 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-1/11f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Repeatedly differentiating

f^{''} = - f

and using

f (0) = 1

f^{'} (0) = 0

gives

f^{(2 k)} (0) = (- 1)^{k}, f^{(2 k + 1)} (0) = 0.

Also,

\frac{d}{d x} (f (x)^{2} + f^{'} (x)^{2}) = 2 f^{'} f + 2 f^{'} f^{''} = 0,

f^{2} + (f^{'})^{2} = 1

. Every derivative is one of

\pm f, \pm f^{'}

, and hence has absolute value at most one.

Taylor's theorem at zero through degree

2 N + 1

gives

f (x) = \sum_{k = 0}^{N} (- 1)^{k} \frac{x ^{2 k}}{( 2 k )!} + R_{N} (x),

where

∣ R_{N} (x) ∣ \leq \frac{∣ x ∣ ^{2 N + 2}}{( 2 N + 2 )!} ⟶ 0

for each fixed

x

. Therefore

f (x) = k = 0 \sum \infty (- 1)^{k} \frac{x ^{2 k}}{( 2 k )!} .

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