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For $x \in \Bbb C$, let $f(x)=i^x = \exp(i\pi x)$, where $i^2=-1$. Then find the fixed points for $f$.

EDIT: Let for all $n\geq 1$ $$\large a_n=\underbrace{i^{i^{\cdots i}}}_{\text{$n$ times}}$$

My question is, does the sequence of tetrations $\{a_n\}_{n\geq1}$ converge to some complex number? If yes, then what is it?

Lord_Farin
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Samrat Mukhopadhyay
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  • How do you define $i^x$ for $x$ in $\mathbb C$? – Did Jun 27 '13 at 20:51
  • Sorry @Did for not answering before. I define $i^x$ in $\mathbb{C}$ as $i^x=\exp(i\pi (Re(x)+iIm(x))=\exp(-Im(x))e^{(i\pi Re(x)))}$ – Samrat Mukhopadhyay Jun 29 '13 at 11:24
  • There are several mistakes in the formula you suggest. Once they will be corrected, you could try to solve the system $\Re(i^x)=\Re(x)$, $\Im(i^x)=\Im(x)$ and see what happens. – Did Jun 29 '13 at 11:39
  • Oops! The edit brings a different question to the fore... Samrat: please do not do that; instead, try to focus on some well-defined question and ask just this question (not several badly defined ones). – Did Jun 29 '13 at 11:40
  • @Did I had this question in mind that whether infinite tetration can converge to some point. Maybe I have not correctly represented that problem. Could you just put some light what the correct question should be? – Samrat Mukhopadhyay Jun 29 '13 at 11:52
  • It seems that you are in fact interested in the function $F:\mathbb R^2\to\mathbb R^2$ defined by $F(x,y)=(\cos(x\pi/2)\mathrm e^{-y\pi/2},\sin(x\pi/2)\mathrm e^{-y\pi/2})$ and that you are asking for the asymptotics of the sequence $(x_n,y_n){n\geqslant0}$ defined by $(x{n+1},y_{n+1})=F(x_n,y_n)$ for some given $(x_0,y_0)$. Not sure even that makes for a good question though... – Did Jun 29 '13 at 12:00
  • @Did I worked a little on my question and I actually found this function to be the one giving the points $(x,y)$ to which the tetration converges(if). I think, from the continuity and monotonicity of this function and due to its boundedness, there is exactly one fixed point $(x,y)$ for this function and if the tetration I'm asking for converges, it converges to this point. Well, maybe my question was not a good one, but I was curious to know it. Thanks for helping. – Samrat Mukhopadhyay Jun 29 '13 at 12:12
  • OK. (But $F$ is not bounded.) – Did Jun 29 '13 at 12:37
  • @Did Oh, yes, my bad, $F(x,y)$ is not bounded. I was just thinking for $\mathbb{R^+}\times \mathbb{R^+}$ – Samrat Mukhopadhyay Jun 29 '13 at 14:07
  • Fortunately, $F(K)\subset K$, where $K=[0,1]^2$ is bounded. – Did Jun 29 '13 at 14:09

1 Answers1

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We had this question quite recently, as I recall. But I cannot now find it. Anyway, here we see $a_n$ for $n$ from $1$ to $50$. They are converging, right?

enter image description here

The limit is: $$ \frac{2i}{\pi}W\left(\frac{-i\pi}{2}\right) \approx .4382829366+.3605924718 i $$ Of course $W$ is the Lambert W function.

GEdgar
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