 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/18/d/solution

The given identity immediately yields

V_{2} = V_{1} - V_{0} sin ω t

. Thus only the fundamental sine coefficient changes:

V_{0} /2

becomes

- V_{0} /2

. No new integral is needed. The first five nonzero terms are

V_{2} (t) = \frac{V _{0}}{π} - \frac{V _{0}}{2} sin ω t - \frac{2 V _{0}}{3 π} cos 2 ω t - \frac{2 V _{0}}{15 π} cos 4 ω t - \frac{2 V _{0}}{35 π} cos 6 ω t + \dots .

This also follows from translation of a Fourier series: a half-period delay multiplies the

n

(- 1)^{n}

, leaving the constant and even harmonics unchanged and reversing the only odd harmonic present.

 New to topics? Read the docs here!