 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/b/solution

Apply Parseval's identity in the same coefficient convention:

\int_{- 1}^{1} g (x)^{2} d x = \frac{a _{0}^{2}}{2} + \sum_{n = 1}^{\infty} (a_{n}^{2} + b_{n}^{2}) .

The left-hand side is

\int_{- 1}^{1} x^{4} d x = 2/5

. Thus

2/5 = 2/9 + (16/ π^{4}) \sum_{n \geq 1} n^{- 4}

n = 1 \sum \infty \frac{1}{n ^{4}} = \frac{π ^{4}}{90} .

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