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06/20/2011, 01:36 PM
(This post was last modified: 06/20/2011, 02:39 PM by Gottfried.)
(06/20/2011, 05:27 AM)sheldonison Wrote: My conjecture is for bases on the Shell Thron boundary, there is an analytic superfunction with a real period, whose structure depends on what the continued fraction representation of the real period is. As long as the period is a real number (with an infinite continued fraction representation), then I suspect the superfunction is analytic. If the period is a rational number, then I don't think there is an analytic superfunction. For example, this base, with a real period=3, probably doesn't have an analytic superfunction, developed from the neutral fixed point, because starting with a point near L, and iterating the function x=B^x three times, doesn't get you back to the initial starting point.
Base= 0.030953557167612060 + 1.7392241043091316i
L= 0.39294655583435517 + 0.46203078407110528i Hi Sheldon 
I've inserted your baseparameter and got the following plot for the orbit/for the three partial trajectories in the same style of my previous plots. I seem to have problems to understand your comment correctly. For instance, isn't that fixpoint attracting instead of neutral?
Having seen this I assume, that also with a startingpoint near the fixpoint we get something converging to the fixpoint, however slow. But, well, that would be now another job to prove.
In my initial plot it seemed, that there is only one base b0, whose orbits are between converging to the fixpoint and diverging, and because the base at 1.71290*I is such a base I assume, that we get either convergence here or divergence to a triplett of cumulation points.
What do you think?
Gottfried
[Update]
A startingpoint x0=0.41*(1+I)=b^^0, even nearer at the fixpoint L, exhibits now repelling properties of the fixpoint. So I think, that in fact there are three "oscillating" fixpoints in the near of the orbit of the last experiment and the trajectories of the first picture do not approach the fixpoint L but that triplett of accumulation(?) points.
Gottfried Helms, Kassel
Posts: 684
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06/20/2011, 02:46 PM
(This post was last modified: 06/20/2011, 04:11 PM by sheldonison.)
(06/20/2011, 01:36 PM)Gottfried Wrote: (06/20/2011, 05:27 AM)sheldonison Wrote: My conjecture is for bases on the Shell Thron boundary, there is an analytic superfunction with a real period, whose structure depends on what the continued fraction representation of the real period is. As long as the period is a real number (with an infinite continued fraction representation), then I suspect the superfunction is analytic. If the period is a rational number, then I don't think there is an analytic superfunction. For example, this base, with a real period=3, probably doesn't have an analytic superfunction, developed from the neutral fixed point, because starting with a point near L, and iterating the function x=B^x three times, doesn't get you back to the initial starting point.
Base= 0.030953557167612060 + 1.7392241043091316i
L= 0.39294655583435517 + 0.46203078407110528i Hi Sheldon 
I've inserted your baseparameter and got the following plot for the orbit/for the three partial trajectories in the same style of my previous plots. I seem to have problems to understand your comment correctly. For instance, isn't that fixpoint attracting instead of neutral?
The definition for the ShellThron region boundary is \( \log(L)=1 \), which is the case. But when the period \( =2\pi i/\log(\log(L))=3 \) is an integer (or a fraction), the equations misbehave. At the ShellThron boundary, the period is always a real number, and the fixed point is neither attracting nor repelling. At first, I thought the idea of a superfunction with a real period was nonsense in this post, but then I was able to get it to work, except for the cases when the period was an integer, or a fraction with a small denominator. So that experimentation is where my conjecture came from.
For example, here is another case, on the ShellThron boundary, that should work fine because the period is a real number, with a period just a little bit bigger than 3. With a sufficient number of iterations, it generates a very nice plot, that appears to lead to an analytic superfunction. But, in the plot, you can see the influence of the base being just a litle bit bigger than an integer. By the way, these ShellThron boundary bases are easy to generate. \( L=\exp(\exp(2 \pi i/\text{period})) \) and \( \text{base}=L^{1/L} \)
base= 0.036314759343852642170871708751 + 1.7435957010705633826865464522i
L= 0.39309905520386861718874315414 + 0.46286165860913191074862913970i
Period= 3.0019951097271885263233102180
 Sheldon
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06/20/2011, 10:22 PM
(This post was last modified: 06/20/2011, 10:50 PM by tommy1729.)
what if b^^0 is large ?
im not sure if we get the same behaviour.
if b^^0 is large we might reach divergence ?
or are we suppose to restrict b^^0 to the ShellThron region or their fixpoints ?
did anyone conjecture or prove bounds on b^^0 ?
i wonder how b = 1.71290 i , b^^0 = 1729 + 1729 i looks like.
the area within the cycles is also of interest.
perhaps not usefull , but i always try to map a cycle to a unit circle and then back again.
this can be done because of the riemann mapping theorem.
then i try to see how fast the iterations cycle on the unit circle ;
i (try to) study the complex angle theta of RIEMANN [superfunction(inversesuper(x_0) + n)] with respect to n.
