Taylor series of cheta
#11
(08/28/2016, 03:13 PM)Xorter Wrote:
(08/24/2016, 03:32 PM)sheldonison Wrote: ....
There is a formal asymptotic series for the Abel \( \alpha(z) \) function solution for the parabolic case. Iterating \( f(z)=\exp(z)-1 \), is congruent through a simple linear transformation to iterating cheta \( f(y)=\eta^y \;\;\; \eta=\exp(\frac{1}{e}) \), by mapping \( z \mapsto \frac{y}{e}-1
\)

\alpha(z) = \frac{-1}{2z} + \frac{\ln(z)}{3} - \frac{z}{36} + \frac{z^2}{540} + \frac{z^3}{7776} + \frac{-71z^4}{435456} + ....
\alpha(\exp(z)-1) = \alpha(z)+1
[/tex]

The formal series will work in either half plane, by changing the ln(z) to ln(-z). But it will not work in both at the same time. It helps to iterate (exp(z)-1) or ln(z+1) a few times to get closer to the fixed point of zero before evaulating the asymptotic series. See G Edgar's post in mathoverflow for some theoretical background on the parabolic case. http://mathoverflow.net/questions/45608/...x-converge

Hello, Sheldon!

I tried your series for abel function. But it does not seem it would be the inverse of cheta function. Why not? Did I make a mistake or what?
(Here is the picture of the graphs.)

I only posted four terms of the series. Also, the series is an asymptotic series so the optimal number of terms varies. But if you iterate \( z \mapsto \ln(z+1) \) enough times until z is fairly closed to the fixed point of zero; say within 0.1 with a 25 term series, then the series gives excellent results. Below is a graph of 25 terms series, with zk=2.02591209868586388250227776560583118127388887; plotting from 0.02 to 0.1. Notice the Kneser.gp invcheta gives nearly identical results within 10^-32 of the formal parabolic series, even though the kneser.gp code uses a completely different algorithm since I didn't even know about the formal parabolic series when I wrote the Kneser.gp program. The equation for the function I graphed is:
\( \left(\frac{\text{cheta}(\text{AbelSeries}(z)+zk)}{e}\right)-1-z\;\; \) where \( zk=-\left(\text{AbelSeries}\left(\frac{\text{cheta}(-100)}{e}-1\right)+100\right) \)

   
Code:
zk=2.02591209868586388250227776560583118127388887;
newabel(z) = { (1/3)*log(z) + subst(parabolic,x,z); }
ploth(t=0.02,0.1,z=cheta(newabel(t)+zk)/exp(1)-1-t)
{parabolic=
(1/x)* -2
+x^ 1* -1/36
+x^ 2*  1/540
+x^ 3*  1/7776
+x^ 4* -71/435456
+x^ 5*  8759/163296000
+x^ 6*  31/20995200
+x^ 7* -183311/16460236800
+x^ 8*  23721961/6207860736000
+x^ 9*  293758693/117328567910400
+x^10* -1513018279/577754311680000
+x^11* -1642753608337/3355597042237440000
+x^12*  3353487022709/1689531377909760000
+x^13* -11579399106239/40790114695249920000
+x^14* -254879276942944519/137219514685385932800000
+x^15*  13687940105188979843/14114007224782553088000000
+x^16*  215276054202212944807/100956663443150497382400000
+x^17* -2657236754331703252459529/1203529624071657866919936000000
+x^18* -146435111462649069104449/50302321749125019205632000000
+x^19*  715411321613253460298674267/135588231530708185101474201600000
+x^20*  16634646784735044775309724063/3702250880735601413534515200000000
+x^21* -104353470644496360229598950087621/7332274212470670094037711585280000000
+x^22* -1026800310866887782669304706891/145015557324117535367532380160000000
+x^23*  10532451718209319314810847524219487/239106170881428081691713129676800000000
+x^24*  426818206492321153424287945331450731/55748747292256998858987528725200896000000
+x^25* -209820349077359397909291778326518401351/1340114117602331703341046363586560000000000
}
- Sheldon
#12
(08/28/2016, 06:20 PM)sheldonison Wrote: I only posted four terms of the series. Also, the series is an asymptotic series so the optimal number of terms varies. But if you iterate \( z \mapsto \ln(z+1) \) enough times until z is fairly closed to the fixed point of zero; say within 0.1 with a 25 term series, then the series gives excellent results. Below is a graph of 25 terms series, with zk=2.02591209868586388250227776560583118127388887; plotting from 0.02 to 0.1. Notice the Kneser.gp invcheta gives nearly identical results within 10^-32 of the formal parabolic series, even though the kneser.gp code uses a completely different algorithm since I didn't even know about the formal parabolic series when I wrote the Kneser.gp program. The equation for the function I graphed is:
\( \left(\frac{\text{cheta}(\text{AbelSeries}(z)+zk)}{e}\right)-1-z\;\;
zk=\text{cheta}^{-1}((0.02+1)\cdot e )-\text{AbelSeries}(0.02)
\)


