• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 TPID 4 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 06/15/2014, 07:35 PM (This post was last modified: 06/15/2014, 07:38 PM by sheldonison.) (06/15/2014, 07:09 PM)tommy1729 Wrote: z + theta(z) takes on all values in the strip -1=0. Now locally that gives sexp(ln^[m](x)). Now sexp(complex oo) IS a singularity as can be clearly seen from sexp(-n) , sexp(+ oo i) and sexp(+oo). Hence sexp(ln^[m](0)) remains a singularity. examples : exp(sqrt(x)), ln(x)^2. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 06/16/2014, 09:21 PM More about that : Let f(z) be a holomorphic function in all relevant domains. At complex infinity f(z) is either undefined or it has a singularity. ( now clearly f(z) is not a polynomial ) Let g(z) be a holomorphic function in all relevant domains apart for z=0 and ignoring branches. Let h(z) be a meromorphic function in all relevant domains. Also g ' (z) =/= h(z). To prove under these conditions : f ( g(z) ) is not analytic at z = 0. Proof : If f ( g(z) ) is analytic at z = 0 then so is D f ( g(z) ) = f ' ( g(z) ) g ' (z). ( use the chain rule ) Since f ' ( g(z) ) is either analytic or not , and g ' (z) is not ( g ' (z) =/= h(z) ) then if f ' ( g(z) ) g ' (z) needs to be analytic then there are only a few options : A) f ' ( g(z) ) is analytic : A1) h is a small number , g ' (h) is bounded ==> A1*) if g ' (h) =/= close to 0 => f ' ( g(h) ) is bounded. A1**) if g ' (h) = close to 0 => f ' ( g(z) ) * g'(z) = 0. THUS a product of an analytic function with a singularity = a singularity OR 0. Hence we require f ' ( g(z) ) * g'(z) = 0. But also f ' ( g(h) ) * g ' (h) = 0 , thus one of f ' ( g(z) ) or g'(z) must be identical 0. Since f is not a polynomial this cannot be ! Hence if f ' ( g(z) ) is analytic near z = 0 then f ( g(z) ) is not. This is 1/3 of the proof. A2) g ' (h) is unbounded => f ' ( g(h) ) is close to 0. (notice : If g ' (h) is unbounded then so is g(h).) This requires f ' ( g(0) ) = f ' ( complex oo ) = 0. But f ' ( complex oo ) has a singularty because f ( complex oo) does (from def). So 2) is not possible if f ' ( g(z) ) must be analytic. B) f ' ( g(z) ) is not analytic : The product of 2 nonanalytic functions is nonanalytic ?? Not that simple. We reduced the problem f(g(z)) is not analytic near z = 0 => f ' (g(z)) is nonanalytic near z = 0 -> f ' (g(z)) g'(z) is not analytic near z = 0. Lemma ?? : if f ' (g(z)) is not analytic then f ' ( g(z) ) is well approximated by truncated_Taylor_f( g(z) ). By this dubious lemma we get : a0 g'(z) + a1 g(z) g'(z) + a2 g(z)^2 g'(z) + ... By integration ( if integral G dx is analytic then so is G , if integral G dx is not analytic then neither is G ) t(z) = C0 + c1 g(z) + c2 g(z)^2 + c3 g(z)^3 + ... Now if g(z) has no algebraic singularities then this t(z) is not analytic. A step closer to a proof perhaps. ... A(z) * g ' (z) = analytic. show A(z) =/= f ' ( g(z) ) Assume A(z) = f ' ( g(z) ) Let g(Q(z)) = z. A(Q(z)) = f ' (z). => C) Q(z) is not analytic => C1) A(Q(z)) * g ' (Q(z)) is not analytic. ... C2) A(Q(z)) * g ' (Q(z)) is analytic. ... D) Q(z) is analytic ... Im getting tired ... Maybe this is in the textbooks. I have not even considered the special case of sexp or 1-periodic and the question is if that is the right strategy or the wrong strategy. