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 On to C^\infty--and attempts at C^\infty hyper-operations JmsNxn Long Time Fellow Posts: 461 Threads: 85 Joined: Dec 2010 02/08/2021, 12:12 AM (This post was last modified: 02/08/2021, 03:55 AM by JmsNxn.) As Sheldon has thoroughly convinced me of non-holomorphy of my tetration. I thought I'd provide the proof I have that it is $C^{\infty}$ on the line $(-2,\infty)$. I sat on this proof and didn't develop it much because I was too fixated on the holomorphy part. But, I thought it'd be nice to have a proof of $C^{\infty}$. Now, the idea is to apply Banach's fixed point theorem, but it's a bit more symbol heavy now. We will go by induction on the degree of the derivative. So let's assume that, $ \tau^{(k)}(t) : (-2,\infty) \to \mathbb{R}\,\,\text{for}\,\,k \sum_{m=1}^\infty ||\tau^{(k)}_{m+1}(t) - \tau^{(k)}_{m}||_{a \le t \le b} <\infty\\$ Where, $ \tau^{(k)}_0(t) = 0\\ \tau^{(k)}_m(t) = \frac{d^k}{dt^k} \log(1+\frac{\tau_{m-1}(t+1)}{\phi(t+1)})\\$ And $||...||_{a\le t \le b}$ is the sup-norm across some interval $[a,b] \subset (-2,\infty)$. As a forewarning, this is going to be very messy... Now to begin we can bound, $ ||\phi^{(j)}(t)||_{a\le t \le b} \le M\,\,\text{for}\,\,j\le k\\$ And that next, $ \phi^{(k)}(t+1) + \tau_m^{(k)}(t+1) = \frac{d^k}{dt^k} e^{\phi(t) +\tau_{m+1}(t)}\\ = \sum_{j=0}^k \binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)})\\ = \sum_{j=0}^{k-1} \binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)}) + \phi(t+1)e^{-t}(\frac{d^k}{dt^k} e^{\tau_{m+1}(t)})$ Now, $ \frac{d^k}{dt^k} e^{\tau_{m+1}(t)} = e^{\tau_{m+1}(t)}(\tau_{m+1}^{(k)}(t) + \sum_{j=0}^{k-1} a_j \tau_{m+1}^{(j)}(t))\\$ So, we ask you to put on your thinking cap, and excuse me if I write, $ \tau_{m}^{(k)}(t+1) = A_m + C_m\tau_{m+1}^{(k)}(t)\\$ And by the induction hypothesis, $ \sum_{j=1}^\infty ||A_{j+1} - A_j||_{a\le t \le b} < \infty\\ \sum_{j=1}^\infty ||C_{j+1} - C_j||_{a\le t \le b} < \infty\\$ Which is because these terms are made up of finite sums and products of $\tau_m^{(j)}$ and these are said to be summable.  Now the proof is a walk in the park. $ \tau_{m}^{(k)}(t) = \frac{\tau_{m-1}^{(k)}(t+1) - A_{m-1}}{C_{m-1}}... = \sum_{j=0}^{m-1} (\prod_{k=0}^{m-1-j} C_{m-1-k}^{-1}) A_j\\$ Where, we've continued the iteration and set $\tau_0 = 0$ and $\tau_1 = 0$ for $k>1$, and $\tau_1 = 1$ for $k=1$ (but we're tossing this away because we know it's differentiable). Therefore, $ \sum_{m=1}^\infty ||\tau_{m+1}^{(k)}(t) - \tau_m^{(k)}(t)||_{a \le t \le b} < \infty\\$ Of which, I've played a little fast and loose, but filling in the blanks would just require too much tex code. EDIT: I'll do it properly as I correct my paper and lower my expectations of the result. *********************** As to the second part of this post--now that we have $C^\infty$ out of the way, we ask if we can continue this iteration and get pentation.  Now, $\text{slog}$ will certainly be $C^{\infty}$ and $\frac{d}{dt}e \uparrow \uparrow t > 0$ so it's a well defined bijection of $\mathbb{R} \to (-2,\infty)$. So, first up to bat is to get another phi function, $ \Phi(t) = \Omega_{j=1}^\infty e^{t-j} e \uparrow \uparrow x \bullet x = e^{t-1} e \uparrow \uparrow (e^{t-2}e \uparrow \uparrow (e^{t-3} e \uparrow \uparrow ...)) $ This will be $C^\infty$ (it'll be a bit trickier to prove because we aren't using analytic functions, but just bear with me). And it satisfies the equation, $ \Phi(t+1) = e^t (e \uparrow \uparrow \Phi(t))\\$ By now, I think you might know where i'm going with this. $ e \uparrow^3 t = \lim_{n\to\infty} \text{slog} \text{slog} \cdots (n\,\text{times})\cdots\text{slog} \Phi(t+\omega_1 + n)\\$ And now I'm going to focus on showing this