 Holomorphic semi operators, using the beta method - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Hyperoperations and Related Studies (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=11) +--- Thread: Holomorphic semi operators, using the beta method (/showthread.php?tid=1386) Pages: 1 2 3 4 5 6 7 8 RE: Holomorphic semi operators, using the beta method - JmsNxn - 04/08/2022 Okay, so this definitely requires more detail, but I see the solution. I made a mistake earlier. We cannot treat $$\theta$$ alone. We actually have to consider $$\boldsymbol{\theta} = (\theta_1,\theta_2,\theta_3)$$, for each occurrence of tetration in the Goodstein equation. If we force these $$\theta$$'s into one, we get the trivial solution $$\theta(s) = 1-s$$. It's a little trickier than I originally thought. But I have it now. I'm just polishing off a proof of concept paper; that is no where near perfect rigor. But describes a good project outline. This means, we're solving: $$1 \oplus_{s-1,\theta_1,F(1\oplus_{s,\theta_2,F(y)} y)} \left(1\oplus_{s,\theta_2,F(y)} y\right) = 1\oplus_{s,\theta_3,F(y+1)} y+1\\$$ Where, $$F(y) = y^{1/y}$$. And then we are relating $$\theta_1 = \theta_2 = \theta_3$$, when we make the appropriate change of variables in each case. Where $$\theta_1$$ is a function in $$s-1, 1\oplus_{s,\theta_2,F(y)} y$$; $$\theta_2$$ is a function in $$s,y$$; and $$\theta_3$$ is a function in $$s,y+1$$. But after changing the variables they are all equivalent. I forgot to pay close attention to the Jacobian above, once you pay attention to the Jacobian this forms a more natural solution; and actually takes care of all the trouble. It ensures we can extend indefinitely in any direction as well--upto branch cuts. This then creates a function $$\theta(s,y)$$, such that: $$1 y = 1 \oplus_{s,\theta(s,y), F(y)} y = \exp_{F(y)}^{\circ s + \theta(s,y)}\left(\log_{F(y)}^{\circ s+\theta(s,y)}(1) + y\right)\\$$ And $$\theta(s,y)$$ has some very weird functional equations... It is not 1-periodic, that was a mistake. It just satisfies $$0 = \theta(0,y) =\theta(1,y) = \theta(2,y)$$. This had me erroneously assume 1-periodic; it's close to it, but not exactly. Which explains how this will get us in between exponentiation and tetration (the value $$\theta$$ will start to explode near here). This is very difficult to talk about though, so I'm going to go silent mode until I have a working project out line--then I'll start rigorously proving the claims in the outline, which aren't very hard (I know how to prove all the claims I make, and I'm making sure they can be shown rigorously). This also works for $$\alpha y$$, for $$y \in \mathbb{C}/\{0,1\}$$, and up to branch cuts which move for $$\alpha,y$$. This becomes a branch cut nightmare. I'm going to try and program in a protocol for $$\alpha,s,y$$ for $$\alpha ,y > 1$$, $$0 \le s \le 2$$. This should be fairly easy to program, it'll be slow, but all we really need is a Newton root finder algorithm. I'm hesitating though, because I think there should be a more efficient manner of calculating $$\theta$$, it shouldn't be this hard, I'm just missing something... EDIT: It's also simpler to realize: $$1 y = \exp_{F(y)}^{\circ s}\left(\log_{F(y)}^{\circ s}(1) + y + \varphi(s,y)\right)\\$$ Which is a slightly less convoluted functional equation. We can call: $$1_\varphi y = \exp_{F(y)}^{\circ s}\left(\log_{F(y)}^{\circ s}(1) + y + \varphi\right)$$ For any $$\varphi \in \mathbb{C}$$. We are then solving: $$1_{\varphi_1} 1 _{\varphi_2} y = 1 _{\varphi_3} y+1\\$$ We choose the appropriate path, using the variable change, which allows us $$\varphi(s,y)$$. This gives us a super easy proof for existence, that works for all $$0 \le \Re(s) \le 2$$; but the construction is definitely not discovered. So just existence is feasible at the moment. We have a surface $$\boldsymbol{\Phi} \subset \mathbb{C}^3$$, such that, for all $$\boldsymbol{\varphi} \in \boldsymbol{\Phi}$$ we get: $$1_{\varphi_1} 1 _{\varphi_2} y = 1 _{\varphi_3} y+1\\$$ Now, we choose a path in $$s$$ and in $$y$$, which reduces us to $$\varphi(s,y)$$, which is our solution. We are literally tracing a path on a surface, Mphlee. That's literally! All we are doing! RE: Holomorphic semi operators, using the beta method - MphLee - 04/08/2022 Damn, I'm really too busy rn. Can't keep up with this amount of info. I hope I can put some time in during the upcoming Easter holidays. I was divided between studying this post or writing something to share... but damn... 24 hours are not enough for