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

Table of contents
- Solution 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.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version