# Infinite powering by $i$ [duplicate]

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.

• 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? Mar 21, 2013 at 8:40
• Oh it's just a question that I have come across. But I will take that into mind Mar 21, 2013 at 8:42
• 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. Mar 21, 2013 at 8:43
• 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 Mar 21, 2013 at 23:00

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.

• Thank you for the help, it is very detailed! Mar 21, 2013 at 8:58
• Doesn't the branch-ambiguity exactly mirror the fact that complex exponentiation of complex numbers is ambiguous? Mar 21, 2013 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 $\log$. Mar 21, 2013 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... Mar 22, 2013 at 7:48
• Isn't it just $x=W(-i\pi/2)$? Or are you looking at a different branch of the $W$ function? Jul 18, 2013 at 19:54

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 $\Bbb{C} \setminus (-\infty, 0]$.

• Can you share a Mathematica Notebook? Mar 21, 2013 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. Mar 21, 2013 at 10:56
• @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];. Mar 21, 2013 at 11:14
• @Fabian, that's a nice method! Thank you. Mar 21, 2013 at 11:17

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

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


Another way is to take natural logarithms:

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

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

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

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

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

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

• Thank you, very comprehensive. Mar 21, 2013 at 8:59
• 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. Mar 21, 2013 at 14:07

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

• Isn't it just $z=W(-i\pi/2)$? Or are you looking at a different branch of the $W$ function? Jul 18, 2013 at 19:55
• @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, 2013 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 $z=W(-i\pi /2)$? Jul 18, 2013 at 20:41
• Wait, I see it. Thanks! Jul 18, 2013 at 20:43

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

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

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

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

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

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).