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closed form for regular superfunction expressed as a periodic function
#21
(08/31/2010, 08:04 AM)mike3 Wrote: Doing some tests, it appears that



has only one occurrence of , and no higher powers of it, and it never seems to be multiplied by any sort of n-dependent coefficient. This means that . I don't have a proof at this point, ...

Now I've got that proof. Here it goes:

We have



(note how the connection to the Faà di Bruno's formula is clear)

with the sum taken over all sequences of non-negative integers such that and , and is the sum of the . As should be obvious from the formula, we see that the value in only occurs in the component . What kind of occurrences of are possible there? Using the formulas for the , we see that , which means that all but one must be zero, and that one that is must be 1. The second constraint, , would imply that if all but one are zero, the nonzero one must equal where is its position in the sequence. This would mean must divide . The first constraint, though, said it must be 1, which means and so the only possible sequence of is ( terms). This means , which is obviously just . So contains only one term with , which is just itself. And then follows trivially.
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#22
(08/31/2010, 08:33 PM)sheldonison Wrote: I'm kind of intrigued at actually having a closed form for a real valued superfunction, like sqrt(2). I'm guessing that the number of terms required for convergence gets extremely large as z increases. From the terms I computed a couple of days ago, it looks like the terms are decreasing exponentially, which means the series acts like it has a singularity. However, since the regular superfunction is entire, we must have convergence to infinity, so the terms must eventually decrease faster than exponentially.
- Sheldon
Or it does already, but just a wee bit faster than exponential. This makes me wonder about an interesting place for mathematical exploration: the behavior of entire functions given by a Taylor series whose terms' coefficients decay just a "wee" bit faster than exponential. As this example shows, such functions can have extremely complicated behavior (note the complicated "fractal structure" of the graphs of these superfunctions.).
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#23
Quote:Or it does already, but just a wee bit faster than exponential. This makes me wonder about an interesting place for mathematical exploration: the behavior of entire functions given by a Taylor series whose terms' coefficients decay just a "wee" bit faster than exponential. As this example shows, such functions can have extremely complicated behavior (note the complicated "fractal structure" of the graphs of these superfunctions.).

Yes, very interesting. I notice that the fractals are often very sparse too. Its only growing super-exponentially on a filagree, and most of the rest of the function is not growing nearly as fast. So it looks like spike singularities. Not sure if that helps any.

I think I may have also figured out the closed forms for the abel functions, the inverse superexponential developed from the fixed point, which I'll eventually post, when I have time to verify the equations.

Now I'm stuck on eta, . I'm trying the substitution y=1/z. I think my results so far are bogus though, so I'm editing them out.
-Sheldon
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#24
i think you will be stuck on most parabolic fixpoints with this method.
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#25
(09/03/2010, 08:19 PM)tommy1729 Wrote: i think you will be stuck on most parabolic fixpoints with this method.
Right now, I'm only interested in the parabolic fixed point for (not the general parabolic question), but its definately a challenge. There are a number of posts on on this forum, which together with the limiting behavior, e-2e/z, and my own experiments, leave me somewhat confused.
- Sheldon
http://math.eretrandre.org/tetrationforu...php?tid=13
http://math.eretrandre.org/tetrationforu...php?tid=10
http://math.eretrandre.org/tetrationforu....php?tid=8
http://math.eretrandre.org/tetrationforu...hp?tid=316
http://math.eretrandre.org/tetrationforu...hp?tid=498
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#26
dear sheldon , i think you are confused because exp(x) - 1 doesnt have a holomorphic half-iterate.

regards

tommy1729
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#27
(09/05/2010, 04:45 PM)tommy1729 Wrote: dear sheldon , i think you are confused because exp(x) - 1 doesnt have a holomorphic half-iterate.

regards

tommy1729
I think I'll need to get Peter Walker's paper, Proc. AMS 1990. He proved that the upper superfunction of exp(x)-1 is entire, which corresponds to the upper superfunction of . I would be interested in knowing how that was done.

My understanding is that there is a analytic half-iterate of exp(x)-1, but it isn't entire.
- Sheldon

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#28
it only has a formal powerseries that converges at the integers but has radius 0 hence not analytic.
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#29
(09/07/2010, 07:46 PM)tommy1729 Wrote: it only has a formal powerseries that converges at the integers but has radius 0 hence not analytic.

Ya but it is analytic outside the fixed point. There is one half-iterate to the right of the fixed point and one to the left, which are different functions, i.e. no analytic continuations of each other. Both iterates have the asymptotic power series expansion at the fixed point that one gets when calculating the *formal* powerseries (which has 0 convergence radius).


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#30
thus it is complex continuous ??

ive been thinking that entire functions with parabolic fixpoints with n somewhere analytic solutions for their half-iterate are half-iterates of another related function that has exactly n non-parabolic fixpoints an no other fixpoints or exactly n parabolic fixpoints with analytic solutions at their fixpoints and no other fixpoints.

as the relationship between eta^x and e^x - 1.

that would make parabolic fixpoints more logical imho ...

tommy1729
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