Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 15 / b / iii

Table of contents
- Solution iii

Solution

Use the Maclaurin series of the logarithm and hyperbolic cosine:

ln (1 + x^{3}) = x^{3} - \frac{1}{2} x^{6} + O (x^{9}), \frac{1}{c o s h x} = 1 - \frac{1}{2} x^{2} + \frac{5}{24} x^{4} + O (x^{6}) .

To retain terms through degree six, only the constant and quadratic terms of the reciprocal factor can multiply the leading cubic term. Thus

\frac{ln ( 1 + x ^{3} )}{cosh x} = x^{3} - \frac{1}{2} x^{5} - \frac{1}{2} x^{6} + O (x^{7}) .

The numerator in the original PDF is

ln (1 + x^{3})

; the cubic power is missing from the local TeX transcription.

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