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Is it hard to find out for which $r$ the infinite power tower $r\uparrow r \uparrow r\uparrow \cdots$ converges?

Let $r$ be a real number satisfying $0<r<e^{-e}$. In order to prove that the iteration $x_1=r$ , $x_{n+1}=r^{x_n}$ does not converge (hence the infinite power tower does not converge), I had the following idea which could solve the problem, but I was not successful.

The equation $r^{r^x}=x$ seems to have $3$ solutions and the middle solution (sorted in magnitude) seems to be the solution of $r^x=x$. First of all, is this correct ?

The equation $r^{r^x}=x$ is equivalent to $r^x\ln(r)=\ln(x)$. I have no idea how to show that it has $3$ solutions. And what I have to show next (if this is the case) that the above iteration eventually switches between values close to the smallest and values close to the largest solution.

This would show that the middle solution is not reached.

  • $\begingroup$ There exist attracting and repelling fixpoints, respective preiodic fixpoints. Try to get the "middle" fixpoint by the inverse of the iteration, or perhaps better the Newton-algorithm when started in the near. I think we have recently had such a discussion here, (at the moment just too lazy to search...) $\endgroup$ Dec 9, 2017 at 10:08
  • $\begingroup$ @GottfriedHelms Am I on the right track ? $\endgroup$
    – Peter
    Dec 9, 2017 at 10:12
  • 1
    $\begingroup$ Peter, look at math.stackexchange.com/questions/2490769/… and especially the pictures of Sheldon . I myself can come back to this later today... $\endgroup$ Dec 9, 2017 at 10:18
  • $\begingroup$ @GottfriedHelms. I apologize for not having seen your contour plot in your previous answer (the link given in your comment). $\endgroup$ Dec 10, 2017 at 8:38
  • $\begingroup$ @Claude - thanks for the notice, and please feel comfortable! $\endgroup$ Dec 10, 2017 at 10:30

2 Answers 2


Considering $$f(x)=r^x\log(r)-\log(x)$$ $$f'(x)=r^x \log ^2(r)-\frac{1}{x}$$ $$f''(x)=r^x \log ^3(r)+\frac{1}{x^2}$$ the first derivative cancels for two values of $x$, namely $$x_a=\frac{W_0\left(\frac{1}{\log (r)}\right)}{\log (r)}\qquad \text{and}\qquad x_b=\frac{W_{-1}\left(\frac{1}{\log (r)}\right)}{\log (r)}$$ where appear Lambert functions.

By the second derivative test, $x_a$ corresponds to a minimimum (with $f(x_a) <0$) and $x_b$ corresponds to a maximum (with $f(x_b) >0$).

So, for $f(x)=0$, there are three roots $a,b,c$ such that $$0 < a < x_a \qquad x_a < b < x_b \qquad x_b< c < 1$$ If fact, we could even show that the second root is such that $x_a < x <\frac 12(x_a+x_b)$.

So, depending on the starting point, Newton method would converge to one of these roots.

Concerning the middle root, it effectively corresponds to the solution of $r^x=x$ that is to say $x=-\frac{W_0(-\log (r))}{\log (r)}$.

For example, using $\color{red}{r=\frac 1 2 e^{-e}}$ and using $x_0=\frac 12 x_a$, we should get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.06861303857 \\ 1 & 0.06484498458 \\ 2 & 0.06506516116 \\ 3 & 0.06506598462 \\ 4 & 0.06506598464 \end{array} \right)$$ Using $x_0=\frac 12 (x_a+x_b)$, we should get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.3372962715 \\ 1 & 0.3270913322 \\ 2 & 0.3273489201 \\ 3 & 0.3273490811 \end{array} \right)$$ Using $x_0=\frac 12 (1+x_b)$, we should get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.7686832329 \\ 1 & 0.8022363077 \\ 2 & 0.8009427413 \\ 3 & 0.8009410041 \end{array} \right)$$


We could have done almost the same solving for $r$ to get $$r_1=\left(\frac{x \log (x)}{W_0(x \log (x))}\right)^{\frac{1}{x}}\qquad \text{and}\qquad r_2=\left(\frac{x \log (x)}{W_{-1}(x \log (x))}\right)^{\frac{1}{x}}$$ For $r \leq e^{-e}$ there is only one $x$ given by $r_1$, while there are two $x$'s for $r_2$.

