 Solution

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

For period two, write

f \sim a_{0} /2 + \sum_{n \geq 1} [a_{n} cos (nπ x) + b_{n} sin (nπ x)]

. Direct integration gives

a_{0} = 1/2

. For

0 < x < 1

f (x) + f (- x) = 2 - 3 x

and

f (x) - f (- x) = x

, so

a_{n} = \int_{0}^{1} (2 - 3 x) cos (nπ x) d x = \frac{3 ( 1 - ( - 1 ) ^{n} )}{n ^{2} π ^{2}}, b_{n} = \int_{0}^{1} x sin (nπ x) d x = \frac{( - 1 ) ^{n + 1}}{nπ} .

\frac{1}{4} + n = 1 \sum \infty [\frac{3 ( 1 - ( - 1 ) ^{n} )}{n ^{2} π ^{2}} cos (nπ x) + \frac{( - 1 ) ^{n + 1}}{nπ} sin (nπ x)] .

It converges to the values specified in part a. Changing the assigned values at the jumps would not change these integral coefficients.

 New to topics? Read the docs here!