converges... Wish me luck; after being trampled by this holomorphy I thought I'd stick to where things are nice--no nasty dips to zero and the like... JmsNxn Long Time Fellow Posts: 461 Threads: 85 Joined: Dec 2010 02/10/2021, 02:30 AM (This post was last modified: 02/10/2021, 03:38 AM by JmsNxn.) So I'm having trouble coming up with a general proof to give us $C^\infty$ pentation (or any hyper-operators), but certainly getting a continuous one is not that hard. We can start by taking, $ \frac{d}{dt} \text{slog}(t) = \frac{1}{(\frac{d}{dt} e \uparrow \uparrow t) \bullet \text{slog}(t)} \le \frac{1}{t}\\$ So that, $ |\text{slog}(a) - \text{slog}(b)| \le \frac{1}{\min(a,b)}|a-b|\\$ Then the sequence of convergents, $ \tau_{m+1}(t) = \text{slog}(\Phi(t+1) + \tau_m(t+1)) - \Phi(t)\\$ Satisfy, $ |\tau_{m+1}(t) - \tau_m(t)| \le |\text{slog}(\Phi(t+1) + \tau_m(t+1)) - \text{slog}(\Phi(t+1) + \tau_{m-1}(t+1))|\\ \le\frac{1}{\Phi(t+1)}|\tau_{m}(t+1) - \tau_{m-1}(t+1)|\\ \le \frac{|\tau_1(t+m) - \tau_0(t+m)|}{\prod_{j=1}^m \Phi(t+j)}\\ \le \frac{|t+m|}{\prod_{j=1}^m \Phi(t+j)}\\$ Because $\tau_0 = 0$ and, $ \tau_1(t) = \text{slog}(\Phi(t+1)) - \Phi(t)\\ = \text{slog}(e^t e\uparrow \uparrow \Phi(t)) - \Phi(t)\\ \le \text{slog}(e\uparrow \uparrow \Phi(t) + t) - \Phi(t)\\ \le t\\$ This certainly converges uniformly. So we have a continuous function, $ e \uparrow \uparrow \uparrow t = \Phi(t+\omega) + \tau(t+\omega)\\$ We can continue this for arbitrary order hyper-operators. The trouble comes from proving that this solution is $C^\infty$. I haven't had the AHA moment yet to prove this. The way I proved tetration is $C^\infty$ is not very helpful here because we used properties of the logarithm. Anyway, $ e \uparrow^n t : \mathbb{R}^+\to \mathbb{R}^+\\ e \uparrow^n t \,\,\text{is continuous}\\$ God, I suck at real analysis, getting $C^\infty$ might take a while... sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 02/10/2021, 04:09 AM (This post was last modified: 02/10/2021, 04:27 AM by sheldonison.) (02/08/2021, 12:12 AM)JmsNxn Wrote: As Sheldon has thoroughly convinced me of non-holomorphy of my tetration. I thought I'd provide the proof I have that it is $C^{\infty}$ on the line $(-2,\infty)$. I sat on this proof and didn't develop it much because I was too fixated on the holomorphy part. But, I thought it'd be nice to have a proof of $C^{\infty}$. Now, the idea is to apply Banach's fixed point theorem, but it's a bit more symbol heavy now. We will go by induction on the degree of the derivative. So let's assume that, $\tau^{(k)}(t)|(-2,\infty)\to\mathbb{R}\,\,\text{for}\,\,k edit the : didn't work replaced with | $\sum_{m=1}^\infty||\tau^{(k)}_{m+1}(t)-\tau^{(k)}_{m}||_{a \le t \le b}<\infty\\$ Where, $\tau^{(k)}_0(t)=0$ $\tau^{(k)}_m(t) = \frac{d^k}{dt^k} \log(1+\frac{\tau_{m-1}(t+1)}{\phi(t+1)})$ And $||...||_{a\le~t\le~b}$ is the sup-norm across some interval $[a,b]\subset (-2,\infty)$. As a forewarning, this is going to be very messy... Now to begin we can bound, $||\phi^{(j)}(t)||_{a\le~t\le b}\le~M\,\,\text{for}\,\,j\le k\\$ And that next, $\phi^{(k)}(t+1)+\tau_m^{(k)}(t+1)=\frac{d^k}{dt^k}e^{\phi(t)+\tau_{m+1}(t)}$ $=\sum_{j=0}^k\binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)})$ $=\sum_{j=0}^{k-1}\binom{k}{j}(\frac{d^{k-j}}{dt^{k-j}}\phi(t+1)e^{-t})(\frac{d^j}{dt^j}e^{\tau_{m+1}(t)})+\phi(t+1)e^{-t}(\frac{d^k}{dt^k}e^{\tau_{m+1}(t)})$ Now, $\frac{d^k}{dt^k} e^{\tau_{m+1}(t)}=e^{\tau_{m+1}(t)}(\tau_{m+1}^{(k)}(t)+\sum_{j=0}^{k-1} a_j \tau_{m+1}^{(j)}(t))$ So, we ask you to put on your thinking cap, and excuse me if I write, $\tau_{m}^{(k)}(t+1)=A_m+C_m\tau_{m+1}^{(k)}(t)$ And by the induction hypothesis, $\sum_{j=1}^\infty ||A_{j+1}-A_j||_{a\le t \le b}<\infty$ $\sum_{j=1}^\infty ||C_{j+1}-C_j||_{a\le t \le b}<\infty$ Which is because these terms are made up of finite sums and products of $\tau_m^{(j)}$ and these are said to be summable.  