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What are some possible answers to the equation x↑↑x = 1? Must a new type of number be conceptualized similar to the answer to the equation x*x = 1? Or can it be proved that this answer lies within the real and complex planes?
Posts: 638
Threads: 22
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05/28/2014, 03:46 PM
(This post was last modified: 05/28/2014, 08:42 PM by sheldonison.)
(05/28/2014, 04:07 AM)KingDevyn Wrote: What are some possible answers to the equation x↑↑x = 1? Must a new type of number be conceptualized similar to the answer to the equation x*x = 1? Or can it be proved that this answer lies within the real and complex planes?
edit update. It turns out that at least one of the solutions to the Op's question, is in the range that I can deal with numerically. For x~=2.6918099719192 + 0.62660048483655i, sexp_x(x)~=1.
The simplest algorithm I could come up with is to treat x^^x as an analytic function, and then iteratively use Newton's method to solve for x^^x=1. For Newton's method to converge, it helps to get somewhere nearby first, which I did by hunt and peck, looking for bases where x^^x was near 1.
I used my complex base tetration program. I wrote a program that explored complex base tetration, link here, but it is limited in terms of what bases it works for, due to problems at both exp(1/e), and at the Shell Thron boundary, and problems due to figuring out a consistent logarithmic branch to use... The Taylor series below can accurately calculate z=tet(b2). Then iterate z=b^z twice, to get 1.
Code: base = 2.6918099719192010269 + 0.62660048483655230254*I;
{tet= 1.0000000000000000000
+x^ 1* ( 1.1102737979673991396 + 0.13444952486116713626*I)
+x^ 2* ( 0.28425372561063406460 + 0.20168515612498045279*I)
+x^ 3* ( 0.20142378046918946453 + 0.12287970799981736460*I)
+x^ 4* ( 0.056631358539842508401 + 0.082262149871280378960*I)
+x^ 5* ( 0.031626588237768922149 + 0.042664820582471964536*I)
+x^ 6* ( 0.0060514513770401941396 + 0.023085772637730528515*I)
+x^ 7* ( 0.0029937024980130884380 + 0.010708203875456837542*I)
+x^ 8* (0.00043235417834169701508 + 0.0051416902992490886284*I)
+x^ 9* (0.00017877732679188096499 + 0.0021743297255104601779*I)
+x^10* (0.00042253710857487390916 + 0.00095593830941724313184*I)
+x^11* (0.00016984009539643333991 + 0.00036797431151789026196*I)
+x^12* (0.00013598497262565877056 + 0.00014934164052065933115*I)
+x^13* (0.000050768369172865507021 + 0.000050790667372515377858*I)
+x^14* (0.000032617521009029439342 + 0.000018777349770437327073*I)
+x^15* (0.000011146610242226519358 + 0.0000050006190101552048081*I)
+x^16* (0.0000066270855458264950034 + 0.0000015446107355450406250*I)
+x^17* (0.0000020122113800518364907 + 0.000000068654818145175922999*I)
+x^18* (0.0000011963396468454392777  0.000000050931639762404521931*I)
+x^19* (0.00000030238855924614592898  0.00000012539726639748157461*I)
+x^20* (0.00000019689839845097420283  0.000000057637500744298040916*I)
+x^21* (0.000000035153434458995604130  0.000000041641278224258482194*I)
+x^22* (0.000000030197104715047471141  0.000000015975908532546115455*I)
+x^23* (0.0000000019166266427945385066  0.0000000095872390599983285641*I)
+x^24* (0.0000000044548959632689572594  0.0000000032492643102921963649*I)
+x^25* ( 0.00000000051723734539716253138  0.0000000018638631462340237533*I)
+x^26* (0.00000000067021649772138737120  0.00000000054652627375747029680*I)
+x^27* ( 0.00000000024760346059064288247  0.00000000032605780533942069098*I)
+x^28* (1.1228394170747083309 E10  7.6537918519681082716 E11*I)
+x^29* ( 7.1268952289713966228 E11  5.3189341538283539220 E11*I)
+x^30* (2.2369708781921881068 E11  8.0625867943287913385 E12*I)
+x^31* ( 1.7517436693666143597 E11  8.3668572196268659348 E12*I)
+x^32* (5.1777236965884120904 E12  2.5187031197558940487 E13*I)
