semi-group homomorphism and tommy's U-tetration
#1
Recently I showed, proved and assertively stated that the semi-group homomorphism is a natural uniqueness condition and it is satisfied on the positive real line for my 2sinh method.

I want to point out that the fixpoint of 2sinh or similar methods namely ZERO , does not appear as an image of the exp map.
That is a cruxial point and makes the methods work. 
And it remains cruxial as I will show below.

Now maybe Im not very vigilant and formal here , and please correct me if im wrong but it seems consistant to me ... so lets continue :

The post is combination of ideas I have had for a long time : semi-group homomorphism , 2sinh method , base change with base eta to base e and the modified U-tetration aka tommy's U-tetration. It will be clear why they are related at the end. 

recall that U-tetration or decremented tetration is the study of interations of dexp(x) = exp(x) - 1.
This obviously related to base eta.
And also to base e.
What already explains the base change connection idea.

exp(x) - 1 has a PARABOLIC fixpoint at 0.

remember 0 is not the range of the exp map.

so U-tetration uses the fixpoint at 0 to get to the super and abel or half-iterates of exp(x) - 1.

HOWEVER two " issues " : 

1) the noninteger iterations are not analytic at x  = 0 !! ( and we get 2 solutions near 0 )

2) the fixpoints is parabolic so the semi-group homom might not hold.

However the half-iterate or noninteger positive iterates for x > 0 ( or real(z) > 0 ) is analytic near the positive real line.

These things are discussed before.

So what is tommy's U-tetration ?

Well the problem with U-tetration is the point x  = 0.
but 0 is not an image of exp.

So we modify the 2sinh method with a parabolic fixpoint analogue ;

ln^[n] ( dexp ( exp^[n](x) ) )

and notice the point x = 0 has dissapeared ( by the exp^[n] ) !

This clearly relates to normal U-tetration and base eta.
So it might even be equal to base change.

Ok so that is the tommy U-tetration.
but does it have the semi-group homomorphism ?
Is it equal to the 2sinh method ? ( notice those last 2 questions are identical by the uniqueness of semi-group homom !! )

Well lets see

let f(x) = x + d x^2 + ...

and

g(f(x)) = f(g(x)) = x.

then

f^[n](x) = x + d n x^2 + ...

From that we can conclude that near x = 0 :

f^[R](x) = lim_n   g^[n]( f^[n](x) + d R (f^[n](x))^2 + ... ) = lim_n   g^[n]( f^[n](x) + d R (f^[n](x))^2 )

It is now easy to show that the semi-group homom works for positive real iterations ( use the distributive property ! )
EXCEPT WHEN X = 0.

however 

ln( f( exp(x) ) ) 

no longer has that zero.

So the property is carried over from hyperbolic fixpoints to parabolic fixpoints ( apart from the fixpoint itself * here x = 0 * and two petals )
And from the U-tetration to tommy's U-tetration ... to finally by uniqueness :

the 2sinh method.

It is also possible to arrive at this conclusion by using the " generalized fixpoint method " by which I mean methods that combine or generalize the hyperbolic and the parabolic.

Such as the julia equation =

J(f(x)) = f ' (x) J(x)

then arriving at

J( f^[R](x) ) = ( f ' (x) )^R  J(x).

and 

f^[R](x) = J-inv ( f ' (x)^R  J(x) )  

which even already starts to looks like the proof for the hyperbolic.


(  A small assumption I made is that the relevant functions are close enough to their linear or quadratic approximation near their fixpoint , so that the semi-group homom clearly works up to iteration values of 1 and then by induction to all positive real iterations, but taylors error term theorems make sure that in some positive radius this MUST be true !! So that is no real issue )


Ok that is alot for just one post.

In conclusion

tommy's U-tetration = tommy's 2sinh method = semi-group homom solution of tetration.

further 

they might be equal to base change but that is still a conjecture ... 

I suspect they are NOT equal to kneser nor the gaussian method.

I suspect they satisfy the Walker/andydude equations.