Now the proof is a walk in the park. $\tau_{m}^{(k)}(t)=\frac{\tau_{m-1}^{(k)}(t+1)-A_{m-1}}{C_{m-1}}...=\sum_{j=0}^{m-1}(\prod_{k=0}^{m-1-j}C_{m-1-k}^{-1})A_j$ Where, we've continued the iteration and set $\tau_0 = 0$ and $\tau_1 = 0$ for $k>1$, and $\tau_1 = 1$ for $k=1$ (but we're tossing this away because we know it's differentiable). Therefore, $\sum_{m=1}^\infty||\tau_{m+1}^{(k)}(t)-\tau_m^{(k)}(t)||_{a\le~t\le b}<\infty$ Of which, I've played a little fast and loose, but filling in the blanks would just require too much tex code. EDIT: I'll do it properly as I correct my paper and lower my expectations of the result. *********************** As to the second part of this post--now that we have $C^\infty$ out of the way, we ask if we can continue this iteration and get pentation.  Now, $\text{slog}$ will certainly be $C^{\infty}$ and $\frac{d}{dt}e \uparrow \uparrow t > 0$ so it's a well defined bijection of $\mathbb{R} \to (-2,\infty)$. So, first up to bat is to get another phi function... Hey James, I reposted your post, given the limitation that the forums Tex support has bugs with spaces.  I don't know if it was always this bad  I might write a oneline perl script to remove spaces or substiture a ~ and replace new lines with a new tex /tex pair and remove \\ at the end of a line.  Of course if I do all that, then I could also have perl trivially replace $...$ combos with a tex /tex pair too  and then I'd automatically convert Latex to math.eretandre Tex .... Anyhow, I'm reading your equations now; are you satisified with proving $C^\infty$ for the Tetration case?   I've seen Walker's proof of $C^\infty$, but I've always been curious about how to do it for other functions. - Sheldon JmsNxn Long Time Fellow Posts: 461 Threads: 85 Joined: Dec 2010 02/10/2021, 08:10 AM (This post was last modified: 02/10/2021, 08:18 AM by JmsNxn.) Hey, Sheldon I had written a page long reply to what you just said; but somehow my computer deleted it and it didn't post (I don't know what happened there). I'm a little upset because I had written a fair amount of math, but in the meantime I'll give you the rundown. I am absolutely certain I can prove that this tetration is $C^\infty$ (I went on to relay how this is really no different than the construction of $\phi$ using infinite compositions). However, I can't quite prove it for hyper-operations; and I'm trying to abstract how I can do it for tetration, so that it works for pentation and the sort. So I don't have an essay-like proof for $C^{\infty}$ nature of $\text{tet}_\phi$. But trust me; it's not much more than the construction of $\phi$. Just a whole load of infinite composition stuff. And I'm busy on generalizing the proof so it works for pentation, hexation, etc... Give me a couple weeks. I'm going to pull more all nighters than I should, and I'll have this. Like how every presidential secretary should respond--give me a week and I'll circle back to your question. Regards, James We'll circle back. EDIT: I also had written about how a perl script won't fix your problems. This forum needs to run on MathJax. Which is the javascript version of LaTeX. For god's sakes Henryk. Lol. JmsNxn Long Time Fellow Posts: 461 Threads: 85 Joined: Dec 2010 02/16/2021, 08:40 AM So I posted a proof of $C^\infty$ before I started working today--I noticed a small expansion error, which I hastily fixed--so I deleted the post. I figured I'd do a couple more run throughs of the result before I posted. It was a dumb mistake towards the end where I got ahead of myself, but everything still works; there