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 Some slog stuff tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/12/2015, 09:55 PM First i want to say that the equation Slog(ln(x)) = slog(x) - 1 is sometimes better then Slog(exp(x)) = slog(x) + 1. Basically because slog is NOT periodic as the exp suggests. Also because we almost always only consider the fixpoint x = ln(x) rather then all the fixpoints x = exp(x). Let the fixpoints be L and L*. First question Im fascinated by the fact that f(g(x)+1) can be periodic while g is not. Are there elementary nonpolynomial f,g that satisfy this ? Second question How does slog behave around the singularities at L and L* ?? 3rd question Does there exist An entire function E such that, Slog(x) = E( ln(x - L) + ln(x - L*) ) ? Or something similar ? Regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/12/2015, 10:16 PM Notice that the 3rd question seems to relate to the Tommy-kouznetsov expansion. Not sure what that implies. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/12/2015, 10:45 PM (This post was last modified: 05/12/2015, 10:53 PM by tommy1729.) I think |slog(z)| - slog(|z|) is bounded for |z|>|L|+1. I think most agree. Regards Tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 05/12/2015, 11:45 PM (05/12/2015, 09:55 PM)tommy1729 Wrote: First i want to say that the equation Slog(ln(x)) = slog(x) - 1 is sometimes better then Slog(exp(x)) = slog(x) + 1. Basically because slog is NOT periodic as the exp suggests. .... Let the fixpoints be L and L*. ... How does slog behave around the singularities at L and L* ?? What is slog(-5)? What is the slog(z) as z goes to minus infinity? slog(-5)=slog(exp(-5))-1=slog(0.0067)-1 = -1.993817.... as z goes to minus infinity slog(z) goes to -2. Since sexp(z) has a fairly straightforward logarithmic singularity at z=-2, $\text{sexp}(z-2) \approx \log(\text{sexp\,}'(-1)\cdot x)$, then slog(z) as z goes to minus infinity is approximated by the inverse: $\lim_{\Re(z) \to -\infty}\; \text{slog}(z) \approx -2 + \frac{\exp(z)}{\text{sexp\,}'(-1)}$ So if the cutpoints are drawn correctly, then for real(z) |re|. Tommy-sheldon disputes are always intresting. Students of tetration should take notes. Regards Tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 05/14/2015, 12:37 PM (This post was last modified: 05/14/2015, 01:29 PM by sheldonison.) (05/13/2015, 11:49 PM)tommy1729 Wrote: If slog was periodic even in the limit, then 1) sexp(slog(x)+1/2) is periodic too. 2) the cutlines and singularities due to L,L* are COPIED infinitely often because of the periodicity. Resulting in slog having infinitely many singularities. By definition; slog(z) = slog(exp(z))-1. So if exp(z) is a small number, close to zero, than the behavior of slog(x) is governed by he behavior of slog(z) near zero, where slog is a nicely defined analytic function in the neighborhood of z=0; where slog(z)~=-1+0.915946x. So now consider a path for slog(z) for z=-5 to z=-5+pi i. slog(-5+pi i) ~= -2.0062. Notice that the pi i values for slog(z) are real valued again, numbers between -2 and -3. So we can do a Schwarz reflection about the pi i line as well.... If you look in the right places, than exp^{1/2}=sexp(slog(z)+1/2) has 2pi i periodicity as well, and exp^{1/2}(z+pi*I) for real(z)~<-0.3624, than exp^{1/2}(z) is real valued, and tends to sexp(-1.5)~=-0.696 as z goes to minus infinity; There is a singularity for slog(z)=-2.5. And exp^{1/2} has singularities at L,L* Based on these arguments, additional singularities would be expected for slog(z) for z=L+2n pi i, L*+2n pi i - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/14/2015, 01:56 PM (This post was last modified: 05/14/2015, 02:35 PM by tommy1729.) Cake is something to turn Tommy into conjectures. ( i dont drink coffee ) Enough jokes. I conjecture there is a nonzero constant c such that C f(x) = f(exp(x)) Where f is entire and Clog f is real-analytic. Some motivation Take Clog on both sides: We then get a NEW F : F(x) + 1 = log(f(exp(x)) / log C. This is similar to the slog or slog equation. We know the branch cut difference should be approx. the pseudoperiod P. So 2pi i / log C = P i assume. Then C = exp(2 pi i / P). That Logic is a bit handwaving but it was the motivation. Related is the idea - not to be confused with question 3 from the OP - ; There exist nontrivial entire functions E_1,E_2 such that E_1(slog(x)) = E_2(x). If the exp only had one primary fixpoint these questions would all be easy. It inspires to ask these question anyway as if we were unaware of the other fixpoint. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/14/2015, 02:28 PM Indeed Sheldon , if sexp(z) is never equal to L + 2 pi n i then all these values are singularities of slog. Unless the functional equations no longer hold there. Hmm. Reminds me of An old question i asked : Are all solutions to exp(exp(s)) = s singularities for slog ? I wonder where things are periodic , I bet plots would be Nice. I do not recall the fake semi-exp to mimic periodicity but i might be wrong, or that may be normal for fakes. Thank you for your reply. Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/14/2015, 02:43 PM If there are additional singularities then question 3 needs revision. Not sure how , it seems complicated due to not exactly periodic. Suggestions ? Regards Tommy1729 « Next Oldest | Next Newest »

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