Power Series

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.8 – Power Series

• Section 11.9 – Representations of Functions as Power Series

Expected Educational Results

• Objective 24–01: I can use the Geometric Series to rewrite functions as power series.

• Objective 24–02: I can differentiate a power series written in summation notation.

• Objective 24–03: I can explain why the index changes when differentiating a power series written in summation notation.

• Objective 24–04: I can integrate a power series written in summation notation.

• Objective 24–05: I can use the geometric series to find the radius of convergence, R, of a power series.

• Objective 24–06: I can use the geometric series to find the interval of convergence, I, of a power series.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Power Series

Radius, $R$$R$, and Interval of Convergence, $I$$I$, of a Power Series

Method 1: Geometric Series Test

Recall: A geometric series converges when: ${\displaystyle \left|r\right|<1}$$\displaystyle \left|\,r\,\right|<1$.

NOTE: When the geometric series is used to find a power series, the radius, $R$$R$, and interval of convergence, $I$$I$, are determined by the solution to: ${\displaystyle \left|\text{common ratio}\right|<1}$$\displaystyle \left|\text{common ratio}\right|<1$.

NOTE: When the geometric series is used to find a power series, the interval of convergence, $I$$I$, is always an open interval: ${\displaystyle (a,b)}$$\displaystyle (a,b)$.

Example 01: Find the radius, $R$$R$, and interval of convergence, $I$$I$, of the power series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}(3x{)}^n}$$\displaystyle \sum_{n=1}^{\infty}{(3x)^n}$.

Solution:

The power series is a geometric series since $3x$$3x$ is raised to the nth-power.

According to the definition of a geometric series, the common ratio of the power series is $r=3x$$r=3x$.

By the Geometric Series Test, the power series converges when

${\displaystyle \left|r\right|<1\Rightarrow \left|3x\right|<1\Rightarrow -1<3x<1\Rightarrow -{\displaystyle \frac{1}{3}}<x<{\displaystyle \frac{1}{3}}}$$\displaystyle \left|\,r\,\right|<1\Rightarrow \left|\,3x\,\right|<1\Rightarrow -1<3x<1\Rightarrow -\dfrac{1}{3}<x<\dfrac{1}{3}$.

In interval notation, the interval of convergence is $I=(-{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}})$$I=\left(-\dfrac{1}{3},\dfrac{1}{3}\right)$.

The radius of convergence is $R={\displaystyle \frac{{\displaystyle \frac{1}{3}}-(-{\displaystyle \frac{1}{3}})}{2}}={\displaystyle \frac{1}{3}}$$R=\dfrac{\dfrac{1}{3}-\left(-\dfrac{1}{3}\right)}{2}=\dfrac{1}{3}$

Use Technology to Solve Absolute Value Inequalities

NOTE: On all Assessments, you may use technology to solve Absolute Value Inequalities without showing any work.

Mathematica:

1
(* Example 01: Find the Interval of Convergence *)
2
Reduce[RealAbs[3x^2]<1,x]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. You may need parens, $($$(\,$ and $)$$\,)$, to group multiple terms in the numerator and denominator.

3. Reduce[ ] has at least two arguments:

   1. The absolute value inequality to be solved;

   2. The independent variable, $x$$x$.

4. For help on using the Reduce[ ] function:

   1. In Mathematica, execute the code: $\text{?Reduce}$$\text{?Reduce}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the Reduce[ ] function – you will be able to copy-paste code.

5. In Mathematica, Abs[3x^2] is $\left|3x^2\right|$$\left|\,3x^2\,\right|$; RealAbs[3x^2] considers only real values of the argument.

6. Remember, correct Mathematica code will be all black except for variables.

7. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Mathematica solution:

${\displaystyle -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}}$$\displaystyle -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Investigation 06

Use the geometric series to find the radius, $R$$R$, and interval of convergence, $I$$I$, of the following power series:

1. ${\displaystyle f(x)=\frac{2}{1+x}}$$\displaystyle f(x)=\frac{2}{1+x}$

2. ${\displaystyle f(x)=\frac{3}{1-x}}$$\displaystyle f(x)=\frac{3}{1-x}$

3. ${\displaystyle f(x)=\frac{1}{x-1}}$$\displaystyle f(x)=\frac{1}{x-1}$

4. ${\displaystyle f(x)=\frac{1}{1-3x^2}}$$\displaystyle f(x)=\frac{1}{1-3x^2}$

5. ${\displaystyle f(x)=\frac{1}{1+2x^3}}$$\displaystyle f(x)=\frac{1}{1+2x^3}$

6. ${\displaystyle f(x)=\frac{x}{1-x}}$$\displaystyle f(x)=\frac{x}{1-x}$

7. ${\displaystyle f(x)=\frac{x^3}{1-x^2}}$$\displaystyle f(x)=\frac{x^3}{1-x^2}$

Theorem: Radius, $R$$R$, of Convergence of a Power Series:

• When a power series converges at $x=a$$x=a$, then $R=0$$R=0$.

• When a power series converges for all $x$$x$, then $R=\mathrm{\infty }$$R=\infty$.

• When a power series converges for $R\in {\mathbb{R}}^{+}$$R\in\mathbb{R}^{+}$ then the radius of convergence is $R$$R$.

Homework

At this time, you should be able to complete the following assignments:

• Section 11.8: # 3, 5, 7, 9, 11, 13, 15, 19, 23, 25.

CC BY-NC-SA 4.0

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Created: Tuesday, 1 December 2020 6:40 EDT Last Modified: Saturday, 02 July 2022 - 09:13 (EDT)