# Power Series is Termwise Differentiable within Radius of Convergence

## Theorem

Let $\displaystyle f \left({x}\right) := \sum_{n \mathop = 0}^\infty a_n \left({x - \xi}\right)^n$ be a power series about a point $\xi$.

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

Then:

$\displaystyle \frac \d {\d x} f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac \d {\d x} a_n x^n = \sum_{n \mathop = 1}^\infty n a_n x^{n - 1}$

## Proof

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

From Power Series Converges Uniformly within Radius of Convergence, $f \left({x}\right)$ is uniformly convergent on $\left\{{x: \left|{x - \xi}\right| \le \rho}\right\}$.

From Polynomial is Continuous, each of $f_n \left({x}\right) = a_n x^n$ is a continuous function of $x$.

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

and again from Polynomial is Continuous, each of $\dfrac \d {\d x} f_n \left({x}\right) = n a_n x^{n-1}$ is a continuous function of $x$.

 $\displaystyle \frac \d {\d x} f \left({x}\right)$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac \d {\d x} a_n x^n$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty n a_n x^{n - 1}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty n a_n x^{n - 1}$ as $n a_n x^{n - 1} = 0$ when $n = 0$

$\blacksquare$