 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/iii/solution

Here

t_{0} = 0

and

t_{2} = t_{1}^{2}

, so the constant and quadratic terms vanish, while the linear coefficient remains

t_{1} \neq = 0

. A further derivative gives

f^{(3)} (x) = \frac{t ^{(3)} ( x )}{1 + t ( x )} - \frac{3 t ^{'} ( x ) t ^{''} ( x )}{( 1 + t ( x ) ) ^{2}} + \frac{2 t ^{'} ( x ) ^{3}}{( 1 + t ( x ) ) ^{3}} .

Consequently

f^{(3)} (0) = t_{3} - 3 t_{1} t_{2} + 2 t_{1}^{3} = t_{3} - t_{1}^{3}

. Since

t_{3}

and

t_{1}

have opposite signs,

t_{3}

and

t_{1}^{3}

also have opposite signs, and their difference is nonzero. The first two nonzero terms are therefore

f (x) = t_{1} x + \frac{t _{3} - t _{1}^{3}}{6} x^{3} + O (x^{4}) .

This is where the full sign condition

t_{3} t_{1} < 0

rules out a further cancellation in the Taylor expansion.

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