Taylor/Mclaurin Series of an Inverse
\[f(x)\]
about a point by collecting powers of \[(x-x_0)\]
or \[x\]
in the equation \[f^{-1}(f(x)) \]
by assuming a Taylor series for \[f^{-1}(x)\]
of the form \[\sum^{\infty}_{n=0}a_n(x-x_0)^n\]
then using the identity \[x-x_0= f^{-1}(f(x))\]
or \[x=f^{-1}(f(x))\]
for a Mclaurin series, where \[f(x)\]
is replaced by the Taylor or Mclaurin series of \[f(x)\]
, then solving iteratively for \[a_0, \; a_1, \; a_2, \; a_3,...\]
Example: Find the Mclaurin series for
\[sin^{-1}x\]
up to \[x^3\]
.\[sinx=\sum^{\infty}_{n=0} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}=x- \frac{x^3}{3!}+-...\]
\[\begin{equation} \begin{aligned}x &= a_0+a_1(x- \frac{x^3}{3!}+...)+a_2(x- \frac{x^3}{3!}+...)^2+ a_3(x- \frac{x^3}{3!}+...)^3+...\\ &= a_0+a_1(x)+(a_2)x^2+(- \frac{a_1}{6} +a_3)x^3 \end{aligned} \end{equation}\]
We have then
\[a_0=0, \; a_1=1, \; a_2=0, \; a_3= \frac{1}{6}\]
and \[sin^{-1}x=x + \frac{x^3}{6}+...\]
. 