this. RE: Holomorphic semi operators, using the beta method - JmsNxn - 04/09/2022 Ya, don't worry about going through these posts. I'd much rather have an update on your research. I'm nearly done a 12 page quick write up, which outlines this method much more clearly than the posts here. It's a little trickier than I originally thought; but it does give an implicit solution. I think I can make it into a differential equation too, so that we can construct the solution--using PDE theory. I'm having a bit of trouble with this, but I'm nearly there. Regards, James RE: Holomorphic semi operators, using the beta method - JmsNxn - 04/14/2022 Here's a quick write up, which implicitly creates the construction.  It's pretty rough around the edges. We don't need the $$\beta$$ method as much as I thought; but how I've coded small solutions used it--I think haphazardly though. This is definitely an existence of holomorphic semi-operators though. I just need to read more about PDEs and create a better construction method... Scratching my head When_Bennet_becomes_Goodstein.pdf (Size: 277.81 KB / Downloads: 64) I'm currently trying to program in a method, but it requires an efficient protocol for $$\exp_b^{\circ s}(1)$$, as we vary $$b$$ and $$s$$. So I can program in the solution. But it'd be like a week to make a single significant graph. Programming this is difficult because most protocols fix b upon initialization (Sheldon, myself, and Kouznetsov). I think I know how to do it, but it's tough. RE: Holomorphic semi operators, using the beta method - MphLee - 04/15/2022 Great, I find this really easy to follow, you did a great job. I just barely had time to read half of it and is already connecting all the dots in my brain. I hope to get some time very soon, I really need to share something vaguely related and have to do it asap... since I believe that I've something that that in your hands could give us the holy grail. RE: Holomorphic semi operators, using the beta method - JmsNxn - 04/19/2022 Alright, so I did a bunch of coding at the cottage this easter weekend. It was shitty horrible weather, so I spent most of the time inside on the laptop grinding out the code. The program is nowhere near done, but I got a first glimpse of what to expect. Essentially I've successfully coded in: $$\exp_{y^{1/y}}^{\circ s}(\log^{\circ s}_{y^{1/y}}(a) + y)\\$$ In limited scenarios... but it should be good enough for semi-operators. I've managed to make a close first order approximation to the solution: $$0 2\\$$ So, this isn't the exact graph, but expect it to look something like this over $$0 \le s \le 2$$: The protocol I've written is still very beta beta beta, so I'm not going to post it. And it mostly works for $$\alpha \approx 0$$, so I can for example do: $$0.5 2\\$$ Again, we need to slightly perturb these solutions to get the correct solution. But it should look something like this. The trouble I'm having now, is figuring out how to solve the root of a three variable equation. I have to read more pari-gp protocols. Beware, that when I release this program it's going to be slow as fuck. This is going to be a very slow program. And I really can't think of a work around As a better explanation of these graphs; they are precisely: $$\alpha 2 = \text{these graphs} + \mathcal{O}(s)\,\,\text{for}\,\,s \approx 0\\$$ And similarly for $$s \approx 1,2$$; but now it's $$\mathcal{O}(s-1),\mathcal{O}(s-2)$$. FUCK YA! I figured it out for more anomalous values. Here are some more graphs. As we know: $$2 2 =4$$ I was having trouble accessing the repelling fixed point to do this. But I can do it now. It produces a straight line at 4. If I take something weird like: $$3.2 2\\$$ We get this graph: Again, we have to perturb these graphs slightly to get the right answer... but my code's almost there!!!!! I'm having trouble with  $$3.2 4$$ at the moment; so dealing with the repelling fixed point when it's in the s-exponent is a little tricky. But that's probably enough for tonight. This is definitely just a dumb coding error though, nothing mathematical. I just need to call it a night and look at it tomorrow ,lol.  But here is the attracting fixed point of $$b = \sqrt{3}$$, let's call it $$u_0$$. $$2 u_0\\$$ EDIT!!!! Alright, I figured out how to add repelling fixed points much better. Now I can have repelling fixed points for all the variables.  