## Infinite power tower of i?

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

dhokarena56
Posts: 179
Joined: Fri Mar 27, 2009 11:52 pm UTC

### Infinite power tower of i?

I was browsing the OEIS (oh, you magnificent time-waster, you!) and came upon Sequence A077589: the digital expansion of ii[sup]i...[/sup], or, if you like, i.

So apparently this is a finite, real, positive number. Anyone care to explain how?

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Infinite power tower of i?

dhokarena56 wrote:I was browsing the OEIS (oh, you magnificent time-waster, you!) and came upon Sequence A077589: the digital expansion of ii[sup]i...[/sup], or, if you like, i.

So apparently this is a finite, real, positive number. Anyone care to explain how?

That’s the decimal expansion of the real part of the complex number i.

By definition of complex exponentiation, ab = eb Log(a), where “Log” is the principle value of the complex log function. So ix = exiπ/2. You can do some work with limits to show the power tower converges. After that, finding the value is easy. It must satisfy z = i = iz = eziπ/2.

But z = eziπ/2 can be solved with the Lambert W function:
1/z = e-ziπ/2
1 = ze-ziπ/2
-iπ/2 = (-ziπ/2)e-ziπ/2

So by definition of W,
-ziπ/2 = W(-iπ/2)
z = (2i/π) W(-iπ/2)
wee free kings

z4lis
Posts: 767
Joined: Mon Mar 03, 2008 10:59 pm UTC

### Re: Infinite power tower of i?

I actually fiddled with the convergence a bit, thinking geometrically, and couldn't get anywhere. Anyone have an easy way to see it?
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Infinite power tower of i?

Well, the goal is to show that the map z→ez·iπ/2 has an attractive fixed point, with i in the basin of attraction. We’ve already found the fixed point, but I don’t know/recall enough about iterated mappings from ℂ→ℂ off the top of my head to come up with a good way to find its basin of attraction.

The best I’ve come up with in a few (sleep-deprived) minutes are some bounds on the orbit of z→iz. Writing ni = rncis(θn), it’s easy enough to show that e-π/2≤rn≤1 and 0≤θn≤π/2 for all n.

This comes from the expressions:
rk+1 = e^(-(π/2) rk sin(θk))
θk+1 = (π/2) rk cos(θk)

Combining those, we get a recurrence involving only the polar angle:
θk+1 = (π/2) cos(θk-1) e^(-θk tan(θk-1))

And an expression for modulus in terms of angle:
rk = (2/π) θk+1 / cos(θk)

It’s easy to see from the angle-recurrence that the arg always remains in [0, π/2] (and only hits the end-points for n = 1 and 2). Going back to the first equation for modulus, we then find the min and max values there. I just hope I didn’t mess anything up, since I’m really tired right now.
wee free kings

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Infinite power tower of i?

All right, I’ve made a little progress. The Brouwer fixed-point theorem guarantees that f(z)=iz=ei½πz has a fixed-point in the quarter-annulus 0≤θ≤½π, e-π/2≤r≤1. That follows since the set is homeomorphic to a closed ball, while f is continuous and maps that set to itself.

Spoiler:
To see that f maps the quarter-annulus 0≤θ≤π/2, e-π/2≤r≤1 to itself, consider a point z in that annulus. Then, z=re, so:
f(z) = ei½π(r·cos(θ) + ir·sin(θ))
f(z) = e-½πr·sin(θ)ei½πr·cos(θ)
f(z) = ρe

Where ρ=e-½πr·sin(θ) and φ=ei½πr·cos(θ). But θ is in the first quadrant, so both sin(θ) and cos(θ) are between 0 and 1, and r is also between 0 and 1. Thus, e-½π≤ρ≤1, and 0≤φ≤½π.

Now, if we can find a compact set homeomorphic to the closed unit ball on which |f′(z)|<1, then the contraction mapping theorem implies the function has a unique and attractive fixed-point in that set. (Note to self: use the phrase “unique and attractive” more often).

The derivative f′(z) = i½πei½π(a+bi), so |f′(z)|=½πe-½πb. That is less than 1 when b>(2/π)ln(½π). So it will suffice to find a set with all imaginary parts greater than that threshold, homeomorphic to the closed unit ball, that f maps into itself. I’m still working on that part. Once we prove such a set exists, then:

Spoiler:
A bit of symbol manipulation shows that the unique fixed-point, call it z0=re, has angle in the first quadrant satisfying:
θeθ·tanθ - ½π·cosθ = 0

And modulus between e-π/2 and 1 satisfying:
ln(r) + ½πrsin√((½πr)2 - (ln(r))2) = 0

Similarly, writing z0=x+iy, the real and imaginary parts satisfy, respectively:
xe½πx·tan(½πx) - cos(½πx) = 0
ye½πy - sin(½πe-½πy√(1 - y2eπy)) = 0

From looking at the graphs, it is clear that each of those functions has only one root in the appropriate region (0≤θ≤π/2, e-π/2≤r≤1). Indeed, the x and y equations have only one root each in 0≤x≤1, 0≤y≤1. Some derivative tests or suchlike should be able to prove those things rigorously.

