double functional equation , continuum sum and analytic continuation

[tommy1729]

double functional equation , continuum sum and analytic continuation
**********************************************************


Ok I think I got misunderstood in the past so I will clarify with a very specific example.

First , the semi-group iso iteration of x^2 is x^(2^t) where t is the amount of iterations.
So that is a logical and defendable choice rather than random super functions.
This also relates to the unit circle as you will understand as I explain.

That lies at the heart for the motivations of the following 


Double functional equation

Many functions even analytic functions or even families of analytic of functions satisfy this 

f(x^2) = f(x) - x.

One example is 

let g(x^2) = 2 g(x)

and - x = sum h(n) g(x)^n

such as g(x) = log(x) and h(n) = -1/n! but there are others too.
( hence a family )

[ does a given h(n) define g(x) uniquely ?? maybe subtopic ]

then 

f(x) = sum  (g(x)^n) / [h(n) (2^n - 1)]


proof :

f(x^2) = sum  (g(x^2)^n) / [h(n) (2^n - 1)] = sum 2^n (g(x)^n) / [h(n) (2^n - 1)]

f(x^2) - f(x) =  sum (2^n - 1) (g(x)^n) / [h(n) (2^n - 1)] =    - x

so f(x^2) - f(x) = - x

thus

f(x^2) = f(x) - x. qed


So are all solutions of this type up to a choice of g(x) and h(n) ??

No.

Consider the simple

f(x) =  t(x) = x + x^2 + x^(2^2) + x^(2^3) + ...

this also satisfies 

f(x^2) = f(x) - x.

However !!

t(x^4) = t(x) - x - x^2

which is NOT satisfied by any " g-type " construction.


Say we want an analytic function within |z| < 1 that satisfies 

f(z^2) = f(z) - z

and !!

f(z^4) = f(z) - z - z^2

and do not forget that logs give singularties btw !!

Then 

f(x) =  t(x) = x + x^2 + x^(2^2) + x^(2^3) + ...

is the natural choice.


So that is the double functional equation.

Keep in mind that the equation

f(z^4) = f(z) - z  - z^2

alone is much weaker and does not make the other equation follow from it.



Next step :

combine continuum sum and semi-group isom , while respecting analytic and natural boundary :


the semi-group iso iteration of x^2 is x^(2^t) where t is the amount of iterations.

so intuitively we write :


f(z^(2^t)) = f(z) - z - z^2 - z^4 - ... - z^(2^(t-1))

this means for x  real and > 0 and t real and > 1 ;

f(x^(2^t)) = f(x) - *some* continuum sum x^(2^(t-1))


continuum sum or CS is here with respect to t and keeps x fixed ofcourse.

But keep in mind we can solve

f( x^(2^t) ) = f ( y^(2^v ) ) 

by simply solving

x^(2^t) = y^(2^v)

This might get useful for extending ranges and domains of variables.

Notice I wrote *some* continuum sum, since it is not shown to be equivalent to other CS or well-defined or anything.
On the other hand the equation itself defines it !


Now since we have all of the above and the equation x^(2^t) = y^(2^v) ;

and we know continuum sum is not in essense restricted , we could perhaps extend our range and domains by taking the CS outside of the unit radius.

Indeed if this * some * CS agrees with traditional CS and/or we can consistantly use x^(2^t) = y^(2^v) then we can go beyond the unit radius.

This would mean we could then do a continuation of 


 t(x) = x + x^2 + x^(2^2) + x^(2^3) + ...

beyond the unit circle !!

And if it is analytic there ( it should because CS usually is ! ) we have made a good extension.



Do you see where I am getting at ?


Ofcourse we also want 

 t(x) = x + x^2 + x^(2^2) + x^(2^3) + ...

"="  LIM  t to +oo  CS  x^(2^t)

for all x , with some kind of continuation or summability method ( related to the method described ) to hold.

It should not be a contradition ofcourse , that would be " weird ".



We might need a UNIQUE summabilty method to make things work out nicely but once that is done and works for all z not on the unit circle this is arguably 
the best way to extend.