+x^33* ( 4.0336539933777435031 E12  1.3265926389475846122 E12*I)
+x^34* (1.2900297482939488422 E12 + 1.8193947317104394501 E13*I)
+x^35* ( 9.0792695347242281651 E13  2.2446482170229203341 E13*I)
+x^36* (3.2656612422347855209 E13 + 7.4097030312727709098 E14*I)
+x^37* ( 2.0437092576263104990 E13  4.2548590115227614080 E14*I)
+x^38* (8.1938431592790255774 E14 + 2.0924876810646870153 E14*I)
+x^39* ( 4.6518922470943806673 E14  9.0908340973673717718 E15*I)
+x^40* (2.0262143014199117716 E14 + 5.2297859307372668790 E15*I)
+x^41* ( 1.0743467122240305126 E14  2.1121380807455872169 E15*I)
+x^42* (4.9476280628213598972 E15 + 1.2446927622317468161 E15*I)
+x^43* ( 2.5145731619654922803 E15  5.1105870278086611369 E16*I)
+x^44* (1.1976391479208810646 E15 + 2.9141474535196681368 E16*I)
+x^45* ( 5.9468603024404147363 E16  1.2519578605776942842 E16*I)
+x^46* (2.8845734316656989916 E16 + 6.8168421844852586540 E17*I)
+x^47* ( 1.4166399462777367792 E16  3.0654575399052642127 E17*I)
+x^48* (6.9322821884199600055 E17 + 1.6036264175328628801 E17*I)
+x^49* ( 3.3908470426705261004 E17  7.4731310274704053355 E18*I)
+x^50* (1.6653357776046968983 E17 + 3.7997687551447424638 E18*I)
+x^51* ( 8.1416335448137121457 E18  1.8140427842389372934 E18*I)
+x^52* (4.0032414617874811484 E18 + 9.0620279617655147611 E19*I)
+x^53* ( 1.9590024443115400557 E18  4.3905474199233700091 E19*I)
+x^54* (9.6344349916460340144 E19 + 2.1720467909141856696 E19*I)
+x^55* ( 4.7210606401205367545 E19  1.0610621969521044634 E19*I)
+x^56* (2.3217937825693896625 E19 + 5.2248674247269975946 E20*I)
+x^57* ( 1.1392068860099503909 E19  2.5632761528515730279 E20*I)
+x^58* (5.6028249121768971146 E20 + 1.2600230734782307737 E20*I)
+x^59* ( 2.7520884846767807057 E20  6.1943235049722404547 E21*I)
+x^60* (1.3537732547522302047 E20 + 3.0441749210418233788 E21*I)
}
 Sheldon
Posts: 1,370
Threads: 335
Joined: Feb 2009
So why dont we have Re(x) << 0 ?
Well probably because Re(x) << 0 implies taking many logs , semilogs etc hence our value has " growth difficulties " in the sense that it gets too small.
Similarly for Re(x) >> 0 then we have the supergrowth problem , it gets too big.
I therefore guess all solutions have 0.5 < Re(x) < 4.
However for x with very (or sufficiently ) large Imaginary parts this might no longer hold as a good argument ...
Notice that log interations have stronger growth issues than exp iterations , because of the chaotic nature of exp iterates ... hence the previous sentense.
Since For x~=2.6918099719192 + 0.62660048483655i, sexp_x(x)~=1 apparantly , I wonder about 2 things
 apart from the fact that I need to learn more about complex base iterations ! 
1) Is this x~=2.6918099719192 + 0.62660048483655i the only solution ??
2) What happens with x~=2.6918099719192  0.62660048483655i ??
Also there should be a proof that x^^x = y Always has a complex solution.
Since every nonconstant analytic function takes every value apart from possibly one value the REAL ( ahum ) question is there such an y such that x^^x has no complex solution ?
Is there something intresting about oo ^ ^ oo for complex infinity ??
But the main question is : is x^^x analytically continuable to the entire complex plane or does it have a natural boundary ?
And as suggested by question nr 2 :
x^^x = conj( conj(x)^^conj(x) ) or similar ??
But thats not all , maybe I need to learn more about complex base iterations , but I know there are multiple solutions right ?
So what answers do the other methods give ?
Many questions ...
I understand the kneser method for real bases > eta , but not for complex.
for real bases we map the reals to the real line , but what do we do for the complex bases ??
Or is that an irrelevant question ?
Sorry complex bases still confuse me.
regards
tommy1729
Posts: 638
Threads: 22
Joined: Oct 2008
05/28/2014, 11:18 PM
(This post was last modified: 05/30/2014, 03:59 AM by sheldonison.)