was one download on the file, so whoever downloaded it I apologize in advance. (Unless the one download was my own download, not sure how it works here.) I am in the process of rewriting my entire paper to focus on $C^\infty$ hyper-operations; whereby it's a long proof by induction. But the initial step is to prove that this tetration is $C^\infty$. Now I can most definitely show this tetration is $C^\infty$; the trouble I'm having is making the proof as general as possible; so that we can create a proof by induction showing $e \uparrow^k t$ is $C^\infty$. The proof I'm posting now is intended to be abstract because I intend to use it as a template for the inductive process. Now, the proof is a little rough around the edges. I haven't fleshed out everything, but honestly it'd only take more words, not more work. Everything definitely works. This isn't much more than what my initial post in this thread was; but it's far better explained. I'm posting this theorem here, without the full paper; essentially to gauge how well explained it is. If anyone has any questions, or any comments or hangups, I'm beyond happy to answer them (and they'll help because they'll teach me how to better write the paper). Despite my errors at assuming this construction would be analytic; I promise no mistakes were made in this circumstance. Somethings may be unclear though, and if they are, please tell me so I can correct myself. Attached Files   just_c_infty-3.pdf (Size: 154.48 KB / Downloads: 61) sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 02/21/2021, 01:38 AM (02/16/2021, 08:40 AM)JmsNxn Wrote: So I posted a proof of $C^\infty$ before I started working today... I am in the process of rewriting my entire paper to focus on $C^\infty$ hyper-operations; whereby it's a long proof by induction. But the initial step is to prove that this tetration is $C^\infty$. Now I can most definitely show this tetration is $C^\infty$; the trouble I'm having is making the proof as general as possible; so that we can create a proof by induction showing $e \uparrow^k t$ is $C^\infty$. Hi James,   I like your paper.  I would suggest generating an infinite sequence of entire $\phi_n$ functions, perhaps defined as follows; this is slightly modified from your approach where this $\phi_2(s)$ = JmsNxn phi(s+1) we could start with  $\phi_1(s)=\exp(s)$ $\phi_2(s)=\exp(\phi_2(s-1)+s);\;$ this $\phi_2(s)$ asymptotically approaches exp(s) as $\Re(s)$ gets arbitrarily negative,  $\phi_n(s)=\phi_{n-1}(\phi_{n}(s-1)+s);\;$ $\phi_n(s)$ also asymptotically approaches exp(s) as $\Re(s)$ gets arbitrarily negative James has proven that $\phi_2(s)$ is entire, and I think each of these phi functions is also entire, and each $\phi_n(s)$ would probably lead to an $e\uparrow^n(s)\;$ function which is also $C^\infty$ only defined at the real axis; details tbd... - Sheldon tommy1729 Ultimate Fellow Posts: 1,438 Threads: 349 Joined: Feb 2009 02/27/2021, 12:08 AM (This post was last modified: 02/27/2021, 12:09 AM by tommy1729.) (02/21/2021, 01:38 AM)sheldonison Wrote: (02/16/2021, 08:40 AM)JmsNxn Wrote: So I posted a proof of $C^\infty$ before I started working today... I am in the process of rewriting my entire paper to focus on $C^\infty$ hyper-operations; whereby it's a long proof by induction. But the initial step is to prove that this tetration is $C^\infty$. Now I can most definitely show this tetration is $C^\infty$; the trouble I'm having is making the proof as general as possible; so that we can create a proof by induction showing $e \uparrow^k t$ is $C^\infty$. Hi James,   I like your paper.  