I was just wondering about this. I searched about it on the net and found that it is called tetration and after this comes pentation and then hexation and so on so forth.

I don't really understand tetration but I'm curious about the answer. What is $i$ raised to itself $i$ times? Also if it makes any sense, what is $i$ 'pentated' or 'hexated' to itself $i$ times? Is there even an answer to this? And if yes, till which 'stage' can we do this ('stage' in the sense 'tetration', 'pentation')? Can we do this indefinitely? Is there a formula for it?

I will apologise if this question has been asked earlier.

I think my question is a little unclear. Here's what I'm saying:

$i$ exponentiated to itself $1$ time: $i^1$

$i$ exponentiated to itself $2$ times: $i^{i^{1}}$

$i$ exponentiated to itself $3$ times: $i^{i^{i^{1}}}$

$i$ exponentiated to itself $i$ time(s): ?

Edit: The 'possible duplicate' links provided are not asking the same thing. Here's another try:

From Wikipedia

Now consider the case where $a = i$ and $n = i$.

I'm not asking what is $i$ multiplied by itself $i$ time(s) (which is equal to $i^i$). I'm asking what is $i$ to the power of $i$ to the power of $i$... this is done '$i$ times'.

  • 3
    $\begingroup$ I think what you are asking about is the extension of exponentiation towers with an integer count of the argument, to real or even complex "counts." There has been no success defining such a thing. See original sources at zakuski.utsa.edu/~jagy/other.html $\endgroup$
    – Will Jagy
    Jan 16 '13 at 19:22
  • $\begingroup$ see this link as well math.stackexchange.com/q/275075/8581 $\endgroup$
    – Mikasa
    Jan 16 '13 at 19:25
  • 1
    $\begingroup$ @GitGud : I'd have written $i^i=e^{i\log i}$, with \log, with a backslash. This not only prevents italicization of $\log$, but also provides for proper spacing before and after $\log$ in expressions like $a\log b$. $\endgroup$ Jan 16 '13 at 19:26
  • 3
    $\begingroup$ @MJD Please ready my question again. I'm asking what is $i$ 'tetrated' to $i$ (not sure if 'tetrate' is a word though). $\endgroup$
    – Alraxite
    Jan 16 '13 at 20:08
  • 1
    $\begingroup$ @GottfriedHelms Right, I'll change that. $\endgroup$
    – Alraxite
    Jan 17 '13 at 20:11

As Will Jagy already pointed out, there is no accepted solution for this. There is a formal procedere which can sometimes lead to a meaningful/approximate answer; but this is then dependent on the convergence of some series, which occur in that procedere, and also, in which way we want to make sense of noninteger powers of negative or complex numbers.
I mean the procedere which applies the Schröder-function on a recentered powerseries.
The formal way to do this is to

  • determine a fixpoint t for the exponentiation with base b such that $b^t = t$ (here our base is $b=i$ ). Also denote the log of $t$ as $\log(t)=u$
  • define the helper function $f(x) = t^x-1 $
  • compute the (lower) triangular Carleman-/Bell-matrix for $f(x)$ , call it for instance $U$
  • diagonalize the matrix $U$ into $M$, $D$, $W (=M^{-1})$ . Then in the diagonal of $D$ are the consecutive powers of $u$
  • define the Schröder-function $s$ from the second column of $M$ such that $ s(x) = \sum_{k=1}^\infty m_{k,1} x^k $ and its inverse from the second column of $W$ such that $ s^{-1}(x) = \sum_{k=1}^\infty w_{k,1} x^k $
  • compute the schröder-value $\sigma = s(x/t-1) $ for some argument $x$. (For instance $x=1$ -this is also tacitly assumed if we write b^^h for the h'th tetrate with base b)
  • compute the function value for the h'th tetration-height by $x_h = (s^{-1}(\sigma u^h)+1) \cdot t$

I've just tried this with the base $b=i$ in Pari/GP (that gave also $ \small t \sim 0.438282936727 + 0.360592471871 i $ and $ \small u \sim -0.566417330285 + 0.688453227108 i$) and for small integer heights h this gives good approximations (although we have complex powerseries involved(!)). However, we see in the last step, that the log $u$ of the fixpoint has to be taken to the h'th power - and this means for your question a complex value to the $i$'th power. This is not unique and the powers of this ("incidentally" selected) value occur then in the formula for the inverse Schröder-function.
Now, after I simply inserted $h=i$ and let Pari/GP determine the result using $\sigma u^i $ for the $i$'th tetrate (beginning from $x_0=1$) with basis $i$ then I arrive at something like $x_i = i^{\text{^^}i} \sim 0.500129061734+0.324266941213 i $ .