Here is: $$33\\$$ Such that this is just the first order expansion in $$s$$ of the $$\varphi$$ argument. I've done all the preliminary math at this point. Now I just have to efficiently solve for a 3 dimensional root equation. This will definitely take me a while. As an italian Mphlee, I'm sure you can appreciate this.  Romans had a perfect way of building bridges. But people could never build bridges as well as the romans. No one could make a bridge like the romans. And it all hinged on the roman keystone. They used to make bridges, with a strange brick in the center that just held everything together. And that was the legacy of roman thought. The $$\varphi$$ I'm looking for, is the keystone of a roman bridge... I've got a shoddy looking bridge here. I just need the italian roman keystone and it's perfect! lmao! RE: Holomorphic semi operators, using the beta method - JmsNxn - 04/20/2022 WOW! This is even easier than I expected. These values line up really really well! If I take: $$2 \oplus_{0.1, \sqrt{2}} 4 = 6.1242048680713775690019097380798646675079851431412...\\$$ It's already correct to two digits of $$2 \oplus_{1.1,\sqrt{3}} 3 = 6.1408969266137790131714605868589172539150029963802...$$ So in this instance $$\varphi_{1},\varphi_2$$, are going to be very very small. This tends to hold for all these values.  It's easy to work with this equation because we can assume that $$2 2 = 4$$ (this will definitely screw up for $$s \approx -1$$; but I'm already assuming we will have branching problems on the negative real axis). So we get a solution of values $$\varphi_1,\varphi_2$$, which solve this equation exactly. The trouble is rectifying it to a single function as we move the other variables. $$2_{\varphi_1} 4 = 2_{\varphi_2} 3$$ The trouble now is figuring out how to properly construct a function: $$\varphi(\alpha,y,s)\\$$ such that: \begin{align} \varphi(2,4,s) &= \varphi_1\\ \varphi(2,3,s+1) &= \varphi_2\\ \end{align} That will be holomorphic in all three variables. ----okay..., I think I can do this for this scenario only. Everywhere else it will fuck up. But since this is a two variable equation, I think I know how.  This would solve: $$2+u 4 + v = 2 + u 3+w\\$$ for $$u,v,w \approx 0$$. And at least do so locally. God I need to go to bed -_-...... Here is a graph of \begin{align} f(s) &= 2 \oplus_{s, \sqrt{2}} 4\\ g(s) &= 2 \oplus_{1+s,\sqrt{3}} 3\\ \end{align} For $$0 \le s \le 1$$.  We're essentially trying to zipper this solution together. We're trying to close this small gap. RE: Holomorphic semi operators, using the beta method - sheldonison - 04/22/2022 (03/24/2022, 09:35 AM)JmsNxn Wrote: ... Again, I haven't worked out the details, but the final expression should look something like this (remember, $$b = e^\mu$$): $$x\omega = \exp_b^{\circ s + \theta(s,x,\omega)}(\log_b^{\circ s + \theta(s,x,\omega)}(x) + \omega)$$ .... Hopefully this makes sense, I'm kind of just shooting at the wall and seeing what sticks, lol. But I do believe this holds some weight. Think of it as taking all the Bennet hyperoperations, and finding a path along them where the typical Goodstein functional equation works.Hi James, Mphlee, I was trying to make sense of what has been posted.  In Kneser's original paper, he talks about the half iterate o , which might be written as: $$f(z)=e^{[0.5]}(z)=\text{tet}_e(\text{slog}_e(z)+0.5);\;\;\;f(f(z))=e^z$$ The base e could be any real or complex base for which Kneser's tetration is defined, and the fractional iterate of 0.5 could be replaced by any arbitrary fractional iterate like .  Such a function would also be defined for non-integer values of n, and would be straightforward to compute with fatou.gp $$f(z)=b^{[\frac{1}{n}]}(z)=\text{tet}_b(\text{slog}_b(z)+\frac{1}{n});\;\;\;f^{[\circ n]}(z)=b^z$$ Then perhaps one might want to think about the super function of f(z), which could be generated surprisingly simply as the following: $$f^{[\circ z]}=\text{tet}_b(\frac{z}{n})$$ And would about the function g whose iterated superfunction is f(z)?  Could we define that as well?  How about the following, which should also be reasonably easy to compute with fatou.gp $$g(z)=f(f^{-1}(z)+1);\;\;\;f(z)=g^{[\circ z]}$$ $$f^{-1}(z)=\text{tet}_b(\text{slog}_b(z)-\frac{1}{n})$$ I hope this is somewhat relevant to the current discussion; this equation might even be closely related to the equation in James PDF which I just downloaded.  