 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-1/3e/d/solution

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

The removable definition gives the entire series

f (z) = \sum_{n = 0}^{\infty} (- 1)^{n} \frac{z ^{2 n}}{( 2 n + 1 )!} .

Therefore

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

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