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?
How to compute the infinite tower of the complex number $i$, that is$ ^{\infty}i$
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0@Rahul, thank you very much for the link. Edit to the title is in order. – 2011-01-15
2 Answers
In case you really mean $ I^x - x = 0 $ you can find this by the iteration
x = repeat 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) $
then
$i^x = x$
$1 = x* \exp( -x \lambda) $
$-\lambda = -x \lambda * \exp( -x \lambda) $
$ W(-\lambda) = - x \lambda $
so
$ x = \frac{W(-\ln(i))}{-\ln(i)} $
(Whether this is more "exact" is rather a question "exact in terms-of-what?")
[end update]
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0Thanks. I am interested in the exact solution. However, I have found it interestin$g$ to know that iteration can also be used for complex numbers. – 2011-01-15
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.
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0@Chulumba You should review branch cuts and multivalued functions. – 2011-03-27