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11/08/2007, 02:16 AM
(This post was last modified: 11/08/2007, 02:18 AM by jaydfox.)
Actually, Henryk, if you graph even a rough mockup of the regular slog developed for the fixed point z=2, the corrugation effect as we wrap around the fixed point at 4 appears, and furthermore, it looks like it should be there. In other words, it makes sense that they should be there, due to a switch from rings around the fixed point at 2 to cyclic wavy curves.
Start with a bunch of concentric rings around z=2, with a relatively small radius. This is just a mockup, so accuracy isn't required. Just make sure to get enough rings, so that when you take the logarithm of each, you don't get any major gaps.
Now, continue to take more and more logarithms of the set of concentric rings. The rings will become ovals, stretching out to the left and getting squished to the right. As the rings pass the origin, the next iteration swing out to minus infinity, and then become curves stretching from between the lines with imaginary parts , and of course, when we include all branches, they become infinitely long wave curves, cyclic at intervals of .
Further iterations will be show the corrugation, due to the change from rings to wavy lines, a switch that occurred when the first ring touched the origin.
~ Jay Daniel Fox
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jaydfox Wrote:Actually, Henryk, if you graph even a rough mockup of the regular slog developed for the fixed point z=2, the corrugation effect as we wrap around the fixed point at 4 appears, and furthermore, it looks like it should be there. Mm, "should" has perhaps no meaning in the mathematical context. But good illustrations of a fact are highly welcome. So any pictures?
My conjecture is that whenever there is an analytic function for which the regular iterations at two fixed points coincide, then this function is already the identity.
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11/09/2007, 04:59 AM
(This post was last modified: 11/09/2007, 05:01 AM by jaydfox.)
bo198214 Wrote:jaydfox Wrote:Actually, Henryk, if you graph even a rough mockup of the regular slog developed for the fixed point z=2, the corrugation effect as we wrap around the fixed point at 4 appears, and furthermore, it looks like it should be there. Mm, "should" has perhaps no meaning in the mathematical context. But good illustrations of a fact are highly welcome. So any pictures? I wouldn't go so far as to say "no meaning", but certainly heuristic analysis can impute more meaning in a situation than is warranted at times.
As for pictures, I'm working on some. I was going to have a go at making some pictures with SAGE, because they look better than those in Excel, but also take longer to get the code set up. I was doing my initial analysis in Excel, until I had a clear enough mental picture to extrapolate the more of the details in my head, much as I had done for base e.
And the quick and easy predication is that, in addition to the singularities at 2 and 4 and their images at imaginary offsets, there would be additional singularities stretching out towards positive infinity, with a relatively calm strip between them, much like the ribs off the backbone of the base e slog. And, as with that slog, there would be additional branches off those branches, between each set of singularities, though without the alternating "logarithmic" and "exponential" types. I still need to create a decent graph for this, because as yet I've only been able to picture it in my mind.
Quote:My conjecture is that whenever there is an analytic function for which the regular iterations at two fixed points coincide, then this function is already the identity.
Not quite sure what you meant by that. At any rate, I can already tell from a series of thought experiments that the rslogs at the upper and lower fixed points (4 and 2 for base sqrt(2)) are quite different and introduce the "corrugations" as you called them in different places, and yet I would imagine that both satisfy the infinite matrix solution. I suppose this is an example where a nonfirst order differential equation can have multiple solutions.
~ Jay Daniel Fox
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11/09/2007, 05:12 AM
(This post was last modified: 11/09/2007, 05:15 AM by jaydfox.)
jaydfox Wrote:And the quick and easy predication is that, in addition to the singularities at 2 and 4 and their images at imaginary offsets, there would be additional singularities stretching out towards positive infinity, with a relatively calm strip between them, much like the ribs off the backbone of the base e slog. And, as with that slog, there would be additional branches off those branches, between each set of singularities, though without the alternating "logarithmic" and "exponential" types. I still need to create a decent graph for this, because as yet I've only been able to picture it in my mind.
By the way, the corrugations introduced by the cyclic shift are actually "desired", in the sense that they help create the singularities at the logarithms of the locations of other singularities.
In fact, I had started to make this connection earlier, but hadn't put it together until Henryk described this very effect of "corrugations".
To see this in action, let's go back to the slog for base e. It has singularities at equally spaced intervals from the two primary fixed points at 0.318 +/ 1.337i.
So, the value of the slog along the imaginary axis (the line with real part 0) is a simple, wellbehaved, complexvalued function. The value along the line with real part 1 is even more wellbehaved, and so on as we move to the left.
If we were to develop a complex fourier series along the imaginary axis, it should behave very nicely, and yet introduce singularities at the fixed points.
So far, this is already interesting, and I had figured out this much about a week ago. But what's more interesting to me are the singularities inside the logarithmic branches. After Henryk's description of "corrugations" (a beautifully descriptive word for this situation), I thought it out, and immediately began piecing together the conneciton. In my mind, these additional fractally branching singularities are due to corrugations as we rotate around any of the singularities.
For the upper primary fixed point, for example, as we rotate clockwise around it, the singularities get closer and closer and the function gets more and more chaotic. On the other hand, if we rotate counterclockwise, it becomes more and more wellbehaved, more and more like the regular slog we would predict if we used only the upper primary fixed point.
