24

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.

Martin Sleziak
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Sy123
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    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? – Marc van Leeuwen Mar 21 '13 at 08:40
  • Oh it's just a question that I have come across. But I will take that into mind – Sy123 Mar 21 '13 at 08:42
  • 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 – Gottfried Helms Mar 21 '13 at 23:00
  • 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. – Damien L Mar 21 '13 at 08:43

8 Answers8

20

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.

Fabian
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  • Thank you for the help, it is very detailed! – Sy123 Mar 21 '13 at 08:58
  • Doesn't the branch-ambiguity exactly mirror the fact that complex exponentiation of complex numbers is ambiguous? – Tobias Kienzler Mar 21 '13 at 15:32
  • @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 $\log$. – Fabian Mar 21 '13 at 20:17
  • @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... – Tobias Kienzler Mar 22 '13 at 07:48
  • Isn't it just $x=W(-i\pi/2)$? Or are you looking at a different branch of the $W$ function? – Adrian Keister Jul 18 '13 at 19:54
16

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

Sangchul Lee
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  • Can you share a Mathematica Notebook? – m0nhawk Mar 21 '13 at 10:54
  • @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. – Sangchul Lee Mar 21 '13 at 10:56
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    @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];. – Fabian Mar 21 '13 at 11:14
  • @Fabian, that's a nice method! Thank you. – Sangchul Lee Mar 21 '13 at 11:17
7

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
Gottfried Helms
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6

Another way is to take natural logarithms:

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

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

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

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

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

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

Mhenni Benghorbal
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Inceptio
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6

$$ e^{i\pi z/2}=z\Rightarrow-\frac{i\pi}2ze^{-i\pi z/2}=-\frac{i\pi}2\tag{1} $$ 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.

robjohn
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  • Isn't it just $z=W(-i\pi/2)$? Or are you looking at a different branch of the $W$ function? – Adrian Keister Jul 18 '13 at 19:55
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    @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 $$ z=\frac{2i}{\pi}w=\frac{2i}{\pi}\mathrm{W}\left(-\frac{i\pi}{2}\right) $$ – robjohn Jul 18 '13 at 20:36
  • 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 $z=W(-i\pi /2)$? – Adrian Keister Jul 18 '13 at 20:41
  • Wait, I see it. Thanks! – Adrian Keister Jul 18 '13 at 20:43
4

Hint:let $ x = {i^{i^{i^{.^{.^{\infty}}}}}}$

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

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

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

Tobias Kienzler
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kalpeshmpopat
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2

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

MAPLE

GEdgar
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1

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).

sds
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