 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/18/c/solution

x = 0

in the convergent Fourier series from part (b). Since

f (0) = 0

and every cosine equals one,

0 = \frac{L}{2} - \frac{4 L}{π ^{2}} \sum_{k = 0}^{\infty} \frac{1}{( 2 k + 1 ) ^{2}} .

Rearranging gives

k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8} .

Convergence at the evaluation point is essential when extracting a numerical series identity from a Fourier series. Here the function is continuous there and the coefficients are absolutely summable.

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