But my equations here don't lead to a generalized Goodstein/Ackermann functions for non integer values of n. RE: Holomorphic semi operators, using the beta method - JmsNxn - 04/23/2022 (04/22/2022, 06:14 PM)sheldonison Wrote: (03/24/2022, 09:35 AM)JmsNxn Wrote: ... Again, I haven't worked out the details, but the final expression should look something like this (remember, $$b = e^\mu$$): $$x\omega = \exp_b^{\circ s + \theta(s,x,\omega)}(\log_b^{\circ s + \theta(s,x,\omega)}(x) + \omega)$$ .... Hopefully this makes sense, I'm kind of just shooting at the wall and seeing what sticks, lol. But I do believe this holds some weight. Think of it as taking all the Bennet hyperoperations, and finding a path along them where the typical Goodstein functional equation works.Hi James, Mphlee, I was trying to make sense of what has been posted.  In Kneser's original paper, he talks about the half iterate o , which might be written as: $$f(z)=e^{[0.5]}(z)=\text{tet}_e(\text{slog}_e(z)+0.5);\;\;\;f(f(z))=e^z$$ The base e could be any real or complex base for which Kneser's tetration is defined, and the fractional iterate of 0.5 could be replaced by any arbitrary fractional iterate like .  Such a function would also be defined for non-integer values of n, and would be straightforward to compute with fatou.gp $$f(z)=b^{[\frac{1}{n}]}(z)=\text{tet}_b(\text{slog}_b(z)+\frac{1}{n});\;\;\;f^{[\circ n]}(z)=b^z$$ Then perhaps one might want to think about the super function of f(z), which could be generated surprisingly simply as the following: $$f^{[\circ z]}=\text{tet}_b(\frac{z}{n})$$ And would about the function g whose iterated superfunction is f(z)?  Could we define that as well?  How about the following, which should also be reasonably easy to compute with fatou.gp $$g(z)=f(f^{-1}(z)+1);\;\;\;f(z)=g^{[\circ z]}$$ $$f^{-1}(z)=\text{tet}_b(\text{slog}_b(z)-\frac{1}{n})$$ I hope this is somewhat relevant to the current discussion; this equation might even be closely related to the equation in James PDF which I just downloaded.  But my equations here don't lead to a generalized Goodstein/Ackermann functions for non integer values of n. Hey, Sheldon. I'll give a quick summary here. We do not need Kneser for any of this, because we're only going to be dealing with bases in the Shell-Thron region (where tetration is considerably easier). I'll start with the real case, which is much easier to digest (the complex plane will work similarly, a little more finicky though). If I take an arbitrary $$y >1$$, then the value $$y^{1/y} = b \in [1,e^{1/e}] \subset \mathfrak{S}$$, the Shell-Thron region. Call: $$\exp^{\circ s}_{b}(x)$$ The regular iteration, for $$x \in \mathbb{R}$$. So for example, if $$b = \sqrt{2}$$, then we'd use the Schroder iteration about $$2$$ for $$x \in (-\infty,4)$$, and for $$x \in [4,\infty)$$, we use the Schroder iteration about the repelling fixed point $$4$$. This can be done for all $$b = y^{1/y}$$ for $$y > 1$$, excepting at $$y=e$$--where you have to switch to solving the Abel equation in the neutral scenario. This would require using the bounded iteration, and the unbounded iteration (for $$e^{1/e}$$, this would be the normal tetration, and cheta solution). Now, we start by preliminarily defining: $$x \oplus_{s} y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y\right)\\$$ Ignoring where the singularities are for $$\log^{\circ s}$$ are momentarily, we get the following interpolation. \begin{align} x \oplus_{0} y &= \exp^{\circ 0}\left(\log^{\circ 0}(x) + y\right) = x+y\\ x\oplus_1 y &= \exp_{y^{1/y}}\left(\log_{y^{1/y}}(x) + y \right) = x \cdot \exp_{y^{1/y}}(y) = x \cdot y\\ x \oplus_2 y &= \exp_{y^{1/y}} \left(\exp_{y^{1/y}}(\log^{\circ 2}_{y^{1/y}}(x) + y)\right) = \exp_{y^{1/y}}\left(\log_{y^{1/y}}(x)\cdot y\right) = e^{\log(x) y} = x^y\\ \end{align} This gives us our preliminary operators, which we want to perturb slightly to satisfy goodstein's equation. We do this by inserting a term $$\varphi(x,y,s)$$, and we define the new operators: $$x _{\varphi} y = \exp_{y^{1/y}}^{\circ s}(\log_{y^{1/y}}^{\circ s}(x) + y + \varphi)\\$$ Initial tests show that $$\varphi$$ is relatively small. For example, we start by knowing $$\varphi = 0$$ when $$s =0,1,2$$, and has a slight bump as you move between. A good way to visualize this was in the graph of my last post. If we stick to the idea that $$2 2 =4$$ then we want to solve for the correct varphi such that: $$2 4 = 2 3\\$$ And if we graph these two functions next to each other over $$(0,1)$$, we get: So essentially, we want to hunt for a varphi that zippers these two functions together. I can do this in a very rough manner, but it's increasingly frustrating to solve the implicit equation. The implicit equation we are solving is tricky to write out, but consider $$\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)$$, and we look at: $$x _{\varphi_1} (x _{\varphi_2} y) = x _{\varphi_3} y+1\\$$ Under the constraint that: \begin{align} \varphi_2(x,y+1,s) &= \varphi_3(x,y,s)\\ \varphi_2(x,x _{\varphi_2(x,y,s+1)} y,s) &= \varphi_1(x,y,s)\\ \end{align} These implicit functions always exist, because the derivative in $$\varphi$$ is non-singular--and we always have existence of points which solve the equation. Then, the desired $$\varphi$$ we'd want would be $$\varphi(x,y,s+1) = \varphi_2$$, so long as it satisfies the above constraints. This then becomes a unique solution (3 variables, 2 constraints). This varphi would then be the solution such that: \begin{align} x y &= \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi(x,y,s)\right)\\ x (x y) &= x (y+1)\\ \end{align} You can also prove this function exists by constructing a first order differential equation in three variables, but that was making my head hurt so I abandoned it. It's hard to do tests on this, because the code I have is just too slow. I'd use fatou.gp, but we have to consistently reinitialize the base of the super-exponential, and there's no work around in fatou.gp that I know of. So instead, I whipped up some code which managed to go as fast as I could get. So we don't need Kneser for any of this, it's just regular iteration or neutral iteration, at least for $$y > 1$$ case. I'm expecting volatile behaviour for $$y < 1/e$$, because here, $$y^{1/y} \not \in \mathfrak{S}$$. This idea only works so far when $$y^{1/y} \in \mathfrak{S}$$. Technically it can be extended further. I'm not too worried about the complex plane at the moment. But, the code still seems to be working. Regards, James Also, this notation is just a place holder. I wanted to distance from Knuth's up arrow, and conway chained notation. I meant to write it like: $$x \langle s\rangle y\\$$ Like the braket angle. But I was lazy and didn't want to type the extra characters for the moment, lol. If you can direct me to a manner of using fatou.gp such that it's holomorphic in the base, we could use that program. But as I understand fatou.gp, we have to fix the base upon initialization. And I don't know of a work around.  So everytime we moved y we'd have to reinitialize, which is obviously too taxing. Unless you can think of a fast fatou.gp way of graphing: $$x \oplus_s y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y\right)\\$$ In the variable $$y$$, I'm happy to adopt fatou.gp protocols. For the moment I'm using the program semi_operators.gp, which is far from being complete. It doesn't work for the neutral case. So it's ideally meant for $$1 < ye$$ and solely real variables. It does not solve for $$\varphi$$ yet, but it allows you test which $$\varphi$$ work. The main protocols, that you use are Iexp(s,z,y) Which is precisely $$\exp^{\circ s}_{y^{1/y}}(z)$$, and Ibennet(s,z,y,{phi = 0}) Which is precisely $$z \langle s\rangle_{\phi} y$$ for $$\phi$$ a constant. You can treat these objects as taylor series too, so: Iexp(0.5+s,2,3) Iexp(0.5,2+w,3) Iexp(0.5,2,3+u) Are all valid operations, and they'll produce a Taylor series in the appropriate variable. Note, that the first one is fast, second one is medium speed, and third one is very slow. The same idea works with Ibennet. semi_operators.gp (Size: 5.65 KB / Downloads: 53) Please note it is very very much still a beta. There are still some small glitches. But it's working for everything I've been using it for. I've run a few example inputs: Code:\r semi_operators Iexp(0.1+s,Iexp(0.1+s,1,2),2) - Iexp(0.2+2*s,1,2) %39 = 1.0560015667790773041 E-47 - 9.874014683335685726 E-48*s + 1.0117887154347293316 E-47*s^2 - 9.769866529528564435 E-48*s^3 + 9.205049268878957323 E-48*s^4 - 8.555693781692655296 E-48*s^5 + 7.887702353919229131 E-48*s^6 - 7.235526068815923254 E-48*s^7 + 6.616389800744812899 E-48*s^8 - 6.038051040511804261 E-48*s^9 + 5.503079393175880247 E-48*s^10 - 5.011230997188426702 E-48*s^11 + 4.560781494644109592 E-48*s^12 - 4.149279553778102565 E-48*s^13 + 3.773972813243431893 E-48*s^14 - 3.432046517242785078 E-48*s^15 + 3.120754229436710630 E-48*s^16 - 2.837486129351430481 E-48*s^17 + 2.5798012395080594446 E-48*s^18 - 2.3454389566002173251 E-48*s^19 + 2.1323189023009714368 E-48*s^20 - 1.9385343945431778221 E-48*s^21 + 1.7623426514180968579 E-48*s^22 - 1.6021535466821381767 E-48*s^23 + 1.4565179655867155578 E-48*s^24 - 1.3241163530764392936 E-48*s^25 + 1.2037477741091062452 E-48*s^26 - 1.0943196448441181181 E-48*s^27 + 