 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/17/e/solution

Evaluate the Fourier sine series of part (d) at the interior point

x = 1/2

. Only odd indices remain, with alternating signs:

\frac{1}{2} = \frac{2}{π} (1 - \frac{1}{3} + \frac{1}{5} - \dots), π = 4 \sum_{k = 0}^{\infty} \frac{( - 1 ) ^{k}}{2 k + 1} .

Thus

A = 4

For the second identity, apply Parseval's identity to

h (x) = x^{2}

. From part (c),

a_{0} = 2/3

a_{n} = 4 (- 1)^{n} / (n^{2} π^{2})

and

b_{n} = 0

, so

\int_{- 1}^{1} x^{4} d x = \frac{1}{2} (\frac{2}{3})^{2} + \frac{16}{π ^{4}} \sum_{n = 1}^{\infty} \frac{1}{n ^{4}} .

The left side is

2/5

\frac{16}{π ^{4}} \sum n^{- 4} = 2/5 - 2/9 = 8/45

. Consequently

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

The first series is conditionally convergent, while the second converges absolutely by the P-series criterion.

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