# Tetration Forum

Full Version: twice a superfunction
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i have been thinking about the following often :

how and when is a function a 'superfunction for two functions' and what degrees of freedom do we have ?

for instance

assume the following equation

f(x+1) = A(f(x))
f(x+i) = B(f(x))

and A and B are somewhat related : they commute of course and satisfy some equation ( e.g. A = sin(log(B)) )

i assume there is no freedom for f(z) apart from choosing f(0).

( once again , riemann mapping theorem and double periodic functions convinced me of that )

but its not so clear when we have a solution and when not.

also , could this be a usefull uniqueness criterion ?

its seems we need at least 3 criterions for a solution to exist :

1) A and B commute
2) A and B share the same fixpoints
3) its clear the superfunction of A and B is nonparadoxal at complex oo.

maybe 4 : 4) the superfunction becomes periodic or semi-periodic near complex oo. ( although that might follow from 3 )

regards

tommy1729
(08/11/2010, 04:28 PM)tommy1729 Wrote: [ -> ]i have been thinking about the following often :

how and when is a function a 'superfunction for two functions' and what degrees of freedom do we have ?

for instance

assume the following equation

f(x+1) = A(f(x))
f(x+i) = B(f(x))

and A and B are somewhat related : they commute of course and satisfy some equation ( e.g. A = sin(log(B)) )

i assume there is no freedom for f(z) apart from choosing f(0).

( once again , riemann mapping theorem and double periodic functions convinced me of that )

but its not so clear when we have a solution and when not.

also , could this be a usefull uniqueness criterion ?

its seems we need at least 3 criterions for a solution to exist :

1) A and B commute
2) A and B share the same fixpoints
3) its clear the superfunction of A and B is nonparadoxal at complex oo.

maybe 4 : 4) the superfunction becomes periodic or semi-periodic near complex oo. ( although that might follow from 3 )

regards

tommy1729

$A(x)=f(f^{\circ -1}(x)+1)$
and
$B(x)=f(f^{\circ -1}(x)+i)$

but this doesn't mean that $A^{\circ i}(x)=B(x)$ should hold always?

On the same branch I think it does.

See it like a vector space on a manifold.

The vector +a followed by the vector +b is the same as the vector +b followed by the vector +a.

Hence they commute.

On the complex field : +1 + i = +i + 1.

Or said even in another way : its like a square :
start at (0,0) going up and left = going left and up. ( you end in (1,-1) )

regards

tommy1729
(08/11/2010, 04:28 PM)tommy1729 Wrote: [ -> ]i have been thinking about the following often :

how and when is a function a 'superfunction for two functions' and what degrees of freedom do we have ?

for instance

assume the following equation

f(x+1) = A(f(x))
f(x+i) = B(f(x))
Then the A and B always commute. I think that this is the criterion.

But maybe the the fact that they commute is too general maybe.

example

$A=h^{\circ s}$
$B=h^{\circ t}$

and

$A\circ B=B \circ A$
the functions commute but their superfunction ($H$) should be defined with

$H(x+s) = A(H(x))$
$H(x+t) = B(H(x))$

then we have $h(x)=H(1+H^{-1}(x))$.

But if we don't know such $s$, $t$ and $h$ and we only know that A and B commute how can we know that $f$ exist?

$f(x+1) = A(f(x))$
$f(x+i) = B(f(x))$

By the way I don't think that we can define a new superfunction from two functions A and B when they don't commute... if it is possible is really weird and interesting...

Have you found some example where we dont need that they commute and the superfunction exist?

PS: I replied to your private message.

http://math.eretrandre.org/tetrationforu...hp?tid=844

It seems to follow that from the (probably) fact that sexp is not pseudounivalent then a superfunction cannot be twice a superfunction on a half-plane.

The proof sketch is as follows : Let f be a nonspeudounivalent superfunction.

Lef f(a)= b.

Then there must exist for most such a and b :

f(a2)=f(a3)=f(a)=b

where a2 and a3 are lineair indep. (real lineair indep of course )

then from the 2 functional equations that make up twice the superfunction it follows that f must be double periodic.

( the superfunctions equations are in the directions a-a2 and a-a3 )

Since double periodic functions are not entire , then the superfunction cannot be analytic on a halfplane.

regards

tommy1729
(03/26/2014, 01:23 PM)tommy1729 Wrote: [ -> ]Considering the thread

http://math.eretrandre.org/tetrationforu...hp?tid=844

It seems to follow that from the (probably) fact that sexp is not pseudounivalent then a superfunction cannot be twice a superfunction on a half-plane.

The proof sketch is as follows : Let f be a nonspeudounivalent superfunction.

Lef f(a)= b.

Then there must exist for most such a and b :

f(a2)=f(a3)=f(a)=b

where a2 and a3 are lineair indep. (real lineair indep of course )

then from the 2 functional equations that make up twice the superfunction it follows that f must be double periodic.

( the superfunctions equations are in the directions a-a2 and a-a3 )

Since double periodic functions are not entire , then the superfunction cannot be analytic on a halfplane.

regards

tommy1729
Sorry but you dragged me where I can not swim... I don't know what entire, linear independence, analyticity and pseudounivalent functions are...
I need alot of time to understand this..

following your post and the definition of univalent function whe have that an u. function is injective (and holomorphic from wiki) thus invertible.

Anyways every invertible function it is the superfunction of another functions and then it is twice a superfunction too since it defines in an unique way the iterates of its subfunction.

So back to my poor knowledge:
I don't know if your weaker assumption (pseudounivalent) for a function $f$ is enough to ensure the existences of 2 ("sub")functions that commute (or other) and to prove that f is twice a superfunction of them.