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 Searching for an asymptotic to exp[0.5] tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/15/2014, 08:22 PM (This post was last modified: 07/15/2014, 08:59 PM by tommy1729.) Conjecture T0 : Let $f(z) = a_0 + a_1 z + a_2 z^2 + ...$ be a real entire function where $a_i > 0$. If for all $n$ we have $a_{n+1} < a_n/n$ then $f(z)$ is stable. This has probably been proved already. (edit) More specificly : Conjecture T1 : Let $f_1(z) = A_0 + A_1 z + A_2 z^2 + ...$ be a real entire function where $0. Let $f_2(z) = B_0 + B_1 z + B_2 z^2 + ...$ be a real entire function where $B_i >= 0$ such that at least $B_i$ is not equal to $0$ for 2 values of $i$. Let $f_3(z) = C_0 + C_1 z + C_2 z^2 + ...$ be a real entire function where $C_i = [f_2({A_i}^{-1})]^{-1}$. Then if $f_1(z)$ is stable , then so is $f_3(z)$. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/15/2014, 09:43 PM NOTE : For those who want to conjecture the inverse of conjecture T1 ; If f_3(z) is stable , then so is f_1(z). THIS IS FALSE ! For instance if f_2(z) = eps + z^2. Notice 1 + z + z^2/sqrt(2) + z^3/sqrt(6) + z^4/sqrt(24) = 0 has the zero : 0.2026 + 1.5304 i and 1 + z + z^2/sqrt(2) + z^3/sqrt(6) + z^4/sqrt(24) + z^5/sqrt(120) = 0 has the zero : 0.545368 + 1.61261 i Hence disproving the inverse of conjecture T1. Btw taking more terms to avoid polynomials : g(z) = 1 + z + z^2/sqrt(2) + z^3/sqrt(6) + z^4/sqrt(24) + z^5/sqrt(120) + ... only makes the case more Obvious. g(z) has zero's with an unbounded large real part. ( or so it appears , anyway the trends seems growing , unbounded is a " mini conjecture / exercise " for those who are intrested ). regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/15/2014, 09:48 PM (This post was last modified: 07/15/2014, 09:51 PM by tommy1729.) Also to warn others , I want to mention gottfriend " dream of a sequence " again. It " pseudofactors " the exp(z). Why does that matter ? Well because theorems about stable polynomials assume factorizations to hold everywhere. Hence the analogue theorems for (stable) Taylor series might fail strongly ! To refresh your memory : http://www.go.helms-net.de/math/musings/...quence.pdf regards tommy1729 sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 07/17/2014, 05:46 AM (This post was last modified: 08/02/2014, 02:47 PM by sheldonison.) (07/14/2014, 12:20 AM)tommy1729 Wrote: ..... This leads to 3 ideas and a remark. remark : you say " oscillating between positive and negative ". Now I know that if a closed jordan curve path has all arguments n times then we have n zero's within that closed path. So that suggests that all real roots have multiplicity 1 and you are in the possesion of a proof ? Also you have claimed that there are no zero's off the real line ? Does that have a proof ? I think it should be provable for Re(z) > 0.I'm getting close to a proof that there are no zeros off the real line, all zeros multiplicity 1. It seems to me that the details come to me very slowly, since I keep thinking about it, and keep making steady progress. Basically, it comes down to how Kneser's half iterate behaves in the complex plane, along with the definition used for the entire asymptotic function. Interestingly, to understand the negative real axis, you mostly have to understand how Kneser's half iterate behaves for positive $z+\pi i$, due to the equation $\exp^{0.5}(z)=\exp(\exp^{0.5}(\log(z)))$. And the definition I'm using for negative axis comes down to $\text{asymptotic}(-z) \approx \exp(\exp^{0.5}(\log(z)+\pi i))+\exp(\exp^{0.5}(\log(z)- \pi i))$ plus stuff, where stuff can be shown too small to create zeros anywhere else, since halfk(z) can be shown to have a large absolute value at the negative axis, so stuff winds up acting like a small perturbation of the zeros of the approximation. More later. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/24/2014, 12:10 PM Im getting close to a proof of zero-free regions. I assume the ideas will be different from sheldon. It is funny , when I talked about tetration on math forums I got alot of "resistance". It was called " nonsense " or " useless " and people who were into it were insulted. And a few of those strong " opponents " where professors in ... complex analysis. And now here we are using complex analysis in the majority of posts of the tetration forum. And it seems that the use of complex analysis here is just beginning , its use is accelerating. Im just a few lemma's away from a proof , so that is what " close " means. How easy those lemma's are is not yet clear to me though. Im Always fighting against the clock. Its a shame that society pulls people away from doing research without financial benefit. