 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/e/solution

Here

f (x)^{2} = x^{2}

, so

\frac{1}{L} \int_{- L}^{L} f (x)^{2} d x = \frac{2 L ^{2}}{3} .

Substituting the coefficients from part (b) into Parseval's identity gives

\frac{2 L ^{2}}{3} = \frac{L ^{2}}{2} + \frac{16 L ^{2}}{π ^{4}} \sum_{k = 0}^{\infty} \frac{1}{( 2 k + 1 ) ^{4}} .

Hence the sum over all positive odd integers is

π^{4} /96

. The requested series starts at

n = 1

3^{4}

, so it excludes the contribution

1^{- 4}

n = 1 \sum \infty \frac{1}{( 2 n + 1 ) ^{4}} = \frac{π ^{4}}{96} - 1.

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