How can one calculate $i^{i^{i^{i^{.^{.^.}}}}}$?

Question

I was reading an article that said about the value of $i^i$, and I thought to do the following $$\left(i^{i^{i^{i^{.^{.^.}}}}}\right)$$i.e, infinite powers of $ i $.

See equation 18 here: http://mathworld.wolfram.com/PowerTower.html — apnorton, Jan 22 '14 at 17:37
See also Lambert W function on Mathworld. Evidently Eisenstein (1844) thought of it too. Upvoted — Alan, Jan 22 '14 at 19:26
You should first specify which definition of the power tower you are using since the function is multivalued. — Moishe Kohan, Jan 22 '14 at 19:43

score 5 · Answer 1 · answered Jan 22 '14 at 17:38

5

Suppose $z=i^{i^{i^{i^{.^{.^.}}}}}$. Then, $i^z=z$. And now...

answered Jan 22 '14 at 17:38

Martín-Blas Pérez Pinilla

41,546
4
46
89

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I don't see how this helps solve the problem... (but I could just be overlooking something obvious) – apnorton Jan 22 '14 at 17:41
And $i^{i^z} = z.$ And $i^{i^{i^z}} = z.$ And ... – John Jan 22 '14 at 17:50
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$z$ is a solution of the equation $i^z=z$. – Martín-Blas Pérez Pinilla Jan 22 '14 at 18:14
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You should of course check that the sequence $s_n = i^{i^{i^{i^{\cdot^{\cdot^{\cdot}}}}}}$ $n$ times converges. If it doesn't, you could still have $i^z = z$ giving a solution but it might not be sensible to say that it is the limit of the sequence of powers. – Cameron Williams Jan 22 '14 at 18:50
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And now...... ? – user85798 Jan 22 '14 at 19:13

score 4 · Answer 2 · answered Jan 22 '14 at 20:39

4

Let's play fast and loose with things. Assume the answer is $x+iy$. Then

$$x+iy=i^{x+iy}=e^{i{\pi\over2}(x+iy)}=e^{-\pi y/2}(\cos(\pi x/2)+i\sin(\pi x/2))$$

from which we can conclude

$$x^2+y^2=e^{-\pi y}\quad\text{and}\quad {y\over x}=\tan(\pi x/2)$$

It follows that $x$ is a solution of $$(x\sec(\pi x/2))^2=e^{-\pi x\tan(\pi x/2)}$$

I get $x\approx 0.438282936727$ for this equation, which agrees with what Matthew Conroy gave in a comment under Foo Barigno's answer.

answered Jan 22 '14 at 20:39

Barry Cipra

79,832

Cool, +1. This GP code "z=I;for(i=1,5000,z=I^z);print(z)" spits out $0.4382829367270321116269751636 + 0.3605924718713854859529405269*I$, and I'm confident in all save the least significant digits. – Matthew Conroy Jan 23 '14 at 01:06
Whoa!, I'll generally require a lot of amazement to say this, but... Whoa! – TheVal Jan 26 '14 at 19:27

Foo Barrigno · Answer 3 · 2014-01-23T15:35:08.187

3

If you iterate over a few towers, you'll see that it does converge towards a point in the complex plane. This point is approximately $0.438283+0.360592i$

edited for significance

edited Jan 23 '14 at 15:35

answered Jan 22 '14 at 18:14

Foo Barrigno

935

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A better approximation is $0.438283 + 0.360592 i$. – Matthew Conroy Jan 22 '14 at 19:34
@MatthewConroy: Interesting, I was using a fairly simple python script to generate the numbers, I didn't expect to see that big of a discrepancy. – Foo Barrigno Jan 22 '14 at 19:40
It looks like you only went to 21 iterations. Try going for longer (like 1000, or so). – Matthew Conroy Jan 22 '14 at 19:47
What is this number? What is its significance/properties if any? – zerosofthezeta Jan 22 '14 at 19:51
It's just the point where $i^z=z$ – Foo Barrigno Jan 23 '14 at 15:36

How can one calculate $i^{i^{i^{i^{.^{.^.}}}}}$?

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