(05/28/2014, 10:34 PM)tommy1729 Wrote: ....
Since For x~=2.6918099719192 + 0.62660048483655i, sexp_x(x)~=1 apparantly , I wonder about 2 things
 apart from the fact that I need to learn more about complex base iterations ! 
1) Is this x~=2.6918099719192 + 0.62660048483655i the only solution ??
2) What happens with x~=2.6918099719192  0.62660048483655i ??
....
And as suggested by question nr 2 :
x^^x = conj( conj(x)^^conj(x) ) or similar ??
The conjugate of x would also be a solution, with x^^x = 1. The f=conj(f(conj(x))) equation is correct since the function is real valued at the real axis. The solution at hand for sexp_x(conj(x))~=3000+5500i Though that probably wasn't your question...
I would expect an infinite number of solutions. Seeing as individually each tetration base solution takes on all values, then probably x^^x=y always has a solution, but finding them isn't easy.
Quote:....
But the main question is : is x^^x analytically continuable to the entire complex plane or does it have a natural boundary ?
We might want to move this to the tetcomplex.gp thread. This complex base tetration stuff was originally inspired by Mike3 solving one particular complex base. I took it a little farther, to try and explore the branch point at exp(1/e), and understand how tetration would work for real bases<exp(1/e), which also turns out to be very interesting; the solution extended from real tetration is not real valued there. So, there's this boundary, called the Shell Thron boundary, where the two primary tetration fixed points switch from two repelling fixed points, to one repelling and one neutral. Its hard to calculate tetration on the boundary, since one of the fixed points is neutral, and has no superfunction, but it turns out the bipolar complex tetration solution is still analytic there, except of course for at eta=exp(1/e). But, if you circle around eta almost one full iteration, you will eventually cross the ShellThron a second time, and that appears to be the natural boundary for complex tetration, so I would expect that boundary to be the boundary for x^^x as well. Starting with a real base>eta, you can go almost one full revolution in either direction. Adrien Douady conjectured a similar boundary for iterations of: See post#17, http://math.eretrandre.org/tetrationforu...416&page=2 Beyond that, there are other known singularities at b=1 and b=0. I got as far as conjecturing a solution for e^^e, but I have never solved a a base with real(b) negative.
Quote:But thats not all , maybe I need to learn more about complex base iterations , but I know there are multiple solutions right ?
So what answers do the other methods give ?
Many questions ...
I understand the kneser method for real bases > eta , but not for complex.
for real bases we map the reals to the real line , but what do we do for the complex bases ??
Or is that an irrelevant question ?
Sorry complex bases still confuse me.
As for as we can tell, Kouznetsov's method agrees with Kneser. The Kneser numeric method is a lot easier to extend to arbitrarily high precision levels. For complex tetration, we are interesting in analytically continuing Kneser so there is theoretically only one solution. The complex base solution is pretty weak, theoretically, though Henryk says you can rigorously show it should exist using the measurable Riemann mapping Theorem. Theoretically, research in complex dynamics might be applicable here too.
Basically, you still have two fixed points. I use a pair of theta mapping for the two primary fixed points. The solution you're looking for is
1) analytically continuable from real base tetration, as you slowly modify
the base, avoiding the singularity at eta.
2) sexp(1)=0, sexp(0)=1, sexp(1)=b
3) converges to the superfunction for one fixed point in the upper half of the complex plane (theta mapping number 1)
4) converges to the superfunction for the other fixed point in the lower half of the complex plane (theta mapping number 2)
The programming for tetcomplex.gp is not nearly as rock stable as the real base kneser.gp program. There are lots of numerical problems at or near the ShellThron boundary, for example, and with using the wrong logarithmic branch, and at the singularity at eta, and with the algorithm sometimes not converging as precision is increased.
 Sheldon
Posts: 638
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05/29/2014, 01:31 PM
(This post was last modified: 05/29/2014, 02:47 PM by sheldonison.)
The op skipped an operation. What about ?
taking ln of both sides, for all ln(1)
where and W is the LambertW function
Solution with the LambertW function
for n=1, I get x=1.6903867571636 + 1.8699079640268i, which I solved numerically with Newton's method. Oh, and then x=1 is the obvious solution as well! I missed the obvious solution the first time; where . Maybe that trivial solution of x=1 is why the Op skipped asking about , but the complex solutions are analogous to the infinite number of complex solutions of
 Sheldon
Posts: 1,370
Threads: 335
Joined: Feb 2009
05/29/2014, 04:37 PM
(This post was last modified: 05/29/2014, 04:41 PM by tommy1729.)
