 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/15/b/i/solution

Write

t_{j} = t^{(j)} (0)

and

s = 1 + t_{0} > 0

. The chain rule gives

f (0) = ln s, f^{'} (0) = \frac{t _{1}}{s} .

The condition

t_{3} t_{1} < 0

ensures

t_{1} \neq = 0

. Since

t_{0} \neq = 0

, also

s \neq = 1

and

ln s \neq = 0

. Hence the first two nonzero terms are the constant and linear terms:

f (x) = ln (1 + t_{0}) + \frac{t _{1}}{1 + t _{0}} x + O (x^{2}) .

The inequality

s > 0

, together with continuity of

t

, makes the logarithm defined on a neighbourhood of zero. The remainder is written in Big O notation.

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