 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v-resit/question-4/b/i/solution

a_{0} = \frac{2}{π} \int_{0}^{π} (1 + x) d x = 2 + π,

and, for

n \geq 1

a_{n} = \frac{2}{π} \int_{0}^{π} (1 + x) cos (n x) d x = \frac{2}{π} [\frac{( 1 + x ) sin ( n x )}{n} + \frac{cos ( n x )}{n ^{2}}]_{0}^{π} = \frac{2 ( ( - 1 ) ^{n} - 1 )}{π n ^{2}} .

Only odd Fourier harmonics remain, so

1 + x = 1 + \frac{π}{2} - \frac{4}{π} m = 0 \sum \infty \frac{cos (( 2 m + 1 ) x )}{( 2 m + 1 ) ^{2}} (0 \leq x \leq π) .

b_{n} = \frac{2}{π} \int_{0}^{π} (1 + x) sin (n x) d x = \frac{2}{π} [- \frac{( 1 + x ) cos ( n x )}{n} + \frac{sin ( n x )}{n ^{2}}]_{0}^{π} = \frac{2 [ 1 - ( 1 + π ) ( - 1 ) ^{n} ]}{πn} .

Consequently

1 + x = \frac{2}{π} n = 1 \sum \infty \frac{1 - ( 1 + π ) ( - 1 ) ^{n}}{n} sin (n x) (0 < x < π) .

x = 0

and

x = π

, the Fourier sine series instead has value zero.

To describe the requested graphs precisely, take

t \in [- π, π)

congruent to

x

2 π

. The Fourier cosine series converges to the continuous triangular periodic extension

F_{c} (x) = 1 + ∣ t ∣ .

Its minima occur at the even multiples of

π

, with value

1

, and its maxima at the odd multiples, with value

1 + π

F_{s} (x) = ⎩ ⎨ ⎧ 1 + t, t - 1, 0, 0 < t < π, - π < t < 0, x \in π Z .

The jump discontinuities at even multiples of

π

go from

- 1

1

; those at odd multiples go from

1 + π

- (1 + π)

. Each jump is assigned its midpoint, zero, by the Fourier series.

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