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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0Shouldn't this be $i^x-x=0 $ ? – 2011-01-15
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0@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 $sqrt{2-sqrt{2-sqrt{2-sqrt{2-\ldots}}}}$ – 2011-01-15
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0@Helms, yes, thanks. – 2011-01-15
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0since 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! – 2011-01-15
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0Ok, now its seem to be fine. – 2011-01-15
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0I don't see how applying Euler's identity gave you the impression that you could find only one solution. Can you expand on that? (By the way, a closed form for the solutions is given here: http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights -- note that the complex logarithm and the Lambert W function are multivalued.) – 2011-01-15
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0I have also seen the solution here: http://www.wolframalpha.com/input/?i=i^x+-x+%3D0. But here the imaginary and real parts are not explicitly expressed. – 2011-01-15
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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 interesting 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$i^i$ is just $e^{-\frac{\pi}{2}}$ which is a proven transcendental. How do you include the $k$? – 2011-03-27
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0if $i^i=e^{- {\pi \over 2}}$ 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. – 2011-03-27
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0@Chulumba You should review branch cuts and multivalued functions. – 2011-03-27