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(06/17/2014, 06:16 PM)sheldonison Wrote: . 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.
Another criteria ?
I dont think so ; if " tommy's theorem " is correct
http://math.eretrandre.org/tetrationforu...hp?tid=890
then analytic tet(z) IMPLIES tet ' (z) > 0.
So its not a new condition , its an implication.
Well for bases > eta ... dont know about the other cases.
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tommy1729
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06/17/2014, 10:48 PM
(This post was last modified: 06/17/2014, 10:49 PM by tommy1729.)
And btw that  tommy's theorem  completes the proof of the " generalized TPID 4".
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tommy1729
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06/17/2014, 11:43 PM
(This post was last modified: 06/18/2014, 04:05 AM by sheldonison.)
(06/17/2014, 09:48 PM)mike3 Wrote: (06/17/2014, 06:16 PM)sheldonison Wrote: 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....
Yes, it would be interesting to see a graph. is real valued at the real axis, where the slope=1 at the integers. I generated the initial seed for tet_alt using , which is approximately what this graph looks like.
Contour plot of at , . Notice how it conveniently loops around the origin. This is really important because when z+theta(z) loops around 2, you go around a logarithmic branch of , which is necessary to get the secondary fixed point as real(z) goes to minus infinity.
Here we have a complex plot of at , , where you can clearly see the influence of the singularity is at z=0.51389+0.38009i. Red is real, and green is imaginary.
tommy Wrote:I dont think so ; if " tommy's theorem " is correct
...
then analytic tet(z) IMPLIES tet ' (z) > 0. Does your theorem imply that the secondary fixed point tet_alt(z) function is a numeric fiction? I think it can be made as rigorous as Tetration from the primary fixed point, and like the primary fixed point tet(z) solution, tet_alt(z) is analytic in the upper and lower halves of the complex plane, except for the integers 2,3,4.... And at every integer>2, the 1st and 2nd derivatives are zero, tet_alt'(n)=0 and tet_alt''(n)=0, which contradicts your theorem. At noninteger point for Re(x)>2, tet_alt'(x)>0.
 Sheldon
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(06/17/2014, 11:43 PM)sheldonison Wrote: Does your theorem imply that the secondary fixed point tet_alt(z) function is a numeric fiction? I think it can be made as rigorous as Tetration from the primary fixed point, and like the primary fixed point tet(z) solution, tet_alt(z) is analytic in the upper and lower halves of the complex plane, except for the integers 2,3,4.... And at every integer>2, the 1st and 2nd derivatives are zero, tet_alt'(n)=0 and tet_alt''(n)=0, which contradicts your theorem. At noninteger point for Re(x)>2, tet_alt'(x)>0.
Then how do you explain the violation of the chain law ?
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tommy1729
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06/18/2014, 12:59 PM
(This post was last modified: 06/18/2014, 02:01 PM by sheldonison.)
(06/18/2014, 12:23 PM)tommy1729 Wrote: Then how do you explain the violation of the chain law ?
regards
tommy1729
The flaw in your proof is that k is an integer, but you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". For k as a fraction, the chain rule does not apply. As I stated, the initial approximation used for tet_alt(z) as a seed before iterating the theta(z) mapping from the secondary fixed point is:
f(z) is not tet_alt(z), but merely an approximation using an analytic entire theta mapping from the primary fixed point tetration solution. But most important it shows the flaw in your theorem. This approximation also has f'(n) and f''(n)=0 at all integers>=2, just like tet_alt, and f(z+1)=exp(f(z)), just like tet_alt. And it is analytic at the real axis. So f(z) is also an example of a function that shows the chain law can only be used for integers.
Unlike the tet_alt(z) function, f(z) has an infinite number of singularities in the upper and lower halves of the complex plane, where is a negative integer<=2. Tet_alt is generated from the equivalent of a Riemann mapping from the secondary fixed point, which guarantees it is analytic in the upper and lower halves of the complex plane, see the alternate fixed point post, #19. for details. But the important thing for our discussion of the generalized tpid#4 is the below is an example showing that need not be entire....
