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Find the value of:

$i^{i^{i^{i^{i^{i^{....\infty}}}}}}$

Simply infinite powering by i's and the limiting value.

Thank you for the help.

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    $\begingroup$ Taking a complex power of a complex number is not uniquely defined. Taking infinite towers of exponentiations rarely converges. What is the application you have in mind? $\endgroup$ Mar 21, 2013 at 8:40
  • $\begingroup$ Oh it's just a question that I have come across. But I will take that into mind $\endgroup$
    – Sy123
    Mar 21, 2013 at 8:42
  • $\begingroup$ To what wrote Fabian, I add that $x$ satisfies also $i^{x} = x$. So $x = 1$ is not possible. So maybe $x$ doesn't exists. $\endgroup$
    – Damien L
    Mar 21, 2013 at 8:43
  • $\begingroup$ I promote again the notation from the top, where we begin with some $x$ and exponentiate by a base $b$ which is written below: $$ \huge x,{\ _{ b} x }, {\ _{ \ _{ b} b} x }, {\ _{ \ _{\ _b b} b} x }, \ldots , {\ _{ \ _{\ _{\ _\infty \ldots b} b} b} x }$$ (which is admittedly awfully typeset...) where we can then begin with $x=0$, $x=1$, $x=b$ or some $x$ on the trajectory (In our case we had $b=x=i$). I think, this is a more realistic and instructive notation because it mimics th top-down computation $\endgroup$ Mar 21, 2013 at 23:00

8 Answers 8

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Let us denote $x=i^{i^{i^{i^\cdots}}}$. Then we have $$i^x=x.$$ It looks like the solution is $x= \frac{2i}{\pi} W(-i\pi/2)$ with $W$ Lambert's $W$ function. Now, $W$ is multivalued. You have to figure out which of the different branches $x$ converges to (and if it converges at all). Numerically, you find (using the principal branch of the logarithm to define the exponentiation) that $x= 0.438283 + 0.360592 i$ which corresponds to the principal branch.

Knowing that you should be able to prove the result by some kind of fixed point theorem.

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  • $\begingroup$ Thank you for the help, it is very detailed! $\endgroup$
    – Sy123
    Mar 21, 2013 at 8:58
  • $\begingroup$ Doesn't the branch-ambiguity exactly mirror the fact that complex exponentiation of complex numbers is ambiguous? $\endgroup$ Mar 21, 2013 at 15:32
  • $\begingroup$ @TobiasKienzler: you can check that $x= -1.86174 - 0.4108 i$ which corresponds to another branch of $W$ still fulfils $i^x=x$ (with the principal branch of the logarithm). So the fix point is not unique, even when you choose a particular branch of $\log$. $\endgroup$
    – Fabian
    Mar 21, 2013 at 20:17
  • $\begingroup$ @Fabian Sorry, I meant the ambiguity in $W$'s multivalued-ness. But your comment states that even for a fixed log-branch there are still multiple fix points, although that doesn't say whether one of them is the "actual" $i^{i^{i^...}}$ if that is even well-defined... $\endgroup$ Mar 22, 2013 at 7:48
  • $\begingroup$ Isn't it just $x=W(-i\pi/2)$? Or are you looking at a different branch of the $W$ function? $\endgroup$ Jul 18, 2013 at 19:54
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Here is a numerical result supporting Fabian's argument.

Here, the complex logarithm

$$ z^{w} := \exp (w \operatorname{Log} z) $$

is defined via the principal value $\mathrm{Log}$ of the logarithm, defined on $\Bbb{C} \setminus (-\infty, 0]$.

enter image description here

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  • $\begingroup$ Can you share a Mathematica Notebook? $\endgroup$
    – m0nhawk
    Mar 21, 2013 at 10:54
  • $\begingroup$ @m0nhawk, I'm sorry, but I did not save it :( Fortunately, the code is short and you can resurrect it by just re-typing it. $\endgroup$ Mar 21, 2013 at 10:56
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    $\begingroup$ @monhawk: maybe this belongs more to Mathematica.SE but the first part (Numerical Iteration) can be made much faster by using iter=100;l = NestList[Power[I, #] &, N[I], iter]; Print["After ", iter," iterations, the result is: ", Last@l];. $\endgroup$
    – Fabian
    Mar 21, 2013 at 11:14
  • $\begingroup$ @Fabian, that's a nice method! Thank you. $\endgroup$ Mar 21, 2013 at 11:17
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This is rather another comment than an answer but contains a picture, so...
If we display the 3-step-like trajectory separated into 3 single trajectories, we get an improvement of imagination of the convergence. See this one enter image description here

Similar improvements can be made with other bases. The idea is, to use this for convergence-acceleration procedures like Euler-sums and similar.

