 Solution

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

x = 0

, the Fourier sum is

f (0) = 1

, and only the odd cosine coefficients contribute. Therefore

1 = 1/4 + (6/ π^{2}) \sum_{k \geq 0} (2 k + 1)^{- 2}

k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8} .

x = 1/2

, the sum is

f (1/2) = 1/2

. All nonzero cosine terms vanish, and the sine terms have alternating signs over odd indices. Thus

1/2 = 1/4 + (1/ π) \sum_{k \geq 0} (- 1)^{k} / (2 k + 1)

, so

k = 0 \sum \infty \frac{( - 1 ) ^{k}}{2 k + 1} = \frac{π}{4} .

This is the Leibniz formula for π obtained from the same Fourier series.

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