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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=<Re(z)=<1 apart from possibly one value.
This follows from picard's little theorem and the periodicity of theta(z).
So since in the strip we take on all complex values (apart from 1 possible value) it follows that the range of sexp in that strip is the same range as sexp.
since the range of sexp is unbounded , than so is the range of sexp(strip).
Q.e.d.
The similarity with the unboundedness of the theta in the OP is striking.
Hope that clarifies.
regards
tommy1729
I'm aware that an entire 1cyclic will take on all values in a unit strip, but I'm unsure of how to prove that will take on all values in the strip. I've been struggling with how to show that, forgive my ignorance here... It would also suffice to show that in the strip where is unbounded.
Second, how do we prove that has to be entire? Is there the possibility that theta(z) has singularities, but sexp(z+theta(z)) is still analytic? Those are the two issues I've struggled whenever I think about theta(z) mappings and uniqueness.
 Sheldon
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(06/15/2014, 07:35 PM)sheldonison Wrote: how do we prove that has to be entire? Is there the possibility that theta(z) has singularities, but sexp(z+theta(z)) is still analytic? Those are the two issues I've struggled whenever I think about theta(z) mappings and uniqueness.
Well if sexp is analytic and z+theta(z) is not analytic we get the composition of an analytic function of a function with a singularity.
Now that has to remain a function with a singularity UNLESS MAYBE the singularities of z+theta(z) get cancelled by functional inverse.
Thus z+theta(z) must have slog singularities.
If I recall correctly the slog singularities are similar to the sexp singularities : of type ln^[m](x) , m>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
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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 1periodic and the question is if that is the right strategy or the wrong strategy.
regards
tommy1729
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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
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This brings us to the related equation
sexp ' (z) = 0.
Which is intresting by itself.
regards
tommy1729
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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
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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=<Re(z)=<1 apart from possibly one value.
This follows from picard's little theorem and the periodicity of theta(z).
So since in the strip we take on all complex values (apart from 1 possible value) it follows that the range of sexp in that strip is the same range as sexp.
since the range of sexp is unbounded , than so is the range of sexp(strip).
Q.e.d.
The similarity with the unboundedness of the theta in the OP is striking.
Hope that clarifies.
regards
tommy1729
I'm aware that an entire 1cyclic will take on all values in a unit strip, but I'm unsure of how to prove that will take on all values in the strip. I've been struggling with how to show that, forgive my ignorance here... It would also suffice to show that in the strip where is unbounded.
Second, how do we prove that has to be entire? Is there the possibility that theta(z) has singularities, but sexp(z+theta(z)) is still analytic? Those are the two issues I've struggled whenever I think about theta(z) mappings and uniqueness.
The answer to the last question would appear to be yes. The mappings taking the tetrational to the regular superfunction of exp, and taking it to the "alternative fixed point bipolar solution" would seem to qualify. In particular, in the first case there will be singularities in the mapping whenever hits a singularity of the regular Abel function. Also, between regular superfunctions, the mappings taking one developed at one fixed point to another developed at another would have this property as well.
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(06/17/2014, 09:30 AM)mike3 Wrote: (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=<Re(z)=<1 apart from possibly one value.
This follows from picard's little theorem and the periodicity of theta(z).
So since in the strip we take on all complex values (apart from 1 possible value) it follows that the range of sexp in that strip is the same range as sexp.
since the range of sexp is unbounded , than so is the range of sexp(strip).
Q.e.d.
The similarity with the unboundedness of the theta in the OP is striking.
Hope that clarifies.
regards
tommy1729
I'm aware that an entire 1cyclic will take on all values in a unit strip, but I'm unsure of how to prove that will take on all values in the strip. I've been struggling with how to show that, forgive my ignorance here... It would also suffice to show that in the strip where is unbounded.
Second, how do we prove that has to be entire? Is there the possibility that theta(z) has singularities, but sexp(z+theta(z)) is still analytic? Those are the two issues I've struggled whenever I think about theta(z) mappings and uniqueness.
The answer to the last question would appear to be yes. The mappings taking the tetrational to the regular superfunction of exp, and taking it to the "alternative fixed point bipolar solution" would seem to qualify. In particular, in the first case there will be singularities in the mapping whenever hits a singularity of the regular Abel function. Also, between regular superfunctions, the mappings taking one developed at one fixed point to another developed at another would have this property as well.
Intresting.
Not sure how it relates to what I posted today & yesterday.
regards
tommy1729
Posts: 684
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06/17/2014, 06:16 PM
(This post was last modified: 06/17/2014, 07:58 PM by sheldonison.)
(06/17/2014, 09:30 AM)mike3 Wrote: ....
The answer to the last question would appear to be yes. The mappings taking the tetrational to the regular superfunction of exp, and taking it to the "alternative fixed point bipolar solution" would seem to qualify. In particular, in the first case there will be singularities in the mapping whenever hits a singularity of the regular Abel function. Also, between regular superfunctions, the mappings taking one developed at one fixed point to another developed at another would have this property as well.
That is a wonderful observation Mike. There is an analytic real valued 1cyclic mapping going from Kneser's solution to the alternate fixed point solution. That is not entire, but is analytic if . This seriously complicates the effort to prove the more generalized version of the TPID#4; that any other tet_primary(z+theta(z)) is unbounded, or has singularity in the vertical strip. We need to add at least one more uniqueness criteria: that the Kneser tetration is the only one that has an slog that is analytic at the real axis, or equivalently, that tet'(z)>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 ... 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. is a 1cyclic 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. is analytic if , since tet_alt(0.380080.51390)~=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 . 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
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(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 ... 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. is a 1cyclic 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. is analytic if , since tet_alt(0.380080.51390)~=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 . 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.