How does tetration work for base 3 + i ?
I looked at some posts and there was talk about parabolic fixpoints and perturbated (parabolic) fixpoints. And stuff about merging in some way. The number of people involved in those threads and/or the number of posts was low, so I suspect im not the only one with questions ...
Apart from the ShellTron region I fail to see the connection of base 3 + i with parabolic fixpoints or perturbations... and maybe there is none ( intended ).
There was some stuff mentioned for complex bases involving riemann mappings and fourier series.
It seems related to Kneser , but its unclear what is mapped where , and why.
pseudocode is nice , but without explainations its " mystic " to the big audience I think.
Also a dead link does not help ofcourse.
W're are talking about analytic tetration , not just continu right ?
I know having more than one fixpoint prevents an analytic solution that agrees on both fixpoints. I dont know how that makes sense then, for kneser we mapped to the real line ... because we wanted realvalued tetration for real bases. But what do we want and do with complex bases ?
Sorry if I have forgotten stuff posted many years ago, I feel a bit like a noob now.
Then again I cannot imagine a young new member/reader without a degree in dynamics too fully understand this from the first time (s)he reads it ...
regards
tommy1729
Posts: 638
Threads: 22
Joined: Oct 2008
05/29/2014, 08:05 PM
(This post was last modified: 06/04/2014, 01:19 PM by sheldonison.)
(05/29/2014, 04:37 PM)tommy1729 Wrote: How does tetration work for base 3 + i ?
...
Start with a base>eta, and rotate counterclockwise till you get to 3+i, or any other base. Lets say we have a complex tetration for b=3+i.
As we rotated, we follow the two fixed points, L_1, L_2, which started out as L and L*. 1 refers to the upper half of the complex plane 2 refers to the lower half.
fixed point, upper
fixed point, lower
derivative at the fixed point, upper
derivative at the fixed point, lower
Let as assume the most straightforward case, that both and , so both fixed points are repelling. This is true at the real axis for bases>exp(1/e). On the ShellThron boundary, , and the Schroeder function does not converge if for some integer n. For some mysterious reason, complex tet(z) is still defined and analytic for these bases; see this question on math overflow. But it is easier to start by assuming that the magnitude of both lambda's is greater than 1.
formal Schroeder equation, upper
formal Schroeder equation, lower
formal Abel function, upper
formal Abel function, lower
So the conjecture is if Tet(z) is the bipolar complex analytic tetration function, then in the upper half of the complex plane,
And in the lower half of the complex plane
If both and , then the conjectured tetration solution will have analytic in the upper half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to +infinity. Also, is analytic in the lower half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to infinity. Then tet(z) decays to L1 as imag(z) goes to +infinity, and decays to L2 as imag(z) goes to infinity. The pseudo periodicity in the upper half of the complex plane will be
as tet(z) decays to
And in the lower half of the complex plane the pseudo periodicity will be
as tet(z) decays to
The tetcomplex.gp program calculates the two theta(z) functions which can it turns out can be represented with an analytic Taylor series, as well as a Taylor series for tet(z) centered at z=0, where the tet(z) Taylor series has a radius of convergence of 2, since there is a singularity at z=2. These three analytic functions, along with the inverse Abel functions, allows calculating tet(z) anywhere in the complex plane. The tet(z) taylor series at z=0 is important, since it allows getting accurate numeric results at the real axis, where the two theta(z) functions have a singularity, and it is difficult to get accurate results for tet(z) at the real axis using just the theta(z) functions, even though the three analytic representations are equal to each other at the real axis. Unlike real tetration, there is no Riemann mapping equivalent for determining the theta(z) functions, and all I can do is conjecture based on numeric evidence, so tetcomplex is fairly weak from a theoretical point of view. Moreover, as you rotate further you get to the ShellThron boundary, and there the Schroder function doesn't even converge, and then you continue on to where the upper lambda_1<1, and you have an attracting fixed point, which further complicates things... Things get even more complicated at the real axis for bases<exp(1/e), and even more complicated as you rotate further, and apparently run up against an analytic barrier the second time you cross the ShellThron boundary. My intent was to publish the tetcomplex.gp code and the results in the computation section of this forum because I don't have the background to make them mathematically rigorous; my best guess is that the techniques used in parabolic implosion might help using Ecalle cylinders to develop a rigorous approach for bases in the neighborhood of
If you load up tetcomplex. and init(3+I), you get the following. The Taylor series is accurate to a little better than 13 decimal digits inside a circle of radius=0.85.
