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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).

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    $\begingroup$ You certainly wouldn't have just $a^n$ in the $x^n$ term. For example, for $n=2$, $$\sum_{m=1}^2 (-1)^{m-1} (m-1)! S(2,m) a^m = a - a^2$$ $\endgroup$ Sep 2, 2015 at 19:20

3 Answers 3

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Changing the order of summation, we get $$ \sum_{n=1}^k \frac{x^n}{n!} \sum_{m=1}^n (-1)^{m-1} (m-1)!S(n,m) a^m$$ $$ = \sum_{m=1}^k (-1)^{m-1} (m-1)! a^m \sum_{n=m}^k \frac{x^n}{n!} S(n,m).$$
Using formula (13) from http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html , we further get $$= \sum_{m=1}^k (-1)^{m-1} (m-1)! a^m\ [y^k]\ \frac{(e^{xy}-1)^m}{m!(1-y)} $$ $$= [y^k]\ \frac{1}{1-y}\sum_{m=1}^k (-1)^{m-1} \frac{((e^{xy}-1)a)^m}{m}.$$ Since $e^{xy}-1 = xy + O(y^2)$, we can extend the summation to $+\infty$ without affecting the result (the extra terms have $y$ in powers $>k$). So, the original expression equals the coefficient of $y^k$ in the generating function: $$\frac{1}{1-y}\sum_{m=1}^{\infty} (-1)^{m-1} \frac{((e^{xy}-1)a)^m}{m}$$ $$ = \frac{\log(1+(e^{xy}-1)a)}{1-y}.$$

P.S. This also implies a closed-form expression for the infinite sum: $$\sum_{n=1}^\infty \frac{x^n}{n!} f^{(n)}(a) = \log(1+(e^{x}-1)a).$$

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Mathematica says that $$f^{(n)}(a) = (-1)^{n+1} \text{Li}_{1-n}\left(1-\frac{1}{a}\right).$$

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I don't know if this will help, but it seems that $f^{(n)}(a)$ consists of the terms in positive powers of $a$ for the asymptotic series of $$- (n-1)!\; (\ln(1-1/a))^{n}$$

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