 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/b/i/solution

Use the periodic extension of

f

when the translated argument falls outside the original interval. With

k_{n} = 2 πn / L

, the sine subtraction identity gives

g (x) = n = 1 \sum \infty b_{n} [sin (k_{n} x) cos (k_{n} ℓ) - cos (k_{n} x) sin (k_{n} ℓ)] .

This translation of a Fourier series generally introduces cosine terms: a translated odd function need not be odd about the original origin. Equality is understood wherever the original Fourier series represents its periodic extension, with midpoint values at jumps.

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