Gottfried Wrote:Hmm, the fixpoints-shifts resulted in similarity scalings with the matrix-method (using powers of the pascalmatrix). Then the eigenvalues should be unchanged.
I dont know whether they should, but actually there are different. I would say that this has to do with that you need an infinite matrix to do a fixed point shift. While you are actually dealing with a finite matrix.
Ok, but perhaps examples are more convincing:
Let us first consider the Carleman matrix (wow I see Andrew has done righteous work in writing this wikipedia article) of \( \sqrt{2}^x \) at 0:
truncated to 5:
\( \begin{pmatrix} 1&0&0&0&0\\1&\frac{1}{2}\ln \left( 2 \right) &\frac{1}{8} \left( \ln \left( 2 \right) \right) ^{2}&\frac{1}{
48} \left( \ln \left( 2 \right) \right) ^{3}&{\frac {1}{384}}
\left( \ln \left( 2 \right) \right) ^{4}\\1&\ln
\left( 2 \right) &\frac{1}{2} \left( \ln \left( 2 \right) \right) ^{2}&\frac{1}{
6} \left( \ln \left( 2 \right) \right) ^{3}&\frac{1}{24} \left( \ln
\left( 2 \right) \right) ^{4}\\1&\frac{3}{2}\ln \left(
2 \right) &{\frac {9}{8}} \left( \ln \left( 2 \right) \right) ^{2}
&{\frac {9}{16}} \left( \ln \left( 2 \right) \right) ^{3}&{\frac {
27}{128}} \left( \ln \left( 2 \right) \right) ^{4}
\\1&2\ln \left( 2 \right) &2 \left( \ln
\left( 2 \right) \right) ^{2}&\frac{4}{3} \left( \ln \left( 2 \right)
\right) ^{3}&\frac{2}{3} \left( \ln \left( 2 \right) \right) ^{4}
\end{pmatrix} \)
The eigenvalues of this matrix are:
1.0000000000
0.6269572883
0.2495195174
0.0482258803
0.0033138436
I think they are not powers of anything. And they change when increasing the matrix size.
Now consider the matrix at the lower fixed point 2, i.e. the matrix of \( \sqrt{2}^{x+2}-2=2(\sqrt{2}^x-1) \).
\( \begin{pmatrix}
1&0&0&0&0\\
0&\ln \left( 2 \right) &\frac{1}{4} \left( \ln \left( 2 \right) \right) ^{2}&\frac{1}{
24} \left( \ln \left( 2 \right) \right) ^{3}&{\frac {1}{192}}
\left( \ln \left( 2 \right) \right) ^{4}\\
0&0& \left( \ln \left( 2 \right) \right) ^{2}&\frac{1}{2} \left( \ln \left( 2
\right) \right) ^{3}&{\frac {7}{48}} \left( \ln \left( 2 \right)
\right) ^{4}\\
0&0&0& \left( \ln \left( 2 \right) \right) ^{3}&\frac{3}{4} \left( \ln \left( 2 \right) \right) ^{4}\\
0&0&0&0& \left( \ln \left( 2 \right) \right) ^{4}
\end{pmatrix} \)
The eigenvalues are the entries on the diagonal, they are powers of \( f'(0)=(2(\sqrt{2}^x-1))'(0)=\ln(2)\approx \)0.6931471806. Numerically
1.0000000000
0.6931471806
0.4804530140
0.3330246520
0.2308350986
When increasing the matrix size only new powers become added, so these eigenvalues are invariante to matrix increase.