Posts: 1,918
Threads: 414
Joined: Feb 2009
(07/02/2011, 10:32 PM)tommy1729 Wrote: in particular i still dont know why 1.729 i ^1.729 i ^ ... 0.5 + 0.5 i gives a circle and that bothers me.
nobody is telling me ...
anyway you might be intrested in the formula for the assumed invariant :
for real z > 0
tommysexp_e(z,x) = tommysexp_e(z,inv(x))
then inv_e(z,x) = tommyQ_e^[2](z,x)
tommyQ_e(z,x) = tommysexp_e(  z,  tommysexp_e(z,x))
Posts: 684
Threads: 24
Joined: Oct 2008
07/05/2011, 01:09 AM
(This post was last modified: 07/05/2011, 01:53 AM by sheldonison.)
(07/04/2011, 10:56 PM)tommy1729 Wrote: (07/02/2011, 10:32 PM)tommy1729 Wrote: in particular i still dont know why 1.729 i ^1.729 i ^ ... 0.5 + 0.5 i gives a circle and that bothers me.
nobody is telling me ...
I can't follow your latest tommysexp equations right now, but the part about 1.729i is an approximation for b=1.712936...i, which is the base on the Shell Thron boundary region with real(b)=0. All bases on the Shell Thron boundary with an irrational period are going to give a circle (as long as the irrational period is a Brjuno number). The definition is the abs(log(fixed_point))=1, which leads to a neutral fixed point. Look up Siegel Disc on Wikipedia, for some background. Also, recent discussion has some really good papers referenced in it (on the Brjuno number). Each iteration circles around the h(z) function (inverse Schroder function), so how circular the path is depends on the magnitude of the initial point (0.5+0.5i)L, which for this case is L~= 0.390842217 + 0.4586753i, so 0.5+0.5i is pretty close to the fixed point, so the path is fairly circular.
1.729*I is a very very slightly repelling fixed point, that will eventually diverge, abs(log(L))=1.00244, but it would take a long time.
 Sheldon
Posts: 1,918
Threads: 414
Joined: Feb 2009
(07/05/2011, 01:09 AM)sheldonison Wrote: (07/04/2011, 10:56 PM)tommy1729 Wrote: (07/02/2011, 10:32 PM)tommy1729 Wrote: in particular i still dont know why 1.729 i ^1.729 i ^ ... 0.5 + 0.5 i gives a circle and that bothers me.
nobody is telling me ...
I can't follow your latest tommysexp equations right now, but the part about 1.729i is an approximation for b=1.712936...i, which is the base on the Shell Thron boundary region with real(b)=0. All bases on the Shell Thron boundary with an irrational period are going to give a circle (as long as the irrational period is a Brjuno number). The definition is the abs(log(fixed_point))=1, which leads to a neutral fixed point. Look up Siegel Disc on Wikipedia, for some background. Also, recent discussion has some really good papers referenced in it (on the Brjuno number). Each iteration circles around the h(z) function (inverse Schroder function), so how circular the path is depends on the magnitude of the initial point (0.5+0.5i)L, which for this case is L~= 0.390842217 + 0.4586753i, so 0.5+0.5i is pretty close to the fixed point, so the path is fairly circular.
1.729*I is a very very slightly repelling fixed point, that will eventually diverge, abs(log(L))=1.00244, but it would take a long time.
 Sheldon
ah , so its not a perfect circle ? it just approximates one ?
didnt gottfried say it was a perfect circle ?
i find 0.5 + 0.5 i  L still relatively large to give such a result , but it must be visual illusion and just a circle approximation then  if i understand you .
thanks.
tommy1729
Posts: 1,918
Threads: 414
Joined: Feb 2009
11/18/2012, 11:41 PM
(This post was last modified: 11/18/2012, 11:43 PM by tommy1729.)
For those who are confused , let me explain the next big step about this.
If we want a superfunction sexp(z) that is analytic for Re(z) > C >> 0 for some C then we need tommysexp(z) or sheldontommysexp(z) to have the same property. Lets call this C property.
tommysexp(z) and sheldontommysexp(z) agree on the reals (converge speed may differ though ).
Now tommysexp(z) seems to be only defined for the real z so it requires analytic continuation ( assuming it is possible ).
However sheldontommysexp(z) IS periodic and might be analytic in that period. However to go beyond that period we will need analytic continuation because the C property forbids nonreal periodicity. ( again assuming it is possible ).
The analytic continuation of tommysexp(z) and sheldontommysexp(z) are identical function.
Hence either/both tommysexp and sheldontommysexp have property C.
To prove property C of tommysexp(z) = sheldontommysexp(z) one can choose one of those 2 and investigate its analytic continuation/singularities/discontinu on the complex plane within the periodic strip for Re(z) > C.
