Differentiation of Power Series

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

Let $\sequence {a_n}$ be a sequence in $\R$.

Let $\ds \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m$ be the power series in $x$ about the point $\xi$.

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

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

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

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

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

By hypothesis $x$ is within the interval of convergence.

It follows from Abel's Theorem that:
 * $\ds \frac {\d^n} {\d z^n} \sum_{m \mathop \ge 0} a_m z^m = \sum_{m \mathop \ge n} a_m m^{\underline n} z^{m - n}$

Then from Derivative of Identity Function and others, we have:
 * $\map {\dfrac \d {\d x} } {x - \xi} = 1$

The result follows from the Chain Rule for Derivatives.