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/24/2014, 10:47 PM The conclusion is that if | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be , then there are no nonreal zero's for f(z). Also the converse : If there are no nonreal zero's for f(z) then | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be. I assume this was already known by sheldon but he did not post the proof. I assumed him to be correct but double checked it. Now its certain. Its intresting , f(z) is now unique by the region where | f(f(z)) - exp(z) | << 1. Hmm well probably. Seems like another mission. It is unique (by that criterion above) if f(z) is unique by the condition : no nonreal zero's and the zero's at the real line are the only possible. Im already considering pro and contra arguments for that mission. The majority of pro arguments are based on the periodicity of exp(z). The contra arguments are based on small perturbations of the zero's. All proofs are based on arg(f(z)). And NO use is made of ln(f(z)) is close to f^[-1](z). Generalization ideas and conjectures are underway. However the initial problem is reduced to an equivalence statement : | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be <=> there are no nonreal zero's for f(z). But a full solution is still not here. Although Im happy with this new insight. Not sure how kneser is suppose to be related as sheldon talks about. We seem to be getting a strong foundation for " fake function theory " regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/24/2014, 10:54 PM At the moment I dream about the zero's of fake exp^[1/4](z) and the zero's of its integral. This could be hard and might then be considered as the analogue of the Riemann Hypothesis for tetration. Or maybe not regards tommy1729 sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 07/25/2014, 02:46 PM (This post was last modified: 08/02/2014, 02:48 PM by sheldonison.) (07/24/2014, 10:47 PM)tommy1729 Wrote: The conclusion is that if | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be , then there are no nonreal zero's for f(z). Also the converse : If there are no nonreal zero's for f(z) then | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be. I assume this was already known by sheldon but he did not post the proof. I assumed him to be correct but double checked it. Now its certain. I like the approach. It might be an elegent simplification of the last step of my proof. The boundary of convergence of |f(f(z))/exp(z)| would be where $f^{-1}(-z)$. Here, the amplitude of $|f(f(z))-\exp(z)|\approx\exp(z)$, but it is smaller everywhere else. Here is a sample calculation, where f(z) is halfway between the 2nd and 3rd zeros. Here, $f(f(z))\approx 2\exp(z)$. If f(z) is one of the zeros, then f(f(z))=0 by definition, and $f(f(z))-\exp(z)=-\exp(z)$. $z \approx -0.183411676 + 6.28413080i$ $f(z) \approx -8.38653002$ $f(f(z)) \approx 1.66174551$ $\exp(z) \approx 0.832425023 + 0.000787053836i$ For my proof, I am focused on the error terms for the assymptotic approximation, and a proof that all of the Taylor series terms (and Laurent series terms) converge. I have a mostly complete version of of the proofs, but its a bit involved; requires some pictures and quite a few equations. Also, there are two different dueling error terms, both of which get arbitrarily small as |z| gets larger. One of the error terms is the 1/x part of the Laurent series for the entire half iterate, which is approximately 0.003898/z-0.00172/z^2 and the other is the number of terms required in the approximation. Then I got side tracked into investigating this Laurent series, which I calculated accurate to 32 decimal digits for terms>=-20. The terms>=0 for the Laurent series are exactly the same as the entire half iterate approximation. All of the terms of the Laurent series converge. But the infinite Laurent series itself probably does not converge anywhere. Anyway, except for all of the pictures required for the proofs, that is where I'm at. Would you be interested in the partial details so far? The last step would be to show that the approximation error term are too small to allow zeros anywhere else other than the real axis for, $\text{asymptotic}(-z) \approx \exp(\exp^{0.5}(\log(z)+\pi i))+\exp(\exp^{0.5}(\log(z)- \pi i))$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/26/2014, 12:21 PM (This post was last modified: 07/26/2014, 12:24 PM by tommy1729.) Im considering that f(f(z)) - exp(z) is small in a region without holes. The idea of holes fascinates me. Many conjectures come to my mind though the quality of them is not yet certain. For instance CONJECTURE HOLE A If g(z) is a fake (entire) half-iterate of exp(z) on the positive real line and all the derivatives are positive then g(g(z)) - exp(z) is small in a region without holes. Problem is also that things are only defined intuitive. For instance the presence of a hole or not depends on your " definition of small ". Only depending on your definition of small is if there is 1 hole or 2 holes close together. ( perhaps name that a pseudohole or fake hole ?) Im not sure if the concept of holes is relevant to the location of zero's of f(z) , some (conjectured) uniqueness conditions , equivalence principles and other stuff discussed sofar , or if it is just a new subject. Maybe I did to much topology lately. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/27/2014, 08:37 AM Another thing. Maybe trivial sorry. But maybe not. In sheldon's picture of f(z) he has colored negative blue and positive red. It seems that between the zero's on the negative real line , we get for every red/blue line on the real line an intersection with another red/blue line coming from the upper resp lower complex plane. This intersection seems to Always be perpendicular !!? Is that an optical illusion or is it true ? Why is that true ? Is it a property of polynomials that carries on because of the hadamard product ? Or is this some complex analysis 101 that I forgot about ? regards tommy1729 « Next Oldest | Next Newest »

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