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 Taylor series of cheta Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 08/22/2016, 08:28 PM I am interested in that how we could expand cheta function and its inverse to reals and complexes. cheta(x) = e[x]e It is important for me because of the rational and complex hyper-operators. Please, help me. Xorter Unizo JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 08/22/2016, 09:41 PM (08/22/2016, 08:28 PM)Xorter Wrote: I am interested in that how we could expand cheta function and its inverse to reals and complexes. cheta(x) = e[x]e It is important for me because of the rational and complex hyper-operators. Please, help me. Well these complex hyper-operators will not be analytic. So there isn't much use to them. Furthermore the cheta function is not e [x] e. Here's an appropriate link to work done on it http://math.eretrandre.org/tetrationforu...ight=cheta Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 08/23/2016, 08:37 PM (08/22/2016, 09:41 PM)JmsNxn Wrote: Well these complex hyper-operators will not be analytic. So there isn't much use to them. Furthermore the cheta function is not e [x] e. Here's an appropriate link to work done on it http://math.eretrandre.org/tetrationforu...ight=cheta I am sorry, but I could not find anything to make the hyper-problem completed. What is cheta function? Because I do not know. What is your method to evaluate it? Please, help me! Xorter Unizo sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 08/24/2016, 03:32 PM (This post was last modified: 08/24/2016, 07:19 PM by sheldonison.) (08/22/2016, 08:28 PM)Xorter Wrote: I am interested in that how we could expand cheta function and its inverse to reals and complexes. cheta(x) = e[x]e It is important for me because of the rational and complex hyper-operators. Please, help me. 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$ 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 - Sheldon Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 08/26/2016, 06:44 PM (08/24/2016, 03:32 PM)sheldonison Wrote: (08/22/2016, 08:28 PM)Xorter Wrote: I am interested in that how we could expand cheta function and its inverse to reals and complexes. cheta(x) = e[x]e It is important for me because of the rational and complex hyper-operators. Please, help me. 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$ 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 I don't really understand it. What is the connection between Abel and Cheta functions? Xorter Unizo sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 08/27/2016, 12:08 AM (This post was last modified: 08/27/2016, 05:15 AM by sheldonison.) (08/26/2016, 06:44 PM)Xorter Wrote: (08/24/2016, 03:32 PM)sheldonison Wrote: (08/22/2016, 08:28 PM)Xorter Wrote: I am interested in that how we could expand cheta function and its inverse to reals and complexes. cheta(x) = e[x]e It is important for me because of the rational and complex hyper-operators. Please, help me. 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$ 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 I don't really understand it. What is the connection between Abel and Cheta functions? The cheta function was named by Jay Daniels in this post. It is the upper superfunction for base $b=\eta=\exp(1/e)$. When Jay made his post, he did not know about the formal asymptotic series; but numerical results for his algorithm as well as an earlier algorithm of mine are identical to the formal asymptotic series. Jay's cheta function would be the inverse of the Abel function; with an additional arbitrary additive constant k. Then the superfunction base eta would be defined as follows using the inverse of the Abel function defined in my earlier post. $S_\eta(z) = \text{superfunction}_\eta(z) = e \cdot \left(\alpha^{-1}(z+k) + 1\right)$ $S_\eta(z+1) = \eta^{\left(S_\eta(z)\right)}$ As far as the function e[x]e, where x is the operator sequence (1=addition, 2=multiplication, 3=exponentiation, 4=tetration), no analytic version of that is known since piecemeal definitions are almost never analytic. So all of this is probably not relevant to your post. The only problem I understand is using tetration to get to the fractional exponentiation operators like $f(z)=\exp^{(1/n)}(z)\;\;\; f^{\circ n}(z)=\exp(z)\;\;f(z)=\text{sexp}(\text{slog}(z)+1/n)$. This is directly analogous to $f(z)=\exp(\log(z)+1/n)=z\cdot \exp(1/n)$. But this does not lead to an analytic e[x]e. - Sheldon Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 08/27/2016, 07:05 PM (08/22/2016, 12:36 AM)JmsNxn Wrote: Search for the cheta function on the forum and get its power series. Take $f(t,x) = cheta(cheta^{-1}(x) + t)$ define $x [t] y = f(t,f(-t,x) + f(-t,y))\,\,t<1$ $x [t] y = f(t-1,yf(1-t,x))\,\,t \ge 1$ continuous solution which is analytic for t < 1 and t >1 with a singularity at t=1 oddly enough $x [t] e$ is analytic everywhere. How can it help me to evaluate 3[0.5]3 and 3[1.5]3? Xorter Unizo Sergo Newbie Posts: 1 Threads: 0 Joined: Aug 2016 08/28/2016, 01:04 PM (This post was last modified: 08/28/2016, 01:08 PM by Sergo.) (08/27/2016, 07:05 PM)Xorter Wrote: How can it help me to evaluate 3[0.5]3 and 3[1.5]3? Hello there, Xorter, to calculate hyperops with fractional ranks, you can use PARI/GP, and kneser.gp, plus small code which uses cheta function for that. First, download PARI/GP: http://pari.math.u-bordeaux.fr/download.html Then, download kneser.gp: http://math.eretrandre.org/tetrationforu...hp?tid=486 And finally, download attachment from my post. After installation of PARI/GP, put kneser.gp and hyper.gp in the same folder as PARI/GP's (gp.exe), then drag'n'drop hyper.gp onto gp.exe, and it will start both kneser.gp and hyper.gp. Now you will be able to calculate all you want, but be aware, that I've shifted original hyperops by one, so addition is [1], instead of [0] (you can change it back in the code). To calculate a[s]b in PARI/GP now, you only need to write Code:h(a,s,b) and that's all. Also, this code allows you to calculate some roots: Code:hr(a,s,b)For example: h(2,3,5) = 32, hr(32,3,5)=2 Attached Files   hyper.gp (Size: 748 bytes / Downloads: 264) Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016   08/28/2016, 03:05 PM (08/28/2016, 01:04 PM)Sergo Wrote: (08/27/2016, 07:05 PM)Xorter Wrote: How can it help me to evaluate 3[0.5]3 and 3[1.5]3? Hello there, Xorter, to calculate hyperops with fractional ranks, you can use PARI/GP, and kneser.gp, plus small code which uses cheta function for that. First, download PARI/GP: http://pari.math.u-bordeaux.fr/download.html Then, download kneser.gp: http://math.eretrandre.org/tetrationforu...hp?tid=486 And finally, download attachment from my post. After installation of PARI/GP, put kneser.gp and hyper.gp in the same folder as PARI/GP's (gp.exe), then drag'n'drop hyper.gp onto gp.exe, and it will start both kneser.gp and hyper.gp. Now you will be able to calculate all you want, but be aware, that I've shifted original hyperops by one, so addition is [1], instead of [0] (you can change it back in the code). To calculate a[s]b in PARI/GP now, you only need to write Code:h(a,s,b) and that's all. Also, this code allows you to calculate some roots: Code:hr(a,s,b)For example: h(2,3,5) = 32, hr(32,3,5)=2 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? Xorter Unizo Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 08/28/2016, 03:13 PM (08/24/2016, 03:32 PM)sheldonison Wrote: (08/22/2016, 08:28 PM)Xorter Wrote: I am interested in that how we could expand cheta function and its inverse to reals and complexes. cheta(x) = e[x]e It is important for me because of the rational and complex hyper-operators. Please, help me. 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$ 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.) Attached Files Image(s)     Xorter Unizo « Next Oldest | Next Newest »

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