bo198214 Wrote:Yes, but how do you compute the infinite case. You can not simply do
\( \sum_{n=0}^\infty (-1)^{n+1} \frac{(A-I)^n}{n} \) as this does not converge. So do you compute it as described here via the Jordan form?
Hmm, I observed that I'm not good in answering correctly in speedy discussions. But may be I've got it right in this case.
If the sequence of corresponding entries of the powers of (A-I)^n/n do not converge well with finite dimension, but at least alternate in sign, I try to approximate the sums \( s^{^{(\infty)}}_{i,k} \) of individual entries \( b^{^{(n)}}_{i,k} \) of the n'th powers of the matrix B=(A-I) by Euler-summation via partial sums, which is sometimes an option.
If the growthrate of the entries is below a certain constant limit \( b^{^{(n+1)}}_{i,k} / b^{^{(n)}}_{i,k} =q_{i,k} \) constant, or \( | q^{^{(n)}}_{i,k} | \) is a decreasing sequence (which is for instance the case in the scalar series for the logarithm) and no further exotic behaviour has to be expected, then Euler-summation of order q can give the final approximation to arbitrary precision for each individual entry for \( s^{^{(\infty)}}_{i,k} \) of the intended result-matrix by finitely many partial sums (where "exotic" has to be precisely specified,well).
Is it that what you meant?
Gottfried
Gottfried Helms, Kassel