MATH 152 Chapter 10.5

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Power Series

Definition

A series in the form of $\textstyle \sum _{n=0}^{\infty }{c_{n}\left(x-a\right)^{n}}$ such that

$s=c_{0}+c_{1}\left(x-a\right)+c_{2}\left(x-a\right)^{2}+c_{3}\left(x-a\right)^{3}+\dots$

$c_{n}$ represents coefficients

$a$ is the "center of power series"

A Step Further

Given a power series, determine the values of $x$ for which the series is convergent.

Every power series is convergent at $x=a$

Result: a function with input $x$ and output is the sum, whose domain is all values of $x$ for which the series is convergent.

domain is known as interval of convergence.

Example

Find the interval of convergence for $\sum _{n=0}^{\infty }{x^{n}}$

Almost always guaranteed to use the ratio test:

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {x^{n+1}}{x^{n}}}\right|\\&=\lim _{n\to \infty }\left|x\right|\\\left|x\right|&<1\\x&\in \left(-1,\,1\right)&{\mbox{(non-inclusive)}}\end{aligned}}$

Check x = ±1 separately ( Note: the ratio test is inconclusive when $\textstyle {\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=1}$)

both diverge by test for divergence in this example

The center of this interval $x\in \left(-1,\,1\right)$ is 0; $a$ just happens to be zero as well. Coincidence? I think not (and Descartes vanished from existence).

Note: When $x\in \left(-1,\,1\right)$, $\textstyle \sum _{n=0}^{\infty }{x^{n}}$ is a convergent geometric series

${\begin{aligned}a&=1\\r&=x\\f\left(x\right)&=1+x+x^{2}+x^{3}+\dots \\&={\frac {1}{1-x}}\end{aligned}}$

Wednesday, November 3, 2010

Interval and Radius of Convergence

There are three possible results when finding the interval of convergence:

$\sum _{n=0}^{\infty }{c_{n}\left(x-a\right)^{n}}{\begin{cases}{\mbox{1. converges only when}}&x=a\\{\mbox{2. converges for all}}&x\in \mathbb {R} \\{\mbox{3. converges on interval}}&x\in \left(a-r,\,a+r\right)\ {\mbox{or}}\ \left[a-r,\,a+r\right]\end{cases}}$

The radius of convergence of a power series is the largest number $r$ such that $\left(a-r,\,a+r\right)\subseteq$ (is a subset of) the interval of convergence:

1. Radius of Convergence is 0
2. Radius of Convergence is ∞
3. Radius of Convergence is $r$

Example 1

Find the radius and interval of convergence of $\sum _{n=0}^{\infty }{n!\,x^{n}}$

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {\left(n+1\right)!\,x^{n+1}}{n!\,x^{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {\left(n+1\right)\,{\cancel {n!}}\,x\,{\cancel {x^{n}}}}{{\cancel {n!}}\,{\cancel {x^{n}}}}}\right|\\&=\lim _{n\to \infty }\left(n+1\right)\,\left|x\right|<1\\&=\infty \;{\mbox{unless}}\;x=0\end{aligned}}$

Therefore, the interval of convergence is $x\in \left[0,\,0\right]$ and radius of convergence is 0.

Example 2

Find the radius and interval of convergence of $\sum _{n=1}^{\infty }{\frac {\left(x+1\right)^{n}}{\left(n+1\right)!}}$

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {\left(x+1\right)^{n+1}}{\left(n+2\right)!}}\,{\frac {\left(n+1\right)!}{\left(x+1\right)^{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {\left(x+1\right){\cancel {\left(x+1\right)^{n}}}}{\left(n+2\right){\cancel {\left(n+1\right)!}}}}\,{\frac {\cancel {\left(n+1\right)!}}{\cancel {\left(x+1\right)^{n}}}}\right|\\&=\lim _{n\to \infty }{\frac {\left|x+1\right|}{n+2}}\\&=0\;{\mbox{for all}}\;x\end{aligned}}$

Therefore the interval of convergence is $x\in \mathbb {R}$ and the radius of convergence is ∞

Example 3

Find the radius and interval of convergence of $\sum _{n=0}^{\infty }{\frac {2^{n}\left(x-3\right)^{n}}{\sqrt {n+3}}}$

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {2^{n+1}\left(x-3\right)^{n+1}}{\sqrt {n+4}}}\,{\frac {\sqrt {n+3}}{2^{n}\left(x-3\right)^{n}}}\right|&=\lim _{n\to \infty }2\left|x-3\right|{\frac {\sqrt {n+3}}{\sqrt {n+4}}}\\&=2\left|x-3\right|<1\\\left|x-3\right|&<{\frac {1}{2}}\\x-3&<{\frac {1}{2}}\\x-3&>-{\frac {1}{2}}\end{aligned}}$

Check each of the endpoints separately:

${\begin{aligned}\sum _{n=0}^{\infty }{\frac {2^{n}\left({\frac {5}{2}}-3\right)^{n}}{\sqrt {n+3}}}&=\sum _{n=0}^{\infty }{\frac {\left(-1\right)^{n}}{\sqrt {n+3}}}\quad {\mbox{Convergent by AST}}\\\sum _{n=0}^{\infty }{\frac {2^{n}\left({\frac {7}{2}}-3\right)^{n}}{\sqrt {n+3}}}&=\sum _{n=0}^{\infty }{\frac {1}{\sqrt {n+3}}}\quad {\mbox{Divergent by LCT with}}\;\textstyle \sum {\frac {1}{\sqrt {n}}}\end{aligned}}$

Therefore, the center is 3, the radius of convergence is ${\tfrac {1}{2}}$, and the interval of convergence is $x\in \left[{\tfrac {5}{2}},\,{\tfrac {7}{2}}\right)$