Find the value of $\large i^{i^{.^{.^.}}}$ ?

How should we start to solve it ?

Also you can see this one if it helps.


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    $\begingroup$ Hint: If there is a solution $x$, it has the property $i^x=x$ $\endgroup$ Jul 18, 2013 at 19:21
  • $\begingroup$ @MahdiKhosravi: Are rast migam. :) Emtehan kon albate khode man ham jozweshun hastama chon untor soalato ghablan ha didam. :D $\endgroup$
    – Mikasa
    Aug 18, 2013 at 7:27

2 Answers 2


Let $z=i^{i^{.^{.^{.}}}}$. Then, as Hagen von Eitzen pointed out, $i^{z}=z$. Then $1=z\,i^{-z}=z \, e^{-i\pi z/2} $. It follows that $$-\frac{i\pi}{2}=-\frac{i\pi z}{2}\,e^{-i\pi z/2}.$$ Using the notion of Lambert's W function, we see that $$-\frac{i\pi z}{2}=W(-i\pi/2),$$ or $$z=\frac{2i}{\pi}\,W(-i\pi /2).$$

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    $\begingroup$ Just gonna do that. Nice answer. $\endgroup$ Jul 18, 2013 at 19:30
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    $\begingroup$ Now, is this a solution? That is, does $i,i^i,i^{i^i},\dots$ converge, so $z$ exists? $\endgroup$ Jul 18, 2013 at 19:31
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    $\begingroup$ Just because $z=x^z$ exists doesn't mean that it makes sense that $x^{x^{x^{\dots}}}=z$. $\endgroup$ Jul 18, 2013 at 19:32
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    $\begingroup$ Since $i$ is not simply just $\exp\left(i\frac{\pi}{2}\right)$ but also given by $\exp\left(i\frac{\pi}{2}+2k\pi i\right)$, shouldn't there be a family of fixed points? $\endgroup$ Jul 18, 2013 at 19:32
  • $\begingroup$ @Andres Caicedo: Is it true that $\left|a^{b}\right| \le |a|^{|b|}$? If so, it seems to me that the sequence $y_{0}=i, y_{n+1}=y_{n}^{i}$ is bounded. If bounded, then it must converge. $\endgroup$ Jul 18, 2013 at 19:45

by maple its possible prove this tower is convergence but its numerically proof

enter image description here

  • $\begingroup$ This is interesting as a "proof." $\endgroup$
    – Emily
    Jul 18, 2013 at 20:37

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