I've been thinking about the calculation of inverse functions through Taylor series expansions.

My hypothesis was that if we had: $$\ f(x) =\sum_{n=0}^\infty \frac{(x-x_0)}{n!}f^{n}(x_0),$$ then $$\ f^{-1}(x) =\sum_{n=0}^\infty \frac{(x-x_0)}{n!}({f^{-1}})^{(n)}(x_0).$$ Here it'd be possible to calculate the derivatives $\ ({f^{-1}})^{(n)}(x_0)$ through the inverse function derivation theorem but for this to work we should have the derivative of the inverse of$\ f$ equal the inverse of the derivative of $\ f$, i.e. $$\ (f^{-1})' = (f')^{-1} \quad (1).$$ In general this does not hold; take for example $\ f(x) = x^2 $.

To formulate my question, assume we have a differentiable bijective map $\ f:A\rightarrow B$ with bijective derivative. Its inverse $\ f^{-1}$ is also a differentiable bijection. Assume the derivative of the inverse is also bijective.

To try and find a function for which$\ (1) $ holds, I was able to deduce (from$\ (1) $) that if such a function exists, we must have $$\ f(x) = (f' \circ f')(x) \quad (2).$$

Clearly this does not hold for any polynomial nor (I'm pretty sure) for any other elementary functions. So, does there exist a function for which this condition holds?

Deriving (2): $\ (f')^{-1}(x) = (f^{-1})'(x) = \frac{1}{(f' \circ f^{-1})(x)} \iff (f' \circ f^{-1})(x) = \frac{1}{(f')^{-1}(x)}$. Then mapping by $\ f$ from the right gives $\ f'(x)=\frac{1}{(f')^{-1}(f(x))}$ and mapping by $\ \frac{1}{f'}$ from the left $\ \frac{1}{f(x)} = (\frac{1}{f'} \circ f')(x) \iff f(x)= \frac{1}{(\frac{1}{f'} \circ f')(x)}=\frac{1}{\frac{1}{(f' \circ f')(x)}}=(f' \circ f')(x)$

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    $\begingroup$ How did you derive the fact $f(x)=(f' \circ f')(x)?$ $\endgroup$ – JasonM Jun 17 '16 at 8:43
  • $\begingroup$ @JasonM I edited the post now. $\endgroup$ – JTervonen Jun 17 '16 at 9:26
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    $\begingroup$ Actually, I'm feeling a bit uneasy about the step where you mapped by $\frac{1}{f'}$ from the left. Are we sure $(\frac{1}{f'}) \circ \frac{1}{((f')^{-1})\circ f}=\frac{1}{f}$? $\endgroup$ – JasonM Jun 17 '16 at 10:01
  • $\begingroup$ Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more. $\endgroup$ – JTervonen Jun 17 '16 at 10:36

Let $f(x)$ be such invertible function s.t. $(f^{-1})'={f'(x)}^{-1}$ which gives a non-linear differential equation $f'^2 +f^2=0$. Clearly, $f=0$ is the only solution to it, which contradicts that $f(x)$ is invertible.

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    $\begingroup$ Could you explain to me how you got the expression $f'^2 + f^2 = 0$? $\endgroup$ – Fabian Schn. Aug 8 '19 at 11:01
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    $\begingroup$ $LHS=(f^{-1})'=(\frac{1}{f})'=-\frac{f'(x)}{f^2(x)}$ and $RHS=\frac{1}{f'(x)}$. $\endgroup$ – Nitin Uniyal Aug 9 '19 at 0:44
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    $\begingroup$ Why is $f^{-1}=\frac{1}{f}$? Wouldn't that imply that $f(x)=\frac{1}{x}$? Also, I don't see how $(f')^{-1}=\frac{1}{f'}$. $\endgroup$ – Void Apr 10 '20 at 3:03

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