Code: base 3 + I
fixed point 0.324536386411255842495384 + 1.00148180609593401223717*I
fixed point2 0.0976763924712228440288378  1.39768129114469821694239*I
Pseudo Period 4.01614979115113383262106 + 0.603487054263112033534298*I
Pseudo Period2 4.02129656594167739617471  1.51597598902672933652589*I
{xsexp=tet= 1.0000000000000
+x^ 1* ( 1.1926785812984 + 0.17528884740114*I)
+x^ 2* ( 0.40358253612197 + 0.28625728248352*I)
+x^ 3* ( 0.26562261086389 + 0.20485344401535*I)
+x^ 4* ( 0.092800207979617 + 0.14734488382204*I)
+x^ 5* ( 0.043931082052940 + 0.086658293838264*I)
+x^ 6* ( 0.0089141331239816 + 0.050085771440674*I)
+x^ 7* ( 0.0013765234983335 + 0.025862318761102*I)
+x^ 8* (0.0026978278190898 + 0.013091395423508*I)
+x^ 9* (0.0021889141827255 + 0.0060620692068208*I)
+x^10* (0.0017806650949948 + 0.0027403191348693*I)
+x^11* (0.0010193966934607 + 0.0011263617029755*I)
+x^12* (0.00061278177551249 + 0.00044428387601631*I)
+x^13* (0.00030810246280557 + 0.00015018324665746*I)
+x^14* (0.00016198268279436 + 0.000043995580522468*I)
+x^15* (0.000074383803479140 + 0.0000057695847336563*I)
+x^16* (0.000035741097977676  0.0000033130217201198*I)
+x^17* (0.000015084052059994  0.0000045728595765211*I)
+x^18* (0.0000067097776111741  0.0000032101686314687*I)
+x^19* (0.0000025610246202369  0.0000019969970162471*I)
+x^20* (0.0000010474967643254  0.0000010720401888591*I)
+x^21* (0.00000033928597452058  0.00000056043396403234*I)
+x^22* (0.00000012145775338563  0.00000026821193818062*I)
+x^23* (0.000000023811900891984  0.00000012760404264929*I)
+x^24* (0.0000000043697811069697  0.000000056319084723659*I)
+x^25* ( 0.0000000042897637038548  0.000000024958463872075*I)
+x^26* ( 0.0000000028416125324423  0.000000010212818024552*I)
+x^27* ( 0.0000000024140647541527  0.0000000042331800324864*I)
+x^28* ( 0.0000000011384968436910  0.0000000015818126036243*I)
+x^29* ( 0.00000000070022678367848  0.00000000060550440289534*I)
+x^30* ( 0.00000000029288747440984  1.9489230025928 E10*I)
+x^31* ( 1.6092471358310 E10  6.5057020624591 E11*I)
+x^32* ( 6.1162851441054 E11  1.3371618168009 E11*I)
+x^33* ( 3.2016640087582 E11  2.2008772123192 E12*I)
+x^34* ( 1.0907314138845 E11 + 1.9162142144252 E12*I)
+x^35* ( 5.6894113787455 E12 + 1.3353362140338 E12*I)
}
{slog=
0
+x^ 1* ( 0.82072096410869  0.12062196311092*I)
+x^ 2* (0.27792404008017  0.050331398522933*I)
+x^ 3* ( 0.0099844371889999 + 0.081302317559299*I)
+x^ 4* ( 0.072917158113986  0.034696957686467*I)
+x^ 5* (0.055207824049220  0.014636137984400*I)
+x^ 6* ( 0.010934755327320 + 0.030216373778745*I)
+x^ 7* ( 0.016694897870269  0.017087641369113*I)
+x^ 8* (0.018533204870406  0.0022511105402448*I)
+x^ 9* ( 0.0063831341544067 + 0.011376158628586*I)
+x^10* ( 0.0046681402819256  0.0083624255733312*I)
+x^11* (0.0074056837820971 + 0.00062628589568203*I)
+x^12* ( 0.0035996797143795 + 0.0042963441872732*I)
+x^13* ( 0.0012030633005558  0.0040593255707054*I)
+x^14* (0.0031337449497880 + 0.00096770025493956*I)
+x^15* ( 0.0019876908480234 + 0.0015804135819437*I)
+x^16* ( 0.00015342020688831  0.0019561030704026*I)
+x^17* (0.0013336988665964 + 0.00074991609693227*I)
+x^18* ( 0.0010715769299693 + 0.00053629516362396*I)
+x^19* (0.00012188121444182  0.00093078481724362*I)
+x^20* (0.00055353208997318 + 0.00048813100945728*I)
+x^21* ( 0.00056344346210701 + 0.00014671872510879*I)
