@Henryk
Did you forget a binomial coefficient somewhere? More specifically:
Andrew Robbins
Did you forget a binomial coefficient somewhere? More specifically:
Quote:Now we know that \( \eta^{\cdot j}(z)=\el^j (e^z-1)^j=\el^j\sum_{k=0}^j C_{jk} e^{kz}(-1)^{j-k} \)
\( \left(\eta^{\cdot j}\right)_n=\el^j \sum_{k=0}^j C_{jk}\frac{k^n}{n!}(-1)^{j-k}=\el^j \sum_{k=1}^j C_{jk} \frac{k^n}{n!}(-1)^{j-k} \)
Andrew Robbins

