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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2/9/i/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 9 / i / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

First expand the exponential function and divide by

x

:

\frac{1 - e ^{- x}}{x} = 1 - \frac{x}{2} + \frac{x ^{2}}{6} - \frac{x ^{3}}{24} + O (x^{4}) .

Seek its reciprocal as

1 + A x + B x^{2} + C x^{3} + O (x^{4})

. Equating coefficients in the product with this expression gives

A - \frac{1}{2} = 0, B - \frac{A}{2} + \frac{1}{6} = 0, C - \frac{B}{2} + \frac{A}{6} - \frac{1}{24} = 0.

Hence

A = 1/2

,

B = 1/12

,

C = 0

. The first three nonzero terms of the Taylor expansion are

\frac{x}{1 - e ^{- x}} = 1 + \frac{x}{2} + \frac{x ^{2}}{12} + O (x^{4}) .

Although the quotient is undefined at zero as written, its limit is one, and this Taylor expansion belongs to its continuous, real analytic extension.

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