+x^22* (0.00015300546130942  0.00043383398182400*I)
+x^23* (0.00021704994178488 + 0.00029242445852136*I)
+x^24* ( 0.00028866200089469 + 0.000013074120388320*I)
+x^25* (0.00011969221735822  0.00019599226576383*I)
+x^26* (0.000076057083217636 + 0.00016620627876529*I)
+x^27* ( 0.00014375133018786  0.000023812355775862*I)
+x^28* (0.000079929211332977  0.000084507323038456*I)
+x^29* (0.000020273535559295 + 0.000090732377903920*I)
+x^30* ( 0.000069266956208391  0.000027147847439129*I)
+x^31* (0.000049060572982138  0.000033844682085868*I)
+x^32* (0.00000050160026643473 + 0.000047795604376386*I)
+x^33* ( 0.000032033848322450  0.000021119852610199*I)
+x^34* (0.000028460773991873  0.000011850056791980*I)
+x^35* ( 0.0000048975013159976 + 0.000024303573536754*I)
+x^36* ( 0.000014012482440201  0.000014205326886274*I)
+x^37* (0.000015796825938335  0.0000029592226504526*I)
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}
 Sheldon
Posts: 1,370
Threads: 335
Joined: Feb 2009
05/29/2014, 11:15 PM
(This post was last modified: 05/29/2014, 11:23 PM by tommy1729.)
Quote: (05/29/2014, 04:37 PM)tommy1729 Wrote: How does tetration work for base 3 + i ?
...
Start with a base>eta, and rotate counterclockwise till you get to 3+i, or any other base. Lets say we have a complex tetration for b=3+i.
As we rotated, we follow the two fixed points, L_1, L_2, which started out as L and L*. 1 refers to the upper half of the complex plane 2 refers to the lower half.
fixed point, upper
fixed point, lower
derivative at the fixed point, upper
derivative at the fixed point, lower
Let as assume the most straightforward case, that both and , so both fixed points are repelling. This is true at the real axis for bases>exp(1/e). On the ShellThron boundary, , and the Schroeder function does not converge if for some integer n. For some mysterious reason, complex tet(z) is still defined and analytic for these bases; see this question on math overflow. But it is easier to start by assuming that the magnitude of both lambda's is greater than 1.
formal Schroeder equation, upper
formal Schroeder equation, lower
formal Abel function, upper
formal Abel function, lower Ok. So i use the koenigs function to solve the schroeder equation and from that I compute the abel function.
But that gets me 2 abel functions ...
What to do with them ?
I cannot simply use one for the upper plane and one for the lower plane , can i ??
Quote:So the conjecture is if Tet(z) is the bipolar complex analytic tetration function, then in the upper half of the complex plane,
Should that not be ?
Quote:And in the lower half of the complex plane
Now I still have 2 abel functions. Not 1.
And I have neither tet_b(z) or theta(z).
This equation just shows how tet_b(z) and theta(z) are related.
Since I have the abel function I can just plug in some theta(z) and get some tet_b(z).
But still 2 of those and I see no reason to prefer some theta resp tet_b over another.
Why dont we just take the inverse of the abel as superfunction(s) ?
hmm.
I see no motivation or uniqueness condition.
And picking one function for the upper and one for the lower seems partially arbitrary ?
Yeah ok that matches the fixpoints at imaginary infinity , but there are others too right.
How do we merge the two theta functions ??
We do merge , to avoid ill defined parts of the complex plane right ?
