 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/4/a/solution

The function is even, so every sine coefficient is zero. Its mean coefficient is

a_{0} = \frac{2}{π} \int_{0}^{π} sin x d x = \frac{4}{π} .

For

n \geq 1

, compute

a_{n} = (2/ π) \int_{0}^{π} sin x cos (n x) d x

. The product-to-sum identity makes this integral zero for odd

n

and gives

- 4/ (π (n^{2} - 1))

for even

n

. Thus the missing Fourier coefficients are $a_{n} = 0$ for odd $n$ and $b_{n} = 0$ for every $n$ , with

a_{0} = 4/ π

. Equivalently,

∣ sin x ∣ = \frac{2}{π} - \frac{4}{π} \sum_{m = 1}^{\infty} \frac{c o s ( 2 m x )}{4 m ^{2} - 1} .

The Fourier series converges to this continuous periodic function; there is no jump value to average.

 New to topics? Read the docs here!