Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 8 / a

Multiply the Taylor series

cos x = 1 - x^{2} /2 + O (x^{4})

(1 - x)^{- 1} = 1 + x + x^{2} + O (x^{3})

. Counting the constant as the first nonzero term gives

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

ii

The identity

sin^{3} x = (3 sin x - sin 3 x) /4

makes the lower-order cancellation explicit. Expand the two sines through degree seven:

sin^{3} x = x^{3} - \frac{x ^{5}}{2} + \frac{13 x ^{7}}{120} + O (x^{9}) .

These are the first three nonzero terms; there is no linear term.

For

F (x) = e^{e^{x}}

, the chain rule gives

F^{'} = e^{x} e^{e^{x}}

and

F^{''} = (e^{x} + e^{2 x}) e^{e^{x}}

. Thus

F (0) = e

F^{'} (0) = e

and

F^{''} (0) = 2 e

, yielding

e^{e^{x}} = e + e x + e x^{2} + O (x^{3}) .

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