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 Infinite tetration of the imaginary unit Gottfried Ultimate Fellow Posts: 789 Threads: 121 Joined: Aug 2007 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 + 0.46203078407110528iHi Sheldon - I've inserted your base-parameter 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 starting-point 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 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 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 + 0.46203078407110528iHi Sheldon - I've inserted your base-parameter 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 Shell-Thron 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 Shell-Thron 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 Shell-Thron 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 Shell-Thron 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 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 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 Shell-Thron 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 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 06/21/2011, 01:58 AM (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 Shell-Thron 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 Gottfried Ultimate Fellow Posts: 789 Threads: 121 Joined: Aug 2007 06/21/2011, 12:13 PM (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 Shell-Thron-description are connected. (I'Ve also put it in the hyperop-wiki). 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 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 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 Shell-Thron-description 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 E-11*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 E-11*I a64=  0.0000000005719223588070705693565 + 1.963669989355466718298 E-10*I a65=  0.0000000003925541915150384351025 + 1.423200907055391980274 E-11*I a66=  1.876665420232897808715 E-10 - 2.041099568060325883782 E-10*I a67=  2.138545944166912423998 E-10 - 1.491616932082952807018 E-10*I a68=  1.057114697606634964409 E-10 - 1.330106086409156802997 E-10*I a69= -2.245771490160356215933 E-11 - 1.199192549169948489927 E-10*I a70=  2.412064012682073922868 E-12 - 1.125620287250565717984 E-10*I a71= -1.925135162098333576978 E-11 - 7.098313998094113622532 E-11*I a72= -4.819517928618243903621 E-11 - 2.365349936790643533816 E-11*I a73= -3.871371903062367023679 E-11 - 2.951358803573412657219 E-11*I a74= -2.971987439383358491286 E-11 - 1.148400477824120990266 E-11*I a75= -2.102115223518953224343 E-11 + 1.078339509214737316490 E-11*I a76= -2.019396751549899274826 E-11 + 6.026259356931675413551 E-12*I a77= -1.163396908506564200969 E-11 + 7.445550617031170199083 E-12*I a78= -1.716154888316130092496 E-12 + 1.025476040032673922023 E-11*I a79= -3.036635954368037678981 E-12 + 8.613530463400133051522 E-12*I a80= -3.752862064975309936775 E-13 + 5.980193219987777616114 E-12*I a81=  3.172166960395143993046 E-12 + 3.298059914357276464051 E-12*I a82=  2.248284621116987511317 E-12 + 3.262575232338994828013 E-12*I a83=  1.999543732722426455567 E-12 + 1.663406533225183694885 E-12*I a84=  1.997678205175370980314 E-12 - 2.563711028673351106773 E-13*I a85=  1.720034892688777377316 E-12 + 4.239470389705304190096 E-14*I a86=  1.095569514705621826147 E-12 - 2.755807745754857768578 E-13*I a87=  4.327405733493612596926 E-13 - 7.737101605156168079589 E-13*I a88=  4.538806176142139212252 E-13 - 5.942948240331469394435 E-13*I a89=  1.836221482061442061104 E-13 - 4.553107307228522798696 E-13*I a90= -1.605165844309677626103 E-13 - 3.556796760694313558042 E-13*I a91= -9.344945167389884815526 E-14 - 3.115596977506452337354 E-13*I a92= -1.130977149566327841894 E-13 - 1.810443673536964897811 E-13*I a93= -1.679204373998674274001 E-13 - 3.619539889304637767143 E-14*I a94= -1.337060593554487395913 E-13 - 4.659807725069687201088 E-14*I a95= -9.263704449807770513808 E-14 - 6.497426500663588784562 E-15*I a96= -5.670632585997269746189 E-14 + 5.002869038220030918858 E-14*I a97= -5.058776182826300737460 E-14 + 3.528909585714982918781 E-14*I a98= -2.596942498401231015211 E-14 + 3.097159125854603958149 E-14*I a99=  2.925281234940225845834 E-15 + 3.316299012434386946739 E-14*I a100 = 2.925281234940225845834 E-15 + 3.316299012434386946739 E-14*Iedit By the way, these coefficients also work for other initial values, such as superfunction(x-0.2364561612687660028188 + 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 Gottfried Ultimate Fellow Posts: 789 Threads: 121 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 integer-period, 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/sin-terms . 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 E-10*I x^11   + ...Now we want to see more for the four-times 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(x-t)+t = b_p^x . Now the fourth iteration means the fourth power of the associated Bell-matrix 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 four-step-period 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 four-step-period 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 + 0where 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 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 06/21/2011, 11:19 PM @ 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 ? Gottfried Ultimate Fellow Posts: 789 Threads: 121 Joined: Aug 2007 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 tex-notation 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_{h-1}) \\$ 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_{h-1}) \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_{h-1}}$ 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 non-integer parameters, so.... Hmmm.... Gottfried Added another picture: (Hmm, now it would be interesting to trace the trajectories backwards...) Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 06/22/2011, 12:25 PM 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 one-periodic 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 « Next Oldest | Next Newest »

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