04/30/2013, 01:04 AM

Simple method for the half iterate of a taylor series.

if

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)

Consider

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

( hence g(0) = g0 )

Define

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

regards

tommy1729

if

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)

Consider

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

( hence g(0) = g0 )

Define

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

regards

tommy1729