📂Analysis

Differentiation of Power Series

Theorem¹

Suppose the power series $\sum\limits_{n = 0}^{\infty} c_{n}x^{n}$ converges for $\left| x \right| \lt R$. And define the function $f$ as follows.

$$ f(x) = \sum\limits_{n = 0}^{\infty} c_{n}x^{n} \qquad \left| x \right| \lt R \tag{1} $$

Then the function $f$ is continuous and differentiable on $(-R, R)$, and its derivative is as follows.

$$ f^{\prime}(x) = \sum\limits_{n = 1}^{\infty} nc_{n}x^{n-1} \qquad \left| x \right| \lt R \tag{2} $$

Moreover, the radii of convergence of $f$ and $f^{\prime}$ are the same.

Explanation

$(2)$ gives a result just like differentiating the infinite terms of $(1)$ term by term. In other words, when differentiating a power series, one may do it as if differentiating a polynomial function.

$$ \dfrac{d}{dx}\left[ \sum\limits_{n = 0}^{\infty} c_{n}(x - a)^{n} \right] = \sum\limits_{n = 0}^{\infty} \dfrac{d}{dx} c_{n}(x - a)^{n} $$

What must be noted is that the radius of convergence of $f^{\prime}$ is the same as that of $f$. This does not mean that the interval of convergence of $f$ and $f^{\prime}$ is the same; at the endpoints of the interval, the convergence may differ.

Proof

Since $\lim\limits_{n \to \infty}\sqrt[n]{n} = 1$,

$$ \limsup\limits_{n \to \infty} \sqrt[n]{n |c_{n}|} = \limsup\limits_{n \to \infty} \sqrt[n]{|c_{n}|} $$

Therefore, the radius of convergence of the series $\sum\limits_{n = 0}^{\infty} nc_{n}x^{n-1}$ is the same as that of $f$. Then, for a positive number $\epsilon \gt 0$, the series in $(2)$ converges uniformly on $[-R + \epsilon, R - \epsilon]$.

Uniform Convergence and Differentiability

Suppose a sequence of functions $\left\{ f_{n} : f_{n} \text{ is differentiable on } [a, b] \right\}$ differentiable on the interval $[a, b]$ converges pointwise at a point $x_{0} \in [a, b]$. If $\left\{ f_{n}^{\prime} \right\}$ converges uniformly on the interval $[a, b]$, then $\left\{ f_{n} \right\}$ also converges uniformly to a function $f$ differentiable on the interval $[a, b]$, and the following holds.

$$ \dfrac{d}{dx} \lim\limits_{n \to \infty} f_{n} (x) = \lim\limits_{n \to \infty} \dfrac{d}{dx} f_{n} (x) \quad a \le x \le b $$

Setting $f_{N}(x) = \sum\limits_{n = 0}^{N} c_{n}x^{n}$ and $f_{N}^{\prime}(x) = \sum\limits_{n = 1}^{N} nc_{n} x^{n-1}$, the conditions of the above lemma are satisfied. Therefore, for $ \left| x \right| \lt R$,

$$ \dfrac{df}{dx} = \dfrac{d}{dx} \lim\limits_{N \to \infty} \sum\limits_{n = 0}^{N}c_{n}x^{n} = \lim\limits_{N \to \infty} \dfrac{d}{dx} \sum\limits_{n = 0}^{N}c_{n}x^{n} = \lim\limits_{N \to \infty} \sum\limits_{n = 1}^{N}nc_{n}x^{n-1} = \sum\limits_{n = 1}^{\infty}nc_{n}x^{n-1} $$

Also, since differentiability implies continuity, $f$ is continuous.

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