Infinite tetration, convergence radius

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 infinnite serie, and therefore coul not perform algebraic manipulations on it before it is know wheter 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 belive this fomula 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 with the inital computational work.
Now I think, that the minima of $\displaystyle \large a^{1/a}$ is zero, by looking at its graph. And study its derivative and endpoints.

So that my “guess” or work shows that a 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 “magicaly”
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 worng? How can I find the correct lower bound?

Solutions Collecting From Web of "Infinite tetration, convergence radius"

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.

Define

$$f(t)=x^t$$

In order to have this converge, we want

$$|f'(a)|<1$$

where $a=f(a)$. If there are multiple solutions, then $a=\min\{t=f(t)\}$. Likewise, it diverges if

$$|f'(a)|>1$$

To find the boundaries of convergence, we are interested in finding

$$|f'(a)|=1$$

We proceed to find that

$$f'(a)=x^a\ln(x)=a\ln(x)$$

Setting this equal to $1$, we get

$$\frac1a=\ln(x)$$

We also have the relationship

$$x=a^{1/a}\implies\ln(x)=\frac1a\ln(a)$$

and thus,

$$\frac1a=\frac1a\ln(a)\implies a=e$$

And so one boundary lies at

$$x=e^{1/e}$$

The other boundary is found by setting the derivative equal to $-1$,

$$-\frac1a=\ln(x)=\frac1a\ln(a)\implies a=1/e$$

And thus,

$$x=1/e^e$$

It is then easy to show convergence at these boundaries, and so we have

$$x\in[1/e^e,e^{1/e}]$$

as was claimed.

(To be completely rigorous, note that $f”(t)\ge0$, and so $|f'(t)|<1$ for $x\in[1/e^e,e^{1/e}]$ and $|f'(t)|>1$ everywhere else (for $x>0$.))

The domain for x is [ (1/e)^e , e^(1/e) ].

For x>e^(1/e), a fails to converge to any finite value because of a straightforward shoot-up to +∞.

But for x<(1/e)^e, a fails to converge to a single specific value because the tetration returns bimodal output for a that goes back-and-forth: a1, a2, a1′, a2′, a1”, a2”…

This bimodalism extends for a short way into the convergent part of the domain, also, but when it does the gap between a1 and a2 slowly closes itself as the tetration goes on for a while. It eventually converges when a1 and a2 meet somewhere in the middle of where they were to start with. Near the left edge of the convergent interval, and about where the bimodal nonsense stops, you find this point: (x=0.06638, a=0.36828074).

There’s also an inflection point at (x=0.39440125, a=0.58192445), or somewhere close to there. It’s the part of the curve that has the most nearly constant slope, so finding exactly where the inflection is can be difficult.

The only special points (x,a) that I know about are:

(1/4, 1/2)
(1, 1)
(√2, 2)