In any case, numerical approximation of the roots gives r≈0.567555163306958, θ≈0.688453227107702, or equivalently, x≈0.438282936727032, y≈0.360592471871386.
wee free kings

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Infinite power tower of i?

Okay, I’ve got a proof of convergence now.

Brouwer guarantees a fixed-point in the quarter-annulus mentioned above. Symbol manipulation and numerical estimation shows there is only one fixed-point in that region, and gives its location as z0 from the previous post.

The norm of the derivative is less than one, |f′(u)| < 1, on any closed disk D centered at z0 with radius less than Im(z0) - (2/π)·ln(½π), for reasons shown above. Numerically that critical radius is R≈0.07310599 as per Wolfram Alpha.

So for any z≠z0 in such a disk, |f(z) - z0| = |f(z) - f(z0)| = |∫[imath]_{z_0}^z[/imath] f′(u)·du| ≤ ∫[imath]_{z_0}^z[/imath] |f′(u)|·|du| < |z - z0| where we use first the triangle inequality and second the fact that |f′(u)|<1 on a straight line from z0 to z. Thus f(z) is closer to z0 than z was.

Additionally, for any two distinct points u, v ∈ D∖{z0}, we have |f(u) - f(v)| = |∫[imath]_{v}^u[/imath] f′(q)·dq| ≤ ∫[imath]_{v}^u[/imath] |f′(q)|·|dq| < |u - v|. Therefore f is a contraction mapping on D. Now the contraction mapping theorem shows the orbit under f of each point in that disk converges to the fixed point z0.

In particular, that means if at any time the orbit of i under the map f gets within distance R of z0, it will converge to z0. As it turns out, the 20th iteration is within such a disk, as per Wolfram Alpha. Hence the orbit of i under f(z)=iz converges to z0, and therefore i=(2i/π)W(-iπ/2).
wee free kings

eSOANEM
:D
Posts: 3652
Joined: Sun Apr 12, 2009 9:39 pm UTC
Location: Grantabrycge

### Re: Infinite power tower of i?

I'm not sure as to the rigour of this proof, but I think that, assuming a value exists, I can show what that value is:

let i=z.

zi=∞+1i=i=z

zi-1=1

(i-1)ln(z)=0

i-1=/=0

so ln(z)=0

so z=1

All the other values posted here are complex unless I'm mistaken but OP stated that the solution was real. I'm by no means sure that my argument holds, but it seems to be the first answer satisfying all the properties we were given for the solution.
my pronouns are they

Magnanimous wrote:(fuck the macrons)

Ben-oni
Posts: 278
Joined: Mon Sep 26, 2011 4:56 am UTC

### Re: Infinite power tower of i?

eSOANEM wrote:I'm not sure as to the rigour of this proof, but I think that, assuming a value exists, I can show what that value is:

let i=z.

zi=∞+1i=i=z

This is wrong. (ab)ba+1b.

Timefly
Posts: 56
Joined: Sun Nov 14, 2010 7:30 pm UTC

### Re: Infinite power tower of i?

eSOANEM
:D
Posts: 3652
Joined: Sun Apr 12, 2009 9:39 pm UTC
Location: Grantabrycge

### Re: Infinite power tower of i?

Ben-oni wrote:
eSOANEM wrote:I'm not sure as to the rigour of this proof, but I think that, assuming a value exists, I can show what that value is:

let i=z.

zi=∞+1i=i=z

This is wrong. (ab)ba+1b.

Really? Doesn't that follow from the definition of tetration? Or is it because a^b=/=b^a?

Edit: I think I see it, it's because exponentiation isn't commutative and because of the direction of its nesting.

Still, OP said the solution was real but the product log solution is complex, I take it that OP and OEIS is wrong in this regard then.
my pronouns are they

Magnanimous wrote:(fuck the macrons)

Mynameisjosh
Posts: 2
Joined: Tue May 29, 2012 6:27 pm UTC

### Re: Infinite power tower of i?

I think there is an easier way to do this:

first assume that i^(i^(i^... converges to some L

L=i^(i^(i^...

We can see that this is equivalent to L=i^L

this is the same as i=L^(1/L)

using the product log function where W(z) is the inverse function of f(W)=We^W and doing some algebra you get L=W(-ipi/2)*(2i/pi)

Check out the paper nested radicals and other infinitely recursive expressions by Michael McGuffin. That'll give you a nice ground on working with all sorts of weird things like infinite exponent towers, continued fractions and nested radicals in a very easy to floow and understand way

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Infinite power tower of i?

eSOANEM wrote:Still, OP said the solution was real but the product log solution is complex, I take it that OP and OEIS is wrong in this regard then.