This whole procedure has - even for the case of a real base and real fixpoints, the further unsolved disadvantage, that the result will be depending on the selection of the fixpoint $t$ . Moreover, even for simple fractional $h$ , say $h=0.5$ , the occuring series are not or only roughly converging correctly, such that the half-iterate from $x$ to $x_{0.5}$ and then the further half-iterate from $x_{0.5}$ to $x_1$ is only approximate with (practically necessarily) truncated power series. So besides the non-uniqueness at the point, where we raise the complex $u$ to the $i$'th power in our specific example, there is also basically a problem of convergence with that whole procedure.

Note, there is one claim that a general solution for the tetration to complex heights was found; look at citizendium at the article of Dmitri Kouznetsov, but I've not yet seen that it had externally been confirmed (and I cannot do it myself). Also we have in the tetration-forum the claim, that we can uniquely determine fractional tetrates at least for some "nontrivial" bases, which is basically derived from Kneser's ansatz for the fractional iteration with base $b=e$. Look at the posts of user "Sheldonison" (Levenstein) who even provides a set of Pari/GP-procedures to compute fractional tetrates for a range of bases outside the "easy" range $1 \ldots e^{1/e} $ of real bases. Unfortunately I'm not yet capable to judge over that claims of Kouznetsov and of Levenstein.

[update] There has been an attempt today to insert in my answer a statement, that the Kouznetsov-method were "the official" tetration, based on some consideration of reasons. I reject that. Without more qualified echo in the professional mathematical community it is not my part to claim such and generate avoidable obfuscation. Please do not attempt to extend the focus of my answer in such a way - you can always put it in another answer.

  • $\begingroup$ I get I^^I=0.4248013286975477+0.4249736033147313i, which is apparently different than Gottfried. I get the same fixpt=0.4382829367270321+0.3605924718713855i, and the same lambda=-0.5664173302854644+0.6884532271077021i. I also generated the Schroder and inverse Schroder series; presumably the same as Gottfried. However, convergence is limited, so I iterated starting with z=I, z=I^z 20 times, before using the Schroder function of z-fixedpt. Call this y=schroder(z-fixpt). Then y=y*lambda^I. Then calculate the y=inverse_schroder(y)+fixpt. Finally, iterate 20 times, y=log(y)/log(I). $\endgroup$
    – Sheldon L
    Jan 16 '13 at 23:59
  • $\begingroup$ With 50 terms schroder/inverse_schroder functions, and zero iterations, I actually get reasonably good results, I^^I~=0.4248427934801628+0.4250051842670692i. So really, only one iteration required to get 15 digits, I^^I=0.4248013286975477+0.4249736033147313i. $\endgroup$
    – Sheldon L
    Jan 17 '13 at 0:06
  • $\begingroup$ Hmm, I arrive at $0.424801328698 + 0.424973603315 i$ if I start at $x=1$, iterate 20 times; compute $y$ by the Schröder-mechanism and then iterate only 19 times $y=\log(y)/log(b) $. So it seems, our results are one iteration apart of each other? $\endgroup$ Jan 17 '13 at 0:25
  • $\begingroup$ @SheldonL: So I get that value I^^I~ $0.500129061734 + 0.324266941213 i$ and I^^(1+I) ~ $ 0.424801328698 + 0.424973603315 i$ with my matrices... $\endgroup$ Jan 17 '13 at 0:38
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    $\begingroup$ The author hints that I^^1=I^I, where as I agree with Gottfried, that in terms of tetration, I^^2=I^I, so that I^^0=1, I^^1=I, I^^2=I^I. Perhaps that threw me off the trail ... :) $\endgroup$
    – Sheldon L
    Jan 17 '13 at 2:41

Well, why not. The goal of such topics is to take a function $f(z)$ defined in some region of the complex plane, often including the real line or part of it, then defining $f_1(z) = f(z)$ and $f_0(z) = z.$ The desire is to be able to define $f_s(z)$ for real or even complex $s,$ such that we get a group $$ f_{s+t}(z) = f_t(f_s(z)) = f_s(f_t(z)). $$ Now, the only setting where this is an unreserved success is when $f$ is a linear fractional transformation. The traditional topic of fractional iteration is asking about $$ f_{1/2}(z). $$ As Baker found, and his student Liverpool finished up, it is fairly common to have such fractional iterates without being able to extend to irrational $s$ or non-real $s.$

I am actually not completely certain how exponentiation towers are set up in this setting. Still, one way or another, you are asking to extend from $s$ a positive integer to $s=i,$ and there is simply no guarantee that there is any way to do this. Certainly none has been found.

As I said in comment, see BAKER_ECALLE. This topic is most naturally part of complex dynamics, although it may not appear that way now.


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