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fractional iterate of x ?
Since it is a bit off topic , and perhaps not of the same 'level' as most posts i present it here in the ' community ' section.

Fractional iterations and calculus ( I include recursion theory ) are things that we all want to unite , i believe , and somehow that is not just a desire but unavoidable ; how else will we carry out computations.

When i try to think about that - without thinking about tetration - i often end up with the same problem.

Is f(x) a fractional ( or real if you like ) iterate of id(x) ?

( id(x) = x )

Although it is usually not hard to decide if such is the case , e.g. f(x) = 1-x clearly is a fractional iterate of id(x) , the general case seems - to me at least - not so trivial.

For starters , id(x) commutes with every function , so the problem is harder than to determine if 2 functions commute under composition.

One could (naively(?)) use matrix methods when dealing with taylor series , arriving at the matrix equation :

A is the carleman matrix of f(x) and B is the carleman matrix of id(x).

r is a real number to be found.

A ^ r = B

( notice A = B ^ r fails , because e.g. half-iterates of id(x) include id(x) )

And then try to solve it by setting r = log(B)/log(A).

But does that really work ? How do we know r is real and not a nonreal matrix ?

Keep in mind that the matrices are of infinite size !!

What if the logaritms give problems ? Suppose log(B) and log(A) have norm or eigenvalues 0 or 00 , use l'hospital ?? ... and if the matrices are not invertible ?

Or does non-invertable mean that f(x) is not a fractional iterate of x ?

I've had other ideas as well apart from the matrix approach.

A specific series expansion for example.

I have to give credit for inspiration to Euler and God.

Especially Euler's Beta function. (Perhaps you see why)

It goes without saying that the issue is mainly about irrational iterates of x , since for rational , a certain test might be poor but will always terminate.

r is also trivially the period of the superfunction of f(x).

But finding that period without having or needing the superfunction is challaging imho at least.

Maybe this has been settled long ago , or maybe i regret pressing ' post thread ' in a minute , but at the moment i am in doubt ...



dedicated to Euler , God and a friend.
I think a keyword here is "babbage equation".
Its thoroughly described for example in Kuczma's book "Iterative functional equations".
Though I mainly forgot about it, its about functional roots of the identity function.
(04/23/2010, 01:06 PM)bo198214 Wrote: I think a keyword here is "babbage equation".
Its thoroughly described for example in Kuczma's book "Iterative functional equations".
Though I mainly forgot about it, its about functional roots of the identity function.


its intresting , but in essence only deals with half-iterates.

i couldnt find anything free on the internet of any value.

im not an expert in "babbage math" , but i believe he used the matrices too , or am i wrong ?

i would love to learn more about this mysterious babbage equation...


(04/24/2010, 08:27 AM)Ansus Wrote: Any function with symmetric against y=x line plot is half-iterate of x.

Yes, these functions are called involutions. And they are roughly the only solutions of the Babbage equation
(*) .

Theorem 11.7.1 in "Iterative functional equations":
Let a self-mapping of a real interval be a continuous solution of equation (*). Then either itself is the identity mapping or has to be even and is a strictly decreasing involution.

I like to compare this with real numbers:
"Let a real number x be a solution of . Then either is 1 itself or has to be even and is a negative number."
i couldnt find much usefullness in relation to babbage's equation ...

but perhaps you would like to know what i wrote elsewhere :

assuming it is correct : quote :

i wrote :

> potentially unsolved : period of superfunction ?
> this thread reminds me of tetration and related ...
> i have explained how to compute the period of a
> taylor series.
> but how does one compute the period of the
> superfunction of a taylor series , without knowing
> the taylor series of the superfunction ???
> for those unfamiliar :
> superfunction[f(x)] = F(x) <=> F(x+1) = f(F(x))
> regards
> tommy1729

basicly if F(x) has a real period > 1 then :

T( (-1)^(2/P) inverseT(x) ) = f(x)

has a solution for T and a period P.

the smallest period P is then the period.

But is finding T easier than finding F ??



( end quote )



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