 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-v/4/b/i/solution

Put

a = π /2

a_{0} = \frac{2}{π} (\int_{0}^{a} x d x + \int_{a}^{π} a d x) = \frac{3 π}{4} .

For

n \geq 1

\int_{0}^{a} x cos (n x) d x \int_{a}^{π} a cos (n x) d x = \frac{a sin ( na )}{n} + \frac{cos ( na ) - 1}{n ^{2}}, = - \frac{a sin ( na )}{n} .

The

n^{- 1}

terms cancel. Thus

a_{n} = \frac{2}{π n ^{2}} (cos \frac{nπ}{2} - 1), f (x) = \frac{3 π}{8} + \frac{2}{π} n = 1 \sum \infty \frac{cos ( nπ /2 ) - 1}{n ^{2}} cos (n x) (0 \leq x \leq π) .

a_{n} = 0

when

n

is divisible by four,

a_{n} = - 4/ (π n^{2})

when

n \equiv 2 (mod 4)

, and

a_{n} = - 2/ (π n^{2})

when

n

is odd.

Similarly,

\int_{0}^{a} x sin (n x) d x \int_{a}^{π} a sin (n x) d x = - \frac{a cos ( na )}{n} + \frac{sin ( na )}{n ^{2}}, = \frac{a cos ( na )}{n} - \frac{a ( - 1 ) ^{n}}{n} .

Therefore

b_{n} = \frac{2 s i n ( nπ /2 )}{π n ^{2}} - \frac{( - 1 ) ^{n}}{n}, f (x) = n = 1 \sum \infty (\frac{( - 1 ) ^{n + 1}}{n} + \frac{2 sin ( nπ /2 )}{π n ^{2}}) sin (n x) (0 < x < π) .

f (0) = 0

at zero, but has value zero at

π

, where

f (π) = π /2

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