(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?
As Mike pointed out, this is the example that shows need not be entire, yet tet(z+theta(z)) is still analytic in the upper/lower halves of the complex plane.
 Sheldon
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06/18/2014, 10:21 PM
(This post was last modified: 06/18/2014, 10:25 PM by tommy1729.)
(06/18/2014, 12:59 PM)sheldonison Wrote: (06/18/2014, 12:23 PM)tommy1729 Wrote: Then how do you explain the violation of the chain law ?
regards
tommy1729
The flaw in your proof is that k is an integer, but you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". For k as a fraction, the chain rule does not apply.
What ? I never saw any exceptions to the chain rule for analytic functions ?
sexp(w1) chain law works. sexp(w+1) chain law works.
In fact the idea that the derivative of sexp relates to a product is the result of the chain law.
So you say :
1) sexp(w+k) is analytic in w,k,w+k and sexp(w) is analytic in w.
2) sexp(w+k) = exp^[k](sexp(w))
3) exp^[k] is also analytic.
4) Yet the derivative of exp^[k](sexp(w)) IS NOT exp^[k] ' (sexp(w)) * sexp ' (w) despite that all functions involved are analytic and the conditions for the chain law are fullfilled ?
Even for the logarithm the chain rule applies even though it has a singularity :
D ln(f(z)) = f ' (z)/f(z) ... = chain law !
The chain law applies to all analytic functions and even many more.
There are many proofs of the chain law and I never read " an exception " ??
Could I be so so wrong in Calc 101 ??
I can imagine going to 0 superexponentially fast but reaching it seems impossible.
Maybe the functional equation is no longer satisfied in the regions or variants of sexp that you use ? That could explain it.
I would like to note that if f ' (h) as h approaches 0 could have as limit 0 BUT that is not the derivative f ' (0).
So in other words , do not confuse the derivative and infinitesimal with the ball from analysis.
As a concrete example : a very mild singularity that is very flat.
There is something known as the CauchyRiemann equations.
Maybe Im making a fool of myself if im terribly wrong but calculus 101 and analysis 101 make a strong case imho.
In math one cannot simply say : Ok you have proved this but my numeric example is counterproof.
The proofs of the chain rule are the most beautiful short proofs.
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tommy1729
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06/18/2014, 10:38 PM
(This post was last modified: 06/18/2014, 10:57 PM by tommy1729.)
The only weird thing I can come up with is this :
sexp(w+k) = exp^[k](sexp(w)) = exp^[w](sexp(k))
So exp[k] ' (sexp(w)) sexp ' (w) = exp[w] ' (sexp(k)) sexp ' (k)
So if sexp ' (w) = 0
0 = exp[w] ' (sexp(k)) sexp ' (k)
AND that is weird ...
Not sure if to relax or get excited now.
This is not satisfactional right now.
0 = exp[w] ' (sexp(k)) sexp ' (k)
Is it an argument for or against ?? Even that is not clear to me.
k = w+1 , its know that sexp ' (w+1) = 0 thus
0 = exp[w] ' (sexp(w+1)) sexp ' (w+1) is valid.
This is equivalent to sexp ' (w) = 0 => sexp ' (2w+1) = 0
Or so it appears.
But it gets stranger :
By repetition :
sexp ' (w) = 0 => sexp ' (2w+1) = 0 => sexp ' ((2w+1)^[2]) = 0 => sexp ' ((2w+1)^[oo]) = 0
The finite limit (2w+1)^[oo] is the fixpoint of 2z + 1. Thus we solve 2z+1 = z.
2z+1 = z => do z on both sides => z + 1 = 0 => z = 1.
So if sexp ' (w) = 0 then sexp ' (1) = 0
But if sexp ' (1) = 0 then sexp ' (0) = 0. And sexp ' (1) = sexp ' (2) = ... = 0
However Im not finished !