9.948381990067232366 E-49*s^28 - 9.043996992691581784 E-49*s^29 + O(s^30) Iexp(0.1,Iexp(0.1,1+w,2),2) - Iexp(0.2,1+w,2) %40 = 1.0560015667790773041 E-47 - 7.292173087295080608 E-48*w + 1.2378925093140665465 E-48*w^2 + 3.233718357384597705 E-50*w^3 - 9.182632082588929039 E-51*w^4 + 4.431670302409573164 E-52*w^5 - 3.214653653169696949 E-54*w^6 - 1.7306769515615559648 E-53*w^7 - 5.503821799350850526 E-55*w^8 - 1.0694188305787609134 E-55*w^9 - 6.202971279155519914 E-56*w^10 - 1.3568224932426610033 E-56*w^11 - 3.216789057723641103 E-57*w^12 - 8.266527408745018921 E-58*w^13 - 2.0705029304634104792 E-58*w^14 - 5.385364260250267394 E-59*w^15 - 1.4542400553561676742 E-59*w^16 - 3.987068346848031708 E-60*w^17 - 1.1043673344277225868 E-60*w^18 - 3.0821036377983951112 E-61*w^19 - 8.666384254248708309 E-62*w^20 - 2.455366258897947865 E-62*w^21 - 7.006802304860847056 E-63*w^22 - 2.0170573985505660255 E-63*w^23 - 5.844838423025127240 E-64*w^24 - 1.7043707844478603356 E-64*w^25 - 5.002282984467736117 E-65*w^26 - 1.4748828465574770541 E-65*w^27 - 4.368195147161827048 E-66*w^28 - 1.3005076403225357554 E-66*w^29 + O(w^30) Iexp(0.1,Iexp(0.1,1,2+u),2+u) - Iexp(0.2,1,2+u) %42 = 1.0560015667790773041 E-47 - 4.472892251399907988 E-46*u + 9.781424310607690251 E-45*u^2 - 1.4699699204329577284 E-43*u^3 + 1.7054441836546445651 E-42*u^4 - 1.6273367290258708051 E-41*u^5 + 1.32885839335278805727827573527918224978 E-40*u^6 - 9.5421624175376840539197291384460400639 E-40*u^7 + 6.1453650732800893341605643123388578479 E-39*u^8 - 3.60300875093315313408444140476588790020 E-38*u^9 + 1.94566303616455738904299083635851188585 E-37*u^10 - 9.7682500447243861266806423963749980846 E-37*u^11 + 4.59447095746943974844305534541439981805 E-36*u^12 - 2.03745361944985745398967717926235994073 E-35*u^13 + 8.5645578821650349164192801336258673790 E-35*u^14 - 3.42829069968351780819928743259801848743 E-34*u^15 + 1.31196653716817812115272921722451662512 E-33*u^16 - 2.24532381181948620962583236044984348020 E-32*u^17 + 4.8652831676001940743924440319709612865 E-31*u^18 + 2.58292920368640701303391891403464628542 E-31*u^19 - 1.97057852828718112279574806521525784630 E-28*u^20 + 4.64987264628885322866316806150922557652 E-27*u^21 - 6.6578671819241133287445615873692665215 E-26*u^22 + 7.2286681115998680900711068774455519814 E-25*u^23 - 6.58959236390523930535296008625163909249 E-24*u^24 + 5.380403054358272658633240764988271137796 E-23*u^25 - 4.096813673994155747724413220990481811285 E-22*u^26 + 2.962290760331694625789446700972377624571 E-21*u^27 - 2.037121995399768881356559797850718838378 E-20*u^28 + 1.324525396616853415146444872768299980083 E-19*u^29 + O(u^30) Ibennet(s,2,3) %43 = 5.000000000000000000000000000000000000000 + 0.7256473965263415355726335073462804205026*s + 0.2065985475702190684145536780286266211912*s^2 + 0.05266661871645363241880119184483379064661*s^3 + 0.01213316362816163676181025275398619937848*s^4 + 0.002484587441238103973304180437106722969663*s^5 + 0.0004278459187290018743995504020806878181593*s^6 + 5.026411809318944120745729874904715046567 E-5*s^7 - 2.097978055159398963357933638524863447127 E-6*s^8 - 3.917442368124171049853741448942720330313 E-6*s^9 - 1.673872252322174081842751643007073521005 E-6*s^10 - 5.380160204435224676806407097766698263024 E-7*s^11 - 1.487524705412200256434548723407493626568 E-7*s^12 - 3.678696023734425527698440224412572731990 E-8*s^13 - 8.202832634858730364492806468668774363198 E-9*s^14 - 1.625500586548617665288097049141693201050 E-9*s^15 - 2.712835452473239056738364619798262174322 E-10*s^16 - 3.109802441299858169295550487800312326676 E-11*s^17 + 1.112993550172292404283302730216624669840 E-12*s^18 + 2.205213806998927153512145905470116524878 E-12*s^19 + 9.181564593335951454943477288140957403512 E-13*s^20 + 2.865405234180613573417040028805685461075 E-13*s^21 + 7.682382271111932455961316648271579055736 E-14*s^22 + 1.839663374945693682982560446757950010565 E-14*s^23 + 3.961187236713129339547902763974145979214 E-15*s^24 + 7.532772539127867144186612215844980322245 E-16*s^25 + 1.185363162997741398930795760686833152740 E-16*s^26 + 1.174398958754750966498720562489546007851 E-17*s^27 - 1.166242631363682451728692597298433002393 E-18*s^28 - 1.131457387001417994544198130378773996438 E-18*s^29 - 4.315233684769584434984544994515309246649 E-19*s^30 + O(s^31) Ibennet(0.5,2+w,3) %44 = 5.421899688575296614480746492856113676386 + 1.679987795255501710878313093811819181796*w - 0.04347785824344861296284767941944766819858*w^2 + 0.007121822850024879241445586349222213280985*w^3 - 0.001376879116264135970261706566706224093898*w^4 + 0.0002747278105529088737346699108556333963802*w^5 - 5.385495530127409423700620537863985861381 E-5*w^6 + 1.027956560181029784062143507868493210664 E-5*w^7 - 1.953787386336938864390986897159886525278 E-6*w^8 + 3.801122856608207740988701957094181033884 E-7*w^9 - 7.442075910542003411460462084939730473596 E-8*w^10 + 1.381795911488855639716786135263370897231 E-8*w^11 - 2.404357523252995119944397239761751813756 E-9*w^12 + 4.322409520474934946212125005899378412773 E-10*w^13 - 8.273085911368946738291857396698229671168 E-11*w^14 + 1.552146378849620728183763620175586108136 E-11*w^15 - 4.097584081722807165336492806098232157828 E-12*w^16 + 1.553865361705301700532908116171772037207 E-12*w^17 - 4.745983078016714430749197422600885694106 E-13*w^18 + 1.492314260312856494890791108637045324979 E-13*w^19 - 5.525157943025930578994615191443662779547 E-14*w^20 + 1.158791240056365382110768676713892250454 E-14*w^21 - 1.439692051073410982618496311343583847457 E-15*w^22 + 4.419281401473502140597701847447461169202 E-16*w^23 + 3.699151836339054125180220323731004410062 E-16*w^24 - 2.232565330945937203451041494407379362271 E-16*w^25 + 3.710049633269071974817933636635621390823 E-17*w^26 - 4.379579123335190351213815623434490645131 E-17*w^27 + 1.032889062250991431171289843167513217652 E-17*w^28 + 1.834623994329765755744838777728366888052 E-18*w^29 + O(w^30) Ibennet(0.5,2,3+u) %45 = 5.421899688575296614480746492856113676386 + 1.363247806781409234684044836437993931866*u - 0.04311722936335908535861724412668305823598*u^2 + 0.01040877031580076014812191025887569735514*u^3 - 0.002870970160590399097264117058994519195313*u^4 - 0.004048640028837108120807518094271585337259*u^5 + 0.06143126859699205576793407015806617810466*u^6 - 0.6322585826171430068096418163463293403929*u^7 + 5.656622724001871846414674149246241211265*u^8 - 44.95104994643104704300166197623567347138*u^9 + 321.1437920573653539303792189552210292800*u^10 - 2082.077165719160331150645148720138351398*u^11 + 12340.61049164733578677208721587805014740*u^12 - 67254.52158829211488788183365838228670235*u^13 + 338476.1381492849208377159580740559204635*u^14 - 1577673.866929055603970674052567385020675*u^15 + 6819681.694962554307106645688386735043041*u^16 - 27318918.85030331730199608661957231215333*u^17 + 101040851.1534861799100329815154092300636*u^18 - 341946433.5942111206146463685267450947096*u^19 + 1038018931.207984498990741844342172871817*u^20 - 2693398546.208956875783299164627112235087*u^21 + 5105796531.426535288760125870735147671064*u^22 - 723951362.2746718252564545130076117477165*u^23 - 58459329036.77037565556938393671302456607*u^24 + 402801584215.7573768444086763089111345704*u^25 - 2006430155007.520755746675915888637100606*u^26 + 8721118226132.705302480198818708474773798*u^27 - 35305635629397.55267055911945872572029274*u^28 + 138069796341481.8443991453860106846432293*u^29 + O(u^30) Which is just checking the Taylor series of our semi-group works in every variable. This works through out the program excepting some buggy values... We also checked the values of some Ibennet operations, done through each variable. RE: Holomorphic semi operators, using the beta method - JmsNxn - 04/27/2022 So, I've thought of working on a single case rather than working on the more general problem. I started by looking at: $$2 4 = 2 3\\$$ Which births from: \begin{align} 2 \left(2 2\right) &= 2 (2+1)\\ 2 4 &= 2 3\\ \end{align} As a reminder of the notation, we write: $$x _\varphi y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)\\$$ If we omit the $$\varphi$$, it is assumed that it is the unique implicit $$\varphi$$ which solves Goodstein's equation. In the case above, we assume that: $$22 = 4\\$$ Therefore $$\varphi = 0$$, because that's the only value which satisfies the equation. From here, we are discussing the surface of values $$\Phi \subset \mathbb{C}^3$$, such if $$\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)$$, then $$2 _{\varphi_1} \left(2 _{\varphi_2} 2\right) = 2 _{\varphi_3} 3\\$$ As we are assuming that $$\varphi_2 = 0$$, to ensure that $$2 2 = 4$$ everywhere, we are reduced to a path along surface the surface $$\Phi$$, $$\boldsymbol{\varphi} = (\varphi_1,0,\varphi_3)$$. So to ensure we have the correct functional equation we get the following: $$\varphi_1 = \log^{\circ s}_{\sqrt{2}}\left(2_{\varphi_3} 3\right) - 6\\$$ Now, we are allowed to move $$\varphi_3$$ freely, so long as $$\varphi_3 = 0$$ when $$s=0,1$$. There is only one value of $$\varphi_3$$ though, that we actually want. But we can still play around with