Matrix Operator Method

[Gottfried]

Hm, then I somewhere must have an error in my computation.
I wanted to compute the Matrix logarithm via the formula
\( \log(A)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} (A-I)^n \).
For A being the power derivation matrix of \( e^x \) at 0, i.e. truncated to 6x6

Hmm, I can crosscheck such a computation in the evening. But for a quick reply: did you observe the partial sums for the entries for increasing n by s_n_(i,j) = sum (k=1..n) entry_(i,j)? If their sign oscillate we have a candidate for the Euler-summation (for each entry separately!). Another try may be to use the alternate series for logarithm
let f=(a-1)*(a+1) ^-1
log(a) = f/1 + f^3/3 + f^5/5 ...
(don't have it at hand, whether this series actually has alternating signs, but I think, it's correct). The same formula can be used for matrices A, where (A+I) is invertible and *all* the eigenvalues of (A+I) are in the admissible range for this series.

Anyway, it is worth to look at the partial sums; Euler-summation of the low order 2 can sum for instance 1-2+4-8+16-...+... which is even more divergent than many of our sums, requiring only few terms for a good approximation of the final result.

For a more detailed discussion I can do better in the evening.

Gottfried