If \(y=\mathrm{f}(x)\), then the inverse of \(\mathrm{f}\) (when it exists) can be obtained from 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.
Verify Lagrange's identity when \(\mathrm{f}(x)=\alpha x\), \((0<\alpha<2)\).
Show that one root of the equation
\[
\tfrac{1}{2}=x-\tfrac{1}{4}x^{3}
\]
is
\[
x=\sum_{n=0}^{\infty}\frac{\left(3n\right)!}{n!\left(2n+1\right)!2^{4n+1}}
\]
Find a solution for \(x\), as a series in \(\lambda,\) of the equation
\[
x=\mathrm{e}^{\lambda x}.
\]
[You may assume that the series in part \((ii) \)converges,
and that the series in part \((iii) \)converges for suitable
\(\lambda\).]
Solution:
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 \(|1-\alpha| < 1 \Leftrightarrow 0 < \alpha < 2\)
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.
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 \(|\lambda| < \frac1e\)).