The semi-group iso problem and bounded derivatives

[tommy1729]

I have been talking alot lately about the semi-group iso.

However the problem is for tetration that there is no real solution.

Consider the fixpoint expansion at the primary fixpoint or equivalantly the kneser method BEFORE the riemann mapping.

This solution HAS THE CORRECT 1-periodic mapping satisfying the semi-group iso property.

And by uniqueness of the 1-periodic mapping this means it is the ONLY solution that has the semi-group iso property.

But it does not map any large real interval to the reals.
It maps the reals to complex numbers almost everywhere.

So that is not the type of tetration we want.

Im not sure how it works with other fixpoints but those also do not map to the reals.

This also implies that any method for REAL tetration can not have the semi-group iso property.

NOT EVEN the carleman matrices , at least not converging to anything meaningful ( like a taylor with nonzero radius ).

I was away for a while because I did not know how to handle it.

It is a big bummer and makes one wonder what it means to do continu iterations !

This problem does not only occur for tetration but for a very large class of problematic functions !!

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IT seems the semi-group iso only works in 2 non-trivial cases ;

1)
 
f(z) has a fixpoint.

or

2)

f(z) is of the form g( k + inverse_g(z) ).

For instance f(z) = ln( k + exp(z) ) for k a real above or equal to 1.

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Another thing is that this has also SEVERE implications to the whole ackermann .. hyperoperator .. pentation .. etc type topics.

So stuff like 3 <s> ( 3 <s+1> y ) = 3 <s+1> (y+1)  and similar equation are also hard to consistantly define.

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So i considered going to older ideas again.

somewhat against my will.

sigh.

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So it was back to square 1 for formal uniqueness conditions that made sense and not completely arbitrary but rather logical.

So i started to focus again on the derivatives of the semi-exp and the extensions.

I will thus probably delve into old ideas but I feel a bit forced.

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lets focus on the reals now.

LET f(x) be the semi-exp.

We want for all real x :

f '(x) > f '' (x) > f ''' (x) > 0.


This might not be completely a uniqueness critertion but puts strong conditions and all solutions visually look similar IF they agree on specific values
 ( f ( -oo ) in particular )

Another intresting idea is the idea of induction : if this holds on a given interval it holds everywhere.

For instance something such as :  if f '(x) > f '' (x) > 0 for x < 0 then actually it holds for all real x.

proofs by inductions and infinite descent for partitions of the reals into intervals seems logical.

playing around with that seems fun.

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perhaps more later

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sorry for my absense , i had to review alot of things.


regards

tommy1729

[tommy1729]

ok some important nuances !

For starters 

The riemann mapping in general does not preserve the semi-group iso.

So very likely the Kneser method does not satisfy the semi-group iso.

the 2sinh method uses a fixpoints and suggests having the semi-group iso but it might not have a nonzero radius and be nowhere analytic.

This brings us to the essence :

the 1-periodic theta function is unique ... but maybe there are different paths and multiple paths have each a semi-group iso ??

This would imply to having uniqueness ( and the semi-group iso ) by 2 things : the path and the function we iterate.

But the real line or the real line intervals only have one path.

So there cannot be 2 solutions from R to a subset of R.

Hence the 2sinh must be unique in having the semi-group iso IF IT EVEN HAS IT.
And even if it has it might not be analytic.

And carleman matrices dont make me optimistic.


So the focus lies on the paths taken.

this might simply explain what happens when we use different fixpoints for kneser like ideas before the riemann mapping ( as to using branch logic as tool for explaining things ) 

So the idea is to change a path A to a path B while preserving the semi-group iso.

lets say a function t does that

t(A) = B

Do we also have a path C with the iso that is then simply :

t(t(A)) = t(B) = C 

??

Thereby perhaps linking functional iterations ( of t ) to changing paths ??

For a given function f(z) how does one find t or one of its iterates ??

This might seem like a riemann mapping idea but it is slightly different and not necc analytic !!! 

( not analytic for instance mapping the path from kneser ( pre riemann mapping ) to 2sinh )

IT seems the path mapping is more like a conjugate thing ;


g( c^k g_F(inverse(z))  ) 

where F comes from the fixpoint expansion with derivative c at fixpoint.

By using the conjugate thing , we clearly see the preserving of the semi-group iso.

( yes this resembles 2sinh method alot , but those where similar paths and a real fixpoint with a real value c , here fix and c are non-real )

But again - for the same reasons - g might not even be analytic !!

( we are working with schroder equation satisfying context here , as opposed to abel ... i mix em up without mentioning sometimes sorry )


So how we know that there are multiple paths with semi-group iso ?

Im not sure and need more understanding but 

the conjugate fixpoint of exp gives the conjugate path !!

so that are at least 2 paths !

...

what makes me wonder if they use the same path functions to the real line ...

( similar to the riemann mapping what does so )



Short story short : there is still hope for R to R semi-group iso tetration.

we need more research and understanding though.

2sinh method is a candidate.

But it is unclear if we have hope for ALSO analytic.

   

***

As for the ideas with derivatives ; the most general case is the uniqueness criterion 

semi-log is analytic and a bernstein function with semi-exp(- oo ) = given constant.

although that given constant might not be free to choose !

And I only assume existance.

 ***

end nuances ( short version )


regards

tommy1729

[tommy1729]

follow-up :

https://math.eretrandre.org/tetrationfor...p?tid=1678

regards

tommy1729

[tommy1729]

A nice exercise is perhaps this

Let f(f(x)) = exp(x)

for all x :

f ' (x) > 0

f ' ' (x) > 0

what is the lowest possible value of A ;

A = inf f(x)

or equivalent 

A = f( - oo )

Does A have a closed form ?

That is a nice idea.

***

Following the above

consider the family of functions f_r(x) such that :

f_r^[r](x) = exp(x) 

with r between 0 and 1.

And consider 

for all x :

f_r ' (x) > 0

f_r ' ' (x) > 0

what are the lowest possible values of A( r ) ;

A( r ) = inf f_r(x)

or equivalent 

A( r ) = f_r( - oo )

It is assumed that the lowest possible value A®  for a given r is independant of the lowest possible values for other values of a different r.


This implies that there is a unique lowest function A of the variable r without contradictions : A( r ).

A( r ) = f_r(- oo) = sexp( slog(-oo) + r)

f_r ' (x) = d/dx sexp( slog(x) + r)

f_r ' ' (x) = d^2/d^2x sexp( slog(x) + r)

 
We then have a uniqueness condition ; A® uniquely defines sexp resp. slog by the above ( and the trivial sexp(x+1) = exp(sexp(x)) )

Food for thought.



regards

tommy1729