Problem source

If $y=\mathrm{f}(x)$, then the inverse of $\mathrm{f}$ (when it exists) can be obtained from \textit{ Lagrange's identity}. This identity, which you may use without proof, is
	\[
	\mathrm{f}^{-1}(y)=y+\sum_{n=1}^{\infty}\frac{1}{n!}\frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}}\left[y-\mathrm{f}\left(y\right)\right]^{n},
	\]
	provided the series converges. 
\begin{questionparts}
	\item Verify Lagrange's identity when $\mathrm{f}(x)=\alpha x$, $(0<\alpha<2)$. 
	\item Show that one root of the equation 
	\[
	\tfrac{1}{2}=x-\tfrac{1}{4}x^{3}
	\]
	 is
	\[
	x=\sum_{n=0}^{\infty}\frac{\left(3n\right)!}{n!\left(2n+1\right)!2^{4n+1}}
	\]
	\item Find a solution for $x$, as a series in $\lambda,$ of the equation
	\[
	x=\mathrm{e}^{\lambda x}.
	\]
\end{questionparts}
	[You may assume that the series in part $\textbf{(ii) }$converges,
	and that the series in part $\textbf{(iii) }$converges for suitable
	$\lambda$.]

Solution source

\begin{questionparts}
\item If $f(x) = \alpha x$ then $f^{-1}(x) = \frac{1}{\alpha}x$.

\begin{align*}
&& \frac{\d^{n-1}}{\d y^{n-1}} [y - \alpha y]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [(1 - \alpha)^n y^n] \\
&& &= (1-\alpha)^n n! y \\
\Rightarrow && y + \sum_{n=1}^{\infty} \frac{1}{n!}\frac{\d^{n-1}}{\d y^{n-1}} [y - \alpha y]^n &= y +\sum_{n=1}^{\infty} (1-\alpha)^ny \\
&&&= y + y\l \frac{1}{1-(1-\alpha)}-1 \r \\  
&&&= \frac{1}{\alpha}y
\end{align*}

Where we can sum the geometric progression if $|1-\alpha| < 1 \Leftrightarrow 0 < \alpha < 2$

\item Suppose that $f(x) = x-\frac14x^3$. We would like to find $f^{-1}(\frac12)$.

\begin{align*}
&& \frac{\d^{n-1}}{\d y^{n-1}} [y -  (y+\frac14 y^3)]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [\frac1{4^n} y^{3n}] \\
&& &= \frac{1}{4^n} \frac{(3n)!}{(2n+1)!} y^{2n+1} \\
\Rightarrow && f^{-1}(\frac12) &= \frac12 + \sum_{n=1}^{\infty} \frac{1}{4^n} \frac{(3n)!}{n!(2n+1)!} \frac{1}{2^{2n+1}} \\
&&&=  \frac12 + \sum_{n=1}^{\infty}  \frac{(3n)!}{n!(2n+1)!} \frac{1}{2^{4n+1}} \\
\end{align*}

Since when $n = 0$ $\frac{0!}{0!1!} \frac{1}{2^{0+1}} = \frac12$ we can include the wayward $\frac12$ in our infinite sum and so we have the required result.

\item Consider $f(x) = x - e^{\lambda x}$ we are interested in $f^{-1}(0)$.

\begin{align*}
&& \frac{\d^{n-1}}{\d y^{n-1}} [y -  (y-e^{\lambda y})]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [e^{n\lambda y}] \\
&&&= n^{n-1} \lambda^{n-1}e^{n \lambda y} \\
\Rightarrow && f^{-1}(0) &=  \sum_{n=1}^\infty \frac{1}{n!} n^{n-1} \lambda^{n-1}
\end{align*}

We don't care about convergence, but it's worth noting this has a radius of convergence of $\frac{1}{e}$ (ie this series is valid if $|\lambda| < \frac1e$).

\end{questionparts}