Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 15 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

The Maclaurin series of the hyperbolic cosine is

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

. The binomial series with exponent

- 1/2

gives

(1 + u)^{- 1/2} = 1 - u /2 + 3 u^{2} /8 + O (u^{3})

. Substituting

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

yields

(cosh x)^{- 1/2} = 1 - \frac{1}{2} (\frac{x ^{2}}{2} + \frac{x ^{4}}{24}) + \frac{3}{8} (\frac{x ^{2}}{2})^{2} + O (x^{6}) = 1 - \frac{x ^{2}}{4} + \frac{7 x ^{4}}{96} + O (x^{6}) .

These are the first three nonzero terms; the remainder is written in Big O notation.

ii

Solution

Put

h = x - π /2

. The sine addition identity gives

sin x = cos h = 1 - h^{2} /2 + h^{4} /24 + O (h^{6})

. Applying the exponential series gives

e^{s i n x} = e exp (- \frac{h ^{2}}{2} + \frac{h ^{4}}{24} + O (h^{6})) = e [1 - \frac{h ^{2}}{2} + (\frac{1}{24} + \frac{1}{8}) h^{4} + O (h^{6})] .

Thus the first three nonzero terms in the Taylor series about

π /2

are

e^{s i n x} = e - \frac{e}{2} (x - \frac{π}{2})^{2} + \frac{e}{6} (x - \frac{π}{2})^{4} + O ((x - \frac{π}{2})^{6}) .

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