Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 15 / c / ii

Table of contents
- Solution ii

Solution

The Maclaurin series give

e^{- x^{3}} = 1 - x^{3} + O (x^{6})

and

cosh x = 1 + x^{2} /2 + x^{4} /24 + O (x^{6})

. Inverting the latter using the geometric series gives

\frac{1}{c o s h x} = 1 - \frac{x ^{2}}{2} + (\frac{1}{4} - \frac{1}{24}) x^{4} + O (x^{6}) = 1 - \frac{x ^{2}}{2} + \frac{5}{24} x^{4} + O (x^{6}) .

Multiplying the two expansions yields

\frac{e ^{- x^{3}}}{cosh x} = 1 - \frac{x ^{2}}{2} - x^{3} + \frac{5}{24} x^{4} + O (x^{5}) .

The remainder is

O (x^{5})

because the product already generates an

x^{5}

term from

(- x^{3}) (- x^{2} /2)

.

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