# Power Series is Termwise Differentiable within Radius of Convergence

## Theorem

Let $\ds \map f x := \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about a point $\xi$.

Let $R$ be the radius of convergence of $S$.

Then:

$\ds \frac \d {\d x} \map f x = \sum_{n \mathop = 0}^\infty \frac \d {\d x} a_n x^n = \sum_{n \mathop = 1}^\infty n a_n x^{n - 1}$

## Proof

Let $\rho \in \R$ such that $0 \le \rho < R$.

From Power Series Converges Uniformly within Radius of Convergence, $\map f x$ is uniformly convergent on $\set {x: \size {x - \xi} \le \rho}$.

From Real Polynomial Function is Continuous, each of $\map {f_n} x = a_n x^n$ is a continuous function of $x$.

$\dfrac \d {\d x} x^n = n x^{n - 1}$

and again from Real Polynomial Function is Continuous, each of $\dfrac \d {\d x} \map {f_n} x = n a_n x^{n - 1}$ is a continuous function of $x$.

 $\ds \frac \d {\d x} \map f x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac \d {\d x} a_n x^n$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty n a_n x^{n - 1}$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty n a_n x^{n - 1}$ as $n a_n x^{n - 1} = 0$ when $n = 0$

$\blacksquare$