Find good bounds for the inverse of a function The general case:
Let $f : A\subset \mathbb R \to B = f(A)\subset \mathbb R$ be a $\mathcal C^\infty$-diffeormorphism.
Find good polynomial bounds for $f$ at any order (i.e. $P,Q \in \mathbb R_n[X]$ such as $\forall x \in B, P(x)\leq f(x) \leq Q(x)$ for every $n$).
What I mean by that, is, for example, we can show that if $f^{-1}=\cos$, $\sum_{k=0}^{2n+1} \frac{(-1)^kx^{2k}}{(2k)!}\leq \cos(x) \leq \sum_{k=0}^{2n}\frac{(-1)^kx^{2k}}{(2k)!}$ for $x\geq 0$.
A specific case:
Let $f : (2\alpha/\pi,+\infty) \to f((2\alpha/\pi,+\infty)), s \mapsto A(\cos(\frac{\alpha}{s}))^{-s}\sin(\frac{2\alpha}{s})+B(\cos(\frac{\alpha}{s}))^{-s}\sin(\frac{\alpha}{s})$ With $A,B,\alpha >0$ three fixed parameters.
I have to consider this function for some modelisation I'm working on. I want to find good bounds of its inverse.

I don't really know how to do it. I would have consider the Taylor development of the inverse $f^{-1}$ , but I don't know how to deduce bounds.
 A: Concerning your special case, $\lim_{s\to\infty} f(s) = 0$. Also there is a point $s_0$ with $f$ decreasing on $[s_0, +\infty)$. Exactly how low $s_0$ can be taken is difficult to determine, but because of the $\sin \frac{2\alpha}s$ factor, it is going to be somewhere around $\frac {4\alpha}\pi$, well above the asymptote caused by the cosine factor.
If we restrict $f$ to $[s_0,\infty)$, then it is injective and can be inverted. $f^{-1}$ has domain $(0,f(s_0)]$, but $\lim_{x \to 0+} f^{-1}(x) = \infty$, from which it decreases to $s_0$. Again, you have a function with a vertical asymptote. It will not be boundable by any polynomial over this entire domain.
So you cannot use polynomials to bound an inverse of $f$ whose domain extends down to $0$. Because $f$ itself is infinite at $\frac {2\alpha}\pi$ there must be at least two other domains on which $f$ is injective. These domains are bounded, so if you restrict $f$ to them, the corresponding $f^{-1}$ will have bounded codomain - i.e., you can bound it with constants.
I do not know what is appropriate for your situation.
