Initially I isolated the y in $x^y=y$, but I just wanted to expand the infinite power tower to two letters in the tower, but I can't solve for z in the equation $x^{y^z}=z$. I tried to use Lambert W Function but I can't get the form $ze^{z}$ or something like this. I will be thankful if someone help me.
Solve $x^{y^z}=z$
lambert-wpower-towers
Related Solutions
There are a few problems with addition.
$$W(x+a)=?$$
$$\log(x+a)=?$$
The inverse functions of exponential related functions don't like addition on the inside. Addition on the outside is perfectly fine though.
$$\log(x)+a=\log(e^ax)$$
Except even then, the Lambert W Function won't simplify if you have addition on the outside... most of the time. And this question isn't one of the ones that simplifies.
So the best is to go with linear approximation methods.
$$1-x=\frac{\alpha x}{e^{\alpha x}-1}$$ $$(1-x)(e^{\alpha x}-1)=\alpha x$$ $$(1-x)e^{\alpha x}-1+x=\alpha x$$
Your equation is an equation of elementary functions. For rational $\alpha$, the equation is related to an algebraic equation in dependence of $x$ and $e^x$. Because the terms $x,e^x$ are algebraically independent and the equation is irreducible, we don't know how to rearrange the equation for $x$ by only elementary operations (means elementary functions). A theorem of Lin (1983) proves, if Schanuel's conjecture is true, that irreducible algebraic equations involving both $x$ and $e^x$ don't have solutions in the elementary numbers.
$$(1-x)e^{\alpha x}=\alpha x+1-x$$ $$(1-x)e^{\alpha x}=(\alpha-1)x+1$$ $$\frac{1-x}{(\alpha-1)x+1}e^{\alpha x}=1$$ $$\frac{1-x}{x+\frac{1}{\alpha-1}}e^{\alpha x}=\alpha-1$$
We see, your equation cannot be solved in terms of Lambert W but in terms of Generalized Lambert W:
$$\frac{x-1}{x+\frac{1}{\alpha-1}}e^{\alpha x}=1-\alpha$$ $$\frac{x-1}{x-\frac{1}{1-\alpha}}e^{\alpha x}=1-\alpha$$ $$x=W\left(^{\ \ \ 1}_{\frac{1}{1-a}};1-a\right)$$
The inverse relation of your kind of equations is what Mezö et al. call $r$-Lambert function. They write: "Depending on the parameter $r$, the $r$-Lambert function has one, two or three real branches and so the above equations can have one, two or three solutions"
So we have a closed form for $x$, and the representations of Generalized Lambert W give some hints for calculating $x$.
[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mező/Baricz 2017] Mező, I.; Baricz, Á.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018
Best Answer
Series solution:
$\def\B{\operatorname B}$
Using $\ln(x)=a,\ln(y)=b$ and Lagrange reversion we get a series for “Hyper Lambert W”
$$x^{y^z}=e^{ae^{bz}}=z\implies z=\sum_{n=1}^\infty \frac1{n!}\left.\frac{d^{n-1}}{dz^{n-1}}e^{ane^{bz}}\right|_{z=0}$$
$e^y$ series:
$$\left.\frac{d^{n-1}}{dz^{n-1}}(e^{ane^{bz}})\right|_{z=0}=\sum_{m=1}^\infty\frac{(an)^m}{m!}\left.\frac{d^{n-1}}{dz^{n-1}}e^{mbz}\right|_{z=0}=\sum_{m=1}^\infty\frac{(an)^m (bm)^{n-1}}{m!}$$
and finally Bell B$_n(x)$ polynomials to get:
$$\boxed{e^{ae^{bz}}=z=1+\frac1b\sum_{n=1}^\infty \frac{(ae^b)^n}{nn!}\B_n(b n)=\frac1b\sum_{n=1}^\infty\frac{(b e^a)^n}{n!}\B_{n-1}(an)}$$
Shown here
Experimental:
A contour integral representation is:
There are no integral bounds and I have little experience with contour integrals. In case switching the contour integral and sum works:
$$e^{ae^{bz}}=z\mathop=^?1+\frac i{2\pi b}\oint\frac{\ln\left(1-\frac{ae^{be^t}}t\right)}tdt\mathop=^?\frac i{2\pi b}\oint\frac{\ln\left(1-\frac{be^{ae^t}}t\right)}tdt$$