Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 2

Table of contents
- a 2
  - Solution a
- b 2
  - Solution b

a

Solution

Substitute

x^{3}

into the Taylor series for the exponential function:

e^{x^{3}} = n = 0 \sum \infty \frac{x ^{3 n}}{n !} .

The general nonzero term is

x^{3 n} / n!

; the power series coefficients of powers not divisible by three are zero. The radius of convergence is infinite.

b

Solution

The power series for

e^{y^{3}}

has uniform convergence on every compact interval, so termwise integration from zero to

x

is valid. Thus

\int_{0}^{x} e^{y^{3}} d y = \sum_{n = 0}^{\infty} \frac{1}{n !} \int_{0}^{x} y^{3 n} d y = n = 0 \sum \infty \frac{x ^{3 n + 1}}{( 3 n + 1 ) n !} .

This power series converges for every real

x

, including negative

x

with the usual orientation of a definite integral.

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