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 06/16/2014, 10:45 PM Ok so f ' (g(z)) is nonanalytic. f := sexp. g(z) = z + theta(z). Whenever Re(g(z)) > 0 and g(z) is bounded then f ' (g(z)) is bounded as is f(g(z)). And in that case we have another bounded sexp yet that is not an analytic one. But the conjecture was bounded and analytic. Notice that if g(z) is not bounded then f(g(z)) is not bounded and hence the boundedness uniqueness is not broken. so g(z) must be bounded. We continue the quest for this hardcore edition of TPID 4 : 1) f ' ( g(z) ) is nonanalytic. 2) g ' (z) is nonanalytic and 1periodic. 3) g(z) is bounded. to do : proof f(g(z)) is nonanalytic. From 3) => f ' ( g(z) ) is bounded. SO if D f(g(z)) is bounded then so is g ' (z). IF ... D f(g(z)) is bounded/analytic => f(g(z)) is bounded/analytic => g'(z) is bounded. CASE ALPHA : f(g(z)) is not bounded near 0 => f(g(z)) not analytic. Case closed. CASE BETA : f(g(z)) is bounded => g'(z) is bounded. HENCE I) f ' ( g(z) ) is nonanalytic and bounded. II) g ' (z) is nonanalytic and 1periodic and bounded. III) g(z) is nonanalytic and bounded. to do : proof f(g(z)) is nonanalytic. --- sidenote : assume f(g(z)) is not analytic : If ln(f(g(z))) =/= log(0) then f ' (g(z)) g ' (z)/ f(g(z)) is not analytic. hence IF f ' / f (g(z)) is analytic then f(g(z)) is not analytic. now f = sexp f ' = sexp ' thus f ' / f = sexp'(g(z))/ sexp(g(z)) and I could bring out the continuum product again but that would not make it trivial ... --- It seems natural to consider G( f ' (x) ) = f (x) and hoping that G is analytic. Then we get G(sexp ' (g(z))) = sexp(g(z)) QED. But G = f( f ' ^[-1](x)). So it comes down to sexp ' ^[-1](x) being analytic ? Now if sexp ' (T) =/= 0 and Re(T) > -2 then it seems sexp ' ^[-1] (z) is analytic. But what if sexp ' (T) = 0 ? Then D sexp ' ^[-1](T) = oo. Now solve for z : (T* such T that lead to D sexp ' ^[-1](T) = oo) : T* = sexp ' ( g(z) ) then those values z are the only ones where the singularity MIGHT be cancelled If z + theta(z) = g(z) has a singularity there. partial QED. Finally ! regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 06/16/2014, 10:49 PM This brings us to the related equation sexp ' (z) = 0. Which is intresting by itself. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 06/16/2014, 10:57 PM It seems sexp ' (T) = 0 implies sexp ' (T+1) = 0. because sexp(T+1) = exp(sexp(T)) sexp ' (T+1) = exp ' (sexp(T)) * sexp ' (T) = 0 ( chain rule used ) Notice T+1+theta(T+1) = T + 1 + theta(T). Nice. regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/17/2014, 09:30 AM (This post was last modified: 06/17/2014, 09:40 AM by mike3.) (06/15/2014, 07:35 PM)sheldonison Wrote: (06/15/2014, 07:09 PM)tommy1729 Wrote: z + theta(z) takes on all values in the strip -1=0 for real(z)>-2. (11/21/2011, 11:19 PM)in Nov 2011, Sheldon Wrote: ...The algorithm I used to generate a seed value was to start with sexp(z) from the primary fixed point, and use $\text{sexp}(z-\sin(\frac{2\pi z} {2\pi}))$... Then this initial approximation required an additional 42 iterations, generating ... the sexp(z) approximation around z=-1. This gave results accurate to ~32 decimal digits. That's from the alternate fixed point post I made, #19. $\theta(z)=\text{slog}(\text{tet}_{\text{alt}}(z))-z$ is a 1-cyclic analytic