Differentiation of Power Series

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

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

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

Then within the interval of convergence:
 * $\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} \sum_{m \ge 0} a_m \left({x - \xi}\right)^m = \sum_{m \ge n} a_m m^{\underline n} \left({x - \xi}\right)^{m - n}$

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

Corollary
The value of $\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} \sum_{m \ge 0} a_m \left({x - \xi}\right)^m$ at $x = \xi$ is:
 * $\displaystyle \left.{\frac {\mathrm d^n}{\mathrm dx^n} \sum_{m \ge 0} a_m \left({x - \xi}\right)^m}\right|_{x = \xi} = a_n n!$

Proof
First we can make the substitution $z = x - \xi$ and convert the expression into:
 * $\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} \sum_{m \ge 0} a_m z^m$

We then use Nth Derivative of Mth Power:
 * $\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} z^m = \begin{cases}

m^{\underline n} z^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$

from which it immediately follows that:
 * $\displaystyle \frac {\mathrm d^n}{\mathrm dx^n} \sum_{m \ge 0} a_m z^m = \sum_{m \ge n} a_m m^{\underline n} z^{m - n}$

Then from Derivative of Identity Function etc. we have:
 * $\displaystyle \frac {\mathrm d} {\mathrm dx} \left({x - \xi}\right) = 1$

The result follows from the Chain Rule.

Proof of Corollary
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 Mth Power: Corollary 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.