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

Expand the natural logarithm and reciprocal separately:Multiplication gives the first three nonzero terms of the Maclaurin series:The local expansions converge for $|x|<2$.

The derivative of the inverse tangent is $(1+x^2{)}^{-1}$. Expand it as a geometric series and integrate term by term near zero, using $\arctan 0=0$:

Use $\cosh x=1+x^2/2+x^4/24+x^6/720+O(x^8)$. The Maclaurin series $\ln (1+u)=u-u^2/2+u^3/3+O(u^4)$ givesAll three nonzero terms are needed; stopping at fourth order would supply only two.

Assume the needed derivatives exist continuously and $df/dg\ne 0$ on a neighborhood, so the inverse function theorem provides a local inverse. Put $p=df/dg$, $q=d^{2}f/d{g}^2$, and $r=d^{3}f/d{g}^3$. The chain rule gives $d/df=p^{-1}d/dg$, soDifferentiate once more:Thus the first three derivatives of an inverse function areAll forward derivatives on the right are evaluated at the corresponding value $g(f)$.

Since $a_1=f^{\prime }(0)\ne 0$, substitute the proposed inverse series into the forward one. Comparing coefficients of $f,f^2,f^3$ givesSolving successively yields the cubic series reversion coefficientsThe conditions $f(0)=0$ and $a_1\ne 0$ ensure that both local series are centred at zero and that the inverse has a linear leading term.