Math 129 Section 005H Lecture 30: Taylor series (§10.2)¹¹ 1 This document is licensed under a Creative Commons Attribution 3.0 United States License Examples

Note on radius of convergence

(Monday, November 1, 2021)

One of you asked today whether the radius of convergence of the Taylor series depends on its center $a$ . The answer is “it depends.” One can show that if

f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}

for all real $x$ , i.e., the radius of convergence $R$ of the seris on the right is $\infty$ , and $f(x)$ equals its Taylor series, then the radius of convergence is also $\infty$ for the Taylor series

f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^{n}

of $f$ about any other point $a$ .

This is not true for Taylor series with finite radius of convergence. For example, consider

f(x)=\frac{1}{x}.

Its Taylor series about $x=1$ is easy to find without differentiating, by a little trickery:

	$\displaystyle\frac{1}{x}$	$\displaystyle=$	$\displaystyle\frac{1}{(x-1)+1}$
		$\displaystyle=$	$\displaystyle\frac{1}{a}\cdot\frac{1}{1+y}$

where $y=x-1$ . But the last expression is the sum of a geometricq series:

	$\displaystyle\frac{1}{1+y}$	$\displaystyle=$	$\displaystyle 1-y+y^{2}-y^{3}+\cdots$
		$\displaystyle=$	$\displaystyle 1-(x-1)+(x-1)^{2}-(x-1)^{3}+\cdots$

\displaystyle\frac{1}{x}

\displaystyle=

\displaystyle 1-(x-1)+(x-1)^{2}-(x-1)^{3}+\cdots

We know this converges if and only if $|y|<1$ , or

\left|x1\right|<1

so the radius of convergence of the Taylor series of $1/x$ about $x=1$ is just $R=1$ .

Let’s now do this about $x=2$ :

$\displaystyle\frac{1}{x}$	$\displaystyle=$	$\displaystyle\frac{1}{(x-2)+2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\cdot\frac{1}{1+(x-2)/2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\cdot\frac{1}{1+y}$

where now $y=(x-2)/2$ . Again, the last expression is a geometric series:

	$\displaystyle\frac{1}{1+y}$	$\displaystyle=$	$\displaystyle 1-y+y^{2}-y^{3}+\cdots$
		$\displaystyle=$	$\displaystyle 1-\Big{(}\frac{x-2}{2}\Big{)}+\Big{(}\frac{x-2}{2}\Big{)}^{2}-% \Big{(}\frac{x-2}{2}\Big{)}^{3}+\cdots$

\displaystyle\frac{1}{x}

\displaystyle=

\displaystyle\frac{1}{2}\left(1-\Big{(}\frac{x-2}{2}\Big{)}+\Big{(}\frac{x-2}{% 2}\Big{)}^{2}-\Big{(}\frac{x-2}{2}\Big{)}^{3}+\cdots\right)

This converges if and only if $|y|<1$ , or

\left|\frac{x-2}{2}\right|<1

which is equivalent to

|x-2|<2.

So the radius of convergence is $R=2$ .

In general, the Taylor series of $1/x$ about $x=a$ has radius of convergence $R=|a|$ . The reason for this is actually the vertical asymptote at $x=0$ : because the function itself diverges at $x=0$ , there is no way its Taylor series can converge. (What can it possibly converge to?) So even without the analysis above, we know that the radius of convergence of the Taylor series of $1/x$ about $x=a$ cannot possibly be larger than $|a|$ .