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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.

JMCF125
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Just_a_fool
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  • 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! –  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. – abnry Feb 11 '14 at 21:25
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    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
  • If I interpret the question correctly, you do not have to go far to be back at i. – M.B. Feb 11 '14 at 21:26
  • oh let me clarify, I didn't consider this potential hazard – Just_a_fool Feb 11 '14 at 21:27
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    @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
  • @PeterTaylor, it matters because the OP didn't realize the pitfall, and it will change whether or not i^i^i^... gives an interesting pattern (in one case it will be the repetition of four values). – abnry Feb 11 '14 at 21:31
  • Ok, now with parentheses it does not look like a group of people anymore. Still, do you mean the principal value only, or along $k$th branch, or just any chosen at each step? I guess the first, but may be the second is also interesting... don't know yet. – Vadim Feb 11 '14 at 21:33
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    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
  • See http://math.stackexchange.com/questions/647753/how-can-one-calculate-iiii/ – Barry Cipra Mar 08 '14 at 13:19

1 Answers1

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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:

  1. http://en.wikipedia.org/wiki/Lambert%27s_W_function

  2. http://en.wikipedia.org/wiki/Tetration

Kaa1el
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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
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    If the limit $i^{i^{.^{.^.}}}$ exists, its value is $W(-\ln i)/(-\ln i)$. But you haven't proven that the limit exists. –  Feb 11 '14 at 21:52
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    @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
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    @Ant I read these for fun! – Kaa1el Feb 11 '14 at 22:02
  • I don't see a reference for your claim "the limit exists if and only if [the function] is defined". How come the sequence fails to converge for $z=10^{-3}$ while the value of the function is $\approx 0.2195$? –  Feb 11 '14 at 22:09
  • @Rahul Yes you are right. I changed "iff" to "only if". If you like, you can come up with a characterization of such sequences and post it here, I am just not an expert in analysis and I am also curious in such subject. :D – Kaa1el Feb 11 '14 at 22:17
  • On convergence: see the picture http://math.stackexchange.com/a/432339/442 – GEdgar Mar 08 '14 at 14:25