Brute force tetration A_k(s) is analytic ! tommy1729 Ultimate Fellow Posts: 1,852 Threads: 399 Joined: Feb 2009 03/18/2021, 01:08 AM (This post was last modified: 03/18/2021, 01:11 AM by tommy1729.) I propose an analytic tetration here. One very similar to other ideas from this year. Imo very natural and just requiring alot of simple computions. I dare say brute force. Consider for real k > 5 and Re(s) > 0 : $A_k(s)=exp(A_k(s-1) - exp(-ks))$ We can solve this by infinite composition : $A_k(s)=exp( - exp(-ks) + exp( - exp(-k(s-1)) + ...)$ ( yes to be formal we need a " z parameter " too, but that is a detail ) Now consider the functions $A_5(s),A_6(s),A_7(s),...$ The limit k to +oo gives us A_oo(s) = A(s). Now A(s) satisfies for Re(s) > 0 : A(0) = exp( -exp(-oo) + exp(-oo + 0)) = exp(0 + exp(-oo)) = exp(0) = 1. and  A(s+1) = exp(A(s) + 0) = exp(A(s)) So assuming the limit exists this is analytic tetration on the halfplane Re(s) > 1. By analytic continuation ( taking log's ) this is analytic tetration !! In fact I think it has no singularities for Re(s) > 1 making it perhaps satisfy a uniqueness condition ?? The limit is designed to ( when it exists ) grow slowly on the complex plane (in the imaginary direction ) for Re(s) > 1 therefore probably remaining analytic. We could in principle even approximate this in say excel. regards Tom Marcel Raes tommy1729 tommy1729 Ultimate Fellow Posts: 1,852 Threads: 399 Joined: Feb 2009 03/18/2021, 12:55 PM Unfortunately I think it is not analytic afterall. It seems the period of A_k(s) is 2 pi i / k. Hence in the limit we have the period " 0 i " which implies for Re(s) > 1  A(s) = A(Re(s)) = Re(A(s)) and that is clearly not analytic. This phenomenon follows from the periodicity of exp( - k s ) : 2 pi i / k.  hmm Maybe if we use another auxiliary function other than exp(- k s) that behaves similar on the real line ? Is that even possible ? Or maybe we should try a much slower changing function. I considered the uncompleted gamma function too ( although that is faster). Im even starting to wonder if methods based on composition only excluding riemann mappings has hope for analyticity at all. So im looking for another auxiliary function f(k,s). I was considering rational functions and giving up the entire constraint because I dont want too many copies ( like periods or so ). Im doubting. regards tommy1729 JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/18/2021, 09:11 PM (This post was last modified: 03/18/2021, 10:18 PM by JmsNxn.) I'm just as discouraged when thinking of using infinite compositions to construct an analytic tetration, Tommy. I think I've settled on the fact that infinite compositions are incredibly helpful for solving equations of the form $y(s+1) = F(s,y(s))$ when $F(s,z)$ has good decay as $\Re(s) \to -\infty$ but are rather impotent at handling problems of the form $y(s+1) = F(y(s))$ (especially when we start requiring specific criterion). That being said, it's still really nice to handle first order difference equations rather easily. I've moved on into different manners of approaching Tetration. I'm back on the fractional calculus approach, and wondering if there's an equivalent way of constructing Kneser's solution without Riemann's mapping theorem. Mostly because I despised Riemann's mapping theorem when I learned it and that hatred still hasn't gone away, lol . I recently had one I thought might work, but I've written it off. Let $\eta^{\circ s}(z)$ be the bounded iteration of $\eta^z$ with base $\eta = e^{1/e}$, and think of, $ F(s) = \Omega_{j=1}^\infty \exp(\eta^{\circ s-j}(z)) \bullet z\\$ Then, $ F(s+1) = \exp (\eta^{\circ s}(F(s)))\\$ And I was looking at similar equations, and trying to get rid of the pesky $\eta$ which appears, with no such luck. :/ Kneser's solution truly is magical, lol. We could try this way with using $\eta^{\circ \frac{s-j}{k}}(z)$ and taking $k\to\infty$ (there's no period in sight), and formally the limit will be a tetration function, but my money is on non convergence . especially because at negative infinity this will tend to $0$, we'd have to cram a fixed point somewhere in there, and then we'd probably lose real to real. I can make a tetration using infinite compositions, but it amounts to nothing more special than the inverse schroder tetration and will definitely not be real valued. I think you are right in your conclusion that the Riemann mapping theorem is integral, and any work around, will have the riemann mapping theorem as a close neighbor. Regards, James If you want something close to tetration which has hope of converging but probably isn't real valued look at, $ \Lambda(s) = \Omega_{j=1}^\infty e^{\phi(s-j) + z} - \phi(s-j +1)\bullet z|_{z=L}\\$ So that, $\phi(s+1) + \Lambda(s+1) = e^{\phi(s) + \Lambda(s)}$ which is nearly there (but it won't be real-valued, and I think convergence may be spotty to derive (we'd need a strict kind of summability criterion)). Of course, we are assuming $L$ is an exponential fixed point. And $z$ doesn't have to exactly equal $L$, but the sequence of nested compositions $\Omega_{j=1}^n...