06/02/2008, 10:39 AM
Re first part:
Perhaps I just dont understand your explanation but for my taste you work too much with infinite matrices. We already have seen that they may not even have unique inverse, nor a unique diagonalization. We can not carry over the rules of finite matrices to infinite matrices. Also the fixed point choice is not unique, there are infinite many complex fixed points.
Re second part:
The trace is indeed an interesting invariant, here the trace of an infinite matrix is really the limit of the traces of the finite matrices:
\( \text{tr}(M)=\lim_{n\to\infty} \text{tr}(M|_n) \).
So what you are saying is (?): If the eigenvalues of the matrices \( B_b|_n \) do converge to the powers of the logarithm of the lower fixed point \( u^n \) then \( \text{tr}(M)=\frac{1}{1-u} \).
So if \( \text{tr}(B_b) \) has indeed this value \( \frac{1}{1-u} \) then this is a "good sign".
So the question then remains whether
\( \sum_{n=0}^\infty \left(\frac{u*n}{e^u}\right)^n \frac{1}{n!}=\frac{1}{1-u} \)
Right?
Lets transform this a bit:
\( \sum_{n=0}^\infty \left(\frac{u*n}{e^u}\right)^n \frac{1}{n!}= \sum_{n=0}^\infty \frac{n^n}{n!} (ue^{-u})^n = \sum_{n=0}^\infty \frac{(-n)^n}{n!} ((-u)e^{-u})^n \)
This looks already damn like the power series of the LambertW function which is
\( W(x)=\sum_{n=0}^\infty \frac{(-n)^{n-1}}{n!} x^n \).
Hence \( W'(x)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{(n-1)!} x^{n-1}=\sum_{n=0}^\infty \frac{(-n)^{n}}{n!} x^{n} \)
And we want to show that \( W'((-u)e^{-u})=\frac{1}{1-u} \).
Ok, we have \( W(x)e^{W(x)}=x \). Take the derivative:
\( W'(x)e^{W(x)}+W(x)e^{W(x)}W'(x)=1 \)
\( W'(x)=\frac{e^{-W(x)}}{1+W(x)} \)
now is \( (-u)e^{-u}=W^{-1}(-u) \) so
\( W'((-u)e^{-u})=W'(W^{-1}(-u))=\frac{e^u}{1-u} \)
Hm this is not quite \( \frac{1}{1-u} \) so where is the error?
Perhaps I just dont understand your explanation but for my taste you work too much with infinite matrices. We already have seen that they may not even have unique inverse, nor a unique diagonalization. We can not carry over the rules of finite matrices to infinite matrices. Also the fixed point choice is not unique, there are infinite many complex fixed points.
Re second part:
The trace is indeed an interesting invariant, here the trace of an infinite matrix is really the limit of the traces of the finite matrices:
\( \text{tr}(M)=\lim_{n\to\infty} \text{tr}(M|_n) \).
So what you are saying is (?): If the eigenvalues of the matrices \( B_b|_n \) do converge to the powers of the logarithm of the lower fixed point \( u^n \) then \( \text{tr}(M)=\frac{1}{1-u} \).
So if \( \text{tr}(B_b) \) has indeed this value \( \frac{1}{1-u} \) then this is a "good sign".
So the question then remains whether
\( \sum_{n=0}^\infty \left(\frac{u*n}{e^u}\right)^n \frac{1}{n!}=\frac{1}{1-u} \)
Right?
Lets transform this a bit:
\( \sum_{n=0}^\infty \left(\frac{u*n}{e^u}\right)^n \frac{1}{n!}= \sum_{n=0}^\infty \frac{n^n}{n!} (ue^{-u})^n = \sum_{n=0}^\infty \frac{(-n)^n}{n!} ((-u)e^{-u})^n \)
This looks already damn like the power series of the LambertW function which is
\( W(x)=\sum_{n=0}^\infty \frac{(-n)^{n-1}}{n!} x^n \).
Hence \( W'(x)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{(n-1)!} x^{n-1}=\sum_{n=0}^\infty \frac{(-n)^{n}}{n!} x^{n} \)
And we want to show that \( W'((-u)e^{-u})=\frac{1}{1-u} \).
Ok, we have \( W(x)e^{W(x)}=x \). Take the derivative:
\( W'(x)e^{W(x)}+W(x)e^{W(x)}W'(x)=1 \)
\( W'(x)=\frac{e^{-W(x)}}{1+W(x)} \)
now is \( (-u)e^{-u}=W^{-1}(-u) \) so
\( W'((-u)e^{-u})=W'(W^{-1}(-u))=\frac{e^u}{1-u} \)
Hm this is not quite \( \frac{1}{1-u} \) so where is the error?