Power Series Revisited

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Definition (Power Series)

A power series centered at $x=a$ for $a\in \mathbb{R}$ is a series defined as

$$\sum _{n=0}^{\infty }{c}_n(x-a{)}^{n}x\in \mathbb{R},c_n\in \mathbb{C}$$

By the Radius of Convergence, we define

$$\alpha :=\underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{|c_n|}R=\{\begin{array}{ll}1/\alpha & \alpha \ne 0,\alpha \text{ exists}\\ \infty & \alpha =0\\ 0& \alpha =\infty \end{array}$$

We say that $R$ is the radius of convergence for a power series.

Theorem (Differentiation of Power Series)

Let the power series be

$$\sum _{n=0}^{\infty }{c}_n{x}^n$$

and let

$$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n|x|<R$$

The following statements are equivalent:

1. It converges in $(a-R,a+R)$ for $|x|<R$, and define
2. for any small $\epsilon >0$, the power series converges absolutely and uniformly in $[a-R+\epsilon ,a+R-\epsilon ]$.
3. The function $f$ is then continuous and differentiable in $(-R,R)$ and

$$f^{\prime }(x)=\sum _{n=0}^{\infty }n{c}_n{x}^{n-1}|x|<R$$

Proof: Part $(1)$ is proven by the root test.

Part $(2)$ is proven by the fact that $\forall x\in [a-R+\epsilon ,a+R-\epsilon ]$,

$$|c_n(x-a{)}^n|\le |c_n(R-\epsilon )|$$

and so $\sum |c_n(R-\epsilon )|$ converges by the Lemma (Comparison Test). By the Criterion 3 (Comparison Test), the convergence is uniform.

Part $(3)$: WTS $f^{\prime }(x)$ exists $\forall x\in (a-R,a+R)$. Let $\epsilon >0$ such that $x\in [x-R+\epsilon ,x+R-\epsilon ]$. Then $f$ converges uniformly on $[a-R+\epsilon ,a+R-\epsilon ]$ by part $(2)$.

Since $\sqrt[n]{n}\to 1$ as $n\to \infty$, we have

$$\underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{n|c_n|}=\underset{n\to \infty }{\mathrm{limsup}}\sqrt[n]{|c_n|}$$

so that the series have the same interval of convergence. Since the derivative is a power series, it converges uniformly in

$$[-R+\epsilon ,R-\epsilon ]\forall \epsilon >0$$

Then by Theorem (Term-by-Term Differentiation), to the series, we get

$$f^{\prime }(x)=\sum _{n=0}^{\infty }n{c}_n(x-a{)}^{n-1}$$

Corollary (Infinite Differentiablility)

A power series is infinitely differentiable in $(a-R,a+R)$ and $f^{(n)}$ can be calculated by term-wise differentiation. See Theorem (Term-by-Term Differentiation).

Furthermore,

$$f(x)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$$

we have

$$\begin{array}{rl}f(a)=c_0& ⟺c_0=f(a)\\ f^{\prime }(a)=c_1& ⟺c_1=f^{\prime }(a)\\ f^{\prime \prime }(a)=c_2\cdot 2!& ⟺c_2=f^{\prime \prime }(a)/2!\\ f^{(n)}(a)=c_{n}n!& ⟺c_n=f^{(n)}(a)/n!\end{array}$$

In particular, if $f$ is a power series around $x=a$, then

1. $f$ has to be infinitely differentiable
2. $f(x)=\sum f^{(n)}(a)\cdot (x-a{)}^n\cdot (n!{)}^{-1}$

Examples

1. Let $f(x)=x^2+1$. To expand as power series at $x=-1$, $$f(-1)=2f^{\prime }(-1)=-2f^{\prime \prime }(-1)=2$$ such that $f(x)=2-\frac{2}{1!}(x+1{)}^1+\frac{2}{2!}(x+1{)}^2$.
2. Let f(x) = \begin{cases} \exp(-1/x^{2}) & x \neq 0 \\ 0 & x = 0 \\

\end{cases} $$ then $f(0)=f^{\prime }(0)=f^{\prime \prime }(0)=\cdots f^{(n)(0)}=0$

In Relation to Complex-Valued Functions

$$\begin{array}{rl}\text{Power Series}\subset \text{Infinitely Differentiable Functions}& \subset \text{Differentiable Functions}\subset \\ \text{Continuous Functions}& \subset \text{Familiy of R→C and R→R functions}\end{array}$$

Proof is trivial.