 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/15/b/solution

Put

u = ln (1 + x) = x - x^{2} /2 + x^{3} /3 + O (x^{4})

. Since

u = O (x)

, the outer Taylor series requires

u^{2} = x^{2} - x^{3} + O (x^{4})

and

u^{3} = x^{3} + O (x^{4})

. Therefore

ln (1 + ln (1 + x)) = u - \frac{u ^{2}}{2} + \frac{u ^{3}}{3} + O (u^{4}) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \frac{1}{2} (x^{2} - x^{3}) + \frac{x ^{3}}{3} + O (x^{4}) = x - x^{2} + \frac{7 x ^{3}}{6} + O (x^{4}) .

Both logarithms are defined on a neighbourhood of zero, so this composition of power series is valid there.

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