09/07/2007, 02:15 PM
I think what you have done is similar what I describe now:
Given the exponentiation to base s: \( \exp_s \).
We know that it has the (lower) fixed point t with \( t^{1/t}=s \).
Let \( \tau_t(x)=x+t \) then
\( f:=\tau_t^{-1}\circ \exp_s \circ \tau_t \)
is a function with fixed point 0 and with \( f_1=f'(0)=s^{x+t}\ln(s)|_{x=0}=s^t\ln(s)=t\ln(s)=\ln(t) \).
Further it is known that a power series f with \( f_0=0 \), \( f_1\neq 1 \) and \( |f_1|\neq 1 \) is conjugate to the linear function \( \mu_{f_1}(x):=f_1x \). As in our case \( 1<s<e^{1/e} \) and hence \( 1<t<e \) we apply it to our \( f \):
\( f=\alpha\circ \mu_{\ln(t)}\circ \alpha^{-1} \)
\( \exp_s = \tau_t\circ \alpha \circ \mu_{\ln(t)}\circ \alpha^{-1}\circ \tau_t^{-1} \)
\( \exp_s^{\circ x} = \tau_t\circ \alpha \circ \mu_{\ln(t)^x}\circ \alpha^{-1}\circ \tau_t^{-1} \).
Now lets express this with power derivation matrices, let \( P(f) \) be the power derivation matrix of \( f \). Then we have:
\( P(\exp_s)=P(\tau_t)P(\alpha)P(\mu_{\ln(t)})P(\alpha)^{-1}P(\tau_t)^{-1} \).
We have the following correspondences
\( P(\mu_a)={}_dV(a) \)
\( P(\exp)=B^{\sim} \)
\( P(\exp_s)=P(\exp\circ\mu_{\ln(s)})=P(\exp)P(\mu_{\ln(s)})=B^{\sim}{}_dV(\ln(s))=({}_dV(\ln(s))B)^{\sim}=B_s^{\sim} \)
\( P(\mu_{\ln(t)})={}_d\Lambda={}_dV(\ln(t)) \)
\( P(\tau_t\circ\alpha)=(W_s^{-1})^{\sim}=(X_s P^{\sim} V(t))^{\sim}=V(t)^{\sim} P X_s^{\sim} \).
Now lets have a look at the structure of \( P(\tau_t) \). We can decompose \( \tau_t = \mu_t\circ \tau_1\circ \mu_{\frac{1}{t}} \) where \( P(\tau_1) \) is the lower triangular Pascal matrix given by \( (\tau_1)_{m,n}=\left(m\\n\right) \). This is because the \( m \)th row of the power derivation matrix of \( x+1 \) consists of the coefficients of \( (x+1)^m=\sum_{n=0}^m\left(m\\n\right)x^n \). As \( P(\mu_t)=V(t)^{\sim} \) we conclude
\( P(\mu_{\frac{1}{t}}\circ \alpha)=X_s^{\sim} \).
I think this completes the correspondences.
Unfortunately this decomposition works merely if the function under consideration has a fixed point. In so far it is very interesting that it also converges for the non-fixed point case with base \( b>e^{1/e} \).
Given the exponentiation to base s: \( \exp_s \).
We know that it has the (lower) fixed point t with \( t^{1/t}=s \).
Let \( \tau_t(x)=x+t \) then
\( f:=\tau_t^{-1}\circ \exp_s \circ \tau_t \)
is a function with fixed point 0 and with \( f_1=f'(0)=s^{x+t}\ln(s)|_{x=0}=s^t\ln(s)=t\ln(s)=\ln(t) \).
Further it is known that a power series f with \( f_0=0 \), \( f_1\neq 1 \) and \( |f_1|\neq 1 \) is conjugate to the linear function \( \mu_{f_1}(x):=f_1x \). As in our case \( 1<s<e^{1/e} \) and hence \( 1<t<e \) we apply it to our \( f \):
\( f=\alpha\circ \mu_{\ln(t)}\circ \alpha^{-1} \)
\( \exp_s = \tau_t\circ \alpha \circ \mu_{\ln(t)}\circ \alpha^{-1}\circ \tau_t^{-1} \)
\( \exp_s^{\circ x} = \tau_t\circ \alpha \circ \mu_{\ln(t)^x}\circ \alpha^{-1}\circ \tau_t^{-1} \).
Now lets express this with power derivation matrices, let \( P(f) \) be the power derivation matrix of \( f \). Then we have:
\( P(\exp_s)=P(\tau_t)P(\alpha)P(\mu_{\ln(t)})P(\alpha)^{-1}P(\tau_t)^{-1} \).
We have the following correspondences
\( P(\mu_a)={}_dV(a) \)
\( P(\exp)=B^{\sim} \)
\( P(\exp_s)=P(\exp\circ\mu_{\ln(s)})=P(\exp)P(\mu_{\ln(s)})=B^{\sim}{}_dV(\ln(s))=({}_dV(\ln(s))B)^{\sim}=B_s^{\sim} \)
\( P(\mu_{\ln(t)})={}_d\Lambda={}_dV(\ln(t)) \)
\( P(\tau_t\circ\alpha)=(W_s^{-1})^{\sim}=(X_s P^{\sim} V(t))^{\sim}=V(t)^{\sim} P X_s^{\sim} \).
Now lets have a look at the structure of \( P(\tau_t) \). We can decompose \( \tau_t = \mu_t\circ \tau_1\circ \mu_{\frac{1}{t}} \) where \( P(\tau_1) \) is the lower triangular Pascal matrix given by \( (\tau_1)_{m,n}=\left(m\\n\right) \). This is because the \( m \)th row of the power derivation matrix of \( x+1 \) consists of the coefficients of \( (x+1)^m=\sum_{n=0}^m\left(m\\n\right)x^n \). As \( P(\mu_t)=V(t)^{\sim} \) we conclude
\( P(\mu_{\frac{1}{t}}\circ \alpha)=X_s^{\sim} \).
I think this completes the correspondences.
Unfortunately this decomposition works merely if the function under consideration has a fixed point. In so far it is very interesting that it also converges for the non-fixed point case with base \( b>e^{1/e} \).
