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 Continuous Hyper Bouncing Factorial Catullus Fellow   Posts: 210 Threads: 46 Joined: Jun 2022 08/08/2022, 10:05 AM (This post was last modified: 08/08/2022, 10:12 AM by Catullus.) You may know about my Hyper Bouncing Factorial function denoted , it is like the Bouncing Factorial, but with a starting number of two and you use exponentiation. How could the Hyper Bouncing Factorial function be defined for non integers, like how the exponential factorial could possibly be extended to non integers in this way? ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 08/09/2022, 05:01 AM (08/08/2022, 10:05 AM)Catullus Wrote: You may know about my Hyper Bouncing Factorial function denoted , it is like the Bouncing Factorial, but with a starting number of two and you use exponentiation. How could the Hyper Bouncing Factorial function be defined for non integers, like how the exponential factorial could possibly be extended to non integers in this way? Lmao, good luck Catullus. This is way too hard to be solved. Where as the Bouncing factorial looks something like: $$\prod_{c=1}^x \Gamma(c+1)\,$$ Which is analytic in $$x$$, and you'd just be adding an additional step to get the analytic bouncing factorial. The hyper bouncing factorial function is just impossible, because we can't even do this kind of hyper factorial. I spent a good amount of time on the Hyper factorial and the closest I came to was very far away from the hyper factorial. I used it to justify that these things "could" be solvable. I managed to solve: $$\Upsilon(s+1) - \Upsilon(s) = e^{s\Upsilon(s)}\\$$ Which turns out to be an entire function. Trying to even get close to anything near hyper factorial from here is leagues beyond me though. I have no idea. My goal was to solve "some kind of" difference equation involving exponentials. Catullus Fellow   Posts: 210 Threads: 46 Joined: Jun 2022 08/09/2022, 05:34 AM (08/09/2022, 05:01 AM)JmsNxn Wrote: I managed to solve: $$\Upsilon(s+1) - \Upsilon(s) = e^{s\Upsilon(s)}\\$$ Which turns out to be an entire function. Trying to even get close to anything near hyper factorial from here is leagues beyond me though. I have no idea. My goal was to solve "some kind of" difference equation involving exponentials.How did you solve it, and what solution did you obtain? ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 08/09/2022, 07:22 AM (08/09/2022, 05:34 AM)Catullus Wrote: (08/09/2022, 05:01 AM)JmsNxn Wrote: I managed to solve: $$\Upsilon(s+1) - \Upsilon(s) = e^{s\Upsilon(s)}\\$$ Which turns out to be an entire function. Trying to even get close to anything near hyper factorial from here is leagues beyond me though. I have no idea. My goal was to solve "some kind of" difference equation involving exponentials.How did you solve it, and what solution did you obtain? Take the function: $$q(s,z) = z + e^{sz}\\$$ If you take the sequence of functions: \begin{align} &q(s-1,z)\\ &q(s-1,q(s-2,z))\\ &q(s-1,q(s-2,q(s-3,z)))\\ &\vdots\\ &\Omega_{j=1}^\infty q(s-j,z)\bullet z\\ \end{align} Then the limit was $$\Upsilon$$, where $$z$$ acts as an open parameter (the (spectrum) parameters exist for $$\Re(z) > 0$$).  This function satisfies the equation, by construction: $$q(s,\Upsilon(s,z)) = \Upsilon(s+1,z)\\$$ Which by linear substitution satisfies: $$\Upsilon(s+1,z) - \Upsilon(s,z) = e^{s\Upsilon(s,z)}\\$$ These questions get really complicated though. And Catullus, I can't answer them. I can prove general normality theorems, and convergence theorems. But everything to do with the hyper factorial is anomalous and nulls every element of my proofs. Leo.W Fellow   Posts: 79 Threads: 6 Joined: Apr 2021 08/10/2022, 11:07 AM This extension problem can be solved striaghtly if you'd take analytic methods, or if you happen to know some higher-level functions. First we define a function as a semi-solution:$$f(z)=\frac{\Gamma(z+1)^2}{z}$$, or you can call it "a bounce of height z" if you wish, then we extend this hyper bouncing or whatever-you-like-to-call factorial $$z\Lambda=f(1)f(2)f(3)\dots f(z)=\prod_{i=1}^{z}{f(i)}$$ Before so long we have already studied summations and products and their relationship, thus we transform the function by this equation $$\log(z\Lambda)=\log(f(1))+\log(f(2))+\log(f(3))+\dots+\log(f(z))=\sum_{i=1}^{z}{\log(f(i))}$$ It happened that since $$\log(f(i))=2\log(\Gamma(i+1))-\log(i)$$ (not considering