If it turns out there are no singularities in the periodic semistrip near the reals , then both functions have property C and are equal.
If there are singularities , property C is broken however it might still have regions of analyticity. A natural boundary proves nowhere analytic ( because exp has fixpoints ).
The reason why tommysexp and sheldontommysexp are equivalent is because of a base change like trick explained by sheldon in this thread. The periodicity is also explained here by him.
Notice that a log(f(z)) with f(z) periodic remains periodic.
Also notice that the C property is a uniqueness criterion !
To see this think of (my proof of) TPID 4.
Or the 1 period function wobble of sexp(z).
regards
tommy1729
Posts: 684
Threads: 24
Joined: Oct 2008
11/20/2012, 12:01 AM
(This post was last modified: 11/21/2012, 04:04 PM by sheldonison.)
(11/18/2012, 11:41 PM)tommy1729 Wrote: tommysexp(z) and sheldontommysexp(z) agree on the reals (converge speed may differ though ).
Now tommysexp(z) seems to be only defined for the real z so it requires analytic continuation ( assuming it is possible ).
However sheldontommysexp(z) IS periodic and might be analytic in that period. However to go beyond that period we will need analytic continuation because the C property forbids nonreal periodicity. ( again assuming it is possible ).
\( \text{tommysexp}_3(z)=\log^{[3]}(\text{2sinh}^{[z]}(\exp^{[3]}(1))) \)
This is tommysexp generated with three iterated logarithms. This function is reasonably well behaved at the origin, with a radius of convegence of approximately 0.457. For reference, the taylor series coefficients are listed below.
\( \text{tommysexp}_4(z)=\log^{[4]}(\text{2sinh}^{[z]}(\exp^{[4]}(1))) \)
But the next step, for n=4, is not so well behaved in the complex plane. The radius of convergence in the complex plane is about 0.03469, and by my count, there are 587507 singularities within a radius of 0.035. And its not just a matter of extending the function beyond the wall of singularities, because the function misbehaves inside the wall of singularities too. Here is where singularities occur, due to the log(0).
\( \text{2sinh}^{[z]}=n\pi i \) \( \text{2sinh}^{[z+1]}=0 \)
The other interesting thing about the singularities is how quickly they fall off. At 99.999% of the singularity radius, \( \text{tommysexp}_4(z)\text{tommysexp}_3(z)<10^{50} \). Its almost as if one can totally ignore the effect of the fourth iteration. And in fact the 4th iteration has no measurable effect on any taylor series coefficient one would use in normal computation. But, somewhere around the 1352620th taylor series term there is an abrupt transition. The \( \text{tommysexp}_4(z) \) function starts behaving radically differently than the \( \text{tommysexp}_3(z) \), as the taylor series terms finally become dominated by the wall of nearby singularities. By the way, estimating the 1.35 millionth taylor series coefficent for a function is a very delicate calculation involving algorithms to make approximations cauchy integrals at carefully picked radius's of \( s(z)\log(e^{s(z)}e^{s(z)})=\sum_{n\to\infty}e^{2n s(z)} \), where \( s(z)=\text{2sinh}^{[z]} \). After picking an appropriate approximation radius, I can estimate the log of the desired taylor series coefficient. It helps that usually the different values of "n" in the sum can be treated independently. I have a parigp program, but haven't posted the details, which are complicated. The algorithm works for both the 2sinh(z) superfunction, as well as the exp(z1) superfunction.
One final thought. \( \text{tommysexp}_3(z) \) has a radius of convergence of ~0.457 which can be seen in the taylor series coefficients below. It is not as well behaved as sexp(z) in the complex plane. The interesting thing is that \( \text{tommysexp}_4(z) \) has the same first 40 taylor series coefficients listed below, accurate to millions of decimal digits. This would also be true of the first million or so terms, until the transition. So, in addition to misbehaving at the 1.35 millionth taylor series term, \( \text{tommysexp}_4(z) \) still has all smaller taylor series coefficients dominated by the singularities in \( \text{tommysexp}_3(z) \). \( \text{tommysexp}_5(z) \) would have a radius of convergence of less than 10^7, that would effect uncountably large taylor series terms due to a wall of uncountably many nearby singularities.
There's nothing particularly special about doing the approximations for tommysexp(0), and similar misbehavior would be expected for any value of tommysexp(z). I could post the methods I used for these approximations which might lead to a rigorous proof, but the general problem is complicated. You need to show the taylor series terms eventually grow faster than any radius of convergence. The difficulty is coming up with a rigorous language to express appropriate approximations of superexponentially large taylor series terms. Even so, I am convinced that tommysexp(z) is nowhere analytic, even though it is infinitely differentiable at the real axis.
