Inverse of $x\sinh x$

Question

I'm looking for a closed expression for the inverse of $$f(x)=x\cdot\sinh x\,.$$ It looks like there should be some combination of Lambert $W$ functions that would do it, but I can't seem to find a solution.

In the application I'm considering, $x$ is real and positive (but not very large).

Answer 1

There is no closed solution with the known classical functions.

If one defines $\,f(x):=(x+\sqrt{1+x^2})^x\geq 1\,$ for $x\in\mathbb{R}\,$ , $\,f(-x)=f(x)\,$ ,

and $\,f^{-1}(x)\,$ is his inverse for $\,x\geq 0$, then it's possible to solve $\,y=x\sinh x\,$ : $$x_{1,2}=\pm \sinh^{-1} f^{-1}(e^y)$$

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Note:

It's a bit like $\,y=g(x):=1+x^2\ge 1\,$ with $\,x_{1,2}=\pm g^{-1}(y)=\pm\sqrt{y-1}\,$ .

Answer 2

$$f(x)=x\sinh(x)=x\ \left(\frac{1}{2}e^x-\frac{1}{2}e^{-x}\right)$$

$f$ is an algebraic function of $x$ and $e^x$. Liouville and Ritt proved that such kind of functions doesn't have elementary inverses.

$$x\sinh(x)=y$$ $$x\sinh(x)-y=0$$ $$x\ \left(\frac{1}{2}e^x-\frac{1}{2}e^{-x}\right)-y=0$$ $$\frac{1}{2}x\left(e^x\right)^2-ye^x-\frac{1}{2}x = 0$$ Because the degree of this exponential polynomial for $e^x$ is greater than $1$, the equation cannot be solved in terms of Lambert W.

The inverse of $f$ is presented in:
Vazquez-Leal, H.; Sandoval-Hernandez, M. A.; Filobello-Ninoa, U.: The novel family of transcendental Leal-functions with applications to science and engineering. Heliyon 6 (2020) (11) e05418

see also: What are the closed-form inverses of $x \sinh(x), x \cosh(x), x \tanh(x), x\ \text{sech}(x), x \coth(x), x\ \text{csch}(x)$?

Answer 3

Inspired by @IV_’s answer, here is a Lagrange reversion result:

$$y\sinh(y)=x\implies y=x+\sum_{n=1}^\infty\frac1{ n!}\frac{d^{n-1}\left(\frac{xe^{-x}}2-\frac{xe^x}2+x\right)^n}{dx^{n-1}}$$

roughly shown here. The next step is to use the trinomial theorem:

$$\frac1{n!}\frac{d^{n-1}\left(\frac{xe^{-x}}2-\frac{xe^x}2+x\right)^n}{dx^{n-1}}=\sum_{m=0}^n\sum_{k=0}^m\frac{(-1)^k\frac{d^{n-1}}{dx^{n-1}}e^{(2k-m)x}x^n}{(n-m)!(m-k)!k!2^m}$$

Here appears a Laguerre function Rodrigues representation here in $(2)$ to $(5)$, but it is more flexible to use the general Leibniz rule with the confluent hypergeometric function $\,_1\text F_1(a;b;z)$:

$$\frac{d^{n-1}}{dx^{n-1}}e^{(2k-m)x}x^n=\sum_{j=0}^{n-1}\binom{n-1}j \frac{d^{n-1-j}}{dx^{n-1-j}}x^n\frac {d^j}{dx^j}e^{(2k-m)x}=xn!\,_1\text F_1(n+1;2;(2k-m)x)$$

Although it converges slowly, we get:

$$\boxed{y\sinh(y)=x\implies y=\pm x\pm x\sum_{n=1}^\infty\sum_{m=0}^n\sum_{k=0}^m\frac{n!(-1)^k\,_1\text F_1(n+1,2,(2k-m)x)}{(n-m)!(m-k)!k!2^m}}$$

shown here in the substitution section. Additionally, $0\le m\le k\le \infty$ or $0\le k\le m\le \infty$ and if all bounds are $\infty$, then all sums are interchangeable.

Approximate graph of inverse for small $n$:

Is there any way to simplify the boxed result or find other solutions using a series?