Conjecture
Let \( B_n \) be the Bell/Carleman matrix of \( f(x)=b^x \), \( 1<b<e^{1/e} \), truncated to \( n \) rows and columns.
There is an enumeration \( (v_{n,k})_{1\le k\le n} \) of the eigenvalues of \( B_n \) such that \( \lim_{n\to\infty} v_{n,k} = \ln(a)^k \) where \( a \) is the lower real fixed point of \( b^x \).
Explanation
This is a key question for deciding whether the diagonalization/matrix power method is independent of its development point. A more general question would be under which circumstances the eigenvalues of the Carleman matrix converge to the powers of an attracting fixed point. It seems they dont do for \( f(x)=x^2+x-1/16 \) which has an attracting fixed point at \( -1/4 \) and a repelling fixed point at \( 1/4 \). Also the behaviour of the eigenvalues for \( e^{-e}<b<1 \) seems unclear. However there is only one real repelling fixed point in this case.
Notes
\( b=a^{1/a} \)
\( \ln(a) \) is the derivative of \( b^x \) at \( a \): \( \frac{\partial b^x}{\partial x}|_{x=a}=\ln(b)b^x|_{x=a}=\ln(b)a=\ln(b^a)=\ln(a) \)
Let \( B_n \) be the Bell/Carleman matrix of \( f(x)=b^x \), \( 1<b<e^{1/e} \), truncated to \( n \) rows and columns.
There is an enumeration \( (v_{n,k})_{1\le k\le n} \) of the eigenvalues of \( B_n \) such that \( \lim_{n\to\infty} v_{n,k} = \ln(a)^k \) where \( a \) is the lower real fixed point of \( b^x \).
Explanation
This is a key question for deciding whether the diagonalization/matrix power method is independent of its development point. A more general question would be under which circumstances the eigenvalues of the Carleman matrix converge to the powers of an attracting fixed point. It seems they dont do for \( f(x)=x^2+x-1/16 \) which has an attracting fixed point at \( -1/4 \) and a repelling fixed point at \( 1/4 \). Also the behaviour of the eigenvalues for \( e^{-e}<b<1 \) seems unclear. However there is only one real repelling fixed point in this case.
Notes
\( b=a^{1/a} \)
\( \ln(a) \) is the derivative of \( b^x \) at \( a \): \( \frac{\partial b^x}{\partial x}|_{x=a}=\ln(b)b^x|_{x=a}=\ln(b)a=\ln(b^a)=\ln(a) \)