Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 3E (Analysis I)

Table of contents
- a 3E
  - Solution a
- b 3E
  - Solution b
- c 3E
  - Solution c
- d 3E
  - Solution d

a

Solution

For every

z \in C

,

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

b

Solution

Termwise differentiation gives

f' (z) = \sum_{n = 1}^{\infty} n a_{n} z^{n - 1} .

The differentiated series has the same radius of convergence

R

.

c

Solution

Fix

a

g (b) = e^{a + b} e^{- b}

using only the power series. Termwise differentiation gives

(e^{z})' = e^{z}

, and the product rule gives

g' (b) = 0

. Hence

g

is constant, so

g (b) = g (0) = e^{a}

. Multiplication by

e^{b}

yields

e^{a + b} = e^{a} e^{b}

.

d

Solution

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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