( and this has the same period 2pi i / ln(ln(L)) ofcourse )
tommy1729
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(06/20/2011, 10:22 PM)tommy1729 Wrote: what if b^^0 is large ?
im not sure if we get the same behaviour.
if b^^0 is large we might reach divergence ? I don't understand  but we observe what appears to be fractal behavior, (on the ShellThron boundary) which would be the edge of convergence.
Quote:....the area within the cycles is also of interest.
perhaps not usefull , but i always try to map a cycle to a unit circle and then back again.
this can be done because of the riemann mapping theorem.
then i try to see how fast the iterations cycle on the unit circle ;
The picture in the wiki is in the upper half of the complex plane, but it can be contactenated with its complex conjugate to make an enclosed region. Then this region could certainly be Riemann mapped onto a unit circle...
 Sheldon
Posts: 901
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Joined: Aug 2007
(06/20/2011, 02:46 PM)sheldonison Wrote: base= 0.036314759343852642170871708751 + 1.7435957010705633826865464522i
L= 0.39309905520386861718874315414 + 0.46286165860913191074862913970i
Period= 3.0019951097271885263233102180
 Sheldon
Waahh... I must have had a period of really weak thinking. Well, this discussion brought me back to the right track, I suppose. After I got it now (again, I must have had it already earlier) I made a picture, how the fixpoint, the log of the fixpoint, and the base according to the ShellThrondescription are connected. (I'Ve also put it in the hyperopwiki). In my notation I always used u (for the log of the fixpoint), t =exp(u) for the fixpoint and b=exp(u/t) for the base. Here is the picture:
Gottfried
Gottfried Helms, Kassel
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06/21/2011, 03:02 PM
(This post was last modified: 06/21/2011, 09:24 PM by sheldonison.)
(06/21/2011, 12:13 PM)Gottfried Wrote: Waahh... I must have had a period of really weak thinking.
Welcome to the club....
Quote:Well, this discussion brought me back to the right track, I suppose. After I got it now (again, I must have had it already earlier) I made a picture, how the fixpoint, the log of the fixpoint, and the base according to the ShellThrondescription are connected....In my notation I always used u (for the log of the fixpoint), t =exp(u) for the fixpoint and b=exp(u/t) for the base....
Thanks Gottfried, great picture. So, using your notation, we add one more equation: \( \text{period}=2\pi i/\log(u) \). In the original example, with all of the nice plots you made, base=1.712936040374417981826i.
u = 0.5070842426299714169803 + 0.8618964966145228964379i
period = 2.988300627934001489933
Now, if you take one of those plots you made, starting with one of your initial points, you'll find that the period works exactly. For example, if you start iterating with b^^0 = 0.7+0.7i. then
\( \text{Superfunction}(0)=0.7+0.7i \)
\( \text{Superfunction}(x \bmod \text{period})=\exp_b^{o x}(0.7+0.7i) \)
And if you repeat this an infinite number of times, then you get the nice periodic contour curve you made. I'm not sure about what the correct notation is for mod with an irrational number. The SuperFunction is real periodic, where iterating (x mod 2.9883) times gets you back to your starting point.
\( \text{SuperFunction}_{L}(x) = L + \sum_{n=1}^{\infty} a_n\exp(2n\pi ix/\text{period}) \)
Using the exact same coefficients, the circular contour of the function you graphed can be represented as f(z), evaluated on the unit circle, where abs(z)=1.
