{5}],

integration over one period gives

a_{0} = \frac{1}{5} \int_{- 5}^{5} f (x) d x = 6, a_{n} = \frac{6}{5} \int_{0}^{5} cos \frac{nπ x}{5} d x = 0,

and

b_{n} = \frac{6}{5} \int_{0}^{5} sin \frac{nπ x}{5} d x = \frac{6}{nπ} [1 - (- 1)^{n}] .

Thus only odd harmonics occur:

f (x) \sim 3 + \frac{12}{π} j = 0 \sum \infty \frac{sin (( 2 j + 1 ) π x /5 )}{2 j + 1} .

The Dirichlet-Jordan convergence theorem makes the sum equal to the function at its continuity points and to the mean of its one-sided limits at a jump. Consequently the sum is 3 at each of $x = - 5, 0, 5$ , regardless of any separately assigned values of the original function at those endpoints.

Figure 1.
A periodic square wave and its Fourier approximations
. Original plot of the alternating low and high intervals and two partial Fourier sums. The jump values of the Fourier series are their midpoint; finite sums exhibit Gibbs oscillations. Details.

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