Code:
zk=2.02591209868586388250227776560583118127388887;
newabel(z) = { (1/3)*log(z) + subst(parabolic,x,z); }
ploth(t=0.02,0.1,z=\text{cheta}(\text{newabel}(t)+zk)/exp(1)-1-t)
{parabolic=
(1/x)* -2
+x^ 1* -1/36
+x^ 2*  1/540
+x^ 3*  1/7776
+x^ 4* -71/435456
+x^ 5*  8759/163296000
+x^ 6*  31/20995200
+x^ 7* -183311/16460236800
+x^ 8*  23721961/6207860736000
+x^ 9*  293758693/117328567910400
+x^10* -1513018279/577754311680000
+x^11* -1642753608337/3355597042237440000
+x^12*  3353487022709/1689531377909760000
+x^13* -11579399106239/40790114695249920000
+x^14* -254879276942944519/137219514685385932800000
+x^15*  13687940105188979843/14114007224782553088000000
+x^16*  215276054202212944807/100956663443150497382400000
+x^17* -2657236754331703252459529/1203529624071657866919936000000
+x^18* -146435111462649069104449/50302321749125019205632000000
+x^19*  715411321613253460298674267/135588231530708185101474201600000
+x^20*  16634646784735044775309724063/3702250880735601413534515200000000
+x^21* -104353470644496360229598950087621/7332274212470670094037711585280000000
+x^22* -1026800310866887782669304706891/145015557324117535367532380160000000
+x^23*  10532451718209319314810847524219487/239106170881428081691713129676800000000
+x^24*  426818206492321153424287945331450731/55748747292256998858987528725200896000000
+x^25* -209820349077359397909291778326518401351/1340114117602331703341046363586560000000000
}

The more term I use in the graph, the less it similars to the inverse of cheta function (abel is the blue function). I used all your terms, but it does not work. Why cannot I get the inverse of cheta function from it?


Attached Files Thumbnail(s)
   
Xorter Unizo
#13
(08/28/2016, 03:05 PM)Xorter Wrote: Hello, Sergo!
Well, this programme looks really promising. Thank you for sharing it with me. It can calculate hyperops with fractional and complex ranks ... BUT it cannot calculate tetrations like H(2;4;3) which would be 16, but it gives 11.713... Why?
And I am really interested in that how can this programme evaluate Cheta and Inverse-cheta functions for fractional and complex arguments. Can you tell me it?

Here's another interesting code fragment which implements
\( f(z)=a^z;\;\;\; h(a,b,c)=f^{\circ b} ( c ) \)
This code fragment works with the other tetration program I wrote; fatou.gp The function h(a,b,c) is an analytic function in all three variables; (a,b,c); there is a parabolic singularity at \( a=\eta=\exp(1/e) \); analytic continuation explains why the sexp function is no longer real valued for fractional iterates for bases<eta.
\( h(a,b,1) = \text{sexp_a}(b) \)
\( h(a,1,b) = a^b \)
\( h(a,-1,b) = \log_a(b) \)
The half iterates of exp and log would be:
\( h(a,0.5,h(a,0.5,b)) = a^b \)
\( h(a,-0.5,h(a,-0.5,b)) = \log_a(b) \)
Code:
\r fatou.gp
h(a,b,c) = {
quietmode=1;
sexpinit(a);
return(sexp(slog(c)+b));
}
- Sheldon
#14
(08/28/2016, 08:24 PM)Xorter Wrote: The more term I use in the graph, the less it similars to the inverse of cheta function (abel is the blue function). I used all your terms, but it does not work. Why cannot I get the inverse of cheta function from it?

Well its an asymptotic series; the theory is somewhat complicated and that would require dozens of pages to explain. For large values of |z|>1 the asymptotic doesn't converge at all and instead diverges! If you stick to values with |z|<0.1 with 25 terms, you'll get accuracy to 32 decimal digits. So lets say you want \( \alpha(100) \); then you need to iterate mapping \( z \mapsto \log(z+1) \) 20 times before evaluating the Abel function to get 32 decimal digits accuracy (iterating 10 times will give you 21 decimal digits accuracy). You get z~=0.109413515605671; and then increment \( \alpha(z)+20 \) to get the Abel function you want. Of course, before you even start you need to map \( z \mapsto \frac{z}{e}-1 \). Then here is the pari-gp code ...

Code:
invchetaabel(z) = {
local(n);
zk=2.02591209868586388250227776560583118127388887;
parabolic=
(1/x)* -2
+x^ 1* -1/36
+x^ 2*  1/540
+x^ 3*  1/7776
+x^ 4* -71/435456
+x^ 5*  8759/163296000
+x^ 6*  31/20995200
+x^ 7* -183311/16460236800
+x^ 8*  23721961/6207860736000
+x^ 9*  293758693/117328567910400
+x^10* -1513018279/577754311680000
+x^11* -1642753608337/3355597042237440000
+x^12*  3353487022709/1689531377909760000
+x^13* -11579399106239/40790114695249920000
+x^14* -254879276942944519/137219514685385932800000
+x^15*  13687940105188979843/14114007224782553088000000
+x^16*  215276054202212944807/100956663443150497382400000
+x^17* -2657236754331703252459529/1203529624071657866919936000000
+x^18* -146435111462649069104449/50302321749125019205632000000
+x^19*  715411321613253460298674267/135588231530708185101474201600000
+x^20*  16634646784735044775309724063/3702250880735601413534515200000000
+x^21* -104353470644496360229598950087621/7332274212470670094037711585280000000
+x^22* -1026800310866887782669304706891/145015557324117535367532380160000000
+x^23*  10532451718209319314810847524219487/239106170881428081691713129676800000000
+x^24*  426818206492321153424287945331450731/55748747292256998858987528725200896000000
+x^25* -209820349077359397909291778326518401351/1340114117602331703341046363586560000000000;

z=z/exp(1)-1;
for (n=1,20,z=log(z+1));
return((1/3)*log(z) + subst(parabolic,x,z) + zk + 20);
}

The theory gets even more complicated for \( \alpha(z) \) for negative values of z, which would require another dozen or so pages to explain. But besides the upper superfunction analogous to cheta, there is another superfunction for iterating \( z \mapsto \exp(z)-1 \). And the two functions are completely different even though they have the same asymptotic Abel series with log(z) replaced with log(-z)! For that other superfunction, for negative values of z, you iterate \( z \mapsto \exp(z)-1 \) instead!
- Sheldon


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