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

Then within the interval of convergence:

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

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

### Corollary

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

First we can make the substitution $z = x - \xi$ and convert the expression into:

- $\displaystyle \frac {\mathrm d^n} {\mathrm d x^n} \sum_{m \mathop \ge 0} a_m z^m$

We then use Nth Derivative of Mth Power:

- $\displaystyle \frac {\mathrm d^n} {\mathrm d z^n} z^m = \begin{cases} m^{\underline n} z^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$

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

It follows from Abel's Theorem that:

- $\displaystyle \frac {\mathrm d^n}{\mathrm dz^n} \sum_{m \mathop \ge 0} a_m z^m = \sum_{m \mathop \ge n} a_m m^{\underline n} z^{m - n}$

Then from Derivative of Identity Function and others, we have:

- $\displaystyle \frac {\mathrm d} {\mathrm d x} \left({x - \xi}\right) = 1$

The result follows from the Chain Rule.

$\blacksquare$