Quote:If both and , then the conjectured tetration solution will have analytic in the upper half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to +infinity. Also, is analytic in the lower half of the complex plane, with a singularity at integers at the real axis, and decaying to a constant as imag(z) goes to infinity. Then tet(z) decays to L1 as imag(z) goes to +infinity, and decays to L2 as imag(z) goes to infinity. The pseudo periodicity in the upper half of the complex plane will be
as tet(z) decays to
And in the lower half of the complex plane the pseudo periodicity will be
as tet(z) decays to
The tetcomplex.gp program calculates the two theta(z) functions which can it turns out can be represented with an analytic Taylor series, as well as a Taylor series for tet(z) centered at z=0, where the tet(z) Taylor series has a radius of convergence of 2, since there is a singularity at z=2. These three analytic functions, along with the inverse Abel functions, allows calculating tet(z) anywhere in the complex plane. The tet(z) taylor series at z=0 is important, since it allows getting accurate numeric results at the real axis, where the two theta(z) functions have a singularity, and it is difficult to get accurate results for tet(z) at the real axis using just the theta(z) functions, even though the three analytic representations are equal to each other at the real axis. Unlike real tetration, there is no Riemann mapping equivalent for determining the theta(z) functions, and all I can do is conjecture based on numeric evidence, so tetcomplex is fairly weak from a theoretical point of view. Moreover, as you rotate further you get to the ShellThron boundary, and there the Schroder function doesn't even converge, and then you continue on to where the upper lambda_1<1, and you have an attracting fixed point, which further complicates things... Things get even more complicated at the real axis for bases<exp(1/e), and even more complicated as you rotate further, and apparently run up against an analytic barrier the second time you cross the ShellThron boundary. My intent was to publish the tetcomplex.gp code and the results in the computation section of this forum because I don't have the background to make them mathematically rigorous; my best guess is that the techniques used in parabolic implosion might help using Ecalle cylinders to develop a rigorous approach for bases in the neighborhood of
ah so you do claim the 2 theta function agree on the real line.
But then how can they be different ??
By schwarz reflection and analytic continuation it seems weird that 2 analytic functions can agree on the real line , yet be different off the real line.
In short
1) I do not get the motivation and uniqueness.
2) I do not understand the fact that there are 2 functions.
3) I do not know how the theta is computed.
So maybe we should map " something " to the real line with a riemann mapping.
Maybe that something contains : 0,1,b,b^b,b^(b^b),...
However we need to map continu flow to the real line, not just some points.
So maybe the question is : can a riemann mapping merge 2 functions into 1 ?
If so , its probably harder than kneser.
I do not immediately see another possible explication or way.
But I assume there is since I have so many questions and probably misunderstandings about this.
Thanks for your swift reply.
Though Im still in a state of confusion and skepticism.
Maybe temporarily.
regards
tommy1729
Posts: 638
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Joined: Oct 2008
05/29/2014, 11:34 PM
(This post was last modified: 05/29/2014, 11:42 PM by sheldonison.)
(05/29/2014, 11:15 PM)tommy1729 Wrote: ....
ah so you do claim the 2 theta function agree on the real line.
But then how can they be different ??
By schwarz reflection and analytic continuation it seems weird that 2 analytic functions can agree on the real line , yet be different off the real line... For a complex base, tet is not a real valued function at the real axis. But if the base is real, then yes, the two theta functions agree naturally due to Schwarz reflection involving L and L*. So the complex tet function naturally becomes the Kneser solution for a real valued base, which is exactly what we need. I'm not trying to address how to compute the two theta functions; that is a much difficult post to write, but I do get very good convergence for many complex bases. The key is to iterate all three Taylor series functions, plus a lot of magic coding. And magically, all three functions converge into excellent numeric agreement for the unit circle Taylor series, where each representation is used where it can give accurate results.
 Sheldon
Posts: 1,370
Threads: 335
Joined: Feb 2009
(05/29/2014, 11:34 PM)sheldonison Wrote: (05/29/2014, 11:15 PM)tommy1729 Wrote: ....
ah so you do claim the 2 theta function agree on the real line.
But then how can they be different ??
By schwarz reflection and analytic continuation it seems weird that 2 analytic functions can agree on the real line , yet be different off the real line... For a complex base, tet is not a real valued function at the real axis. But if the base is real, then yes, the two theta functions agree naturally due to Schwarz reflection involving L and L*. So the complex tet function naturally becomes the Kneser solution for a real valued base.
Im aware of that.
I mean mapping a continu line on the complex plane to the positive reals.
For instance mapping f(4+3i) to g(1) as a point example.
... it seems weird that 2 analytic functions can agree on the real line , yet be different off the real line...
so f(1) = g(1) = a , f(2) = g(2) = b , ... for all positive reals ... yet f and g are different ??
regards
tommy1729