(2w + 1)^[1] also applies !
so sexp ' (2) = 0 , take a "new" w := 2w + 1 = 2.
then 2w + 1 = 2 => 2w = 1 => w = 1/2.
And then we get the result that sexp ' ( halfinteger > 0) = 0.
By induction this becomes :
sexp ' ( positive real ) = 0.
So sexp(z) is of the form A z + B.
Seems like another strong case for tommy's theorem and the chain rule.
regards
tommy1729
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06/18/2014, 10:41 PM
(This post was last modified: 06/18/2014, 10:57 PM by sheldonison.)
(06/18/2014, 10:21 PM)tommy1729 Wrote: What ? I never saw any exceptions to the chain rule for analytic functions ?
sexp(w1) chain law works. sexp(w+1) chain law works.
In fact the idea that the derivative of sexp relates to a product is the result of the chain law.
So you say :
1) sexp(w+k) is analytic in w,k,w+k and sexp(w) is analytic in w.
2) sexp(w+k) = exp^[k](sexp(w))
3) exp^[k] is also analytic.
4) Yet the derivative of exp^[k](sexp(w)) IS NOT exp^[k] ' (sexp(w)) * sexp ' (w) despite that all functions involved are analytic and the conditions for the chain law are fullfilled ?
The flaw in your proof is that exp^[k] may not be analytic, if k is not an integer, even if sexp(z) is analytic. If k is an integer, exp^[k] is well defined. But you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". A simple counter example to your proof is f(z), which has f'(n)=0 and f''(n)=0 for all integers>2.
The reason why is because has a cube root branch singularity for n>=0, at . This is relevant since the exp^[k](z) function used implicitly assumes is analytic.
Also, the f(z) function is the "seed" value used to generate tet_alt(z) from the secondary fixed point, and is a rough approximation for tet_alt, and has the same tet'(n)=0 and tet''(n)=0 for integers>2.
 Sheldon
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06/18/2014, 11:15 PM
(This post was last modified: 06/18/2014, 11:28 PM by tommy1729.)
(06/18/2014, 10:41 PM)sheldonison Wrote: The flaw in your proof is that exp^[k] may not be analytic, if k is not an integer, even if sexp(z) is analytic. If k is an integer, exp^[k] is well defined. But you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". A simple counter example to your proof is f(z), which has f'(n)=0 and f''(n)=0 for all integers>2.
The reason why is because has a cube root branch singularity for n>=0, at . This is relevant since the exp^[k](z) function used implicitly assumes is analytic.
Also, the f(z) function is the "seed" value used to generate tet_alt(z) from the secondary fixed point, and is a rough approximation for tet_alt, and has the same tet'(n)=0 and tet''(n)=0 for integers>2.
Aha but then its a partial misunderstanding.
You see , I indeed assumed to be analytic.
I believe that Inverse_Kneser is analytic.
SO I guess the next step is
To do : prove that if an sexp function has no singularities of the real line , then its inverse (slog) is analytic.
But I guess that if Kneser does have derivatives equal to 0 we are back at square 1.
 EDIT 
OK I see sheldon already said that in post 19. Sorry.
 EDIT 
Hmm
Its getting complicated.
Need to think more.
regards
tommy1729
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A few comments
sexp(z+wave(z)) has derivative
sexp ' (z+wave(z)) (1 + wave'(z))
Therefore the techniques from the first few posts relate the solution of TPID 4 to the boundedness conjecture under some condition :
CONDITION => If sexp(z) is bounded then so is sexp ' (z) in the relevant strip.
That condition is a conjecture that needs to be proven !!
A related thing
Let CP be the continuum product.
Is it NECC true that
IF
sexp ' (z) = CP [sexp(z)]
THEN
sexp ' (z + wave(z)) = ( CP [sexp(z)] ) (1 + wave ' (z)) = CP[sexp(z+wave(z))]
And how do these concepts of continuum product , derivative , boundedness etc " really " relate ?
Ok that last is a vague question , but still.
regards
tommy1729
