Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $w = z^{z^{z^{...}}}$ converges, we can determine its value by solving $w = z^{w}$, which leads to $w = -W(-\log z))/\log z$. To be specific here, let's use $u^v = \exp(v \log u)$ for complex $u$ and $v$.

Two questions:

  • How do we determine analytically if the tower converges? (I have seen the interval of convergence for real towers.)
  • Both the logarithm and Lambert W functions are multivalued. How do we know which branch to use?

In particular $i^{i^{i^{...}}}$ numerically seems to converge to one value of $i2W(-i\pi/2)/\pi$. How do we establish this convergence analytically?

(Yes, I have searched the 'net, including the tetration forum. I haven't been able to locate the answer to this readily.)

share|improve this question
    
You may look for "Shell-Thron-region", which is the complex extension for the "Euler-interval" for real numbers (where the infinite power tower of a real basis converges), see for instance math.eretrandre.org/hyperops_wiki/… . i is inside that region, so the infinite power-tower converges (I think, only the principal branch is always selected). As far as I recall the original article of D. Shell is online available for more details. –  Gottfried Helms Aug 10 '12 at 6:06
2  
Thron's article and Shell's article. In any event, remember that the power function itself is multivalued... –  J. M. Aug 10 '12 at 6:26
    
"How do we determine analytically if the tower converges" (jstor.org/discover/10.2307/2300357) had some analysis for determining the convergence of infinite exponentials. "Both the logarithm and Lambert W functions are multivalued. How do we know which branch to use?" IIRC, the branches of h, are split-level with-respect-to W, i.e. inside the Shell-Thron region you have to use W branch n, and outside the Shell-Thron region you have to use W branch n+-1 (I don't remember), if you want the real part/imaginary part of the result to fall within some specific range. –  Andrew Robbins Aug 17 '12 at 14:18

2 Answers 2

(Edit-version 2)

If you have a base $b$ such that you look at $b^{b^{b^{...}}}$ then find a solution for $t$ such that $b = t^{1/t}$ If you have such a $t$, then look at its logarithm $u = \ln(t)$. If $|u| \le 1$ then the infinite tower is a convergent expression. Note that the function h(x), such that $t = h(b) \to t^{1/t} = b $ is multivalued and you take the principal value) I've made a picture enter image description hereabout this "Shell-Thron-region" in the tetration-forum (the picture reflects only the upper halfplane, the full picture is symmetric around the x-axis).

The blue curve indicates the complex bases $b$ on the boundary between convergence and divergence of the resp. infinite powertower. Outside of this curve the powertower diverges. To each point on this curve there is another point $t$ in the complex plane associated (which is on the magenta curve). I connected some example points $b=t^{1/t} $ with a grey line. The yellow curve indicates the points $u$, the logs of the points $t$.

Inside (and on) the yellow circle (with radius 1 ) are all points $u$ whose exponentials $t$ are inside (and on) the magenta curve and whose associated bases $b$ are inside (and on) the blue curve -and thus whose infinite powertower with base b is convergent.
For values $b$ outside the blue curve, the corresponding $t$ are outside the magenta and the corresponding $u$ are outside the yellow curve the infinite powertower diverges.
(Note, that due to the multivaluedness we can have values $u$ and $t$ outside their curves which have corresponding $b$ inside the blue curve, but that doesn't matter for the question since there must then another value $u$ and $t$ inside their curved regions)

Your example is $b=i$, and that value is inside the blue curve, so the infinite powertower is convergent.

share|improve this answer

You might try using one of the following series expansions:

${a_1}^{{a_2}^{{.^{.^{a_n}}}}} = {\large\rm T}_{k=1}^{n} a_k = \sum_{k_j \ge 0 \atop 1 \le j \le n} \prod_{i=1}^{n}\frac{{(k_{i-1} \ln(a_i))}^{k_i}}{(k_i)!}$

which Barrow (link above) gives a variant of without logarithms, or

$a^{a^{a^{.^{.^{.}}}}} = \exp_a^{\infty}(1) = \sum_{k=0}^{\infty} \frac{\ln(a)^{k}}{k!} (k+1)^{(k-1)} = \sum_{k=0}^{\infty} \frac{(a - 1)^{k}}{k!} \sum_{j=0}^{k}\left[{k \atop j}\right] (j + 1)^{(j - 1)}$

which is just a substitution of variables in the Lambert-W series, the second series is just the Stirling transform of the first.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.