• 1 Vote(s) - 5 Average
• 1
• 2
• 3
• 4
• 5
 using sinh(x) ? tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/07/2010, 12:34 AM (06/02/2010, 10:56 PM)sheldonison Wrote: After that, do you agree that converting from your f(y) to g(y) by iterating natural logarithms would appear to be exactly the same? Here, the superfunction <=> f(y), and TommySexp <=> g(y). The only reason I threw in the "k" constant was so that the resulting sexp_e(z) can be normalized so that TommySexp_e(0)=1. Otherwise, with the k constant=0, I get TommySexp(0)=0.92715... The value I'm using for k is 0.067838366. $\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} ( \operatorname{superfunc}(z+n+k))$ no , i use iterations of exp(x) ! you only use logaritms and the superfunction of 2sinh(x) ... i use logaritms , superfunction of 2sinh(x) and exp(x). numerically it might not be a big difference for large values ( since 2sinh(x) and exp(x) are close for large x ) , but for small x or properties it might be a huge difference ... maybe that is why you get 0.92715... ? i dont know if i need a correcting 'k'. and btw as for the diff-eq , i made a correction to that post (typo mainly but important ) , maybe it will help. regards tommy1729 sheldonison Long Time Fellow Posts: 623 Threads: 22 Joined: Oct 2008 06/08/2010, 01:59 PM (This post was last modified: 06/09/2010, 12:18 AM by sheldonison.) (06/07/2010, 12:34 AM)tommy1729 Wrote: (06/02/2010, 10:56 PM)sheldonison Wrote: After that, do you agree that converting from your f(y) to g(y) by iterating natural logarithms would appear to be exactly the same? ... The value I'm using for k is 0.067838366. $\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} ( \operatorname{superfunc}(z+n+k))$ no , i use iterations of exp(x) ! you only use logaritms and the superfunction of 2sinh(x) ... i use logaritms , superfunction of 2sinh(x) and exp(x). numerically it might not be a big difference for large values ( since 2sinh(x) and exp(x) are close for large x ) , but for small x or properties it might be a huge difference ... maybe that is why you get 0.92715... ? i dont know if i need a correcting 'k'. and btw as for the diff-eq , i made a correction to that post (typo mainly but important ) , maybe it will help. regards tommy1729The correcting factor is k=$\lim_{n \to \infty}\operatorname{superfunc}^{-1}\exp^{[n]}(0)$ If you try your equations to calculate TommySexp(0.5) with x=0, y=0.5, and k=3 Quote:let g(x,y) be the y'th iterate of exp(x) evaluated at x. let f(x,y) be the y'th iterate of 2 sinh(x) evaluated at x. then g(x,y) = lim k -> oo log log log ... (k times ) [f( g(x,k) ,y)] then g(0,0.5) = (for k=3) log log log (3 times) [f( g(0,3), 0.5)] g(0,3) is $\exp^{[3]}(0)$=3814279.104.... Calculating f( g(3,0) ,0.5) is a a somewhat complex operation, and I'd like to understand if you would calculate that expression the same way I would. My approach would be to use the inverse superfunction, then add 0.5, then calculate the superfunction. $x=\exp^{[3]}(0) ; y=0.5$ $\text{f(x,y)=}\operatorname{superfunc}(\operatorname{superfunc}^{-1} (x) + y)$ Update or alternatively, iterate $\text{f(x,y)=}\lim_{n \to \infty}\operatorname{2sinh}^{[n]}((\operatorname{2sinh^{-1 [n times]}(x))*2^y)$ $\text{f( g(0,3) ,0.5)=}\lim_{n \to \infty}\operatorname{2sinh}^{[n]}((\operatorname{2sinh^{-1 [n times]}(g(0,3)))*\sqrt 2)$ The inverse superfunction gives 3.0678383... which is 3 plus the normalization constant, then adding 0.5 ... then calculate the superfunction of 3.5678383.... 8.10306E+77 Finally, take the logarithm three times, and you get (for Tommy's k=3) TommySexp(0.5) =~ f(0,0.5) =~ 0.4987433... Numerically, this is exactly what the limit equations I've posted give. - Sheldon - Sheldon tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/09/2010, 12:18 PM $\operatorname{TommySexp_e}(z,x)= \lim_{n \to \infty } \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/11/2010, 02:23 AM (This post was last modified: 06/11/2010, 02:25 AM by mike3.) I'm really dubious about this method. The problem is f(x) = 2 sinh(x) is quite a wildly different function to exp(x) on the complex plane, which is really the "best" place to study analytic functions. Thus the idea we can get $\exp^z(w)$ from $f^z(w)$ and still have it analytic (don't we want analytic tetration?) seems absurd, because of the extreme difference in the "fractal structure" of their graphs in the complex $z$-plane. Just as how the "cheta" method failed. The real version may be "smooth" but I don't think it will be analytic anywhere at all. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 06/12/2010, 04:36 AM (06/11/2010, 02:23 AM)mike3 Wrote: Thus the idea we can get $\exp^z(w)$ from $f^z(w)$ and still have it analytic (don't we want analytic tetration?) seems absurd, because of the extreme difference in the "fractal structure" of their graphs in the complex $z$-plane. Just as how the "cheta" method failed. The real version may be "smooth" but I don't think it will be analytic anywhere at all. Well though by some dubious reason the base change could also be analytic. Just want to say that it is still not proved whether analytic or not. Though it doesnt look like... mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/12/2010, 11:14 AM (06/12/2010, 04:36 AM)bo198214 Wrote: Well though by some dubious reason the base change could also be