\( f(z) = L + \sum_{n=1}^{\infty} a_n z^n \)
Here are the 100 coefficients for these equations, which is accurate to 15 decimal digits or so. a0=L. The equations were generated with b^^0=0.7+0.7i
 Sheldon
Code: a0= 0.3920635484599088334159 + 0.4571543197696414316818*I
a1= 0.1926347048245026996218 + 0.2553626064604919992564*I
a2= 0.04367562340340935922713 + 0.02205560982867233582908*I
a3= 0.007530714487061478885377  0.004281576514011399313377*I
a4= 0.04709123530117242894461 + 0.001729440845159316720314*I
a5= 0.01306847282348073176814  0.005728356074317823223406*I
a6= 0.0006614004482465856090804  0.003745545557771203176425*I
a7= 0.009018437522021720431834  0.01057073314544308526135*I
a8= 0.001367493374159604282404  0.005020786666321506753021*I
a9= 0.001093179701013968437283  0.001263132425285266404038*I
a10= 0.0009612888490325652001380  0.004685330087439725794279*I
a11= 0.001197476243914925206686  0.001632842246574827208293*I
a12= 0.0007333426764609148904379 + 0.00002704910834568717554014*I
a13= 0.001585607063510229010734  0.0007749689289491325438067*I
a14= 0.0008122799208911508244806  0.00002786102261464546068704*I
a15= 0.0001912817298718253167176 + 0.0002595615603574707578201*I
a16= 0.0006062049020599662754753 + 0.0003026807801206704776212*I
a17= 0.0002133612201397169986658 + 0.0002546714235074630134220*I
a18= 0.00003743767676364786366434 + 0.0001366516889627626748407*I
a19= 0.00004960231979796590150008 + 0.0002622624914391191734664*I
a20= 0.00002914410828772657407477 + 0.0001343295228982834413090*I
a21= 0.00005740673450802333073318 + 0.00002412575934410344993297*I
a22= 0.00007190414851207875090348 + 0.00007932051866760387225803*I
a23= 0.00005181266163292071488059 + 0.00002460336753897918612847*I
a24= 0.00002388309209425730931165  0.00001336900104575625063695*I
a25= 0.00004334222830648700442252  0.000003957863248238146555857*I
a26= 0.00002144406955373366943567  0.00001097237534884560319531*I
a27= 0.000001798338756957510733360  0.00001189640167775918700159*I
a28= 0.000009381044976804181706355  0.00001521809568526962885314*I
a29= 0.000001814217743399466423905  0.00001000368045094924883848*I
a30= 0.000003653404747697099145868  0.000003824946827096487902828*I
a31= 0.000002722276064735807862073  0.000006885441806388226474380*I
a32= 0.000002908243010370422712642  0.000003178655514749099491204*I
a33= 0.000002312499568376790395264 + 0.0000002460508774499087597595*I
a34= 0.000002980300482220947987573  0.0000008054002380194513555136*I
a35= 0.000001823918641511708691392 + 0.0000001690025918878855565754*I
a36= 0.0000005334119278536951372042 + 0.0000008723944367755511233820*I
a37= 0.000001016497503905268380668 + 0.0000008015125616246374896681*I
a38= 0.0000004112254538462517744914 + 0.0000006638895287291174927516*I
a39= 0.0000001620819496611418767756 + 0.0000004194202258747815345800*I
a40= 0.00000001618007685693369358734 + 0.0000005451229963514326146522*I
a41= 0.0000001209168813396943102704 + 0.0000003110475973992931378252*I
a42= 0.0000001898717133881552152764 + 0.00000005522749966037761739770*I
a43= 0.0000001894442243720459740995 + 0.0000001316472729147078143581*I
a44= 0.0000001376003918043586154671 + 0.00000003856956234095862926094*I
a45= 0.00000006989793638646620143736  0.00000005174316419871478195931*I
a46= 0.00000009272637028549195160717  0.00000003149903997060054611874*I
a47= 0.00000004862361433330243915387  0.00000003733152323435119037092*I
a48= 0.0000000004594161970077067830909  0.00000003824731911971521292835*I
a49= 0.00000001262869999583407154392  0.00000003974147931865195611863*I
a50= 0.0000000003496984553054855241969  0.00000002633575753285570565953*I
a51= 0.00000001329027566459892886883  0.00000001031672759687675209111*I
a52= 0.00000001047177328908874356543  0.00000001439926459715597781841*I
a53= 0.000000009176173196441562162962  0.000000006649128397591598125552*I
a54= 0.000000007145022906860095624727 + 0.000000001929181197324110824839*I
