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Simple method for half iterate NOT based on a fixpoint.
Simple method for the half iterate of a taylor series.

g: R -> R
x>0 => g(x) > x
x,y>0 => g(x+y) > g(x)
g is entire.

To find
f(f(x)) = g(x)


g(x) = g0 + g1 x + g2 x^2 + g3 x^3 + ...
( hence g(0) = g0 )


f(x) = f0 + f1 x + f2 x^2 + f3 x^3 + ...

Now use D^n g(0) = D^n f(f(0)) = n! gn
( this follows from taylor's theorem and the chain rule )

We get

g(0) = f(f(0)) = f(f0) = g0

g'(0) = f'(f(0)) f'(0) = f'(f0) f'(0) = g1

pick f0 with 0 < f0 < g0.

Now pick an analytic function t(x) such that t(f^(n)(f0)) = f^(n)(0).

We arrive at the set of equations :
g^(1)(0) = f^(1)(f0) t(f^1(f0)) = g1
which can be solved.

Now use the solutions of D^(n-1) g(0) = D^(n-1) f(f(0)) = (n-1)! g(n-1) together with t(x) to solve D^n g(0) = D^n f(f(0)) = n! gn.

This is always possible because there are exactly 2 new variables in the n th equation relative to the n-1 th equation who are f^(n)(0) and f^(n)(f0) and by using t(x) we get only 1 new variable per equation. So there is no more a degree of freedom !

Doing so gives us by induction (and the lack of free parameters aka degrees of freedom) all values f^(n)(0) hence we get the values fn. As was desired.

f(x) = f0 + f1 x + f2 x^2 + f3 x^3 + ...

is now the unique solution based on picking f0 and t(x) assuming t(x) is picked *wisely*.

By wisely it is meant that
1) All solution fn are real
2) There are NO branches too choose when we solve for the new variable in each step from equation n-1 to equation n.

Note that if for all n : gn > 0 (*) => fn > 0. ( * = COND 1 )
If that is the case we can repeat the procedure to find g^[1/4](x) or equivalently f^[1/2].

HOWEVER it is not certain if the SAME t(x) can be used again.

Notice a similar method will probably work finding g^[1/3](x).
(with a t1(x) AND t2(x) to remove 2 degrees of freedom)

Also note that COND 1 easily gives a method for g^[1/2^m] for integer m.

Notice that for a real z : 0<z<1 and binary(0 or 1) zn we have z = z1/2 + z2/2^2 + z3/2^3 + ...
WHICH means we could approximate g^[z](x) (for bounded x) if COND 1 holds.
( analytic with respect to z is however a tricky question ! not ? )

Notice this method does not rely on fixpoints !!

It might be intresting to find a connection to fixpoints.

Also of possible intrest is to find a t(x) such that that method is equivalent to another one already discussed.

Thanks for your intrest


tommy1729 Wrote:Simple method for the half iterate of a taylor series

Nice! But I have two questions :

1. Does it always converges?
2. Is the half-iterate analytic?

to question 1)

No it does not always converge.

That is because almost always the half-iterate has singularities.
The half-iterate of exp has singularies.
More precisely EVERY half-iterate of exp has singularities.

See :

Since it does not always converge , we have a limit radius of convergeance WHEREEVER we expand.

In particular the fixpoints have singularities.

to question 2)

clearly related to question 1)

within its radius of convergeance the function is indeed analytic.
even better , analytic continuation is possible.

notice that the taylor series is EXACTLY bounded where there are singularies so when the expansion point is not near a singularity we have a nonzero radius of convergeance !

NOTE that the singularities are determined by the fixpoints of g and the function t(x).



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