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:

$\ds \frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m = \sum_{m \mathop \ge n} a_m m^{\underline n} \paren {x - \xi}^{m - n}$

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

Corollary

The value of $\ds \frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m$ at $x = \xi$ is:

$\ds \valueat {\frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m} {x \mathop = \xi} = a_n n!$

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 $n$th Derivative of $m$th Power:

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

$x$ is by hypothesis 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.

$\blacksquare$

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