bo198214 Wrote:Well, I have it in my "Hütte Mathematische Tafeln":Gottfried Wrote:If their sign oscillate ...No, their sign doesnt oscillate.
The reason is simply: Some of the Eigenvalues are greater than 1 and hence the logarithm sequence does no more converge.
This seems to can be fixed with the series
\( \log(A)=\sum_{n=0}^\infty \frac{2}{2n+1} \left((A-I)(A+I)^{-1}\right)^{2n+1} \)
which then properly converges.
\(
\log( \frac{1+x}{1-x}) =2\left({x + \frac{x^3}{3} + \frac{x^5}{5} +... }\right) \hspace{100} for |x|<1
\\
\log( \frac{x+1}{x-1})=2\left({x + \frac{x^3}{3} + \frac{x^5}{5} +... }\right) \hspace{100} for |x|>1
\)
To apply one of this series to a matrix, all eigenvalues must match the same bound separately.
I think, this settles this question for the most interesting cases.
For matrices with eigenvalues both <1 and >1 , which occurs with the Bs-matrixes for s outside the range 1/e^e ... e^(1/e) we need still workarounds, like the techniques for divergent summation. But this is then a completely different chapter. I'm happy, we have now arrived at a level of convergence of understanding of (one of?) the core points of concepts.
Gottfried
Gottfried Helms, Kassel

