I got this problem from my teacher as a optional challenge. I am open about this being a given problem, however it is not homework.

The problem is stated as follows. Assume we have an infinite tetration as follows

$$x^{x^{x^{.^{.^.}}}} \, = \, a$$

With a given $a$ find $x$. The next part of the problem was to discuss the convergence radius of a. If a is too big or too small the tetration does not converge.

Below is my humble stab at the problem.

My friend said you would have to treat the tetration as a infinite series, and therefore could not perform algebraic manipulations on it before it is know whether it converges or diverges.

However my attempt is to first do some algebraic steps, then discuss the convergence radius.

I) Initial discussion

At the start it is obvious that the tetration converges when $a=1$ (just set $x=1$) Now after some computer hardwork it seems that the tetration fails to converge when a is roughly larger than 3.

II) Algebraic manipulation

$$ x^{x^{x^{.^{.^.}}}} \, = \, a$$

This is the same as

$$ x^a \, = \, a$$

$$ \log_a(x^a) \, = \, \log_a(a)$$

$$ \log_a(x) \, = \, \frac{1}{a}$$

$$ x \, = \, a^{\frac{1}{a}}$$

Now, if we let $a=2$ then $x = \sqrt{2}$. After some more computational work, this seems to be correct, which makes me believe this formula is correct.

III) Discsussion about convergence

By looking at the derivative of $ \displaystyle \large a^{\frac{1}{a}} $ we see that the maxima occurs when $a=e$, which also seems to correspond to the inital computational work. Now I think, that the minima of $\displaystyle \large a^{1/a}$ is zero by looking at its graph, studying its derivative and the endpoints.

So that my "guess" or work shows that it converges when

$$ a \in [0 \, , \, 1/e] $$

VI) My questions

Can my algebraic manipulations be justified? They seem rather sketchy taking the a`th logarithm and so on . (Although they seem to "magically" give out the right answer)

By looking at Wikipedia it seems that the tetration converge when $$ a \in \left[ 1/e \, , \, e \right] $$

This is almost what I have, why is my lower bound wrong? How can I find the correct lower bound?

  • $\begingroup$ Perhaps you want to take a look at this question : I must say though it is not a duplicate question, but the question you ask is related. (This link is dealing with $a=2$ and maybe other users will think of some other linked questions) math.stackexchange.com/questions/87870/… $\endgroup$ Feb 11, 2012 at 22:50
  • $\begingroup$ I am confident that there is probably an answer to this question already written out on this website. $\endgroup$ Feb 11, 2012 at 22:50
  • $\begingroup$ Yes, I actually searched before asking I could not find anything dealing with the generall case. Nor anything dealing with how to find the lower bound. Also this is not a straight duplicate, as I ask if my method is correct. =) $\endgroup$ Feb 11, 2012 at 23:06
  • $\begingroup$ Recall that the logarithm function is a multi-valued function in the complex numbers... "taking the log" might be nasty, as detailed in the link I've added, we had issues with that. Maybe there's a link in my linked question you could take a look at. $\endgroup$ Feb 11, 2012 at 23:13
  • $\begingroup$ Look at $x, x^x, x^{x^x}, \ldots$. In other words, starting with $a_0 = x$ and then $a_{n+1}=x^{a_n}$, you would want $\lim_{n \to \infty} a_n = a$. This does not converge if $x \lt e^{-e}$. $\endgroup$
    – Henry
    Feb 11, 2012 at 23:34

3 Answers 3


First, I would recommend reading "Exponentials Reiterated" by R.A.Knoebel, http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=3087&pf=1. It is by far the most comprehensive resource on infinite tetration. Second, your question might have been answered sooner if you had posted on the Tetration Forum, http://math.eretrandre.org/tetrationforum/index.php. Lastly, I think it is very encouraging to see others interested in this subject, since I've been interested in it for quite a long time now.

  • $\begingroup$ Funny, I linked to that in this answer... whose corresponding questions was what Pat was pointing to in the comments. $\endgroup$ Feb 15, 2012 at 17:03
  • 3
    $\begingroup$ Hi, welcome Andy at MSE! Good to hear of you after a long time... $\endgroup$ Feb 15, 2012 at 17:17



In order to have this converge, we want


where $a=f(a)$ and $t$ is sufficiently close to $a$. If there are multiple solutions, then $a=\min\{t=f(t)\}$. Dividing both sides by $|t-a|$ and rewriting, we see we need to have


By the mean value theorem, we need


for some $\xi$ between $t$ and $a$. Since $f'$ is continuous, this means we must have


We proceed to find that


and thus, we want


leading to $a\in[1/e,e]$. Since $a=x^a$, we have $x=a^{1/a}$, and hence $x\in[1/e^e,e^{1/e}]$.

Convergence on the boundaries is not guaranteed. For $x=1/e^e$ and $a=1/e$, note that $f(f(t))$ is increasing and bounded above by $a$ for $t<a$, and decreasing and bounded below by $a$ for $t>a$. For $x=e^{1/e}$ and $a=e$, note that $f(t)$ is increasing and bounded above by $a$ for $t<a$. Hence it will converge on the boundaries.

Likewise note that it cannot converge outside of this range unless the initial value is equal to $a$, with $x=-a$, for example.

And so we have


as was claimed.

  • $\begingroup$ I would really like to know how this work, why do the fixed points of $f\left( t\right) =x^{t}$ only converge if $\left| f'\left( a\right) \right| =1$? And how do you show convergence at the $e^{-e}$ boundary? $\endgroup$
    – JimSi
    Jul 24, 2018 at 9:42
  • $\begingroup$ @JimSi updated. $\endgroup$ Jul 25, 2018 at 12:50
  • $\begingroup$ If $x=1/e^e$, then $f(f(t))$ is not decreasing anywhere. $\endgroup$
    – 1mik1
    Jan 31, 2021 at 18:03
  • $\begingroup$ Also, if $x=1/e^e$, why don't you consider $f(t)$ instead of $f(f(t))$? $\endgroup$
    – 1mik1
    Jan 31, 2021 at 18:15
  • $\begingroup$ Actually $|f(t)-a|<|t-a|$ is sufficient, but not necessary for the convergence. $\endgroup$
    – cyanide
    Apr 14, 2022 at 6:35

[note: As mentioned in comments, this is actually infinite exponentiation, not infinite tetration.]

I've played around with this. For $e^{-\frac{1}{e}} \le x \le e^{\frac{1}{e}}$ it converges as you insert extra terms at the bottom of the tower, and for $x=e^{-\frac{1}{e}} $ it looks as though it does but I've not proved it.

Call $T_n$ the value of the $n$th tower (eg $T_3 = x^{x^x}$)

For $1<x\le e^\frac{1}{e}$ the convergence happens because for each $n$, $T_n <T_{n+1}<e$ so you've got an increasing sequence bounded above by $e$. (An incressing sequence which is bounded above is guaranteed to converge.)

For $x<1$ things are trickier because you get an alternating sequence and have to prove that it converges to one value rather than odd terms converging to one value and even ones to another.

We can also prove that if all terms of an infinite tower are in the range $[1,e^\frac{1}{e}]$ it will converge: it's not even necessary for the sequence of terms (as opposed to the sequence of towers) to converge as long as no more than finitely many are outside that range.


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