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 matrix function like iteration without power series expansion bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/30/2008, 03:01 PM (This post was last modified: 06/30/2008, 03:40 PM by bo198214.) 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)|, then $|f^{\circ t}(x)|. 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. 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 $b>e^{1/e}$. « Next Oldest | Next Newest »

 Messages In This Thread matrix function like iteration without power series expansion - by bo198214 - 06/30/2008, 03:01 PM RE: matrix function like iteration without power series expansion - by Gottfried - 06/30/2008, 09:23 PM RE: matrix function like iteration without power series expansion - by bo198214 - 07/01/2008, 06:48 AM RE: matrix function like iteration without power series expansion - by Gottfried - 07/01/2008, 02:18 PM RE: matrix function like iteration without power series expansion - by Gottfried - 07/02/2008, 11:23 AM RE: matrix function like iteration without power series expansion - by Gottfried - 07/08/2008, 06:56 AM RE: matrix function like iteration without power series expansion - by andydude - 07/08/2008, 06:01 PM RE: matrix function like iteration without power series expansion - by andydude - 07/08/2008, 09:35 PM RE: matrix function like iteration without power series expansion - by Gottfried - 07/08/2008, 06:23 PM RE: matrix function like iteration without power series expansion - by Gottfried - 07/09/2008, 08:58 AM RE: matrix function like iteration without power series expansion - by andydude - 07/09/2008, 02:38 PM RE: matrix function like iteration without power series expansion - by andydude - 07/09/2008, 02:57 PM RE: matrix function like iteration without power series expansion - by bo198214 - 07/09/2008, 06:18 PM RE: matrix function like iteration without power series expansion - by andydude - 07/14/2008, 06:21 PM RE: matrix function like iteration without power series expansion - by bo198214 - 07/14/2008, 09:55 PM RE: matrix function like iteration without power series expansion - by Gottfried - 07/10/2008, 07:01 AM

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