We basicly have 3 methods to compute \( f^{\circ t} \) for a given analytic function \( f \). This is the natural Abel function, as a generalization of Andrew's slog, then it is the matrix operator method as favoured by Gottfried and there is the old regular iteration method. While the method of natural Abel function works only for developments at non-fixed points, the matrix operator method works for developments on fixed points and on non-fixed points and the regular iteration method works only at fixed points. The matrix operator method and the regular iteration method coincide on fixed points.
Here I want to investigate the application of these methods to the lower operations addition, multiplication and power operation. More precisely what is the outcome of the iteration of \( x+c \), of \( xc \) and of \( x^c \) as supplement to the already much discussed iteration of \( c^x \).
We would expect the methods to yield the following results, lets denote "the" Abel function of \( f \) by \( f^\ast \), which is the inverse of the function \( t\mapsto f^{\circ t}(x_0) \) and is determined up to an additive constant. Vice versa \( f^{\circ t}(x)={f^\ast}^{-1}(f^\ast(x)+t) \).
Let us start in this post with the investigation of the simplest case 1, \( f(x)=x+c \).
This has no fixed points for \( c\neq 0 \) so we apply the natural Abel function method and the matrix operator method. In both cases we need the Carleman-matrix, which has in its \( n \)-th column the coefficients of the \( n \)-th power of \( f \). The \( n \)th power is \( f(x)^n=(x+c)^n =\sum_{k=0}^n \left(n\\k\right) c^{n-k} x^k \). So the N-truncated Carleman-matrix is \( F_{k,n}=\left(n\\k\right)c^{n-k} \), \( 0\le k,n\le N \).
For example with \( N=3 \):
\( F=\begin{pmatrix} 1 & c & c^2 & c^3\\ 0 & 1 & 2c & 3c^2\\ 0 & 0 & 1 & 3c\\ 0 & 0 & 0 & 1\end{pmatrix} \).
The natural Abel method
Let \( \overline{F} \) be the matrix without the first column and without the last row which has N rows and columns and subtract the Identity matrix, this gives for \( N=3 \):
\( G:=\overline{F}-I=\begin{pmatrix} c & c^2 & c^3\\ 0 & 2c & 3c^2\\ 0 & 0 & 3c\end{pmatrix} \). To retrieve the natural Abel function we have to solve the equation system \( G\ast (f^{\ast}_1,f^{\ast}_2,\dots,f^{\ast}_N)^T = (1,0,\dots)^T \). We can easily backtrace that \( f^{\ast}_2,\dots,f^{\ast}_N=0 \) and that \( f_1^\ast = \frac{1}{c} \). Hence is \( f^\ast(s) = \frac{1}{c}s \) plus \( f^\ast_0 \) and we have verified our claim as this is valid for arbitrarily large N.
The matrix operator method
To compute \( F^t \) we can simply apply here
\( F^t=(F-I+I)^t=\sum_{k=0}^\infty \left(t\\k\right)(F-I)^k \). The first (not 0th) column of \( F^t \) are then the coefficients of \( f^{\circ t} \). But we see that the first column of \( (F-I)^k \) has all entries 0 for \( k\ge 2 \). Hence only two terms in the above infinite sum contributes to the first column, this is \( \left(t\\0\right)(F-I)^0=I \) and \( \left(t\\1\right)(F-I)=t(F-I) \). The first column of the first one is \( (0,1,0,\dots) \) and of the second one is \( t(c,0,\dots) \). Hence \( ({f^{\circ t}}_0,{f^{\circ t}}_1,\dots)=(tc,1,0,\dots) \) and so \( f^{\circ t}(x)=tc+x \) fullfilling our claim too.
Hopefully I will continue somewhen with cases 2. and 3. .
Here I want to investigate the application of these methods to the lower operations addition, multiplication and power operation. More precisely what is the outcome of the iteration of \( x+c \), of \( xc \) and of \( x^c \) as supplement to the already much discussed iteration of \( c^x \).
We would expect the methods to yield the following results, lets denote "the" Abel function of \( f \) by \( f^\ast \), which is the inverse of the function \( t\mapsto f^{\circ t}(x_0) \) and is determined up to an additive constant. Vice versa \( f^{\circ t}(x)={f^\ast}^{-1}(f^\ast(x)+t) \).
- \( f(x)=x+c \), \( f^{\circ t}(x)=x+ct \), \( f^\ast(s)=\frac{s}{c} \).
- \( f(x)=xc \), \( f^{\circ t}(x)=xc^t \), \( f^\ast(s)=\log_c s \).
- \( f(x)=x^c \), \( f^{\circ t}(x)=x^{c^t} \), \( f^\ast(s)=\log_c(\ln(s)) \)
Let us start in this post with the investigation of the simplest case 1, \( f(x)=x+c \).
This has no fixed points for \( c\neq 0 \) so we apply the natural Abel function method and the matrix operator method. In both cases we need the Carleman-matrix, which has in its \( n \)-th column the coefficients of the \( n \)-th power of \( f \). The \( n \)th power is \( f(x)^n=(x+c)^n =\sum_{k=0}^n \left(n\\k\right) c^{n-k} x^k \). So the N-truncated Carleman-matrix is \( F_{k,n}=\left(n\\k\right)c^{n-k} \), \( 0\le k,n\le N \).
For example with \( N=3 \):
\( F=\begin{pmatrix} 1 & c & c^2 & c^3\\ 0 & 1 & 2c & 3c^2\\ 0 & 0 & 1 & 3c\\ 0 & 0 & 0 & 1\end{pmatrix} \).
The natural Abel method
Let \( \overline{F} \) be the matrix without the first column and without the last row which has N rows and columns and subtract the Identity matrix, this gives for \( N=3 \):
\( G:=\overline{F}-I=\begin{pmatrix} c & c^2 & c^3\\ 0 & 2c & 3c^2\\ 0 & 0 & 3c\end{pmatrix} \). To retrieve the natural Abel function we have to solve the equation system \( G\ast (f^{\ast}_1,f^{\ast}_2,\dots,f^{\ast}_N)^T = (1,0,\dots)^T \). We can easily backtrace that \( f^{\ast}_2,\dots,f^{\ast}_N=0 \) and that \( f_1^\ast = \frac{1}{c} \). Hence is \( f^\ast(s) = \frac{1}{c}s \) plus \( f^\ast_0 \) and we have verified our claim as this is valid for arbitrarily large N.
The matrix operator method
To compute \( F^t \) we can simply apply here
\( F^t=(F-I+I)^t=\sum_{k=0}^\infty \left(t\\k\right)(F-I)^k \). The first (not 0th) column of \( F^t \) are then the coefficients of \( f^{\circ t} \). But we see that the first column of \( (F-I)^k \) has all entries 0 for \( k\ge 2 \). Hence only two terms in the above infinite sum contributes to the first column, this is \( \left(t\\0\right)(F-I)^0=I \) and \( \left(t\\1\right)(F-I)=t(F-I) \). The first column of the first one is \( (0,1,0,\dots) \) and of the second one is \( t(c,0,\dots) \). Hence \( ({f^{\circ t}}_0,{f^{\circ t}}_1,\dots)=(tc,1,0,\dots) \) and so \( f^{\circ t}(x)=tc+x \) fullfilling our claim too.
Hopefully I will continue somewhen with cases 2. and 3. .