Unusual equation (solving)

Question

I have the following representation of $z$ in terms of $s$, where both $z,s\in \mathbb{R}$ and $|s|<1$:

$$e^{s+2}\sqrt{1-s^2}=z$$

Now, how can I express $s$ in terms of $z$?

I first attacked the problem by taking logarithms:

$$2s + \log (1-s^2) = \log (\frac{z^2}{e^4})$$

Or

$$2s + \log (1-s) + \log(1+s) = \log (\frac{z^2}{e^4})$$

And then developing a Taylor Series expansion of the remaining logarithm, with no success (I got stuck). Changes of variables such as $s+1=t$ did not work for me neither. Besides, I do not imagine any other way to solve this problem.

I do not mind getting something like a Lambert-W function or similar as a result.

Am I missing anything here? What strategy shall I use?

Thanks in advance.

Answer 1

Rearrange:

$$e^{s+2}\sqrt{1-s^2}=z\iff e^{2s}(s^2-1)=-e^{-4}z^2$$

and use Lagrange inversion:

$$e^{2s}(s^2-1)=a\implies s=1-\sum_{n=1}^\infty \frac{a^n}{n!}\frac{d^{n-1}}{ds^{n-1}}\left.\left(\frac{s-1}{e^{2s}(s-1)(s+1)}\right)^n\right|_1$$

General Leibniz rule gives:

$$\frac{d^{n-1}}{ds^{n-1}}e^{-2ns}(s+1)^{-n}=\sum_{m=0}^{n-1}\binom{n-1}m\frac{d^{n-1-m}}{ds^{n-1-m}}e^{-2ns}\frac{d^m}{ds^m}(s+1)^{-n}$$

Summing uses Bessel K:

$$\boxed{s=1-\sum_{n=1}^\infty\frac{(-e^{-4}z^2n)^n}{\sqrt{\pi n}n!}\operatorname K_{\frac12-n}(2n)}$$

shown here. There is another branch of the inverse function too