 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/18/c/i/solution

Apart from its endpoint convention, the ramp is odd, so its constant and cosine Fourier coefficients vanish. For the sine coefficient, evenness of the integrand and integration by parts give

b_{n} = \frac{4 K}{L ^{2}} \int_{0}^{L /2} x sin \frac{2 πn x}{L} d x = \frac{4 K}{L ^{2}} [- \frac{xL}{2 πn} cos \frac{2 πn x}{L} + \frac{L ^{2}}{4 π ^{2} n ^{2}} sin \frac{2 πn x}{L}]_{0}^{L /2} .

Hence

a_{0} = a_{n} = 0, b_{n} = \frac{K ( - 1 ) ^{n + 1}}{πn} .

This is the Fourier series of a periodic sawtooth. At the periodic jump, the series sums to zero rather than the assigned endpoint value

K /2

; changing a value at one endpoint does not change the coefficients.

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