analytic. Just want to say that it is still not proved whether analytic or not. Though it doesnt look like... Ah. OK. It'd be a surprise if it were, but you know, sometimes that can happen tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/21/2010, 09:14 PM (06/09/2010, 12:18 PM)tommy1729 Wrote: $\operatorname{TommySexp_e}(z,x)= \lim_{n \to \infty } \ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ regards tommy1729 as for the ROC i assume a plot for increasing n says more than a thousand words. plot n = 1 -> 100. z = 1/2 $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ --- basicly if the n'th derivate has no pole i guess they are all holomorphic for x > 0. tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/22/2010, 11:18 PM i think i can generalize my idea so that it fits for all bases > 1 instead of only bases > sqrt(e). *********** work in progress ... tommy1729 sheldonison Long Time Fellow Posts: 623 Threads: 22 Joined: Oct 2008 06/23/2010, 06:06 PM (This post was last modified: 06/23/2010, 06:07 PM by sheldonison.) (06/21/2010, 09:14 PM)tommy1729 Wrote: ... as for the ROC i assume a plot for increasing n says more than a thousand words. plot n = 1 -> 100. z = 1/2 $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ --- basicly if the n'th derivate has no pole i guess they are all holomorphic for x > 0. The half iterate function is multi-valued, something that I did't quite realize until I tried understanding Tommy's 2sinh limits in the complex plane. Studying the half iterate shows reasons why extending this limit equation to the complex plane seems difficult. $\operatorname{half-iterate}_e(z) = \lim_{n \to \infty} \ln^{[n]}(\operatorname{2sinh}^{[0.5]}(\exp^{[n]}(z)))$ The full iterate of the sexp function is just $e^x$ which is of course single valued. But the the ln(x) function is multiple valued, and so is the inverse sexp. There are infinitely many half iterates, each corresponding to a different horizontal line in the complex plane graph of the sexp function. If we have sexp(z)=x, then sexp(z+0.5) is an half iterate of x. Consider the behavior of $\exp^{[n]}(0.4986)$. At the real axis, the half iterate of 0.4986 is approximately 1, which also means the half iterate of the ln(0.4986) is near the singularity. Each of the following numbers is the approximate exponent of the previous line. -0.3624 + pi*i -0.6960 => half iterate is near singularity 0.4986 => half iterate is ~1.0000 1.6464 4.4817 88.384 2.42*10^38 .... The other half iterates of 0.4986, correspond to other horizontal lines in the sexp function. Consider ln(0.4986) + 2pi*i 1.8440 + i*1.6811 -0.6960 + 2pi*i 0.4986 => half iterate is complex valued, not equal to ~1 1.6464 4.4817 88.384 2.42*10^38 .... Finally, consider z=ln(0.4986)+i*2pi. Again, the sequence ends the same, and has a different values for the half iterate. -0.3624 + 3*pi*i -0.6960 => half iterate is complex valued, not near singularity 0.4986 => half iterate is complex valued, not equal to ~1 1.6464 4.4817 88.384 2.42*10^38 .... So, in generating the half iterate, we need to include the history of all the full iterate exponentiations that occurred prior to the half iterate! This makes it difficult or impossible to extend Tommy's proposed limit equation for the half iterate based on the 2sinh function to the complex plane. Tommy's equations, like the base conversion equations, iterate the exponentiation function with increasing "n", so that the full iterate of the 2sinh function matches the full iterate of the exponentiation function. The function is well defined (infinitely differentiable) for real numbers. Furthermore, the half iterate of 2sinh is well defined on the complex plane, based on the Superfunction earlier in this thread. But, in the examples I gave, even though the full iterate may be increasing to infinity, the half iterate is complex valued, and the imaginary portion probably never goes to zero with increasing "n". The half iterate will be increasing at times, and then decreasing to near zero, and then increasing again, never growing arbitrarily large. Using the published sexp taylor sereies, one could generate some of the complex half iterates of the half iterate of 0, and iterating $\exp^{[n]}(z)$ probably has a complex value that never goes to zero. Thus, even if we could select a different portion of the superfunction to generate a horizontal line from, its not clear that the sequence of terms for increasing "n" would converge to any particular value. Either Tommy's sexp is not analytic, or if it is analytic, it requires a different approach to extend it to the complex plane. - Sheldon tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/23/2010, 09:11 PM (06/23/2010, 06:06 PM)sheldonison Wrote: (06/21/2010, 09:14 PM)tommy1729 Wrote: ... as for the ROC i assume a plot for increasing n says more than a thousand words. plot n = 1 -> 100. z = 1/2 $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ --- basicly if the n'th derivate has no pole i guess they are all holomorphic for x > 0. The half iterate function is multi-valued, something that I did't quite realize until I tried understanding Tommy's 2sinh limits in the complex plane. Studying the half iterate shows reasons why extending this limit equation to the complex plane seems difficult. $\operatorname{half-iterate}_e(z) = \lim_{n \to \infty} \ln^{[n]}(\operatorname{2sinh}^{[0.5]}(\exp^{[n]}(z)))$ The full iterate of the sexp function is just $e^x$ which is of course single valued. But the the ln(x) function is multiple valued, and so is the inverse sexp. There are infinitely many half iterates, each corresponding to a different horizontal line in the complex plane graph of the sexp function. If we have sexp(z)=x, then sexp(z+0.5) is an half iterate of x. Consider the behavior of $\exp^{[n]}(0.4986)$. At the real axis, the half iterate of 0.4986 is approximately 1, which also means the half iterate of the ln(0.4986) is near the singularity. Each of the following numbers is the approximate exponent of the previous line. -0.3624 + pi*i -0.6960 => half iterate is near singularity 0.4986 => half iterate is ~1.0000 1.6464 4.4817 88.384 2.42*10^38 .... The other half iterates of 0.4986, correspond to other horizontal lines in the sexp function. Consider ln(0.4986) + 2pi*i 1.8440 + i*1.6811 -0.6960 + 2pi*i 0.4986 => half iterate is complex valued, not equal to ~1 1.6464 4.4817 88.384 2.42*10^38 .... Finally, consider z=ln(0.4986)+i*2pi. Again, the sequence ends the same, and has a different values for the half iterate. -0.3624 + 3*pi*i -0.6960 => half iterate is complex valued, not near singularity 0.4986 => half iterate is complex valued, not equal to ~1 1.6464 4.4817 88.384 2.42*10^38 .... So, in generating the half iterate, we need to include the history of all the full iterate exponentiations that occurred prior to the half iterate! This makes it difficult or impossible to extend Tommy's proposed limit equation for the half iterate based on the 2sinh function to the complex plane. Tommy's equations, like the base conversion equations, iterate the exponentiation function with increasing "n", so that the full iterate of the 2sinh function matches the full iterate of the exponentiation function. The function is well defined (infinitely differentiable) for real numbers. Furthermore, the half iterate of 2sinh is well defined on the complex plane, based on the Superfunction earlier in this thread. But, in the examples I gave, even though the full iterate may be increasing to infinity, the half iterate is complex valued, and the imaginary portion probably never goes to zero with increasing "n". The half iterate will be increasing at times, and then decreasing to near zero, and then increasing again, never growing arbitrarily large. Using the published sexp taylor sereies, one could generate some of the complex half iterates of the half iterate of 0, and iterating $\exp^{[n]}(z)$ probably has a complex value that never goes to zero. Thus, even if we could select a different portion of the superfunction to generate a horizontal line from, its not clear that the sequence of terms for increasing "n" would converge to any particular value. Either Tommy's sexp is not analytic, or if it is analytic, it requires a different approach to extend it to the complex plane. - Sheldon 1) i dont think the half-iterate is multivalued for real x > 0. 2) the 'history' as you like to call it for large real x is unique for many steps , hence the limiting method. 3) I(exp[n](z)) with I(z) =/= 0 might not deliver a converging imaginary part for increasing n ... and perhaps not even a converging real part ... ( since the imag and real are related at each step of the iteration ) or at least numerical instability or chaos ( which is prob not complex differentiable ) however i am aware of that. it doesnt need to be a problem. first , because I(exp[n](z)) might diverge doesnt imply that I( $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(0)))$ ) needs to diverge since the ln iterations might cancel that effect. but even so , my method is to find a REAL solution to tetration thus therefore i used 'x' instead of a complex z in $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ to go towards the taylor series you simply take the derivates you need of $\ln^{[n]} (\2sinh^{[z]}(exp^{[n]}(x)))$ and create the taylor series with real coefficients. then just plug in complex Z instead of x and your done. however the question is , what is the radius of that taylor series expanded at x = 0 or even elsewhere. if that radius is nonzero , we could probably extend further by using analytic continuation , or mittag-leffler expansion. and that is the way we go to the complex plane. perhaps an intresting note is that if the ROC is large , we could check if that taylor series has the same period as exp(x) ( 2pi i ). if it is indeed large enough and 1) the period is indeed 2pi i then the limit formula might hold for the complex plane and be equal to the taylor of the limit formula for the reals. 2) the period is not 2pi i then the limit formula will NOT hold for the complex plane and NOT be equal to the taylor of the limit formula for the reals. also , if the radius is 0 everywhere , despite unlikely , then if the limit converges for all complex , then the function must have some local or global fractal or semi-fractal properties. regards tommy1729 « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post exp^[3/2](x) > sinh^[1/2](exp(x)) ? tommy1729 7 5,015 10/26/2015, 01:07 AM Last Post: tommy1729 2*sinh(3^h*asinh(x/2)) is the superfunction of (...) ? Gottfried 5 4,760 09/11/2013, 08:32 PM Last Post: Gottfried zeta and sinh tommy1729 0 1,768 05/30/2011, 12:07 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)