Qaanol wrote:That’s the decimal expansion of the real part of the complex number i.
wee free kings

gmalivuk
GNU Terry Pratchett
Posts: 26765
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here and There
Contact:

### Re: Infinite power tower of i?

Mynameisjosh wrote:I think there is an easier way to do this:
You mean like how it was done in the very first reply? Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
---
If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

(he/him/his)

eSOANEM
:D
Posts: 3652
Joined: Sun Apr 12, 2009 9:39 pm UTC
Location: Grantabrycge

### Re: Infinite power tower of i?

Qaanol wrote:
eSOANEM wrote:Still, OP said the solution was real but the product log solution is complex, I take it that OP and OEIS is wrong in this regard then.

Qaanol wrote:That’s the decimal expansion of the real part of the complex number i.

Sorry, I misread that. my pronouns are they

Magnanimous wrote:(fuck the macrons)

Mynameisjosh
Posts: 2
Joined: Tue May 29, 2012 6:27 pm UTC

### Re: Infinite power tower of i?

gmalivuk wrote:You mean like how it was done in the very first reply? Reading comprehension FTW. Well fortunately my first post is indicative of my normal attention span

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Infinite power tower of i?

I started looking at the analogous situation for an arbitrary base, so fu(z) = uz = ez·Log(u). In particular, I wanted to know for which values of u a fixed point of fu is an attractor. I did not try to find the basin of attraction, just whether it has one.

What I found is that fu has a fixed point at z0 = -W(-Log(u)) / Log(u). That is an attractor when |f′(z0)|<1, meaning |Log(u)·e-W(-Log(u))| < 1.

So I fired up Matlab and plotted which values of u satisfy that. Turns out to look like this, where the black region has attractive fixed points: The pixels are sampled every .02 along both real and imaginary axes. The topmost part has imaginary component 2.10, and the bottom-most -2.10. The leftmost point has real component -0.38, and the rightmost 2.04. Along the real axis (which is the symmetry axis here), the leftmost point is 0.08 and the rightmost is 1.52.

I don’t recognize that shape, but the formula involved lots of “Log(u)” so I plotted the same thing for Log(u), giving this: That looks very much like a cardioid, but I’m not sure if it really is. In any case, the left-most point is -2.70, the rightmost point is at 0.86, and the right-most point on the real axis is 0.36. The max imaginary value is 2.00. I rather suspect the left-most point is actually -e, and the right-most point on the real axis is 1/e, but I don’t know for sure.

It appears that for maps of the form fu = uz, the values of u that have an attractive fixed point correspond to values of Log(u) within a cardioid. Of course, I only looked at the principle branch of the Lambert W function, so there could be other u-values for which additional fixed-points exist and are attractive.
wee free kings

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Infinite power tower of i?

A bit more numerical work, and some pretty pictures. First, the points in white are those u-values for which the orbit of u under f(z) = uz diverges to infinity (or at least has absolute value great enough to overflow Matlab) at or before the 1000th iteration. The images go from -3.5 to 6.5 on the real axis, and -7.5 to 7.5 on the complex axis. This region shows most of the “interesting” behavior. Click to view full-size, where pixels are sampled every 0.01 units. Exponential fractal: white points numerically
calculated to diverge by 1000th iteration,
black points have not diverged  Cycle length after 1000 iterations
Black = Chaotic, or cycle-length greater than 10
Blue = 1-cycle (fixed-point attractor)
Green = 2-cycle (2-point attractor)

Yellow = 10-cycle
White = Numerically observed to diverge

Here are the same things, but drawing the result for z at the point log(z) (technically, drawing at point z the result for exp(z), so these are 2πi-periodic). That is, I have simply applied the map Log(z) to the complex plane, thereby warping the locations of points, but not changing any of the values in the iterated exponential map.

As you can see, it very strongly appears that the locus of fixed-point attractors, shown in blue in the last image, has natural log in the shape of a cardioid. The apparent cusp is at 1/e, and the opposite side is at -e. The top and bottom seem to be at c±2i for some real number c I haven’t tried to identify.

To find the cycle-lengths, I calculated the 1000th through 1010th iterations, and compared them pointwise to see when the orbit returned to its value from the 1000th iteration, within a tolerance of 0.001. To visualize how the cycles go, here are the 1000th through 1010th iterations in order, with arctan-polar coloration:

Edit: it also looks like the red area (period 3) in the log-plot has bounds that approach ±π/2 in the limit of large real component.
wee free kings