Guys! That I didnt see that before!
We have a very simple formula for computing the \( t \)-th iterate of an arbitrary function:
\( f^{\circ t} = \sum_{n=0}^\infty \left(t\\n\right) \sum_{k=0}^n \left(n\\k\right) (-1)^{n-k} f^{\circ k} \)
This series does not always converge but at least if \( f \) has an attracting fixed point reachable from \( x \) then \( f^{\circ t}(x) \) converges, because then \( f^{\circ k}(x) \) is bounded, say \( |f^{\circ k}(x)|<M \), then
\( |f^{\circ t}(x)|<M\sum_{n=0}^\infty \left(t\\n\right) \sum_{k=0}^n \left(n\\k\right)=M\sum_{n=0}^\infty \left(t\\n\right) 2^n = M3^t \). I.e. the sum is absolutely convergent.
I bet my pants that this \( f^{\circ t} \) is the regular iteration at the lower (attracting) fixed point in the case \( f(x)=b^x \), \( 1<b<e^{1/e} \).
Why is this matrix function like?
Well, if \( A \) has eigenvalues all smaller than 1 then, we dont need the Jordanization of \( A \), but the matrix power is also given as a convergent series
\( (A-I+I)^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n=\sum_{n=0}^\infty \left(t\\n\right) \sum_{k=0}^n \left(n\\k\right) (-1)^{n-k} A^k \).
If \( A \) is the Carleman matrix of \( f \), then we just consider the first row in all those matrix powers and add them up.
Dont make the error however to assume that
\( (f-\text{id})^{\circ n}=\sum_{k=0}^n \left(n\\k\right) (-1)^k f^{\circ n} \)
This is not true. You can only do this with matrices as \( A(B+C)=AB+AC \) while mostly \( f\circ (g+h)\neq f\circ g + f\circ h \).
Notes:
We have a very simple formula for computing the \( t \)-th iterate of an arbitrary function:
\( f^{\circ t} = \sum_{n=0}^\infty \left(t\\n\right) \sum_{k=0}^n \left(n\\k\right) (-1)^{n-k} f^{\circ k} \)
This series does not always converge but at least if \( f \) has an attracting fixed point reachable from \( x \) then \( f^{\circ t}(x) \) converges, because then \( f^{\circ k}(x) \) is bounded, say \( |f^{\circ k}(x)|<M \), then
\( |f^{\circ t}(x)|<M\sum_{n=0}^\infty \left(t\\n\right) \sum_{k=0}^n \left(n\\k\right)=M\sum_{n=0}^\infty \left(t\\n\right) 2^n = M3^t \). I.e. the sum is absolutely convergent.
I bet my pants that this \( f^{\circ t} \) is the regular iteration at the lower (attracting) fixed point in the case \( f(x)=b^x \), \( 1<b<e^{1/e} \).
Why is this matrix function like?
Well, if \( A \) has eigenvalues all smaller than 1 then, we dont need the Jordanization of \( A \), but the matrix power is also given as a convergent series
\( (A-I+I)^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n=\sum_{n=0}^\infty \left(t\\n\right) \sum_{k=0}^n \left(n\\k\right) (-1)^{n-k} A^k \).
If \( A \) is the Carleman matrix of \( f \), then we just consider the first row in all those matrix powers and add them up.
Dont make the error however to assume that
\( (f-\text{id})^{\circ n}=\sum_{k=0}^n \left(n\\k\right) (-1)^k f^{\circ n} \)
This is not true. You can only do this with matrices as \( A(B+C)=AB+AC \) while mostly \( f\circ (g+h)\neq f\circ g + f\circ h \).
Notes:
- Yes, this looks like the double binomial formula, note however that the formula should not be rearranged into the form \( \sum_{n=0}^\infty a_n f^{\circ n} \) because this can lead to non-convergence. This is best seen with the formula \( ((x-1)+1)^t \) it must not be rearranged into the form \( \sum_{n=0}^\infty a_n x^n \) because 0 is a singularity for this function (\( t\not\in \mathbb{N} \)) and hence there is no power series development at 0.
- Perhaps Gottfried can jump in to provide summability in the divergent case \( b>e^{1/e} \).