I am currently trying to evaluate a complicated function $f$ at a point $x+a$ using a high order Taylor polynomial about the point $a$. The polynomial has the standard form:
$$\sum_{n=1}^k \frac{x^n}{n!} f^{(n)}(a)$$
However it takes a long time to evaluate because the $n^{th}$ derivative requires me to compute the value of a second summation:
$$f^{(n)}(a) = \sum_{m=1}^n (-1)^{m-1} (m-1)!S(n,m) a^m$$
Here, $S(n,m)$ denotes a Stirling number of the second kind and the integer sequence $(m-1)!S(n,m)$ is a well-documented integer sequence (A028246).
My question is as follows: it possible to simply this Taylor polynomial i.e.,
$$ \sum_{n=1}^k \frac{x^n}{n!} \sum_{m=1}^n (-1)^{m-1} (m-1)!S(n,m) a^m$$
into a single sum that has the form:
$$ \sum_{n=1}^k C_n x^n a^n$$
Here the $C_n$ are constants that do not depend on $x$ or $n$ (and could have closed form expressions or be computed using recursion).