I would suggest generating an infinite sequence of entire $\phi_n$ functions, perhaps defined as follows; this is slightly modified from your approach where this $\phi_2(s)$ = JmsNxn phi(s+1) we could start with  $\phi_1(s)=\exp(s)$ $\phi_2(s)=\exp(\phi_2(s-1)+s);\;$ this $\phi_2(s)$ asymptotically approaches exp(s) as $\Re(s)$ gets arbitrarily negative,  $\phi_n(s)=\phi_{n-1}(\phi_{n}(s-1)+s);\;$ $\phi_n(s)$ also asymptotically approaches exp(s) as $\Re(s)$ gets arbitrarily negative James has proven that $\phi_2(s)$ is entire, and I think each of these phi functions is also entire, and each $\phi_n(s)$ would probably lead to an $e\uparrow^n(s)\;$ function which is also $C^\infty$ only defined at the real axis; details tbd... $\phi_3(s)=\phi_{2}(\phi_{3}(s-1)+s)$ probably grows more like pentation. Notice the similarity to the superfunctions of the previous function in the list. So that would probably fail to get another c^oo solution to tetration but rather a c^oo solution to pentation or higher. Unfortunately probably not analytic either. Generalizing to fractional index n is then probably similar to the classic ' semi-super ' function type questions. So sorry but .. I am not convinced of its usefullness. *** $f_n(s)=f_{n-1}((f_{n}(s-1)+s)/2)$ **could** however converge to f(s) = s + 1 ( the successor function !) for appropriately defined f_1(s). Maybe that could be usefull for some kind of hyperoperator ? However going to negative index n does not seem to give interesting results ( only linear functions ?). Ofcourse many variants of the above can be considered and the question is very vague and open. But it is not certain in what direction we should proceed .. or is it ?? It does not seem simpler than generalizing ackermann , making ackermann analytic etc  Regards tommy1729 MphLee Fellow Posts: 175 Threads: 16 Joined: May 2013 02/27/2021, 11:37 AM I agree with Tommy. That kind of functional equation is structurally analogous to double recursion. I wonder if it is possible to adapt, at least theoretically disregarding radii of convergence, the classic and Nixon's limit formulas to that cases. The problem that I see, as Tommy does, that thing going back tho the successor as the limits goes to infinity. There has to be a deep reason for this. In superfunctions/subfunctions and non-integer-superfunctions (aka non-integer ranks) successor is a fixed point of the "operator". The iteration theory of those kinds of (non-linear)operators on functions spaces has to coincide with the rank theory of the underlying function spaces. As different generating laws define different HOF(amilies) and in some cases a generating law is in some cases a kind of operator on functions, ranks theory reduces to a special kind of iteration theory. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 02/27/2021, 09:57 PM (This post was last modified: 03/02/2021, 10:27 PM by sheldonison.) (02/27/2021, 12:08 AM)tommy1729 Wrote: $\phi_3(s)=\phi_{2}(\phi_{3}(s-1)+s)$ probably grows more like pentation. .... So sorry but .. I am not convinced of its usefullness. ... It does not seem simpler than generalizing ackermann , making ackermann analytic etc  Regards  tommy1729Hey Tommy, I mostly agree with you with one small caveat. Here is my observation. Kneser's tetration function is actually beautifully behaved in the complex plane, or at least as reasonably nicely behaved as any tetration superfunction can be.  The same thing can be said for Jame's $\phi_2$ function, except that it is also entire and 2pi*i periodic. Now, all of the "analytic" higher order functions like pentation generated from the lower fixed point of Kneser's Tetration are actually pretty poorly behaved in the complex plane.  All those singularities at the negative integers for tetration get reflected in pentation as real(z) grows and they eventually show up arbitrarily close to the real axis.  