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Imaginary iterates of exponentiation - jaydfox - 09/12/2007

Okay, first of all, we need a name for performing the ith iteration of exponentiation. The 1st iteration is exponentiation, the -1st iteration is the logarithm. What's the ith iteration? Notice that the -ith iteration is the inverse of the ith iteration, but unlike the logarithm/exponentiation duality, imaginary iterations and their inverses are conjugates of each other for real numbers and real bases. For complex numbers and/or bases, we've picked one side or the other of the real line, so they might not necessarily be conjugates anymore (indeed, they probably aren't).

While I'm on the subject of naming, I'm searching for a better verb for performing logarithms. To perform an exponentiation is to exponentiate. To perform a logarithm is to...??? To perform an exorcism (on an evil spirit) is to exorcize (the spirit). To give criticism (to a person or about a movie) is to criticize (the person or movie). I'm tempted at times to say that to perform a logarithm is to logarithe, but it doesn't quite roll off the tongue...

Okay, back on subject. We need an ith iteration of exponentiation. We could start with a tetration solution and derive one, but I'd very much prefer to derive one from basic principles instead, and use it to derive and/or validate a tetration solution.

First, some observations. The ith iteration of natural exponentiation of base e has two primary fixed points at . If we momentarily call the ith iteration "imaginary exponentiation" and the -ith iteration "imaginary logarithm", the the upper fixed point is an attracting fixed point for imaginary exponentiation, and the lower fixed point is a repelling fixed point. For imaginary logarithms, the upper fixed point is repelling and the lower is attracting.

I'm wondering if we could possibly experiment with exponentiation of complex bases to study this effect. I've experimented and found that complex bases (the few I played with) have both attracting and repelling fixed points of exponentiation, so that logarithms will tend to one and exponentiations will tend to the other, unless they happen to escape first (we could say that infinity is an attracting "fixed point" of exponentiation or logarithms in these cases...).

I had been hoping to find a complex base with a fixed point at 1.337-0.318i, which is -i times the fixed point of natural exponentiation. However, I haven't found a way to find a base that has a specific fixed point.

Even if I found such a base, I'm not sure what I'd do with it, other than play around with it and see if I get any inspiration...


RE: Imaginary iterates of exponentiation - andydude - 01/16/2008

is such a base. If you know a fixed point then


RE: Imaginary iterates of exponentiation - quickfur - 02/21/2008

jaydfox Wrote:[...]
While I'm on the subject of naming, I'm searching for a better verb for performing logarithms. To perform an exponentiation is to exponentiate. To perform a logarithm is to...??? To perform an exorcism (on an evil spirit) is to exorcize (the spirit). To give criticism (to a person or about a movie) is to criticize (the person or movie). I'm tempted at times to say that to perform a logarithm is to logarithe, but it doesn't quite roll off the tongue...
[...]
What about ... logarithmatize?? Big Grin Or, just "log (something)" for short. But "log" is already too overloaded. Hmm. Logarithate, maybe? or logarithmate. Bah, you're right, there is no good term for it. Sad


RE: Imaginary iterates of exponentiation - Ivars - 02/26/2008

jaydfox Wrote:First, some observations. The ith iteration of natural exponentiation of base e has two primary fixed points at . If we momentarily call the ith iteration "imaginary exponentiation" and the -ith iteration "imaginary logarithm", the the upper fixed point is an attracting fixed point for imaginary exponentiation, and the lower fixed point is a repelling fixed point. For imaginary logarithms, the upper fixed point is repelling and the lower is attracting.

Hi Jaydfox,

This may be relevant:
Omega constant =0.5671432904097838729999686622 is defined by 1=Omega*(e^Omega):

So selfroot of Omega:

Omega^(1/Omega) = e^ln(Omega^(1/Omega) = e^((1/Omega)*(ln(Omega))=e^((1/Omega)*(-Omega)) =e^(-1)=1/e=0,367879441

Infinite tetration of selfroot of Omega:

h(Omega^(1/Omega))=h(1/e) = -W(-ln(1/e))/(ln(1/e))= -W(1)/-1=Omega=0,56714329=-ln(Omega),

Square superroot of (Omega^1/Omega) :

ssrt(Omega^(1/Omega) = ln(1/e)/W(ln(1/e))= -1/W(-1)= -1/(-0.318131505204764 +- 1.337235701430689*I) = 0.16837688705553+-0.707755195958823*I.

W(-1) = 0.318131505204764 +- 1.337235701430689*I


So why not call it Omegation?

Ivars


RE: Imaginary iterates of exponentiation - Ivars - 03/20/2008

Just to clarify for myself:

i -th exponentiation of anything (e.g.) 2 would be:

2^i = 2*2*2*2*2............ i times = 0.76923890..+I*0.63896127...=e^(i*ln2)

i-th root of this number would be 2. (1/i=-i )

Learning, thanks.

ith exponentiation of a function is not i-th iteration of a function. See next post by administrator which puts it straight.