Also I only mentioned reals here, Im convinced they are real-analytic ( by induction and boundedness of the iterations )

And I have ideas about their behaviour on the complex plane...

But those are complicated matters. Without proofs for now.

 

regards

tommy1729
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#2
since exp(x) - 1 is not as close to exp(x) as 2sinh(x) I recommend iterating this at most 1 real time when computing tommy U-tetration.

In theory it should always work though.
But it would be unnecessarily complicated and slower. Extra so on the complex plane.

regards

tommy1729
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#3
I'm confused, wasn't this shown to be non-analytic?

I believe this will just construct a smooth solution on \(\mathbb{R}\).

Additionally, the semi-group homomorphism is only unique in the neighborhood of a fixed point. There are plenty of semi-group homomorphisms that are holomorphic, they just can't be holomorphic at the fixed point (At a neighborhood \( p \in \mathcal{N} = \{|z-p| < \delta\}\).

There are countably infinite semi-group homomorphisms. The trouble is, there's only one that is holomorphic about a fixed point. That's the uniqueness statement.
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#4
(08/01/2022, 10:55 PM)tommy1729 Wrote: Such as the julia equation =

J(f(x)) = f ' (x) J(x)

then arriving at

J( f^[R](x) ) = ( f ' (x) )^R  J(x).

and 

f^[R](x) = J-inv ( f ' (x)^R  J(x) )  

which even already starts to looks like the proof for the hyperbolic.

Hi tommy
I'm not feeling like to but have to point out that, these equations aint equivalent, the Julia function denoted as J(z) fits:
$$J(f(z))=J(z)f'(z)$$
and can be arrived at
$$J(f^t(z))=J(f(f^{t-1}(z)))=J(f^{t-1}(z))f'(f^{t-1}(z))=\dot=J(z)\prod_{n=0}^{t-1}{f'(f^n(z))}$$
Regards, Leo Smile
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#5
(08/12/2022, 10:57 AM)Leo.W Wrote:
(08/01/2022, 10:55 PM)tommy1729 Wrote: Such as the julia equation =

J(f(x)) = f ' (x) J(x)

then arriving at

J( f^[R](x) ) = ( f ' (x) )^R  J(x).

and 

f^[R](x) = J-inv ( f ' (x)^R  J(x) )  

which even already starts to looks like the proof for the hyperbolic.

Hi tommy
I'm not feeling like to but have to point out that, these equations aint equivalent, the Julia function denoted as J(z) fits:
$$J(f(z))=J(z)f'(z)$$
and can be arrived at
$$J(f^t(z))=J(f(f^{t-1}(z)))=J(f^{t-1}(z))f'(f^{t-1}(z))=\dot=J(z)\prod_{n=0}^{t-1}{f'(f^n(z))}$$

Right !

thanks.

so switching to julia is overrated here.

But the main idea is still valid.

regards

tommy1729
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#6
(08/12/2022, 08:12 PM)tommy1729 Wrote:
(08/12/2022, 10:57 AM)Leo.W Wrote:
(08/01/2022, 10:55 PM)tommy1729 Wrote: Such as the julia equation =

J(f(x)) = f ' (x) J(x)

then arriving at

J( f^[R](x) ) = ( f ' (x) )^R  J(x).

and 

f^[R](x) = J-inv ( f ' (x)^R  J(x) )  

which even already starts to looks like the proof for the hyperbolic.

Hi tommy
I'm not feeling like to but have to point out that, these equations aint equivalent, the Julia function denoted as J(z) fits:
$$J(f(z))=J(z)f'(z)$$
and can be arrived at
$$J(f^t(z))=J(f(f^{t-1}(z)))=J(f^{t-1}(z))f'(f^{t-1}(z))=\dot=J(z)\prod_{n=0}^{t-1}{f'(f^n(z))}$$

Right !

thanks.

so switching to julia is overrated here.

But the main idea is still valid.

regards

tommy1729

and we see the ghost of continuum sums and products reappear again Smile
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