Let $x = i^{i^{i^{i^{.^{.^{.{^ \infty}}}}}}}$. This is the solution of the equation $i^x - x = 0 $ . I used Euler's identity to find a solution. But I haven't yet found the real and imaginary parts of the solution. Are there more solutions? If so why did I miss them?

  • $\begingroup$ Shouldn't this be $i^x-x=0 $ ? $\endgroup$ – Gottfried Helms Jan 15 '11 at 14:06
  • $\begingroup$ @Loy That is, $i$ raised to $i$ raised to $i$...infinitely many times. This is similar to say the problem of finding the exact value of $sqrt{2-sqrt{2-sqrt{2-sqrt{2-\ldots}}}}$ $\endgroup$ – Chulumba Jan 15 '11 at 14:09
  • $\begingroup$ @Helms, yes, thanks. $\endgroup$ – Chulumba Jan 15 '11 at 14:10
  • $\begingroup$ since ii=exp(−π/2), you can have at most 4 options in which one could be a solution. exp(−π/2), exp(−iπ/2) exp(π/2), and exp(iπ/2) depending upon how many terms you use. But it is going up to inf. I can not figure out limit of this series. But none of these options satisfy your equation. You got a typo in your question at best! $\endgroup$ – Dilawar Jan 15 '11 at 14:13
  • $\begingroup$ Ok, now its seem to be fine. $\endgroup$ – Dilawar Jan 15 '11 at 14:14

In case you really mean $ I^x - x = 0 $ you can find this by the iteration

 x = <some initial value> 
     x = I^x   // where I is the imaginary unit
   until convergence

You'll get approximately $ x = 0.438282936727 + 0.360592471871*I $ (using Pari/GP, for instance)

However, you can also find the value using the lambert-w-function.
[update] using the lambert-w:

let $ \lambda=\ln(i) $
$i^x = x$
$1 = x* \exp( -x \lambda) $
$-\lambda = -x \lambda * \exp( -x \lambda) $
$ W(-\lambda) = - x \lambda $
$ x = \frac{W(-\ln(i))}{-\ln(i)} $

(Whether this is more "exact" is rather a question "exact in terms-of-what?")

[end update]

  • $\begingroup$ Thanks. I am interested in the exact solution. However, I have found it interesting to know that iteration can also be used for complex numbers. $\endgroup$ – Chulumba Jan 15 '11 at 14:24

Solving $i^i=x$ we get $e^{(i \frac{\pi}{2} + i2k\pi)i}=e^{-\frac{\pi}{2} - 2k\pi}$, than the values of that tower are dence in $\mathbb{C}$, or maybe equal (that's a good question I think). In mathematical softwares, one among the values is choosen, usually setting $k=0$ in the expression above.

  • $\begingroup$ $i^i$ is just $e^{-\frac{\pi}{2}}$ which is a proven transcendental. How do you include the $k$? $\endgroup$ – Chulumba Mar 27 '11 at 15:15
  • $\begingroup$ if $i^i=e^{- {\pi \over 2}}$ then it is also $i^i=e^{- {\pi \over 2}+ k*2 \pi i}$ . For integer $k$ the exp-function is periodic by $2 \pi i$ in its argument. $\endgroup$ – Gottfried Helms Mar 27 '11 at 15:53
  • $\begingroup$ @Chulumba You should review branch cuts and multivalued functions. $\endgroup$ – Alex B. Mar 27 '11 at 16:44

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