Root of $x = \tanh(a x)$ Consider the equation:
$$x = \tanh(a x)$$
Given $a\in\mathbb{R}$, let $x=f(a)$ denote the maximum non-negative solution of this equation.
Is there a name for the function $f(a)$? Or can it be expressed in terms of known functions?
This equation comes up in statistical physics, for instance in the Curie-Weiss model. See Eq. 2.12 in https://www.unige.ch/math/folks/velenik/smbook/Curie-Weiss_Model.pdf.
 A: I don't think that there is a suitable function. Lambert's W seems insufficient.
Anyway, notice that the inverse function can easily be expressed:
$$a=\frac{\text{artanh}(x)}{x}.$$

A: You want to solve for $x$ the equation $$x = \tanh(a x)\tag1$$ which, being transcendental, will require some numerical method and, as usual, a reasonable estimate of the solution.
However, from a formal point of view, there is an anlytical solution. Let $ y=ax$  and rewrite $(1)$ as
$$\tanh(y)=\frac{e^{2 y}-1}{e^{2 y}+1}= \frac ya\tag 2$$ that is to say
$$e^{-2y}=-\frac{y-a}{y+a} \tag 3$$ which has solutions in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).
This being said, it is not very useful from a practical point of view.
For iterative methods, you could use, as an approximation, a $[2n,2n]$ Padé approximant. For example
$$\frac{\tanh (y)}{y}\sim \frac{y^4+105 y^2+945}{15 \left(y^4+28 y^2+63\right)}=\frac 1a$$ is more than decent for $0.2 \leq \frac 1a \leq 1.0$ and let you with solving a quadratic equation in $y^2$.
For sure, you could have better but it will require at least to solve a cubic or quartic in $y^2$.
