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Could be tetration if this integral converges - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Could be tetration if this integral converges (/showthread.php?tid=847) |
Could be tetration if this integral converges - JmsNxn - 04/03/2014 Hi everyone. I've been fiddling with this for a while and it looks like we should have some kind of convergence on an expression for tetration I've found. I'm not certain if i'm making mistakes or not but any help would be great. My mathematica is broke. Lets take The mellin inversion theorem. If where Then we get the result that, for We also have the imaginary asymptotics of the Gamma function: and the real asymptotics: We almost have everything, lets add one more condition. We are going to analytically continue tetration in a few steps. Let us say that Then let us take the function for Then we know since this is a modified mellin transform: for So what right? We need an extension of tetration to solve this... Nope. Lets do some more magic with the Gamma function and you'll see that's why it's in the kernel. It is known that: So we see that the Gamma function has poles at the negative integers. The rightmost term is entire as well, so its only contribution asymptotics. Now lets look at the modified mellin transform and talk about the contour Then we know, by Cauchy's theorem: And by the asymptotics of And we are left with: But wait! This relies only on.... the discrete values. Okay okay, does this satisfy recursion though? Let's see, let us define, for It is not difficult to show that by our restrictions on It is not difficult to show that: But using a little math!: We are allowed to take Therefore, Since the modified mellin transform is a modified fourier transform as well... It is one to one and so we just showed it satisfied recursion. There we are. Analytic tetration for This will work on probably all real numbers. Without the fixpoint theorems it has no problem with ![]() RE: Could be tetration if this integral converges - sheldonison - 04/30/2014 (04/03/2014, 02:14 PM)JmsNxn Wrote: Hi everyone. I've been fiddling with this for a while and it looks like we should have some kind of convergence on an expression for tetration I've found. I'm not certain if i'm making mistakes or not but any help would be great. That was my first reaction, for complex z, the inverse of the super-exponential is really badly behaved, and gets arbitrarily large near the real axis. For convergence testing, would it be better to focus on large real values of z, or small real values of z, sexp(0)=1, sexp(1)=e? I assume its the integral that causes problems? Do we know what the values of any of these integrals are? If you have generated them, could you post them? update It looks like this equation is different than the one in the thread entire function close to sexp, which has an additional exponential term in integral and may be entire, whereas I don't think the version above, without the exponential term converges. Here is the version that converges. Presumably, James is aware of the problems... - Sheldon RE: Could be tetration if this integral converges - tommy1729 - 04/30/2014 Aha this is where I have seen the formula before ![]() RE: Could be tetration if this integral converges - JmsNxn - 05/02/2014 (04/30/2014, 11:17 AM)sheldonison Wrote: That was my first reaction, for complex z, the inverse of the super-exponential is really badly behaved, and gets arbitrarily large near the real axis. For convergence testing, would it be better to focus on large real values of z, or small real values of z, sexp(0)=1, sexp(1)=e? I assume its the integral that causes problems? Hey Sheldon. Firstly If Then if If this happens then for we will have a Tetration with the remarkable property of being well behaved as we move the imaginary argument. AND: Where this tetration is well behaved enough so that happens. Once we have that condition 1 is satisfied I can extend the tetration for all Since the gamma function has simple zeroes at the non positive integers we see they cancel out the poles and all that's left is It's very difficult but what that's left to prove is that And then everything that's left is my theorems that I have in the paper I'm writing. My professor has looked over it and everything is right--I'm just working out the final knots of it. It shows a very competent way of iterating superfunctions, however it lacks theorems on convergence of mellin transforms. I.e. the convergence of the mellin transform of I am wary on the convergence of the mellin transform though. I am not sure that it will converge. However. If you can find a tetration such that it satisfies: So we have uniqueness! RE: Could be tetration if this integral converges - mike3 - 05/03/2014 JmsNxn, I am a little suspicious of this method. In particular, I'm