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Hey folks,
until now we mainly considered holomorphy in the second argument of tetration:
.
If we now fix and demand also the function to be analytic on one small interval , the function is already determined for all other bases, particularly for bases by analytic continuation along the real axis (if there is no singularity at ).
For bases we know that the regular iteration at the lower fixed point is analytic there and continuable to 1.
So the question would be to what values the regular iteration continues for . I guess it has a branch point at and continues to complex values beyond .
I will explore this thought more concretely in the following posts.
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11/05/2009, 10:50 PM
(This post was last modified: 11/05/2009, 10:50 PM by BaseAcid Tetration.)
You only seem to mean the tetrapower instead of the tetraexponential.
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I've also been toying with this, too. It appears, however, that it continues to real values, not complex values, for .
Consider the following “regular iteration” limit formula:
http://math.eretrandre.org/tetrationforu...10#pid1610
where is the attracting fixed point, i.e. .
There's also a series formula, apparently (with respect to "w" in , so setting w = 1 yields the tetrational), but I haven't yet figured out how one is supposed to evaluate the general coefficient. How can that be done?
To test the analytic continuation, we can use the Cauchy integral: if is analytic, then we derive the powerseries coefficients at via
and
.
This seems to provide a more efficient algorithm for the recovery of the coefficients, than straight numerical differentiation from the difference quotient (which seems to require more rapidlyescalating levels of numerical precision).
We can now choose some close to , set for some fractional tower where is obtained from the regular formula, and a path that encircles it, but does not leave the kidneybean ("ShellThron" region) of convergence, e.g. a small circle round the point. Then, by increasing n, we obtain the Taylor coefficients. For , expanded about , using a circle of radius 0.01, we get the following estimates for the first 25 coefficients:
Code: a_0 ~ 1.24622003310
a_1 ~ 0.447921100148
a_2 ~ 0.194428566238
a_3 ~ 0.143167873861
a_4 ~ 0.144967399774
a_5 ~ 0.182159301224
a_6 ~ 0.263407426426
a_7 ~ 0.417098477762
a_8 ~ 0.702204147888
a_9 ~ 1.23535078363
a_10 ~ 2.24705897139
a_11 ~ 4.19678269601
a_12 ~ 8.00913024091
a_13 ~ 15.5621057277
a_14 ~ 30.7029302261
a_15 ~ 61.3746543453
a_16 ~ 124.093757768
a_17 ~ 253.427621734
a_18 ~ 522.152429631
a_19 ~ 1084.31773542
a_20 ~ 2267.61147731
a_21 ~ 4772.09902388
a_22 ~ 10098.4528601
a_23 ~ 21464.8938685
a_24 ~ 45852.6753827
For , we can use this get , which is real, not complex. How does that agree with other methods of tetration for bases greater than ? This series should have radius of convergence 0.42, determined by the distance to the nearest singularity/branchpoint, which is at z = 1.
I'm not sure of a formal proof of the "continuability", though one approach may be to try and differentiate the regular iteration formula, then prove that the limit of the derivative as converges  in order for it to switch to nonreal complex values as is passed, that point would have to be some sort of singularity, like a branch point, and so the function would not be differentiable there, and if it is, then that is not the case.
I'll see if maybe I can get some graphs on the complex plane but calculating the regular iteration is a bear as it requires lots of numerical precision, at least for the limit formula. Maybe that series formula would be better?
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11/06/2009, 11:58 AM
(This post was last modified: 11/06/2009, 12:11 PM by mike3.)
Also, fidding around with this some more, it seems that we cannot continue below in one step, suggesting that it is a singularity (likely a branchpoint).Trying to continue into the pseudocircle with this method from some direction in the complex plane also does not seem to work well, at least with only one step, suggesting lots of singularities/branchpoints in this area. This is Weird!!! Yet we can go up the real axis past ...
It seems possible that certain Hopfian bifurcation points (of the integer tetrations) are also branchpoints, but not all (note that if one looks at the tetration fractal, the point is an accumulation point of Hopfian points, yet if they were all singularities it would be impossible to analytically continue to , yet this seems to work OK.). A graph might be useful in figuring out what's going on here, but solving the regular iteration near the edges of the ShellThron region is difficult, esp. if we want a highresolution plot. Perhaps if there was some way to evaluate the regular iteration on the edge of the region? But this is tough  see the thread about regularly iterating ... Note we can parameterize the edge using exponentials, however (see here: http://math.eretrandre.org/tetrationforu...11#pid3011), so obtaining points on it is not a problem.
All these wild things make me think that if tetration is ever developed to the point where it could be a "real" special function like the gamma function, error function, zeta function, hypergeometric, Lambert W, etc. it'd be one of the most exotic yet. Especially considering the "fractal" structure (looks like the dynamical Julia set of the exp. function!) it produces for bases like those greater than when it is plotted for complex values of the tower.
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11/06/2009, 12:12 PM
(This post was last modified: 11/09/2009, 09:36 AM by bo198214.)
(11/06/2009, 04:15 AM)mike3 Wrote: I've also been toying with this, too. It appears, however, that it continues to real values, not complex values, for .
that would be awesome. And this would mean that there is no singularity at ?
Quote:For , we can use this get , which is real, not complex.
with "use this" you mean the powerseries development you just derived at ? But how do you know that it converges and that there is no branchpoint at ?