a55= 0.000000007620140099414333325288  3.665361956240423108017 E11*I
a56= 0.000000004657112126987005058983 + 0.000000001463721624742275648158*I
a57= 0.000000001214340042927719948489 + 0.000000003020802059448936620889*I
a58= 0.000000001945865674907896473438 + 0.000000002626112862490583268957*I
a59= 0.0000000006895093416586537172232 + 0.000000001989304633944696536150*I
a60= 0.0000000007408669730258221080258 + 0.000000001225840780840576126459*I
a61= 0.0000000004162667478872884401106 + 0.000000001341526002164161883378*I
a62= 0.0000000005132292039301019798648 + 0.0000000007502246684352801693042*I
a63= 0.0000000006270046072871590876051 + 6.314128778383910279034 E11*I
a64= 0.0000000005719223588070705693565 + 1.963669989355466718298 E10*I
a65= 0.0000000003925541915150384351025 + 1.423200907055391980274 E11*I
a66= 1.876665420232897808715 E10  2.041099568060325883782 E10*I
a67= 2.138545944166912423998 E10  1.491616932082952807018 E10*I
a68= 1.057114697606634964409 E10  1.330106086409156802997 E10*I
a69= 2.245771490160356215933 E11  1.199192549169948489927 E10*I
a70= 2.412064012682073922868 E12  1.125620287250565717984 E10*I
a71= 1.925135162098333576978 E11  7.098313998094113622532 E11*I
a72= 4.819517928618243903621 E11  2.365349936790643533816 E11*I
a73= 3.871371903062367023679 E11  2.951358803573412657219 E11*I
a74= 2.971987439383358491286 E11  1.148400477824120990266 E11*I
a75= 2.102115223518953224343 E11 + 1.078339509214737316490 E11*I
a76= 2.019396751549899274826 E11 + 6.026259356931675413551 E12*I
a77= 1.163396908506564200969 E11 + 7.445550617031170199083 E12*I
a78= 1.716154888316130092496 E12 + 1.025476040032673922023 E11*I
a79= 3.036635954368037678981 E12 + 8.613530463400133051522 E12*I
a80= 3.752862064975309936775 E13 + 5.980193219987777616114 E12*I
a81= 3.172166960395143993046 E12 + 3.298059914357276464051 E12*I
a82= 2.248284621116987511317 E12 + 3.262575232338994828013 E12*I
a83= 1.999543732722426455567 E12 + 1.663406533225183694885 E12*I
a84= 1.997678205175370980314 E12  2.563711028673351106773 E13*I
a85= 1.720034892688777377316 E12 + 4.239470389705304190096 E14*I
a86= 1.095569514705621826147 E12  2.755807745754857768578 E13*I
a87= 4.327405733493612596926 E13  7.737101605156168079589 E13*I
a88= 4.538806176142139212252 E13  5.942948240331469394435 E13*I
a89= 1.836221482061442061104 E13  4.553107307228522798696 E13*I
a90= 1.605165844309677626103 E13  3.556796760694313558042 E13*I
a91= 9.344945167389884815526 E14  3.115596977506452337354 E13*I
a92= 1.130977149566327841894 E13  1.810443673536964897811 E13*I
a93= 1.679204373998674274001 E13  3.619539889304637767143 E14*I
a94= 1.337060593554487395913 E13  4.659807725069687201088 E14*I
a95= 9.263704449807770513808 E14  6.497426500663588784562 E15*I
a96= 5.670632585997269746189 E14 + 5.002869038220030918858 E14*I
a97= 5.058776182826300737460 E14 + 3.528909585714982918781 E14*I
a98= 2.596942498401231015211 E14 + 3.097159125854603958149 E14*I
a99= 2.925281234940225845834 E15 + 3.316299012434386946739 E14*I
a100 = 2.925281234940225845834 E15 + 3.316299012434386946739 E14*I
edit By the way, these coefficients also work for other initial values, such as superfunction(x0.2364561612687660028188 + 0.4934489697206941114376i), for b^^0 = 0.5+0.5i, which corresponds to abs(z)=0.3543311933142207352068.
And all of this seems to work great as long as the period is an irrational real number. But, as you noticed, from the example with a real period = exactly 3, iterating three times doesn't get you back to your starting point....
Finally, the fractal singularity comes from log(0), when you have b^^0=1, then b^^2= singularity; but then b^^n is also a singularity for all negative integers <=2, which means, since the function is real periodic, with an irrational real period, that there are singularities everywhere, when b^^0=1. For the coefficients above, the fractal behavior appears to be at around abs(z)=1.298, or around imag(x)=0.124i.