not sure the integral converges. I have done some numerical tests and it seems that the sup of In addition to this, I am suspect of the method used to show uniqueness, in particular, the use of the 1-cyclic warping of the tetration. sheldonison had constructed here: http://math.eretrandre.org/tetrationforum/showthread.php?pid=5019 an alternate tetration function which decays to a different, non-principal set of fixed points of the logarithm at I suspect these are related by a 1-cyclic mapping which is such that it is not entire but instead has branch singularities, and so the warping requires a more careful treatment. In particular, you can get the warping by restricting to the real axis, then applying the 1-cyclic map, then analytically continuing again to the plane. The branch nature precludes a simple substitution on the whole plane. Also, are you sure you have that right, that a tetration function ? I believe that any tetration function This can be shown from the chaos of the exponential map RE: Could be tetration if this integral converges - mike3 - 05/03/2014 On the other hand, what I saw here: http://en.wikipedia.org/wiki/Mellin_inversion_theorem suggests that the Mellin inverse transform requires only the right boundedness in vertical strips, not on a whole half-plane. So it should work... ... but then I tried a numerical test to compare the value of with As you can see, This suggests that there is an error in your derivation of RE: Could be tetration if this integral converges - mike3 - 05/03/2014 (04/03/2014, 02:14 PM)JmsNxn Wrote: And by the asymptotics of I believe this is where the problem lies. As I mentioned, the reciprocal tetrational is going to be unbounded. I believe it is also possible with the topological-transitivity thing I mentioned to show that it will also take on almost every complex value infinitely often, on RE: Could be tetration if this integral converges - JmsNxn - 05/03/2014 (05/03/2014, 07:13 AM)mike3 Wrote:(04/03/2014, 02:14 PM)JmsNxn Wrote: And by the asymptotics of YES yes yes yes. Thank you. I don't know very much on how tetration behaves imaginarily. I just meant IF! it satisfied that exponential bound we'd be good. If what typically happens when the function does not have an exponential decay happens then we usually use some asymptotic analysis or an expansion: And on your point about uniqueness. I was being very brief but giving an over view of how we may be able to qualify uniqueness. In technical terms, "it's the only function that in the inverse mellin transform produces an entire function f that is Weyl differintegrable on the right half plane However after seeing what you just posted I have to draw the same conclusion as you. IF we can find some entire function we are back in the game OR IF we can find some entire function we are back in the game. By back in the game I mean I think I can provide an analytic expression for tetration. I'm just finishing the paper I'm working on at the moment and it contains a fair amount of what I'm talking about a lot more rigorously. I'll attach it once I know it's in it's final form. It shows what I am talking about more c learly when I am using fractional calculus on recursion. RE: Could be tetration if this integral converges - mike3 - 05/04/2014 (05/03/2014, 06:12 PM)JmsNxn Wrote: And on your point about uniqueness. I was being very brief but giving an over view of how we may be able to qualify uniqueness. In technical terms, "it's the only function that in the inverse mellin transform produces an entire function f that is Weyl differintegrable on the right half plane So you mean "the inverse Mellin transform of the tetrational", right? Or do you mean of the reciprocal? Also, what do you mean by "Weyl differintegrable"? According to here: http://en.wikipedia.org/wiki/Weyl_differintegral that is something that only functions with a Fourier series, i.e. periodic, can have. Tetration is not periodic (although it does have a pair of "pseudo periods"). (05/03/2014, 06:12 PM)JmsNxn Wrote: However after seeing what you just posted I have to draw the same conclusion as you. Hmm. Given the nasty, chaotic behavior of tetration I've mentioned, it would seem the second kind of function would be more difficult than the first. Actually, I think it might be possible to get a function of the first type (for the series). If we could find an entire function However, it seems you can get a different According to this: http://mathoverflow.net/questions/2944/which-sequences-can-be-extended-to-analytic-functions-e-g-ackermanns-funct there is a method to construct an entire interpolant of any increasing sequence, which would include RE: Could be tetration if this integral converges - mike3 - 05/04/2014 WOW, this is easier than I thought... All we do is take that as an entire interpolant for the tetrational sequence. |