possible values. Upon doing this, we get: $$2 _{\varphi_1} 4 = 2 _{\varphi_3} 3\\$$ For the moment it appears that $$\varphi_3$$ is arbitrary, but that's because we haven't introduced a non constant "exponent". Safe to say, though, that there is some function $$\varphi_3(s)$$ that will be the actual function we want. This will be on a moving surface $$\Phi(s)$$, which changes as $$s$$ moves. To make things more interesting, we look at: $$2 (2 + w) = 4 + A(s)w + \mathcal{O}(w^2)\\$$ Therein $$A = 2$$ for $$s = 0$$. Whereupon $$2 \left( 2 (2 + w)\right) = 2 (4 + A(s)w + \mathcal{O}(w^2))\\$$ Then, $$2(4 + A(s)w + \mathcal{O}(w^2)) = 2 3 + w\\$$ Since we're going to try and only care about the first term in $$w$$, and make some kind of a differential equation, we can drop the pesky $$\mathcal{O}(w^2)$$. And since we've already solved for $$\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)$$ when $$w=0$$ ($$\varphi_2 = 0$$). But in addition to that, we have the functional equation: $$\varphi_1 = \log^{\circ s}_{(4+Aw)^{1/(4+Aw)}}\left(2_{\varphi_3} (3+w)\right) - 6-A(s)w\\$$ And with this equation: $$2 _{\varphi_1} \left(2 _{\varphi_2} 2 + w\right) = 2 _{\varphi_3} 3+w\\$$ Now we consider, \begin{align} \varphi(2,4+A(s) w,s) &= \varphi_1\\ \varphi(2,2+w,s+1) &= \varphi_2 = \mathcal{O}(w)\\ \varphi(2,3+w,s+1) & = \varphi_3\\ \end{align} So the correct answer, and the exact $$\varphi_3$$ we would need, is the value such that, \begin{align} \lim_{s\to 0} \lim_{w\to 0} \frac{\partial}{\partial w}\varphi_3 &= \lim_{s\to 0} \lim_{w\to 1} \frac{\partial}{\partial w} \varphi_2\\ \lim_{s\to 1} \lim_{w\to 0} \frac{\partial}{\partial w}\varphi_3 &= \lim_{s\to 1} \lim_{w\to 1} \frac{\partial}{\partial w} \varphi_2\\ \end{align} We mean this derivative in the complex sense too. This limit formula is mainly important at the $$0,1$$ endpoints, but it should hold everywhere. It's nice to use this limit formula, because we've reduced the question to making sure that we're only worried about the end points (addition and multiplication and exponentiation). Ensuring the derivatives work here is crucial, and with derivatives we get close to almost having our answer. This gives us an avenue of solving for $$A$$, which then gives us an avenue to solve for $$\varphi_2$$, then $$\varphi_1,\varphi_3$$ are related as above, then we are simply choosing the correct value of $$\varphi_3$$, such you can paste them all together with no discontinuities, which is certainly possible, just from looking at it like an initial value problem ODE. This effectively reduces the problem into solving for $$\varphi_2$$ and $$\varphi_3$$, where $$\varphi_1$$ is a function of both. So it effectively reduces the problem by one degree. Which makes sense considering that we're trying to do calculus on a surface $$\Phi \subset \mathbb{C}^3$$, which has complex-dimension 2. I'm having not much luck testing these things because my code is just too damn slow. Unless Sheldon has a speedup, I'm not sure how to make the code faster. Some additional notes, which help visualize the problem is that; there always exists a $$\boldsymbol{\varphi}$$ if we assume that two of the components are zero. So if I assume that $$\varphi_2 = 0$$ and $$\varphi_3 = 0$$, then we always have a solution $$\varphi_1$$ such that: $$2 _{\varphi_1} \left(2_0 2+w\right) = 2 _0 3 + w$$ This would not be the solution we'd want though, as it wouldn't paste together properly. There's only one function that equals the correct equation pasted together. But this is very important. Because if you fix $$s$$, then there exists a value $$\varphi_1$$ such that $$\boldsymbol{\varphi} = (\varphi_1,0,0)$$ which solves the equation. Then, if you move $$\varphi_1$$ by $$\epsilon$$, you move $$\varphi_2$$ and/or $$\varphi_3$$ by $$\epsilon'$$. So we can with $$s$$ fixed, have a surface $$\Phi \subset \mathbb{C}^3$$ (which means $$\Phi$$ is of complex-dimension $$2$$), of possible values. Moving $$s$$ moves the shape of the surface. When $$s=0,1,2$$, we're reduced to a point mass at $$(0,0,0)$$. As we move $$s \sim 0 \to 1$$ the surface grows in a ball shaped mass, and retracts back to the point mass. This is done in $$\mathbb{C}^3$$, so it's a surface in this space... I'm still working heavily on this, but it might be a while before I can sort it out. The code is pretty fucking awful, and I need to find a way to improve it to do more accessible tests. I hope Sheldon can help, but he's moved away from tetration, so I might be on my own, lol. Regards, James