# i^i^i^i^… Is there a pattern? [duplicate]

I was messing around with $i$ and I (haha) noticed that certain progressions arise when I keep on raising $i$ to $i$ to $i$ and so forth. Though, I am not really quite sure what is going on (and I don't have time to explore further).

In other words, is there an interesting pattern in the sequence:

$i$ , $i^i$, $i^{\left(i^i\right)}$, $i^{\left(i^{\left(i^i\right)}\right)}$, etc.

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## marked as duplicate by GEdgar, Claude Leibovici, egreg, Macavity, C-SMar 8 '14 at 15:59

Comment: remember that $i^i$ is defined to be $e^{i\ln i} \cong e^{-\pi/2}$. I write $\cong$ instead of $=$ because this depends on your definition of $\ln$. This is very important, since different definitions of $\ln$ give us different values! –  Mike Miller Feb 11 '14 at 21:25
$i^{i^i}$ is ill-defined just as $3^{3^3}$ is. $3^{(3^3)}=3^{27}\neq (3^3)^3=3^9$. First you got to define how you are raising powers. –  nayrb Feb 11 '14 at 21:25
Yes, it looks more and more like a group of people saluting you... But seriously, $i^i$ has many values... –  Vadim Feb 11 '14 at 21:25
@nayrb, since $(a^b)^c = a^{(bc)}$, the convention is that $a^{b^c} = a^{(b^c)}$. –  Peter Taylor Feb 11 '14 at 21:27
But from the wording of the question it sounds like the OP was inputting the following sequence into something: i ^ i= ^ i= ^i =^i =... This would give $(...(i^i)^i)...)^i)$, which should be 4-periodic up to picking a branch of exponentiation. –  Kevin Carlson Feb 11 '14 at 21:33

Actually the limit exists.

Define $a_0=i$, $a_{n+1}=i^{a_n}$, $\lim_{n\to\infty}a_n=\frac{W(-\ln(i))}{-\ln(i)}\approx0.4383+0.3606i$, where $W(z)$ is the Lambert W function, $\ln(z)$ is the principle branch of $\log(z)$.

More generally, for each $z\in\mathbb{C}$, we can define such sequence $a_n(z)$, the limit exists only if $\frac{W(-\ln(z))}{-\ln(z)}$ is defined and they are equal.

Also the proof isn't hard, just messing with the definitions.

Correct me if there is any mistakes, I am just retrospecting what I read in high school.

Reference:

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Horey shit, those are some beautiful complex plots! Though currently a bit beyond me :P! –  Just_a_fool Feb 11 '14 at 21:51
If the limit $i^{i^{.^{.^.}}}$ exists, its value is $W(-\ln i)/(-\ln i)$. But you haven't proven that the limit exists. –  Rahul Feb 11 '14 at 21:52
@Rahul I didn't prove the limit exists, but I am sure someone did, just read the references in these two wiki pages. –  Kaa1el Feb 11 '14 at 21:59
did you study that in high school? –  Ant Feb 11 '14 at 22:01
@Ant I read these for fun! –  Kaa1el Feb 11 '14 at 22:02