# Since $x^{x^x}$ is one to one on the interval $(0,\infty)$ solve for $x$ if $x^{x^x} = y$ and $y>0$

Proof that $$x^{x^x}$$ is one to one the interval $$(0, \infty)$$ is on prove or disprove that $x^{x^x}$ is one to one on the interval $(0,\infty)$. I have tried using the Lambert-W Function but have been unable to come to a solution. How do I solve this?

Note: I was unsure about what tag this should go under, so I may have put the wrong tag. please correct me if I did

• What do you mean by "solve for $x$"? Just because a function is invertible, does not mean its inverse can be expressed in a nice way (with elementary functions, or the Lambert-W function, or whatever). Jul 13, 2019 at 15:49
• @Servaes beat me to it. There's no reason to believe you can express the solution of that equation as $x=x(y)$ for a function composed of elementary functions and the Lambert-W. Jul 13, 2019 at 16:04
• Jul 13, 2019 at 17:05
• Solution of $x^x=y$ for $x$ can be done in terms of the Lambert W function. There is no reason to think that solution of $x^{x^x}=y$ for $x$ can be done in terms of the Lambert W, or indeed any other known function. Jul 13, 2019 at 19:44
• Perhaps you can use calculus to show that the function is increasing on its domain. Jul 13, 2019 at 19:53

The solution is defined as $$x=\sqrt[3]y_s$$, the 3rd super-root of $$y$$. Multiple numerical methods for computing the super-root is given in Calculating the nth super-root when n is greater than 2? A simple method is shown in my answer and is simply an application of bisection.
Analogous to the Lambert W function, define $$H(x)$$ be the inverse function of $$x e^{x e^x}$$, i.e. $$H(x) e^{H(x)e^{H(x)}} = x$$ for $$x\in\mathbb{R}$$. This is well defined for all real $$x$$, since $$xe^{xe^x}$$ is monotonic. The function you are looking for can be found by observing $$e^{H(\log s)e^{H(\log s)*e^{H(\log s)}}} = s$$ so $$e^{H(\log s)}$$ is the 3rd super-root of $$s$$, analogous to how $$e^{W(\log s)}$$ is the super square root of $$s$$. Just as $$W(t)$$ can be written as the integral of an elementary function$$W(t) = \frac{t}{\pi} \int_0^\pi\frac{(1-x\cot x)^2 + x^2}{t + x\cot x e^{-x\cot x}}dx$$ we will find an analogous formula for $$H$$: $$H(t) = \frac{t}{\pi}\int_{-\infty}^\infty \Im\left(\frac{1 + ((x+i\frac\pi4)^2 + x+i\frac\pi4)e^{x+i\frac\pi4}}{t - (x+i\frac\pi4)e^{(x+i\frac\pi4)e^{x+i\frac\pi4}}}\right) dx$$ (See WolframAlpha if you want to know what the integrand looks like in closed form without $$i$$ or imaginary parts.) See below for a proof. This is not particularly nice, but it means that $$x^{x^x}=y$$ can be solved for $$x$$ using only elementary functions and integration. As a numerical check, suppose we want to solve $$x^{x^x} = 2$$. Then we would take $$x = e^{H(\log 2)}$$. Using WolframAlpha, we compute that $$H(\log 2) \approx 0.389799$$. Exponentiating gives $$x=1.47668$$. This is not far off:$$1.47668^{1.47668^{1.47668}} = 1.99998...$$ Just to show that it also works for at least some values between 0 and 1, lets also solve $$x^{x^x} = \frac12$$. We find $$H(-\log 2) = -1.00137$$, hence we should have $$x \approx 0.36736$$ is the solution. Indeed:$$0.36736^{0.36736^{0.36736}} = 0.499984$$
We prove the integral formula for $$H$$ using contour integration: Fix $$t$$. Observe that the function $$f(x) = \frac{1 + (x+x^2)e^x}{xe^{xe^x} - t}$$ has exactly one pole on the real line, at the point $$x=H(t)$$. By noting that $$\frac{d}{dx} xe^{xe^x} = (1+(x+x^2)e^x)e^{xe^x}$$, we find that the residue at this pole is $$\frac1{e^{{H(t)}e^{H(t)}}} = \frac{H(t)}t$$. Hence for any counterclockwise oriented curve in $$\mathbb{C}$$ that surrounds $$H(t)$$ and includes no other solutions to $$ze^{ze^z}=t$$ in its interior, we have $$\frac{1}{2\pi i}\oint_\gamma f(x)dx = \frac{H(t)}{t}$$ Take $$\gamma$$ to be a rectangular contour: Note that the integral along the top minus the integral along the bottom is equal to $$2\int_{-A}^A \Im(-f(x+i\frac\pi4))dx$$ (negative because we are integrating right to left on the top of the box). Thus we need to show that the integral over the left and right sides of the box goes to 0, and that $$f$$ has no more poles inside the region. To see that $$f$$ goes to 0 on the left and right sides, observe that as $$x$$ goes to infinity, if $$|\Im x|<\pi/2$$, then the denominator goes to infinity doubly exponentially, making the whole fraction go to 0 rapidly. When $$x$$ goes to negative infinity, the denominator goes to infinity linearly, while the numerator is bounded. Since both the right and left legs of the contour have constant length, if $$f$$ goes to 0 on them as $$A$$ goes to infinity, these do not contribute to the integral in the limit.
Thus it remains to show that $$f$$ has no other poles in the interior of the contour. Letting $$z=x+i y$$, we compute $$\arg {z e^{z e^z}} = \arg z + \Im(z e^z) = \arg z + e^x x \sin(y) + e^x y \cos(y)$$ For $$z$$ in the subregion defined by $$\Re z>-1$$, $$\Im z \in(0,\pi/4)$$, all terms are positive, so $$z e^{z e^z}$$ is guaranteed to be nonreal. Note also that for $$\Re z < -1$$, we have $$|z e^{z e^z}| > e^{-|z e^z|} = e^{-|z| e^x} \ge e^{-|x|e^x -\frac\pi4 e^x} \ge c$$ we $$c=\min\limits_{x<-1} e^{-|x|e^x -\frac\pi4 e^x} = \exp\left(\frac{-4-\pi}{4e}\right)\approx 0.5185$$. Hence if $$|t|<0.5185$$, $$f$$ cannot possibly have any poles in the region of integration, so we can solve $$x^{x^x} = y$$ using this method at least for $$|\log x|< c$$. It should also work in a wider range, by analytic continuation in the largest interval where the integral converges it ought to give correct values for $$H$$, since the integrand is an analytic function of $$t$$. Numerically it seems to fail for $$x=\frac18$$.
• Does this mean any super root $\sqrt[n]{y}_s,n\in\Bbb N$ has a single integral representation in terms of elementary functions and complex components? Mar 9 at 23:51