[Update]: Also the process of convergence can be improved over the need to iterate 100 times and more. Just use the Newton-iteration. Here is a code-snippet in Pari/GP:

f(x)  = exp( L *x)    \\ implements x->  b^x where L is the log of te base b
fd(x) = L * exp(L*x)  \\ implements the derivative of f(x)

L = log(I)
x0=0.5+0.5*I       \\ Initialize
[x0=x0 - (f(x0)-x0)/(fd(x0)-1)  , exp(L*x0)-x0]  \\ repeat this, say, 7 times

Result:

x0=0.5+0.5*I    \\ initialize
 %214 = 0.500000000000 + 0.500000000000*I

[x0=x0 - (f(x0)-x0)/(fd(x0)-1)  ,  exp( L*x0)-x0]   \\ repeat this say 7 times
 %215 = [0.429683379978 + 0.358463904092*I, 0.0149144114062 - 0.00263680525658*I]
 %216 = [0.438282449555 + 0.360624709917*I, -0.0000214307236671 - 0.0000508331490807*I]
 %217 = [0.438282936547 + 0.360592471486*I, 0.000000000547853619231 + 0.000000000479209718138*I]
 %218 = [0.438282936727 + 0.360592471871*I, 1.24483565546 E-19 - 2.36342583549 E-20*I]
 %219 = [0.438282936727 + 0.360592471871*I, -1.59860647096 E-39 - 3.49116795082 E-39*I]
 %220 = [0.438282936727 + 0.360592471871*I, 2.79037134755 E-78 + 2.15595352591 E-78*I]
 %221 = [0.438282936727 + 0.360592471871*I, 2.83277459577 E-156 - 9.05172112238 E-157*I]
 %222 = [0.438282936727 + 0.360592471871*I, 5.10320381 E-203 - 2.551601908 E-203*I]
  \\ convergence sufficient, 200 dec digits
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Another way is to take natural logarithms:

$$i^{i^{i^{i^{i^{i^{\dots \infty}}}}}}=y$$

$$\ln y= \ln (i)^y$$

$$y\ln i=\ln y$$

$$\ln i=\dfrac{i \pi}{2}$$

$$\dfrac{y.i\pi}{2}= \ln y$$

$$e^{\frac{iy\pi}{2}}=y$$

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  • $\begingroup$ Thank you, very comprehensive. $\endgroup$
    – Sy123
    Mar 21, 2013 at 8:59
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    $\begingroup$ Your grouping of the exponential is unconventional. More usual is the other one, where you would get $y=i^y$ as in Fabian's answer. $\endgroup$
    – GEdgar
    Mar 21, 2013 at 14:07
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$$ e^{i\pi z/2}=z\Rightarrow-\frac{i\pi}2ze^{-i\pi z/2}=-\frac{i\pi}2\tag{1} $$ Therefore, $$ z=\frac{2i}{\pi}\mathrm{W}\left(-\frac{i\pi}2\right)\tag{2} $$ Which Mathematica gives as N[2 I/Pi LambertW[0, -I Pi/2], 20] $$ 0.43828293672703211163 + 0.36059247187138548595 i\tag{3} $$ Since this is the only value where the derivative of $e^{i\pi z/2}$ has absolute value less than $1$, it is the only stable limit point. In particular, the derivative is $$ 0.89151356577604704289e^{2.25924955390259874973\,i}\tag{4} $$ when close to the limit, the map is a contraction with ratio $0.89151356577604704289$ combined with a rotation of $2.25924955390259874973$ radians. This is seen in the plots supplied in other answers.

Raising $(4)$ to the power $t$ and setting $\theta=2.25924955390259874973\,t$ gives that $$ r=z_0\,e^{-\lambda\theta}\tag{5} $$ where $\lambda=-\dfrac{\log(0.89151356577604704289)}{2.25924955390259874973}=0.05082865892244868531$.

Thus, the iterates lie close to an exponential curve.

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  • $\begingroup$ Isn't it just $z=W(-i\pi/2)$? Or are you looking at a different branch of the $W$ function? $\endgroup$ Jul 18, 2013 at 19:55
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    $\begingroup$ @AdrianKeister: if we let $w=-\frac{i\pi}{2}z$, then $$ -\frac{i\pi}2ze^{-i\pi z/2}=-\frac{i\pi}2 $$ becomes $$ we^w=-\frac{i\pi}{2} $$ which directly says $$ w=\mathrm{W}\left(-\frac{i\pi}{2}\right) $$ then $$ z=\frac{2i}{\pi}w=\frac{2i}{\pi}\mathrm{W}\left(-\frac{i\pi}{2}\right) $$ $\endgroup$
    – robjohn
    Jul 18, 2013 at 20:36
  • $\begingroup$ So, where's my mistake in doing $i^{z}=z$, so $1=zi^{-z}=ze^{-i\pi z/2}$. Therefore, $$-\frac{i\pi}{2}=-\frac{i\pi z}{2}\,e^{-i\pi z/2}$$ implying that $z=W(-i\pi /2)$? $\endgroup$ Jul 18, 2013 at 20:41
  • $\begingroup$ Wait, I see it. Thanks! $\endgroup$ Jul 18, 2013 at 20:43
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Hint:let $ x = {i^{i^{i^{.^{.^{\infty}}}}}}$

hence $x = {i^{x}}$

$\ln x = x\ln(i)$

$\frac{\ln x}{x} = \ln(0+i)$

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Here is the Maple version of the graph from sos440...

MAPLE

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In clisp:

(loop for i upfrom 1 
      and prev = 0 then x 
      and x = #c(0L0 1L0) then (expt #c(0 1) x) 
      while (< long-float-epsilon (abs (- x prev))) 
      finally (return (values x i)))
#C(0.43828293672703211162L0 0.36059247187138548596L0) ;
393

so, in fewer than 400 iterations you get 20 correct decimal digits. (The convergence is quadratic).

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