11/08/2007, 08:16 PM
Let us continue with the 2nd case \( f(x)=xc, c>0, c\neq 1 \) with the expected iteration \( f^{\circ t}(x)=xc^t \) and the expected Abel function \( f^\ast(s)=\log_c(s) \).
This function has one (and only one) fixed point at 0, so here we can directly apply the regular iteration (which we couldnt for our previous function \( f(x)=x+c \), which had no fixed points) and the matrix operator method. However for the natural Abel function we must consider a development at a non-fixed point.
Iterative regular iteration
We compute the principal Schroeder function by
\( \sigma(x)=\lim_{n\to\infty} \frac{f^{\circ n}(x)}{c^n} = \lim_{n\to\infty} \frac{c^nx}{c^n} = x \).
The principal Abel function is \( \alpha(x)=\log_c(\sigma(x))=\log_c(x) \).
Matrix operator method/formal powerseries iteration
The Bell matrix \( F \) of \( f(x)=xc \) has in the \( i \)-th row the coefficients of the \( i \)-th power of \( f \), i.e. \( f(x)^i=(xc)^i=c^ix^i \), which means it consists of 0's except on the diagonal it has the entries \( 1,c^1, c^2,\dots \).
\( F=\begin{pmatrix} 1 & 0 & 0 & \dots \\ 0 & c & 0 &\dots \\ 0 & 0 & c^2 & 0\dots \\ 0 & \dots\end{pmatrix} \).
This however is already in the diagonal form and the \( t \)-th power of \( F \) consists of \( 1,c^t, c^{2t}, c^{3t},\dots \) on the diagonal and 0 otherwhere. The first row contains the coefficients of the \( t \)-th iterate, so \( f^{\circ t}(x)=c^t x \) according to our assumption.
Natural Abel method
First we have to choose a non-fixed point of development, say \( a \). Then we form transposed conjugate \( g(x)=f(x+a)-a=(x+a)c-a=xc+a(c-1) \), knowing that then \( f^\ast(x+a)=g^\ast(x) \), we would expect that \( g^\ast(x)=\log_c(x-a) \).
For simplification write \( g(x)=xc+d, d=a(c-1) \).
The powers of \( g \) are \( g(x)^n=\sum_{i=0}^n \left(n\\i\right) d^{n-i}c^i x^i \) hence the Carleman matrix G of g has the entries \( G_{i,n}=\left(n\\i\right) d^{n-i}c^i \) this is a upper triangular matrix. Truncated to 4x4:
\( G=\begin{pmatrix}1 & d^1 & d^2 & d^3 \\ 0 & c^1 & 2cd & 3cd^2 \\ 0 & 0 & c^2 & 3c^2d \\ 0 & 0 & 0 & c^3\end{pmatrix} \), subtracting the identity matrix and first column and last row removed, we have to solve:
\( \begin{pmatrix}d^1 & d^2 & d^3 \\ c^1-1 & 2cd & 3cd^2 \\ 0 & c^2-1 & 3c^2d\end{pmatrix}\left(g^\ast_1\\g^\ast_2\\g^\ast_3\right)=\left(1\\0\\0 \right) \)
For a moment let the development point \( a=1 \), then \( d=c-1 \). And we can line by line transform the above equation system into a triangular one, first subtracting the first line from the second line
\( \underbrace{\begin{pmatrix}d^1 & d^2 & d^3 \\ 0 & 2cd-d^2 & 3cd^2-d^3 \\ 0 & c^2-1 & 3c^2d\end{pmatrix}}_{H'}\left(g^\ast_1\\g^\ast_2\\g^\ast_3\right)=\left(1\\-1\\0 \right) \)
then \( H'_{2,2}=2cd-d^2=2c(c-1)-(c-1)^2=c^2-1 \). And we subtract the second line from the third:
\( \underbrace{\begin{pmatrix}d^1 & d^2 & d^3 \\ 0 & 2cd-d^2 & 3cd^2-d^3 \\ 0 & 0 & 3c^2d-(3cd^2-d^3)\end{pmatrix}}_{H''}\left(g^\ast_1\\g^\ast_2\\g^\ast_3\right)=\left(1\\-1\\1 \right) \)
This scheme can arbitrarily continued. We can now subtract the third line from the forth, because \( H''_{3,3}=c^3-1 \) and so on.
Generally this would give us a triangular \( N\times N \) matrix
\( H^{(N-1)}_{k,n} = \sum_{i=0}^{k-1} (-1)^{k+i+1}\left(n\\i\right) d^{n-i} c^i,\quad k\le n,\quad 1\le k,n\le N \).
Hm, does the solution of this system converge to \( \log_c(x-1) \), more specifically is \( \lim_{N\to\infty} g^\ast_k = \frac{-(-1)^k}{k} \) for \( c=e \), does it converge at all? My numerical computations show that it is roughly (up to \( 10^{-3} \) precision) the natural logarithm. However it seems not really to converge, or is it just to slow?