Quote:I'm not sure of a formal proof of the "continuability", though one approach may be to try and differentiate the regular iteration formula, then prove that the limit of the derivative as converges  in order for it to switch to nonreal complex values as is passed, that point would have to be some sort of singularity, like a branch point, and so the function would not be differentiable there, and if it is, then that is not the case.
*nods* but at least it is already known that the regular iteration is not analytic at . However it is currently not clear to me what this states about the regularity of at .
Quote:I'll see if maybe I can get some graphs on the complex plane but calculating the regular iteration is a bear as it requires lots of numerical precision, at least for the limit formula. Maybe that series formula would be better?
Ya I will try it with the series formula (or perhaps a mixture with limit formulas).
Actually it seems that noone posted pictures of tetrapowers yet!!!
(yes Bat I mean tetrapowers.)
So it will be time that we have some pictures at least, as the theoretic consideration seems utmost complicated to me.
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(11/06/2009, 12:12 PM)bo198214 Wrote: that would be awesome. And this would mean that there is no singularity at ?
Yes, that would be correct.
Quote:with "use this" you mean the powerseries development you just derived at ? But how do you know that it converges and that there is no branchpoint at ?
That's correct. Though as I've mentioned, there is as of yet no rigorous proof that the series converges even though the first 25 terms given do, hence why I mentioned it suggests, not proves, the hypothesis.
That it takes on a real value here suggests this continuation cannot be interpreted as regular iteration at these bases, which would take on complex values.
Quote:*nods* but at least it is already known that the regular iteration is not analytic at . However it is currently not clear to me what this states about the regularity of at .
Not analytic at ? That's strange. Not analytic at would make more sense. If you look at a graph of for , it is a smooth, monotone curve that has a horizontal asymptote at . If we mirror it to get the shape of the inverse function , we see the singularity is at and is a smooth point. Then we have , and as the slog is smooth at , and the tet is smooth at 1, it should be smooth there, no?
Quote:Ya I will try it with the series formula (or perhaps a mixture with limit formulas).
Actually it seems that noone posted pictures of tetrapowers yet!!!
(yes Bat I mean tetrapowers.)
So it will be time that we have some pictures at least, as the theoretic consideration seems utmost complicated to me.
Can you tell me what this series formula is, by the way? I am only familiar with the limit formulas. Series seems easier to analyze theoretically than a limit with the whole "log log log log log log log ..." thing, and maybe also easier to analyze numerically, in that it may not have the increasing numerical precision requirement.
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11/06/2009, 11:29 PM
(This post was last modified: 11/06/2009, 11:42 PM by bo198214.)
(11/06/2009, 09:16 PM)mike3 Wrote: That's correct. Though as I've mentioned, there is as of yet no rigorous proof that the series converges even though the first 25 terms given do, hence why I mentioned it suggests, not proves, the hypothesis.
That it takes on a real value here suggests this continuation cannot be interpreted as regular iteration at these bases, which would take on complex values.
No, Mike, thats no hint at all. You can take any function real analytic at some point, close to some real singularity. The truncated powerseries always yields real values outside the convergence radius. Better would be a graph of the root test, or at least a graph that shows some convergence of the series.
Quote:Not analytic at ? That's strange. Not analytic at would make more sense.
yes, of course you are right. I correct it in my original post.
Quote:Can you tell me what this series formula is, by the way?
Have a look at the tetration method draft formula 2.5. This is regular iteration at fixed point 0.
You get the formula by coefficient comparison of and , where the index indicates the corresponding coefficient of the powerseries at 0.
For fixed points different from 0 you consider the function which has the fixed point at 0. Then .
I.e. in our case we have .
To work with a simpler formula we consider another conjugation .
, , , .
This gives the following unsimplified coefficients of :
One can see that the coefficients are polynomials in with rational coefficients in .
One needs to investigate whether is analytic in with .
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11/07/2009, 12:23 AM
(This post was last modified: 11/07/2009, 12:31 AM by mike3.)
By a "graph that shows the convergence of the series" you mean like a graph of partial sums? Also, I could try a root test or ratio test but I think to make it really good I'll need more terms... Is there a general algorithm or formula to generate those polynomials in the regular iteration series?
EDIT: I just saw the paper and there appears to be a recurrence formula on page 8 but am a little confused by the notation. Specifically, what's supposed to be? nth Taylor coefficient of m iterations of g? But we have not yet determined g, so how can we iterate?
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(11/07/2009, 12:23 AM)mike3 Wrote: By a "graph that shows the convergence of the series" you mean like a graph of partial sums?
Yes, yes, xaxis: n, yaxis: nth partial sum.
Quote:Also, I could try a root test or ratio test but I think to make it really good I'll need more terms...
With root test the same: xaxis: n, yaxis: . The roottest should converge to something bigger than the distance from the development point to the perhapssingularity.
Quote:I just saw the paper and there appears to be a recurrence formula on page 8 but am a little confused by the notation. Specifically, what's supposed to be? nth Taylor coefficient of m iterations of g? But we have not yet determined g, so how can we iterate?
In this draft integeriteration is always denoted by , while power is as usual . depends only on values , . You can take the polynomial of g truncated to n, then take the mth power and then get the coefficient at n.
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