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How could you evaluate $i$ to the power of $i$ which is the to the power of $i$ with infinitely many $i$ exponents? My initial idea was to set it all equal to x which allowed me to write:

$$i^x=x$$

I tried a couple of different ways to solve this but struggled to get anywhere. When I put this equation in Wolfram alpha they use $W_n$ term that I am unfamiliar with. Could someone please explain to me what this means or if there is any alternative ways to solve this problem?

Xander Henderson
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Tom
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    It is $i^{i \cdot i \cdot i \cdots }$ which does not converge. Do you mean a power tower with $i$ instead? See also here. – Gary Oct 14 '22 at 05:42
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    Hi :) I don't believe, that the sequence $(a_n)n$ with $a{n+1}=a_n^i,\ a_0=i$ converges. It's periodic, since $i^4=1$. – Jochen Oct 14 '22 at 05:43
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    It is not hard to see that the sequence is $i, e^{-{\pi \over 2}}, -i, e^{{\pi \over 2}},...$ – copper.hat Oct 14 '22 at 05:45
  • Sorry I meant each i is to the power of the last i not to the power of the whole thing, I'm not to sure how to format that correctly it gives me an error message when I try to type it like that. – Tom Oct 14 '22 at 05:50
  • Yeah sorry I meant to write it as a power tower but couldn't format it. – Tom Oct 14 '22 at 05:52
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    Generally, equations like $a^x=x$ involve something like "Lambert's W function" and won't have nice solutions that are easily written down. Maybe there is some complex analysis that at least proves one or more solutions exist though. – jdods Oct 14 '22 at 05:52

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From the euler formula we know that $$i=e^{\frac{i\pi}{2}}$$ Which means that $$i^i= \left(e^{\frac{i\pi}{2}}\right)^i=e^{\frac{i^2\pi}{2}}$$ and for n powers of $i$ $$(((i^i)^i)^i)^{\dots}=e^{\frac{i^n\pi}{2}}$$ As you can see the limit does not exist since $i^4$ oscillates between $1$,$-1$, $i$, and $-i$.

Sam
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