 Sheldon
Posts: 901
Threads: 130
Joined: Aug 2007
06/21/2011, 08:00 PM
(This post was last modified: 06/22/2011, 09:39 AM by Gottfried.)
Hi Sheldon,
thanks for the great discussion, btw!
Now I'm looking at the problem of the integerperiod, for instance, a period of four.
First I go one more step behind the '' u'' parameter. Because we want to have it with abs(u) or length =1, its real and imaginary part are just cos and sin of the same angular parameter, call it p. To have a period of four we set p=Pi/2
Now I have u,t and b dependent on the parameter p:
Code: .
p = Pi/2 ~ 1.57079632679
u_p = e^(p*I) = cos(p) + I*sin(p) = 1*I
t_p = exp(u) = e^e^(p*I) ~ 0.540302305868 + 0.841470984808*I
b_p = exp(u/t) = e^(e^(p*I  e^(p*I)) ) ~ 1.98933207608 + 1.19328219947*I
where the expression for b_p can also be written in cos/sinterms .
Then the powerseries for b_p^x, dependent on x, begins
Code: b_p^x=
1
+0.841470984808 + 0.540302305868*I x^1
+0.208073418274 + 0.454648713413*I x^2
0.0235200013433 + 0.164998749433*I x^3
0.0272351508693 + 0.0315334373045*I x^4
0.00799103562219 + 0.00236385154553*I x^5
0.00133356984257  0.000388077080832*I x^6
0.000130354483873  0.000149583780624*I x^7
0.00000360863179089  0.0000245376549262*I x^8
+0.00000113568806559  0.00000251083074814*I x^9
+0.000000231225619785  0.000000149917634174*I x^10
+0.0000000250518630389  1.10873065678 E10*I x^11
+ ...
Now we want to see more for the fourtimes iteration. Series having a constant term are not nice for iteration, so we recenter by the fixpoint.
We get
Code: g(x)=b_p^(x+t)t= 1.00000000000*I x^1
0.270151152934 + 0.420735492404*I x^2
0.151549571138 + 0.0693578060912*I x^3
0.0412496873584  0.00588000033583*I x^4
0.00630668746090  0.00544703017386*I x^5
0.000393975257588  0.00133183927037*I x^6
0.0000554395829760  0.000190509977510*I x^7
0.0000186979725780  0.0000162943104841*I x^8
0.00000272640610291  0.000000400959087876*I x^9
0.000000251083074814 + 0.000000113568806559*I x^10
0.0000000136288758340 + 0.0000000210205108896*I x^11
...
Let's call this function g(x), such that g(xt)+t = b_p^x . Now the fourth iteration means the fourth power of the associated Bellmatrix and this gives then the powerseries
Code: g[4](x)=
1.00000000000 x^1
0 x^2
0 x^3
0 x^4
0.0369804674784 + 0.0209284634084*I x^5
0.0133357749117 + 0.0430512149364*I x^6
0.0314482605408  0.00625863317399*I x^7
0.00325514172007  0.0187262354932*I x^8
0.0111167178521 + 0.00450965240622*I x^9
0.00619602644164 + 0.00863859539483*I x^10
0.00808355752859  0.00636782383183*I x^11
+ ...
which immediately shows, that not all x are mapped to themselves after four iterations. But now the beginning of that series looks *very* strange to me  the first four coefficients are exactly what one would expect for a precise fourstepperiod but with the fifth and the following it looks, as if there were some small, but systematic, distortion in that whole formula. Perhaps I've also an error anywhere, but because we actually shall not get a clean fourstepperiod I think the series is ok.
Hmmm  what does this tell to us now?
Gottfried
[update 1] Well, numerically this agrees with the direct computation.
Even more strange looks the formal powerseries for the sum of g[4](x) and its inverse:
Code: (g[4](x)+g[4](x))/2= 1.00000000000 x^1
0 x^2
0 x^3
0 x^4
0 x^5
0 x^6
0 x^7
0 x^8
0.00232388598570 + 0.00386972180224*I x^9
0.00766787431799 + 0.00722126227129*I x^10
0.0112186194499  0.00878239357562*I x^11
0.00780725843775  0.0123162125380*I x^12
0.0108462124501 + 0.00556062630806*I x^13
0.00254708386547 + 0.00766320056079*I x^14
0.00366236379688 + 0.000506041490944*I x^15
0.00286712784159 + 0.000171130301212*I x^16
0.00308230052060  0.00406930549958*I x^17
...