I believe this is consistant with my summability method despite that required entire functions actually , yet consistant does not mean much here.



Notice we did continuum sums of a sum of iterations , something gottfried and myself and others have been talking about for many years and  gottfried looked at it from a matrix perspective.

Also the continuum sum is far from a new concept.

In that sense it is all just old news.

but it all fits together.

Im not sure how this all fits within the ideas of Caleb and James to extend beyond natural boundaries but notice that a simple plug in does not work here.

How it relates to contour and path integrals I honestly do not know.
I have to think about that.




regards

tommy1729

[tommy1729]

I cant help thinking about this 

https://mathworld.wolfram.com/FalseLogar...eries.html

the famous false log.


regards

tommy1729

[tommy1729]

Another way or at least idea is this

Reflection idea.

Since f(x^2) - f(x) + x = 0 for all x

,it is possible to extend this idea to all z not on the unit circle.

It even makes that equation differentiable.

Lets say

f(x^2) - f(x) = - 2.

then we try to find x.

A solution is garanteed to exist by the fact that it is analytic when not on the unit circle.

And now we compute f(x) = y

Then we simply set

f(x) = y = f(2)

f(x^2) = f(2^2) = f(4) = - 2 + y.



If this is consistant with the previous post is another matter.


also 

Lets say

f(x^2) - f(x) = - 2.

then we find x.

might be an issue because it might not give a unique x , so more care is needed.


deciding which x might use the stuff from the other post though.


It feels a bit like going from the reals to the complex.

just like x^2 = -1 had no solution before the complex numbers,

f(x^2) - f(x) = -2

has no evident solution in the complex.


Just like finding the real numbers as solutions to the cubic equation required complex numbers ,
maybe finding complex solutions to this

f(z^2) - f(z) = -2

requires new numbers.

Just an idea.


Ok in this case the analogue is a bit weird :

afterall f(z^2) - f(z) = - id(z) when considered literal and f(4) = f(2) = oo ( diverges )

so

oo - oo = -2 ??

maybe with limits ...

Maybe a better analogue is this

sqrt(2)^sqrt(2)^... = 2.

x^(1/x) = sqrt(2)

x= 2

however we have on the other side

x = 4 works as well !!



On the other hand solving f(x^2) - f(x) might not be easier than computing f(x)
so the benefit is not proven.

***


I also suggested another reflection idea in the past and recently again.

The reflection formula is simply based on the shape of the boundary :


f( z ) = f ( conjugate (1/z) )

or

f( z ) = f ( 1 / z )

This might give the conditions

f(z^2) - f(z) = - z

f(z^4) = f(z) - z - z^2

f(z^(-1)) = f(z)

a triple functional equation ??

but does this not give contradictions somewhere ???

LETS SEE

if

f(z) = f(1/z)

f( 1/z^2 ) = f(z^2)


so

f(z) = 1/z + 1/z^2 + 1/z^4 + 1/z^8 + ...

for |z| > 1

f(z^2) = f(z) - z
should hold

but now for

f(z^2) = f(z) - 1/z !!!

if |z| > 1.


These lead to conditional functional equations rather than new numbers !

So in this case * if you accept it *

we get

for Re(z) < 1

f(z^2) = f(z) - z

f(z^4) = f(z) - z - z^2

and for Re(z) > 1

f(z^2) = f(z) - 1/z

f(z^4) = f(z) - 1/z - 1/z^2

and for all z not on the unit circle

f(z) = f(1/z)

well assuming all paradoxes have now been removed.

regards

tommy1729

[tommy1729]

...

So in this case * if you accept it *

we get

for Re(z) < 1

f(z^2) = f(z) - z

f(z^4) = f(z) - z - z^2

and for Re(z) > 1

f(z^2) = f(z) - 1/z

f(z^4) = f(z) - 1/z - 1/z^2

and for all z not on the unit circle

f(z) = f(1/z)

well assuming all paradoxes have now been removed.

regards

tommy1729

notice that 

- 1/z - 1/z^2 - 1/ z^3 = -z - z^2 - z^3

has only got solutions on the unit circle !!