Ivars


RE: Imaginary iterates of exponentiation - bo198214 - 03/20/2008

Ivars, please dont post completely unrelated things!
Before you reply you should at least have a base understanding what you are replying to! You can not just throw in your ideas because of similar sounding words like "imaginary exponentiation" that even are explained in the same paragraph to have a different meaning.

Taking powers with complex exponents is a well explored area, there is nothing mysterious about it, its even tought at highschool.
However taking the I-th iteration of a function is a rather unknown area and barely connected to the I-th power. And *this* was discussed by Jay here in this thread for .


RE: Imaginary iterates of exponentiation - Catullus - 06/29/2022

Like before:
Quote:The .png file attached to this post is smaller than the .txt file attached to this.
The .txt file attached might not look right on a smart phone.
The text graphs need a monospaced font.
The image attached to this post looks blurry, because of the size of the image.
.
Here are graphs of the real and imaginary parts of [Image: svg.image?exp%5Ei(x)] and [Image: svg.image?exp%5E%7B-i%7D(x)], using the Kneser method:


RE: Imaginary iterates of exponentiation - Gottfried - 06/29/2022

Perhaps an additionally illustrating discussion in mathoverflow from 2011:

"Do complex iterations of functions have any meaning?"                   

1)  Question by Anixx  (see some graphic of similar style as given here)               
2)  Answer by me (Gottfried Helms)  (as well one picture)               
3)  A long & detailed (recent! 2022) answer by Tom Copeland . (Also referring some unclear points in my answer of 2011) 


=================================================

Here is an excerpt from my answer:
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
quote
This is a picture, where I studied the application of imaginary heights, using the base for exponentiation \(b=\sqrt2\). It has the attracting real fixpoint \(t=2\).       
 
As an example, look at the left side, with \(z_0=1 + 0\cdot î\). Using iteration with real heights (here in steps of \(1/10\) ) we move rightwards to \(z_1=b^{z_0}=b = 1.414...\) and by more iterations more towards the fixpoint \(t =2+ 0 î\). This is indicated by the orange arrows.   

Note that because \(t\) is a *fixpoint*, we cannot arrive at points on the real axis more to the right hand!         

But using imaginary heights, iterations move from \(z_0 \) to \(z_h\) on the indicated circular curve (computed data are in steps of \(0.1 { \pi \over \ln \beta} î\) see legend), which is indicated by the blue arrow.         
This iteration does not go towards the fixpoint, but repeats to cycle around it. On that cycling the trajectory crosses the real axis beyond the fixpoint.   

(Legend: the circular curves which connect the computed iteration-values of imaginary heights are Excel-cubic-splines and thus only very rough approximations of the true continuous iterations) 

[Image: aI3Ry.png]
end quote
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Perhaps a more expanded, systematic and visual presentation would be good for us all ...

Gottfried


RE: Imaginary iterates of exponentiation - MphLee - 06/30/2022

(06/29/2022, 11:04 AM)Gottfried Wrote: Perhaps an additionally illustrating discussion in mathoverflow from 2011:

"Do complex iterations of functions have any meaning?"                   

1)  Question by Anixx  (see some graphic of similar style as given here)               
2)  Answer by me (Gottfried Helms)  (as well one picture)               
3)  A long & detailed (recent! 2022) answer by Tom Copeland . (Also referring some unclear points in my answer of 2011) 

[...]
Perhaps a more expanded, systematic and visual presentation would be good for us all ...

Gottfried

Those are great links, thanks for collecting them. Your material is great and intuitive, also thanks for reminding me of Copeland's great answer, I forgot about it.
Ofc, a more systematic and visual presentations woul be beneficial for everyone... but who has the time for it?


RE: Imaginary iterates of exponentiation - JmsNxn - 07/01/2022

I always like to relate things to the unit disk when doing these iterations.

The value \(\exp^{\circ it}(z)\) traces an almost elliptic path, similar to \(\lambda^it\) for \(\lambda\) the multiplier. Essentially the object looks very much like \(z \mapsto \lambda^i z\), but it's placed in this weird fractally shape (the immediate basin). If you map the immediate basin (using Schroder) to \(\mathbb{C}\), this comparison becomes directly evident. Fun fact, you can use this to prove that the immediate basin about the fixed point is simply connected. There exists a sequence of jordan curves which slowly approximate the boundary.

$$
\lim_{x\to\partial \mathcal{A}} \exp^{\circ it}(x)\\
$$

You can prove this is a jordan curve with some work, but it's doable. So picture a bunch of almost elipses which start to slowly get more and more fractally as \(x\) approaches the boundary, and more and more circular as x approaches the fixed point.

You can copy paste a lot from the mapping \(z \mapsto \lambda^i z\), thanks to the Schroder function. All in all, it's not something inherently that interesting though. The function \(\exp^{\circ it}(z)\) is definitely more interesting--especially for periodic, about the fixed point, solutions. Looking at the boundary of the immediate basin, you would never guess that it is a Jordan curve, and that the immediate basin is simply connected, but it is. Totally weird looking fractally shape that it is. This limit sort of lets you watch as a nice looking ellipse starts to look more and more fractally.