This very behavior resembles the slog itself. To the right, it gets more and more fractal, but to the left, it gets smoother and smoother, not unlike the smoothness of the basic exponential in the left halfplane (negative real part).
If we took the slog, with all its smoothness to the left and all its fractal complexity to the right, and "wrapped" that around a logarithmic singularity, we would get the fractal branching in one direction and smoothness in the other direction.
Of course, this is the vaguest of notions so far, a mere hunch, and I need to brush up on complex fourier analysis before I could even begin to tackle this properly.
~ Jay Daniel Fox
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11/09/2007, 07:04 AM
(This post was last modified: 11/09/2007, 07:08 AM by jaydfox.)
Actually, the fourier analysis might be very beneficial, if I understood it better. I was trying to think of an easy way to derive the terms, without resorting to tedious math, and it hit me.
I was thinking about the fourier series of the along the imaginary axis, which is cyclic with period 2*pi*i. So we would need sines and cosines with of the form , with a_k and b_k complex. These could be expressed as . Well, at 0, the slog is realvalued, so the c_k would have to be real as well.
This is where it hit me. The c_k would look like a power series in terms of e^z, which isn't terribly fascinating, except for the minor detail that , and we have a way of calculating the power series for . Therefore, we have the coefficients c_k already!
This in and of itself essentially validates a few things. First, it validates that the slog will be periodic with period 2*pi*i, though I've already figure out other arguments to demonstrate this. But it's nice to have an independent verification.
Second, it validates the very notion of creating these singularities with the corrugations caused cyclic functions. These corrugations first create ripples, viewed as the smooth sections of various branches, but in some places the ripples add up to singularities.
~ Jay Daniel Fox
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11/12/2007, 08:45 PM
(This post was last modified: 11/12/2007, 08:57 PM by bo198214.)
jaydfox Wrote:I wouldn't go so far as to say "no meaning", but certainly heuristic analysis can impute more meaning in a situation than is warranted at times. I just found it strange to say something "should" be that way if it mathematically (non changably) "is" this or that way. Its strange if it "should" be different from what it "is" and also strange, as in our case, if it "should" be the same that it "is".
Quote:As for pictures, I'm working on some.
Yes, they would be really helpful. Though I can perceive your joy in the imagination of the mathematical situation, the picture remains somewhat vague, if not supported by something directly accessible.
And regarding Fourier analysis ... I was absent when this topic was tought Means I probably have to go back to the basics of it, before I can follow somewhat more sophistaced usage of it.
Quote:yet I would imagine that both satisfy the infinite matrix solution.
Yes the infinite matrix equation is just an other form of (i.e. equivalent to) the Abel equation. Every solution of the Abel equation is a solution to the matrix equation and vice versa.
By an email conversation with Jean Ecalle it turns out that my conjecture was not completely true.
What however is true that
Quote:For each *entire* function the regular iterations at each two different fixed points are different.
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11/13/2007, 01:47 AM
(This post was last modified: 11/13/2007, 01:55 AM by jaydfox.)
Quote:For each *entire* function the regular iterations at each two different fixed points are different.
Do you mean that the function being iterated is entire, or that the continuous iterator is entire? For example, multiplication is entire, as is exponentiation (the continous iterator). However, exponentiation is entire, but tetration has at least the singularities at z=k, k an integer <= 2.
Edit: Since he just left (or is about to leave) on a 2month trip, I might not get the answer directly. So anyone feel free to answer this question if you happen to know.
~ Jay Daniel Fox
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bo198214 Wrote:jaydfox Wrote:As for pictures, I'm working on some. Yes, they would be really helpful. Though I can perceive your joy in the imagination of the mathematical situation, the picture remains somewhat vague, if not supported by something directly accessible. Okay, I've got some pictures of the coming along nicely. I'm evaluating the regular slog at z=2, by first exponentiating 0 a "bunch" of times, where "bunch" is a number sufficient to reach a desired level of precision. For graphing purposes, a "bunch" is only a few dozen.
I think these will be some of my prettiest graphs yet, once they're completed. I'll create a separate thread for them when they're done.
~ Jay Daniel Fox
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11/13/2007, 10:36 AM
(This post was last modified: 11/13/2007, 10:37 AM by bo198214.)
jaydfox Wrote:Do you mean that the function being iterated is entire, Exactly.
Quote:For example, multiplication is entire, as is exponentiation (the continous iterator).
However multiplication has only one fixed point, so the theorem can not apply.
Quote: However, exponentiation is entire,
And thatswhy the can not be the same at any two fixed points. Thatswhy the at one fixed point has singularities at all other fixed points. Every Abel function of an entire function has singularities at every fixed point except at most 1.
Quote:Edit: Since he just left (or is about to leave) on a 2month trip,
Its my last day, I rather should be occupied with packing
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bo198214 Wrote:jaydfox Wrote:Edit: Since he just left (or is about to leave) on a 2month trip, Its my last day, I rather should be occupied with packing
Ah, you'll miss the pictures for then. I won't have them done until Wednesday night at the earliest, probably later depending on my work schedule.
~ Jay Daniel Fox
