 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/8/solution

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

The logarithm must be expanded about two, not about one. Factor

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

Applying the supplied logarithm expansion to the bracket gives

lo g 2 + (x /2 + x^{2} /4) - \frac{1}{2} (x /2)^{2} + O (x^{3})

. Thus, through second order,

lo g (1 + e^{x}) = lo g 2 + \frac{x}{2} + \frac{x ^{2}}{8} + O (x^{3}) .

In fact the cubic coefficient vanishes, since

lo g (1 + e^{x}) = x /2 + lo g [2 cosh (x /2)]

and the second term is even; the Maclaurin series remainder can be sharpened to

O (x^{4})

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