I've been thinking about the calculation of inverse functions through Taylor series expansions.
My hypothesis was that if we had: f(x)=∞∑n=0(x−x0)n!fn(x0),
Here it'd be possible to calculate the derivatives (f−1)(n)(x0) 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(1).
To formulate my question, assume we have a differentiable bijective map f:A→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′∘f′)(x)(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)=1(f′∘f−1)(x)⟺(f′∘f−1)(x)=1(f′)−1(x). Then mapping by f from the right gives f′(x)=1(f′)−1(f(x)) and mapping by 1f′ from the left 1f(x)=(1f′∘f′)(x)⟺f(x)=1(1f′∘f′)(x)=11(f′∘f′)(x)=(f′∘f′)(x)
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