matrix function like iteration without power series expansion

[andydude]

I guess I should be more specific. The key observation to show the equivalence is that
\( n!\left({s \atop n}\right) = \prod_{k=0}^{n-1}(s-k) \)
which means Woon's formula simplifies to:
\(
A^s = w^s
\left(
1 +
\sum_{n=1}^{\infty} (-1)^n \left({s \atop n}\right)
\left[
1 +
\sum_{m=1}^{n} (-1)^m w^{-m} \left({n \atop m}\right) A^m
\right]
\right)
\)
and letting \( w=1 \) this simplifies to
\(
A^s =
\left(
1 +
\sum_{n=1}^{\infty} (-1)^n \left({s \atop n}\right)
\left[
1 +
\sum_{m=1}^{n} (-1)^m \left({n \atop m}\right) A^m
\right]
\right)
\)
and assuming \( 0^0 = 1 \) this simplifies to
\(
A^s =
\left(
\sum_{n=0}^{\infty} (-1)^n \left({s \atop n}\right)
\left[
\sum_{m=0}^{n} (-1)^m \left({n \atop m}\right) A^m
\right]
\right)
\)
and combining (-1) terms, this simplifies to
\(
A^s =
\sum_{n=0}^{\infty} \left({s \atop n}\right)
\sum_{m=0}^{n} (-1)^{n+m} \left({n \atop m}\right) A^m
\)
and that is, as you call it, "Henryk's formula" from Woon's formula.

Andrew Robbins