Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 15 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii

i

Solution

Expand the natural logarithm and reciprocal separately:

ln (2 + x) = ln 2 + \frac{x}{2} - \frac{x ^{2}}{8} + O (x^{3}), \frac{1}{2 - x} = \frac{1}{2} (1 + \frac{x}{2} + \frac{x ^{2}}{4} + O (x^{3})) .

Multiplication gives the first three nonzero terms of the Maclaurin series:

\frac{ln ( 2 + x )}{2 - x} = \frac{ln 2}{2} + \frac{1 + ln 2}{4} x + \frac{1 + 2 ln 2}{16} x^{2} + O (x^{3}) .

The local expansions converge for

∣ x ∣ < 2

.

ii

Solution

The derivative of the inverse tangent is

(1 + x^{2})^{- 1}

. Expand it as a geometric series and integrate term by term near zero, using

arctan 0 = 0

:

arctan x = \int_{0}^{x} (1 - t^{2} + t^{4} + O (t^{6})) d t = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} + O (x^{7}) .

iii

Solution

Use

cosh x = 1 + x^{2} /2 + x^{4} /24 + x^{6} /720 + O (x^{8})

. The Maclaurin series

ln (1 + u) = u - u^{2} /2 + u^{3} /3 + O (u^{4})

gives

ln (cosh x) = \frac{x ^{2}}{2} + (\frac{1}{24} - \frac{1}{8}) x^{4} + (\frac{1}{720} - \frac{1}{48} + \frac{1}{24}) x^{6} + O (x^{8}) = \frac{x ^{2}}{2} - \frac{x ^{4}}{12} + \frac{x ^{6}}{45} + O (x^{8}) .

All three nonzero terms are needed; stopping at fourth order would supply only two.

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