function at the real axis, even though such an alternate fixed point tet(z) doesn't have an analytic inverse at integers.... And, as you might expect, theta(z) is not entire since the fixed points are completely different in the complex plane. $\theta(z)$ is analytic if $|\Im(z)|\approx<0.38008$, since tet_alt(0.38008-0.51390)~=0.3181+1.337, which is the fixed point for the Kneser primary tetration function. And at the theta(z) singularity, tet_alt(z)=tet_primary(z+theta(z)) is analytic. Should I post a graph of theta(z) near the singularity? I could also generate and post the Fourier series for $\theta(z)$. This is a fascinating function, z+theta(z)=slog(tet_alt(z)). It circles around the integer values of tet(z) as you pass by at constant Im(z)>0. If z=-2, this means circling around a logarithmic singularity... - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/17/2014, 09:48 PM (06/17/2014, 06:16 PM)sheldonison Wrote: (11/21/2011, 11:19 PM)in Nov 2011, Sheldon Wrote: ...The algorithm I used to generate a seed value was to start with sexp(z) from the primary fixed point, and use $\text{sexp}(z-\sin(\frac{2\pi z} {2\pi}))$... Then this initial approximation required an additional 42 iterations, generating ... the sexp(z) approximation around z=-1. This gave results accurate to ~32 decimal digits. That's from the alternate fixed point post I made, #19. $\theta(z)=\text{slog}(\text{tet}_{\text{alt}}(z))-z$ is a 1-cyclic analytic function at the real axis, even though such an alternate fixed point tet(z) doesn't have an analytic inverse at integers.... And, as you might expect, theta(z) is not entire since the fixed points are completely different in the complex plane. $\theta(z)$ is analytic if $|\Im(z)|\approx<0.38008$, since tet_alt(0.38008-0.51390)~=0.3181+1.337, which is the fixed point for the Kneser primary tetration function. And at the theta(z) singularity, tet_alt(z)=tet_primary(z+theta(z)) is analytic. Should I post a graph of theta(z) near the singularity? I could also generate and post the Fourier series for $\theta(z)$. This is a fascinating function, z+theta(z)=slog(tet_alt(z)). It circles around the integer values of tet(z) as you pass by at constant Im(z)>0. If z=-2, this means circling around a logarithmic singularity... Yes, it would be interesting to see a graph. « Next Oldest | Next Newest »

 Possibly Related Threads… Thread Author Replies Views Last Post RED ALERT : TPID CONJECTURES GONE ??? tommy1729 4 69 08/12/2022, 10:08 PM Last Post: tommy1729 TPID 6 Catullus 1 131 07/04/2022, 12:55 PM Last Post: tommy1729 Sexp redefined ? Exp^[a]( - 00 ). + question ( TPID 19 ??) tommy1729 0 3,668 09/06/2016, 04:23 PM Last Post: tommy1729 Flexible etas and eulers ? TPID 10 tommy1729 0 3,188 08/19/2016, 12:09 PM Last Post: tommy1729 (almost) proof of TPID 13 fivexthethird 1 5,315 05/06/2016, 04:12 PM Last Post: JmsNxn introducing TPID 16 tommy1729 4 10,232 06/18/2014, 11:46 PM Last Post: tommy1729 TPID 8 tommy1729 0 3,672 04/04/2011, 10:45 PM Last Post: tommy1729 Discussion of TPID 6 JJacquelin 3 10,899 10/24/2010, 07:44 AM Last Post: bo198214 Another proof of TPID 6 tommy1729 0 3,852 07/25/2010, 11:51 PM Last Post: tommy1729 proof: Limit of self-super-roots is e^1/e. TPID 6 bo198214 3 11,748 07/10/2010, 09:13 AM Last Post: bo198214

Users browsing this thread: 1 Guest(s)