Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 15 / b

Assume the needed derivatives exist continuously and

df / d g \neq = 0

on a neighborhood, so the inverse function theorem provides a local inverse. Put

p = df / d g

q = d^{2} f / d g^{2}

, and

r = d^{3} f / d g^{3}

. The chain rule gives

d / df = p^{- 1} d / d g

, so

\frac{d g}{df} = \frac{1}{p}, \frac{d ^{2} g}{d f ^{2}} = \frac{1}{p} \frac{d}{d g} \frac{1}{p} = - \frac{q}{p ^{3}} .

Differentiate once more:

\frac{d ^{3} g}{d f ^{3}} = \frac{1}{p} \frac{d}{d g} (- q p^{- 3}) = - \frac{r}{p ^{4}} + \frac{3 q ^{2}}{p ^{5}} .

\frac{d g}{df} = \frac{1}{df / d g}, \frac{d ^{2} g}{d f ^{2}} = - \frac{d ^{2} f / d g ^{2}}{( df / d g ) ^{3}}, \frac{d ^{3} g}{d f ^{3}} = \frac{3 ( d ^{2} f / d g ^{2} ) ^{2} - ( df / d g ) ( d ^{3} f / d g ^{3} )}{( df / d g ) ^{5}} .

All forward derivatives on the right are evaluated at the corresponding value

g (f)

ii

Since

a_{1} = f^{'} (0) \neq = 0

, substitute the proposed inverse series into the forward one. Comparing coefficients of

f, f^{2}, f^{3}

gives

a_{1} b_{1} = 1, a_{1} b_{2} + a_{2} b_{1}^{2} = 0, a_{1} b_{3} + 2 a_{2} b_{1} b_{2} + a_{3} b_{1}^{3} = 0.

b_{1} = \frac{1}{a _{1}}, b_{2} = - \frac{a _{2}}{a _{1}^{3}}, b_{3} = \frac{2 a _{2}^{2} - a _{1} a _{3}}{a _{1}^{5}} .

The conditions

f (0) = 0

and

a_{1} \neq = 0

ensure that both local series are centred at zero and that the inverse has a linear leading term.

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