Still considering parabolic fixpoints at 0.

And also those "semi-Taylor" expansions that have radius 0.

For instance solving f(f(x)) = x + x^2.

See also : http://math.stackexchange.com/questions/...324#912324

But in general f(f(x)) = x + r x^2 + r_2 x^3 + ... = g(x) (real-entire)

with r > 0.

Considering the semi-Taylor for f(x) ;

Now if we rewrite the logs and sqrt's of x as Taylor series in (x+1) and truncate at A sums and we truncate the remaining Taylor series at A sums , then I think that :

f(x) , for every x > 0 can be given by truncating the semi-Taylor at A sums ( as described above ).

Actually not A , but A(x) , where A is a function of x.

Notice that I did not say A is an integer.

So im talking about continuum sums again.

ALthough the best fitting integer is also intresting ofcourse.

Natural question is ofcourse , for a given g(x) , how to find A(x) ??

It seems easy in a numerical experimental way , so I have hope for this.

( this is the new thread what I talked about in post 5 of http://math.eretrandre.org/tetrationforu...hp?tid=965 )

Experimental math seems easy , considering that the truncation should give values for A(x) such that the truncated f(x) is between x and g(x).

symbolic :

for Q > x > 0

where Q is the smallest value > 0 where g'(Q) = 0 ,

x < f_A(x) < g(x).

I used f_A(x) for f(x) truncated at the A(x) th term.

To give some examples I wonder about :

f_A(x) resp A(x) for

1) x + x^N ( some N > 2 )

2) x exp(x)

...

I think A(x) satisfies some logical things.

for instance : if A(x) belongs to g(x) , then A(x) + 1 belongs to (g(x)+1) x.

Also I think the concept of "growth" (as considered by me and sheldon) is important as are r_2 and r_3.

This will probably give nice Visuals.

regards

tommy1729

And also those "semi-Taylor" expansions that have radius 0.

For instance solving f(f(x)) = x + x^2.

See also : http://math.stackexchange.com/questions/...324#912324

But in general f(f(x)) = x + r x^2 + r_2 x^3 + ... = g(x) (real-entire)

with r > 0.

Considering the semi-Taylor for f(x) ;

Now if we rewrite the logs and sqrt's of x as Taylor series in (x+1) and truncate at A sums and we truncate the remaining Taylor series at A sums , then I think that :

f(x) , for every x > 0 can be given by truncating the semi-Taylor at A sums ( as described above ).

Actually not A , but A(x) , where A is a function of x.

Notice that I did not say A is an integer.

So im talking about continuum sums again.

ALthough the best fitting integer is also intresting ofcourse.

Natural question is ofcourse , for a given g(x) , how to find A(x) ??

It seems easy in a numerical experimental way , so I have hope for this.

( this is the new thread what I talked about in post 5 of http://math.eretrandre.org/tetrationforu...hp?tid=965 )

Experimental math seems easy , considering that the truncation should give values for A(x) such that the truncated f(x) is between x and g(x).

symbolic :

for Q > x > 0

where Q is the smallest value > 0 where g'(Q) = 0 ,

x < f_A(x) < g(x).

I used f_A(x) for f(x) truncated at the A(x) th term.

To give some examples I wonder about :

f_A(x) resp A(x) for

1) x + x^N ( some N > 2 )

2) x exp(x)

...

I think A(x) satisfies some logical things.

for instance : if A(x) belongs to g(x) , then A(x) + 1 belongs to (g(x)+1) x.

Also I think the concept of "growth" (as considered by me and sheldon) is important as are r_2 and r_3.

This will probably give nice Visuals.

regards

tommy1729