Is it Possible to Develop an inverse function using the function it self

Question

Is it Possible to Develop (taylor expansion) of an inverse function by knowing the function it self ? If Yes ,Can you illustrate with a simple function I know that we use the identity formula $$ f(f^-1(y))= x $$

Answer 1

I'm going to assume that the inverse is not easy to compute or find.

Lets call the inverse function $g(x)$

an inverse function is one that

$f(x) = y \iff g(y) = x$

Note that there can be left and right inverses

ie:

$f(x) = y \implies g(y)=x$

or the other way around

$g(y) = x \implies f(x)=y$

$f(x) = x^2$ is an example of a function which can't always be reversed

since $f(2) = 4$ and $f(-2) = 4$

Keep left and right inverses in mind.

For a Taylor series, you just need derivatives of the function at certain points

$g(x) \equiv \sum_{n=0}^{n=\infty} \frac{g^n(0)}{n!}x^n$

As long as you can find the derivatives of $g$ at zero, you're in the clear.

If on the other hand you cant do this, then things get more interesting.

You can find them from the derivatives of $f$

for example, if we are considering a left inverse function

$x = g(f(x))$

differentiate with respect to $x$

$1 = g'(f(x))f'(x)$

This gives us

$1 = g'(y)f'(x)$

$\frac{1}{f'(x)} = g'(y)$ So we have the first derivative

Now differentiate $1 = g'(f(x))f'(x)$ with respect to x again

See here for working

The result is

$g''(y) = \frac{-f''(x)}{[f'(x)]^3}$

You can continue in this fashion or investigate the generalized Faa di Bruno formula which makes my head hurt.