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