Power Series

Tags

Calculus

Cegep/2

Word count

420 words

Reading time

3 minutes

Series with the form

$$\begin{array}{rl}S& =\sum _{n=0}^{\mathrm{\infty }}{C}_n(x-a{)}^n\\ S_N& =\sum _{n=0}^{N-1}{C}_n(x-a{)}^n\end{array}$$

+ Only three possibilities of convergence

1. The series converges only when $x=a$ ($R=0$).
2. The series converges for all $x\in \mathbb{R}$ ($R=\mathrm{\infty }$).
3. The series converges on the interval $(a-R,a+R)$.

$R$ is known as the **radius of convergence**.

Theorem

If $f(x)=\sum _{n=0}^{\mathrm{\infty }}{C}_n(x-a{)}^n$ with radius $R>0$, then $f$ is differentiable on the interval $(a-R,a+R)$.

Also,

1. $f^{\prime }(x)=\frac{\mathrm{d}}{\mathrm{d}x}[\sum _{n=0}^{\mathrm{\infty }}{C}_n(x-a{)}^n]=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{d}}{\mathrm{d}x}[C_n(x-a{)}^n]$
2. $\int f(x)\mathrm{d}x=\int \sum _{n=0}^{\mathrm{\infty }}{C}_n(x-a{)}^{n}dx=\sum _{n=0}^{\mathrm{\infty }}\int C_n(x-a)\mathrm{d}x$

With the same radius of convergence.

Examples

Consider the power series $\sum _{n=0}^{\mathrm{\infty }}n!x^n$.

1. What is the radius of convergence?

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|& =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{(n+1)!x^{n+1}}{n!x^n}\right|\\ & =\underset{n\to \mathrm{\infty }}{lim}|(n+1)x|\\ & =\mathrm{\infty }>1\end{array}$$

Therefore, the series is divergent by RT $\mathrm{\forall }x\ne 0$.

If $x=0$, the limit is

$$\underset{n\to \mathrm{\infty }}{lim}|(n+1)(0)|=\underset{n\to \mathrm{\infty }}{lim}|0|=0<1$$

Therefore, the series is convergent by RT. The radius of convergence is 0.

2. What is the interval on which the series converges?

Since it converges on a point, there is no interval.

3. What is the series centred?

$x=0$

Consider the power series $\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n}}{(2n)!}$.

1. What is the radius of convergence?

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\left|\frac{(-1{)}^n{x}^{2(n+1)}}{(2(n+1))!}\frac{(2n)!}{(-1{)}^n{x}^{2n}}\right|& =\underset{n\to \mathrm{\infty }}{lim}\left|\frac{x^2}{(2n+2)(2n+1)}\right|\\ & =0<1\end{array}$$

Therefore, the series is convergent by RT $\mathrm{\forall }x\in \mathbb{R}$. The radius of convergence is $\mathrm{\infty }$.

2. What is the interval on which the series converges?

$\mathbb{R}$

3. What is the series centred?

$x=0$

Contributors

Ajitani Hifumi

Changelog

Last edited about 2 hours ago

View full history

7a6541d-feat: feet