I have experimented with an analytic hexation generated from a complex conjugate pair of fixed points from such an "analytic" pentation, but it too is really poorly behaved in the complex plane.  So imho, the higher order analytic Ackermann functions after tetration don't seem as interesting to me as tetration. Contrast this with the family of $\phi_n$ functions all of which seem to be entire and 2pi*i periodic.  Since they are 2pi*I periodic with limiting behavior as real(z) grows arbitrarily negative of $\exp(z)$, then each of these functions has a corresponding Schroeder function whose inverse $\Psi^{-1}$ is also entire with $\phi_n(z)=\Psi_n^{-1}(\exp(z))$ $\Psi^{-1}_n(x)=x+\sum_{n=2}^{\infty}a_n\cdot x^n;\;\;$ there is a formal entire inverse Schroeder function for each phi_n function. Moreover, each formal Schroeder function's Taylor series can be generated with surprising ease.  For $\Psi^{-1}_2(x)=x+...$ we initialize a function f=x, and we iterate n times to calculate n+1 Taylor series terms as follows, where each iteration gives one additional exact term in the Taylor series of the inverse Schroeder function. $f_1=x; f_n(x)=x\cdot\exp(f_{n-1}(\frac{x}{e}));\;\;\Psi^{-1}_2(x)=\lim_{n\to\infty}f_n(x);\;\;\Psi^{-1}_2(x)=x+\frac{x^2}{e}+...$  Surprisingly, the iteration for $\Psi^{-1}_3(x);\Psi^{-1}_4(x);\Psi^{-1}_5(x);\;$ are only a little bit more complicated but can also easily be coded in a single lines of pari-gp code or mathematical equations as follows: $g_1=x; g_n(x)=\Psi^{-1}_2(x\cdot\exp(g_{n-1}(\frac{x}{e})));\;\;\Psi^{-1}_3(x)=\lim_{n\to\infty}g_n(x);\;\;$this works for Psi_4,5 etc. So I would assert that this family of iterated entire superfunctions is more well behaved than any other family of iterated superfunctions that I am aware of.  It is also more accessible  and easy to calculate than any other family than I am aware.   In that sense, it is also more accessible than Kneser, which is actually pretty difficult to understand and calculate. - Sheldon MphLee Fellow Posts: 175 Threads: 16 Joined: May 2013 03/01/2021, 11:22 PM (This post was last modified: 03/02/2021, 12:57 AM by MphLee.) (02/27/2021, 09:57 PM)sheldonison Wrote: Since they are 2pi*I periodic with limiting behavior as real(z) grows arbitrarily negative of $\exp(z)$, then each of these functions has a corresponding Schroeder function whose inverse $\psi^{-1}$ is also entire with $\phi_n(z)=\psi_n^{-1}(\exp(z))$ $\psi^{-1}_n(x)=x+\sum_{n=2}^{\infty}a_n\cdot x^n;\;\;$ there is a formal entire inverse Schroeder function for each phi_n function. I'm probably missing some key piece of the puzzle (terminology). Are you talking about a kind of inverse Schroeder-like function right? A confortable abuse of name similar to how we can call $\phi_{n+1}$ inverse Abel-like function of $\phi_{n}$? In a strict sense, I don't see how $\psi_{n}$ is a Schroeder function of $\phi_{n}$ or of $\phi_{n-1}$. For this sequence I see this: define $\sigma_{n}:=\psi_n^{-1}$ in your notation. $\sigma_{n+1}(z)=\phi_n(\ln z+\sigma_{n+1}(\frac{z}{e}))$ $\sigma_{n+1}(z)=\sigma_n(z\cdot e^{\sigma_{n+1}(\frac{z}{e})})$ That is "inverse Schroeder-like" in some sense. $\sigma_{n+1}(ez)=\phi_n(1+\ln z+\sigma_{n+1}(z))$ $\sigma_{n+1}(ez)=\sigma_n(ez\cdot e^{\sigma_{n+1}(z)})$ ADD: it is possible to express phi_2 (an offset of JmsNxn's phi) as an infinite composition. $\phi_2(z)=\Omega_{j=0}^\infty \phi_1(z-j+w)\bullet w$ at $w=0$ Is there a similar "closed form" for the other phi_n? Is it too optimistic to declare $\phi_{n+1}(z)=\Omega_{j=0}^\infty \phi_{n-1}(z-j+w)\bullet w$ at $w=0$? MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ « Next Oldest | Next Newest »

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