  • $\begingroup$ @Peter. I did not notice that, in his answer, Gottfried Helms gave the contour plot of $r^x\ln(r)-\log(x)=0$ for $0\leq r \leq e^{-e}$ and $0 \leq x \leq 1$. $\endgroup$ Dec 10, 2017 at 5:19

A plot of the function $y_2=b^{b^x}$ using $b=0.01$ (for instance in the picture below) together with $y_0=x$ shows the three fixpoints. Looking at the curve for $y_1=b^x$ as well explains also, why there are three fixpoints.

[update - the proof for existence of three fixpoints]

The existence of two more crossing points where $t_k=ff(t_k)$ (fixpoints) is needed by two conditions.

Let's define $f(x)=b^x$, $ff(x)=b^{b^x}$ , and the identity function $g(x)=x$.
We know/observe, that $f(t)=t$ means $b=t^{1/t}$ and writing $u=\log(t)$ means $\log(b)= u / \exp(u)$ .

In the OP we have selected some $b < e^{-e}$ and from this we find a determination for $u$ and $t=\exp(u)$:
$$\begin{array} {} \log(b)&=u / \exp(u) &\lt -1/ \exp(-1) = - 1 e^1 \\ \implies & -u \cdot \exp(-u) &> e \\ \implies & -u &> W(e)=1 \\ \implies & u &< -1 \\ \implies & | u| &> 1 \\ \end{array}$$

So the fixpoint for $f(x)$ is $t=\exp(u)< e^{-1} \approx 0.367879 $ and is of course a fixpoint for $ff(x)$ .

Let now, for example, $b=0.01$. Then $u =-1.28... $ and $t=0.278$.

We look at the first graph and want to argue that for $ff(x)$ besides of the fixpoint $t$ in the middle there need to be two more fixpoints (crossing points of the curve and the line), left and right to that fixpoint $t$.

  • (condition 1): $f(0)=b^0 = 1$ and thus $ff(0)=b$. The y-coordinate of the red curve at $x=0$ is $b$ and thus greater than the y-coordinate of $g(0)=0$ at $x=0$.

  • (condition 2): The function for the derivative of $ff(x)$ is $ff'(x) = \log(b)^2 b^x b^{b^x} $ At the fixpoint $t$ is thus the slope $ \log(b)^2 t^2 = (u/t)^2 t^2=u^2 >1 $ and thus must cross the line $g(x)$ from below. So the values for $ff(x)$ in the near left neighbourhood of $t$ must be smaller than the y-coordinate $g(x)$.

  • Because now at $x=0$ we have $ff(x)> g(x)=0$ but in the near left neighbourhood of $x=t$ we have $ff(t-\delta)< g(t-\delta)$ there must be some other point $t_1$ where $ff(t_{-1}) = g(t_{-1}) = t_{-1}$ and this is thus another fixpoint. Actually we find $t_{-1} = 0.013092... $.

  • The similar consideration for $x$ near $1$ gives a third fixpoint $t_1$ with $ff(t_1)=t_1 = 0.941488... $ Both fixpoints are moreover in the relation $f(t_{-1})=t_1$ and $f(t_1)=t_{-1}$


See this plot: picture The red curve for the function $b^{b^x}$ crosses three times the blue $y=x$ line and by the grey curve for $b^x$ it should be easily recognizable why this must be thus.

Added: There exists this little bonmot: "If you don't really understand, generalize". Things might even be clearer when more iterations are in the picture.


It suggests that even-height ($h=2k >0$) iterates should have three fixpoints and odd-height ($h=2k+1$) iterates should have one fixpoint.


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