Derivative of Complex Power Series

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

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

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

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

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

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

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

We then use Nth Derivative of Mth Power:
 * $$\frac {d^n}{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:
 * $$\frac {d^n}{dx^n} \sum_{m \ge 0} a_m z^m = \sum_{m \ge n} a_m m^{\underline n} z^{m - n}$$

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

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

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