Now we have the first 8 coefficients zero???
[update 2] Now things become really strange. We can build a sum of the weighted iterates which seems to asymptotically gives a powerseries where only one coefficient <>0 remains at x^1. The beginning of the sum is, where the []brackets denote the iterate and the (x)part is left away:
Code: .
s1(p)=((g[4]+g[4])  (g[8]+g[8])/4)  ((g[16]+g[16])/16  (g[32]+g[32])/64 )/16 ...
I tried to formulate this as recursion (hope I've debugged all errors)
Code: .
let g4[k]=g[4*k]
then
s0(p)=(g4[2^p]+g4[2^p]) / 4^p
s1(p)=(s0(2p)s0(2p+1)) / 4^(p*2)
s2(p)=(s1(2p)s1(2p+1)) / 4^(p*2^2)
s3(p)=(s2(2p)s2(2p+1)) / 4^(p*2^3)
..
lim n>inf s_n(0) = a x + 0
where the value of a is not yet clear to me.
This is a very strange sum of the iterates, I'd say...
Gottfried Helms, Kassel
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Joined: Feb 2009
@ sheldon : with superfunction , do you mean the regular superfunction expanded at the fixpoint ?
@ everyone : euh , why does 1.7129 i ^ 1.7129 i ^ ... ( 0.5 + 0.5 i) give a circle and what is its radius ?
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06/22/2011, 09:28 AM
(This post was last modified: 06/22/2011, 10:10 AM by Gottfried.)
(Further looking at the (non) periodicity...)
We can write the iteration of b_p more convenient. Remember my convention
Code: .
p = Pi/2 ~ 1.57079632679
u_p = e^(p*I) = cos(p) + I*sin(p) = 1*I
t_p = exp(u) = e^e^(p*I) ~ 0.540302305868 + 0.841470984808*I
b_p = exp(u/t) = e^(e^(p*I  e^(p*I)) ) ~ 1.98933207608 + 1.19328219947*I
With this we can formulate another recursion, simple and based on the angular parameter p: (I swith to texnotation because of the use of indices)
\(
\hspace{48} p_1 = p*I  \exp(p*I) = \cos(p) + I*(p  \sin(p)) \\
\hspace{48} p_h = p_1 + \exp(p_{h1}) \\
\)
where always
\(
\hspace{48} b_h = b^\text{\^\^h } = \exp(\exp(p_h))
\)
The iteration p_h has here the nested form to the depth h:
\(
\hspace{48} p_h = p_1 + \exp(p_1 + \exp(p_1 + \exp(p_1+ ...))))
\)
and it looks very promising, that this can be periodic or even constant only in few and possible exotic cases. Surely, that does not automatically mean, that \( b^\text{\^\^h }= \exp(\exp(p_h)) \) cannot be periodic, as the example of h and sin(h) shows, but it is a strong hint.
Additionally, we can also insert an alternative startexponent z0 as given in my other examples; we have then a small modification:
\(
\hspace{48} p_1 = p*I  \exp(p*I) \\
\hspace{48} p_2 = p_1 + \log(z0) \\
\hspace{48} p_h = p_1 + \exp(p_{h1}) \text{ // for h>2 } \\
\\
\hspace{24} \text{ where } \\
\hspace{48} b_2 = b^{z_0} \\
\hspace{48} b_h = \exp(\exp(p_h)) = b^{b_{h1}}
\)
Here it seems interesting, that by the recursion the initial value z0 disappears into the tail (notational) of the p1+exp(p1+exp(...)))  expression.
Also I doubt by this now, that there could be a substantial difference between the irrational and the rational periods  but well, we have the difference between binomials of integer and noninteger parameters, so....
Hmmm....
Gottfried
Added another picture:
(Hmm, now it would be interesting to trace the trajectories backwards...)
Gottfried Helms, Kassel
Posts: 1,924
Threads: 415
Joined: Feb 2009
i assume for a real period x in the variable y in sexp_1.7129i(slog_1.7129i(z) + y) that tet_1.7129i(z) is unique up to floor(x)furcations and a oneperiodic theta(z) and that chaos occurs in lim a> oo tet_1.7129i(a+ai).
and believe it or not 1.7129 was known to me because of the commen digits with 1729
tommy1729