Perhaps one has to derive it from the formulas without numerics in some quiet hours.
This function has one (and only one) fixed point at 0, so here we can directly apply the regular iteration (which we couldnt for our previous function \( f(x)=x+c \), which had no fixed points) and the matrix operator method. However for the natural Abel function we must consider a development at a non-fixed point.
Iterative regular iteration
We compute the principal Schroeder function by
\( \sigma(x)=\lim_{n\to\infty} \frac{f^{\circ n}(x)}{c^n} = \lim_{n\to\infty} \frac{c^nx}{c^n} = x \).
The principal Abel function is \( \alpha(x)=\log_c(\sigma(x))=\log_c(x) \).
Matrix operator method/formal powerseries iteration
The Bell matrix \( F \) of \( f(x)=xc \) has in the \( i \)-th row the coefficients of the \( i \)-th power of \( f \), i.e. \( f(x)^i=(xc)^i=c^ix^i \), which means it consists of 0's except on the diagonal it has the entries \( 1,c^1, c^2,\dots \).
\( F=\begin{pmatrix} 1 & 0 & 0 & \dots \\ 0 & c & 0 &\dots \\ 0 & 0 & c^2 & 0\dots \\ 0 & \dots\end{pmatrix} \).
This however is already in the diagonal form and the \( t \)-th power of \( F \) consists of \( 1,c^t, c^{2t}, c^{3t},\dots \) on the diagonal and 0 otherwhere. The first row contains the coefficients of the \( t \)-th iterate, so \( f^{\circ t}(x)=c^t x \) according to our assumption.
Natural Abel method
First we have to choose a non-fixed point of development, say \( a \). Then we form transposed conjugate \( g(x)=f(x+a)-a=(x+a)c-a=xc+a(c-1) \), knowing that then \( f^\ast(x+a)=g^\ast(x) \), we would expect that \( g^\ast(x)=\log_c(x-a) \).
For simplification write \( g(x)=xc+d, d=a(c-1) \).
The powers of \( g \) are \( g(x)^n=\sum_{i=0}^n \left(n\\i\right) d^{n-i}c^i x^i \) hence the Carleman matrix G of g has the entries \( G_{i,n}=\left(n\\i\right) d^{n-i}c^i \) this is a upper triangular matrix. Truncated to 4x4:
\( G=\begin{pmatrix}1 & d^1 & d^2 & d^3 \\ 0 & c^1 & 2cd & 3cd^2 \\ 0 & 0 & c^2 & 3c^2d \\ 0 & 0 & 0 & c^3\end{pmatrix} \), subtracting the identity matrix and first column and last row removed, we have to solve:
\( \begin{pmatrix}d^1 & d^2 & d^3 \\ c^1-1 & 2cd & 3cd^2 \\ 0 & c^2-1 & 3c^2d\end{pmatrix}\left(g^\ast_1\\g^\ast_2\\g^\ast_3\right)=\left(1\\0\\0 \right) \)
For a moment let the development point \( a=1 \), then \( d=c-1 \). And we can line by line transform the above equation system into a triangular one, first subtracting the first line from the second line
\( \underbrace{\begin{pmatrix}d^1 & d^2 & d^3 \\ 0 & 2cd-d^2 & 3cd^2-d^3 \\ 0 & c^2-1 & 3c^2d\end{pmatrix}}_{H'}\left(g^\ast_1\\g^\ast_2\\g^\ast_3\right)=\left(1\\-1\\0 \right) \)
then \( H'_{2,2}=2cd-d^2=2c(c-1)-(c-1)^2=c^2-1 \). And we subtract the second line from the third:
\( \underbrace{\begin{pmatrix}d^1 & d^2 & d^3 \\ 0 & 2cd-d^2 & 3cd^2-d^3 \\ 0 & 0 & 3c^2d-(3cd^2-d^3)\end{pmatrix}}_{H''}\left(g^\ast_1\\g^\ast_2\\g^\ast_3\right)=\left(1\\-1\\1 \right) \)
This scheme can arbitrarily continued. We can now subtract the third line from the forth, because \( H''_{3,3}=c^3-1 \) and so on.
Generally this would give us a triangular \( N\times N \) matrix
\( H^{(N-1)}_{k,n} = \sum_{i=0}^{k-1} (-1)^{k+i+1}\left(n\\i\right) d^{n-i} c^i,\quad k\le n,\quad 1\le k,n\le N \).
Hm, does the solution of this system converge to \( \log_c(x-1) \), more specifically is \( \lim_{N\to\infty} g^\ast_k = \frac{-(-1)^k}{k} \) for \( c=e \), does it converge at all? My numerical computations show that it is roughly (up to \( 10^{-3} \) precision) the natural logarithm. However it seems not really to converge, or is it just to slow?
Perhaps one has to derive it from the formulas without numerics in some quiet hours.