 Sheldon
added pretty graph showing accuracy of tommysexp3 taylor series terms, and switchover to tommmysexp4 taking over near 1.35 millionth taylor series term
Code: Taylor series coefficients for tommysexp(0)
a0= 1.00000000000
a1= 1.09146536077
a2= 0.273334906394
a3= 0.215218479242
a4= 0.0652715037680
a5= 0.0391656564309
a6= 0.0171521314068
a7= 0.0117058806325
a8= 0.00471958861559
a9= 0.00123667589678
a10= 0.00226288150336
a11= 0.00321559868096
a12= 0.00820154271015
a13= 0.00218777707040
a14= 0.0477372146016
a15= 0.117247563749
a16= 0.0788849220733
a17= 0.883038307996
a18= 0.610166815881
a19= 5.10470797278
a20= 7.81612106347
a21= 28.6479580355
a22= 60.0481521891
a23= 173.314801252
a24= 382.323334609
a25= 1156.14068365
a26= 2075.65103517
a27= 8124.15769733
a28= 8589.21772341
a29= 56026.7180961
a30= 10973.8915911
a31= 353646.843999
a32= 264336.726813
a33= 1880963.01782
a34= 3669078.97541
a35= 7012765.98983
a36= 30701470.8277
a37= 1548864.00646
a38= 184151139.583
a39= 254566511.140
Posts: 1,918
Threads: 414
Joined: Feb 2009
(11/20/2012, 12:01 AM)sheldonison Wrote: \( \text{2sinh}^{[z]}=n\pi i \)
What is wrong with that ?
We can compute log of that ?
Posts: 684
Threads: 24
Joined: Oct 2008
11/20/2012, 11:01 PM
(This post was last modified: 11/20/2012, 11:03 PM by sheldonison.)
(11/20/2012, 10:53 PM)tommy1729 Wrote: (11/20/2012, 12:01 AM)sheldonison Wrote: \( \text{2sinh}^{[z]}=n\pi i \)
What is wrong with that ?
We can compute log of that ? if \( \text{2sinh}^{[z]}=n\pi i \) then \( \text{2sinh}^{[z+1]}=e^{n\pi i}e^{n\pi i}=0 \)
and therefore there is a singularity.
\( \log(\text{2sinh}^{[z+1]})=\log(0) \)
 Sheldon
Posts: 1,918
Threads: 414
Joined: Feb 2009
I think the same applies to the base change ...
I do not immediately know a way around it.
Although those singularities maybe vanish from analytic continuation if they were in an undefined region before analytic continuation.
Posts: 684
Threads: 24
Joined: Oct 2008
11/20/2012, 11:46 PM
(This post was last modified: 11/20/2012, 11:53 PM by sheldonison.)
(11/20/2012, 11:32 PM)tommy1729 Wrote: I think the same applies to the base change ...
I do not immediately know a way around it. Yes! The exact same thing applies to the base change. Iterating \( s(z)=\exp(z1) \) is exactly analogous to iterating tetration base \( \eta=\exp(1/e) \). Then, taking the logarithm as the first step in the base change results in singularities whenever \( s^{[z]}=2n\pi i \), where \( \log(s^{[z+1]})=\log(e^{2n\pi i}1)=\log(0) \)
Actually, for every singularity in the base change, there are two singularities for 2sinh(z), and the singularities for \( 2n\pi i \) for both functions are very nearly in the same place in the complex plane!
 Sheldon
Posts: 1,918
Threads: 414
Joined: Feb 2009
11/28/2012, 06:59 PM
(This post was last modified: 11/28/2012, 07:40 PM by tommy1729.)
I would like to point out that tommysexp for bases \( >= e^2 \) is problemfree for real values.
The reason is that for bases \( >=e^2 \) there is a unique (entire) superfunction for the 2sinh ( example base \( e^2 \) thus \( e^{2x}  e^{2x} \)).
And that is true because 2sinh(2x) has only one complex fixpoint : 0.
So for these bases tommysexp is valid , has uniqueness and existance and satisfies all desired properties with the possible exception of analytic. However it is Coo.
While doing basechange I prefer to use tommysexp base exp(2).
Lets call that basechangetommy.
I suspect basechangetommy and tommysexp for bases larger than exp(2) to have similar properties.
I even considered them to be equal once , although I now keep it as a sort of vague closeness idea.
I am considering properties and might even call it my new pet ideas in tetration.
I believe in intresting properties for these functions.
Im also considering replacing the logs with functions that are asymptotic to logs but entire. ( an idea that I came up with together with mick , the guy who posts on MSE )
However that is quite complicated and not well understood at the moment. For instance it is unknown if this is the sought of analytic continuation or a totally different function ?
