 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/15/c/solution

Again the function is even, so all sine coefficients vanish. Its constant term is

a_{0} /2 = sinh 1

. With

k = nπ

\int cosh x cos (k x) d x = \frac{s i n h x c o s ( k x ) + k c o s h x s i n ( k x )}{1 + k ^{2}} .

Thus

a_{n} = 2 (- 1)^{n} sinh 1/ (1 + n^{2} π^{2})

. In the requested factored notation,

A_{0} = 1, A_{n} (x) = \frac{2 ( - 1 ) ^{n} cos ( nπ x )}{1 + n ^{2} π ^{2}} .

This is the period-two specialization of the Fourier series of periodic hyperbolic sine and cosine. The function values match at the endpoints, although its derivative has a jump in the periodic extension.

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