 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/17/c/solution

Factor

1 + e^{x} = 2 e^{x /2} cosh (x /2)

. This gives

ln (1 + e^{x}) = ln 2 + \frac{x}{2} + ln cosh \frac{x}{2} .

The last function is even. Its Taylor series begins

ln cosh (x /2) = x^{2} /8 + O (x^{4})

, since

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

and

ln (1 + w) = w + O (w^{2})

. Therefore

ln (1 + e^{x}) = ln 2 + \frac{x}{2} + \frac{x ^{2}}{8} + 0 x^{3} + O (x^{4}) .

In particular, the cubic coefficient is zero; the identity explains its cancellation before any lengthy differentiation.

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