 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-1/9e/b/solution

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

f

has

n + 1

x_{0}

, then

f (x) = \sum_{k = 0}^{n} f^{(k)} (x_{0}) (x - x_{0})^{k} / k! + f^{(n + 1)} (ξ) (x - x_{0})^{n + 1} / (n + 1)!

for some intermediate

ξ

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