 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/6/a/ii/solution

Use the period-two Fourier series convention

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

Away from its jumps this square wave is an odd function, so

a_{0} = a_{n} = 0

b_{n} = \int_{- 1}^{1} f (x) sin (nπ x) d x = 2 \int_{0}^{1} sin (nπ x) d x = \frac{2 ( 1 - ( - 1 ) ^{n} )}{nπ} .

Only odd Fourier harmonics remain. Therefore

f (x) \sim \frac{4}{π} j = 0 \sum \infty \frac{sin (( 2 j + 1 ) π x )}{2 j + 1} .

f (x)

away from the jumps; at a jump it gives the midpoint of the one-sided limits by Dirichlet's theorem on Fourier series. The printed definition leaves the values at the integers unspecified. Choosing finite values there consistent with period two changes neither the Fourier coefficients nor these limiting values.

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