However

1/z + 1/z^2 + 1/z^4 = z + z^2 + z^4

has solutions off the unit circle !!

This is both an argument for and against the reflection

f(1/z) = f(z)


notice that x + x^2 + x^3 + ... occured in the zeta topics !



regards

tommy1729

[tommy1729]

...

So in this case * if you accept it *

we get

for Re(z) < 1

f(z^2) = f(z) - z

f(z^4) = f(z) - z - z^2

and for Re(z) > 1

f(z^2) = f(z) - 1/z

f(z^4) = f(z) - 1/z - 1/z^2

and for all z not on the unit circle

f(z) = f(1/z)

well assuming all paradoxes have now been removed.

regards

tommy1729

notice that 

- 1/z - 1/z^2 - 1/ z^3 = -z - z^2 - z^3

has only got solutions on the unit circle !!

However

1/z + 1/z^2 + 1/z^4 = z + z^2 + z^4

has solutions off the unit circle !!

This is both an argument for and against the reflection

f(1/z) = f(z)


notice that x + x^2 + x^3 + ... occured in the zeta topics !



regards

tommy1729

maybe the conjugate : (1/z)* works better

[tommy1729]

ofcourse the bad thing is that these kind of ideas might NOT have number theory properties as once desired or hoped for ??

[tommy1729]

Ok so basically I discussed the very simple case where the natural boundary is the unit circle and our special function is a sum of iterations.

And it only had one fixpoint and that was very attracting.

And perhaps I should mention that for |x| < 1 , x^v for real v always maps to smaller values.
So even x^(2^v(t)) where v(t) is a real valued function associated  with the super by t iterations ( or even not associated ) will remain within the unit circle.

Other functions and their random supers might not behave so nice; staying within the bounds of integer iterates ( when doing real iterates ) is a must but not a given.
( although this might be a local uniqueness idea !? )

The boundary of the filled julia exactly matches the natural boundary so they are certainly related.

Also the filled julia of x^2 has only one component , what justifies the reflection idea.


But this clearly is a special case.

Functions like f1(x) = x + x^2 + x^4  + x^8 + ... or f2(x) = x + x^3 + x^9 + x^27 + ...
Fall into this category.

Notice that f3(x) = f1(x) + f2(x) does not satisfy equations unless perhaps f3(x) = f3(1/x).

SO already that is more complicated.

Many special functions with simple natural boundaries can be created by using sums, compositions, products over double sums , summing over primes 1 mod 3 etc

We have recently seen such functions of type lambert, zeta, theta etc.

 sometimes we can just plug in values and sometimes not.

But without further ado, I want to go more into the type of iteration sum functions in the spirit of this thread.

***

let 

f(z) = (z^2 + z^3)/2

Then define the opa function

opa(z) = z + f(z) + f(f(z)) + f(f(f(z))) + ...

The taylor series gets quite complicated.

The function f(z) seems more contracting than z^2 because of the average of different arguments.

So we expect the filled julia set to be larger than the filled unit circle.

Indeed it is.

Ofcourse many questions arise !!

the julia has only one component which is good.

but f(z) has multiple fixpoints and hence also multiple "opinions " about fractional iterations.

This makes the iterations more complicated.

The riemann mapping from the natural boundary of opa(z) to the unit circle is also not trivial I think.

And the functional equations and reflection formulas are more complex or debatable.

YET f(z) is one of the simplest cubics.
And thus opa(z) is one of the simplest iteration sums.

This topic is getting wildly complicated.

analytic number theory or analytic combinatorics might relate or not.

This makes sense though :

opa( (z^2 + z^3)/2 ) = opa(z) - z

but it is probably not a uniqueness case as we have seen from the first post.

So alot to think about.

But I will add a picture of a truncated opa(z) for inspiration.

White is associated to where the function opa(z) " diverges after the boundary ".

Maybe you can see the fixpoints of f(z) at -2,0,1 ...

regards

tommy1729