# Differentiation of Power Series/Corollary

## Theorem

Let $\xi \in \R$ be a real number.

Let $\left \langle {a_n} \right \rangle$ be a sequence in $\R$.

Let $\displaystyle \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m$ be the power series in $x$ about the point $\xi$.

The value of $\displaystyle \frac {\mathrm d^n}{\mathrm d x^n} \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m$ at $x = \xi$ is:

$\displaystyle \left.{\frac {\mathrm d^n}{\mathrm d x^n} \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m}\right|_{x = \xi} = a_n n!$

## Proof

$\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m = \sum_{m \mathop \ge n} a_m m^{\underline n} \left({x - \xi}\right)^{m - n}$

where $m^{\underline n}$ denotes the falling factorial.

When $x = \xi$ all the terms in $\left({x - \xi}\right)^{m - n}$ vanish except when $m = n$.

When $m = n$, from Nth Derivative of Nth Power we have:

$\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} a_m \left({x - \xi}\right)^m = a_m n!$

But then $m = n$ and the result follows.

$\blacksquare$