\bullet z |_{z=z_n}$ must be at least taken in a manner such that $z_n \to L$ in a summable manner. Using a fixed point truly is the missing ingredient to get holomorphic functions close to tetration; and riemann's mapping theorem is needed to get real to real. This equation came from my analysis of $\tau$, upon which I convinced myself of this convergence which convinced me that $\tau$ is holomorphic; the error being I needed $\tau$ to be holomorphic SOMEWHERE before I did this; and as it turns out, $\tau$ is holomorphic nowhere. But the equation should still converge in a reasonable manner. I can write a proof if you want. tommy1729 Ultimate Fellow Posts: 1,852 Threads: 399 Joined: Feb 2009 03/19/2021, 01:04 PM (This post was last modified: 03/19/2021, 01:16 PM by tommy1729.) Ok I put my money on the uncompleted gamma function "G" : $A_k(s)=exp(A_k(s-1) - G(-ks))$ We can solve this by infinite composition : $A_k(s)=exp( - G(-ks) + exp( - G(-k(s-1)) + ...)$ ( yes to be formal we need a " z parameter " too, but that is a detail ) Now consider the functions $A_5(s),A_6(s),A_7(s),...$ The limit k to +oo gives us A_oo(s) = A(s). Now A(s) satisfies for Re(s) > 0 : $A(0)=LIM.exp(-G(-oo)+exp(-oo+0))=LIM.exp(0+exp(-oo))=exp(0)=1.$ and  $A(s+1)=exp(A(s)+0)=exp(A(s))$ I see no clear reason why this should fail. So this deserves some attention. In particular I assume G in the imaginary direction. *** Btw I want to mention that the functional equation f(s+1) = exp(f(s) + s) does a poor job at approximating tetration when you consider the derivative ; we do not get f ' (x) = f(x) ln(f(x)) ln(ln(f(x))) * ... but f ' (x) = f(x) * (1 + ln(f(x)) * ( 1 + ln ln ( f(x) )... The uncompleted gamma function method described above however does ( in the limit ) give f ' (x) = f(x) ln(f(x)) ln(ln(f(x))) * ... and does not suffer the periodicity problem of the exp method solution. Hence I am betting on this uncompleted gamma trick. regards tommy1729 Tom Marcel Raes tommy1729 Ultimate Fellow Posts: 1,852 Threads: 399 Joined: Feb 2009 03/19/2021, 01:22 PM (This post was last modified: 03/19/2021, 01:22 PM by tommy1729.) btw I named G after Gottfriend not so much gamma :p Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 03/19/2021, 05:22 PM (03/19/2021, 01:22 PM)tommy1729 Wrote: btw I named G after Gottfriend not so much gamma :p This is sweet :-)) Gottfried Gottfried Helms, Kassel JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/19/2021, 09:12 PM (This post was last modified: 03/19/2021, 09:13 PM by JmsNxn.) So we're taking, $ G(-ks) = \int_1^\infty e^{-x}x^{-1-ks}\,dx\\$ which has decay to zero as $\Re(s) \to \infty$. However, you then take, $ \Omega_{j=1}^\infty e^{z+G(-k(s-j))}\bullet z\\$ But, $G(-k(s-j)) = G(jk-ks) \to \infty$ as $j \to \infty$. As in, $ \sum_{j=1}^\infty e^{z+G(-k(s-j))} = \infty\\$ So I'm not sure how you are planning to derive convergence of this infinite composition. tommy1729 Ultimate Fellow Posts: 1,852 Threads: 399 Joined: Feb 2009 03/20/2021, 12:34 AM Ok so apart from the comment of James, I think there is another reason why the gamma trick does not work. The thing is on the real line -G(-s) gives large negative values for large negative s as desired.  However - G(- k s ) for complex s with large negative part and increasing k gives arg that are neither positive or negative.  Or they give positive values when we want negative ones. If it even converges. This makes even the convergeance of A_k(s) problematic for almost every nonreal s and every k. Let alone a limiting k to oo. And this implies that many functions fail. But maybe the problem is not so much with the s but with the way we use k. using k s might be a bad idea because that kinda makes s and k similar in contribution. Im thinking about new ways to use k. regards tommy1729 JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/22/2021, 08:58 PM I wonder if taking the limit of $\phi$ in a different manner would work. Let's define, $ \phi_k(s) = \Omega_{j=1}^\infty e^{\frac{s-j}{k} + z}\bullet z\\$ Which satisfies $\phi_k(s+1) = e^{\frac{s}{k} + \phi_k(s)}$. Then we define tetration through the limit, $ F_k(s) = \log^{\circ k}(\phi_k(s+k))\\$ This will avoid the periodic problem, this will still keep a somewhat nice function $\tau_k(s)$, but we're probably going to run into problems somewhere along the way. At least, my gut says so. JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/22/2021, 11:39 PM (This post was last modified: 03/22/2021, 11:50 PM by JmsNxn.) Another way I was thinking, that still has the periodic problem, but is more natural because asymptotically it looks like tetration. $ \Phi(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{j-s} + 1}\bullet z\\$ Where, $ \Phi(s+1) = \frac{e^{\Phi(s)}}{e^{-s}+ 1}\\$ And asymptotically, $ \Phi(s+1)/e^{\Phi(s)} \to 1\\$ So that logarithms may behave better in the complex plane here. I wouldn't worry too much about periodicity because we can always think of the logs collecting $2 \pi i$'s. This way might actually work with holomorphy. Can't be sure though. It definitely constructs $\mathcal{C}^\infty$ tetration on $(-2,\infty)$. But now, we have to deal with poles when $\Im(s) = \pi i$. With this we should expect, $ \log \Phi(s+1) = \Phi(s) + \mathcal{O}(e^{-s})\\$ which looks ripe for convergence in the complex plane. « Next Oldest | Next Newest »

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