multivalued-ity), this sum can be written in closed form with special functions:  $$\log(z\Lambda)=\sum_{i=1}^{z}{\log(f(i))}=2\log(G(z+2))-\log(\Gamma(z+1))$$ where $$G$$ denotes Barnes' G function, or more explicitly it's represented by $$G(z)=\Gamma(z)^{z-1}e^{\frac{z\log(2\pi)+z(1-z)}{2}-\phi^{(-2)}(z)}$$ where $$\phi^{(-2)}(z)$$ is polygamma function and defined originally by the recurrence $$G(z+1)=G(z)\Gamma(z)$$. And all used functions are "entire". This can be more presented with antiderivatives and derivatives of Zeta function, if desired. Regards, Leo  JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 08/11/2022, 12:07 AM (08/10/2022, 11:07 AM)Leo.W Wrote: This extension problem can be solved striaghtly if you'd take analytic methods, or if you happen to know some higher-level functions. First we define a function as a semi-solution:$$f(z)=\frac{\Gamma(z+1)^2}{z}$$, or you can call it "a bounce of height z" if you wish, then we extend this hyper bouncing or whatever-you-like-to-call factorial $$z\Lambda=f(1)f(2)f(3)\dots f(z)=\prod_{i=1}^{z}{f(i)}$$ Before so long we have already studied summations and products and their relationship, thus we transform the function by this equation $$\log(z\Lambda)=\log(f(1))+\log(f(2))+\log(f(3))+\dots+\log(f(z))=\sum_{i=1}^{z}{\log(f(i))}$$ It happened that since $$\log(f(i))=2\log(\Gamma(i+1))-\log(i)$$ (not considering multivalued-ity), this sum can be written in closed form with special functions:  $$\log(z\Lambda)=\sum_{i=1}^{z}{\log(f(i))}=2\log(G(z+2))-\log(\Gamma(z+1))$$ where $$G$$ denotes Barnes' G function, or more explicitly it's represented by $$G(z)=\Gamma(z)^{z-1}e^{\frac{z\log(2\pi)+z(1-z)}{2}-\phi^{(-2)}(z)}$$ where $$\phi^{(-2)}(z)$$ is polygamma function and defined originally by the recurrence $$G(z+1)=G(z)\Gamma(z)$$. And all used functions are "entire". This can be more presented with antiderivatives and derivatives of Zeta function, if desired. Yes, that's the bouncing Factorial. Thanks for doing the work I was too lazy to do, lol. But the bouncing HYPER factorial. Get outta here, that's next century's problem, lmao. tommy1729 Ultimate Fellow     Posts: 1,700 Threads: 374 Joined: Feb 2009 08/12/2022, 03:29 AM (08/08/2022, 10:05 AM)Catullus Wrote: You may know about my Hyper Bouncing Factorial function denoted , it is like the Bouncing Factorial, but with a starting number of two and you use exponentiation. How could the Hyper Bouncing Factorial function be defined for non integers, like how the exponential factorial could possibly be extended to non integers in this way? " in this way " ?  I think those conditions/properties are inconsistant with the desired function. Leo.W Fellow   Posts: 79 Threads: 6 Joined: Apr 2021 08/12/2022, 11:10 AM (08/11/2022, 12:07 AM)JmsNxn Wrote: Yes, that's the bouncing Factorial. Thanks for doing the work I was too lazy to do, lol. But the bouncing HYPER factorial. Get outta here, that's next century's problem, lmao.I think it's easy to do so if we already find an asymptotic term? If we can prove that this hyper bouncing factorial growing fast almost as $$O(_xx)=O(x2)$$ and then recurrent! Regards, Leo  Catullus Fellow   Posts: 210 Threads: 46 Joined: Jun 2022 08/15/2022, 06:29 AM I have an idea for how to define a continuous hyper bouncing factorial! To do it, you would do polynomial interpolation of , and then do  of that. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 08/15/2022, 07:54 AM (08/15/2022, 06:29 AM)Catullus Wrote: I have an idea for how to define a continuous hyper bouncing factorial! To do it, you would do polynomial interpolation of , and then do  of that. I know a lot of people are going to treat your comment as incorrect. But I imagine something like: $$\text{slog}(x \Lambda^+) = x + g(x) + o(g(x))\\$$ As $$x \to \infty$$; this is probably possible. This of course relates to what Leo was talking about. And how he described first order difference equations and their relation to abel solutions. Either way; we are reducing the problem into a linear problem, where $$g$$ grows at worse exponentially. Then, yeah. I think Leo is right, that we can construct the Hyper bouncing factorial. In so far as, we need to remember at best this is an existence proof. It is not a constructive proof, just existential. Which Leo is to credit for a lot of this. « Next Oldest | Next Newest »

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