9.1: Power Series and Functions

Section Preview

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• Solving Inequalities
  • Solving Absolute Value Inequalities: Conceptually, the interval of convergence is the distance-based inequality \(|x-a| \lt R\); the "all points within distance \(R\) of the center" model is what makes the radius-of-convergence picture coherent.

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Identify a power series and provide examples of them.
• Determine the radius of convergence and interval of convergence of a power series.
• Use a power series to represent a function.

A power series is a series with terms involving a variable. More specifically, if the variable is \(x\), then all the terms of the series involve powers of \(x\). As a result, a power series can be thought of as an infinite polynomial. Power series are used to represent common functions and also to define new functions. In this section, we define power series and show how to determine when a power series converges and when it diverges. We also show how to represent certain functions using power series.

Form of a Power Series

Definition: Power Series

A series of the form\[\sum_{n=0}^\infty c_n x^n = c_0+c_1x+c_2x^2+\cdots\nonumber\]is called a power series centered at \(x=0\). A series of the form\[\sum_{n=0}^\infty c_n(x-a)^n = c_0+c_1(x-a)+c_2(x-a)^2+\cdots\nonumber\]is a power series centered at \(x=a\).

To make this definition precise, we stipulate that \(x^0 = 1\) and \((x-a)^0 = 1\) even when \(x=0\) and \(x=a\), respectively. It is incredibly important to understand that \(0^0\) is still considered indeterminate for all other purposes within Mathematics. Defining \(x^0 = 1\) even when \(x = 0\) is a useful but unique convention for power series.

The series\[1+x+x^2+\cdots =\sum_{n=0}^\infty x^n,\nonumber\]\[\sum_{n=0}^\infty \dfrac{x^n}{n!} = 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots,\nonumber\]and\[\sum_{n=0}^\infty n! x^n = 1+x+2!x^2+3!x^3+\cdots\nonumber\]are all examples of power series centered at \(x = 0\). The first of these is a geometric series with common ratio \(r = |x|\), and we know that it converges if \(|x| \lt 1\) and diverges if \(|x| \geq 1\).

The series\[\sum_{n=0}^ \infty \dfrac{(x-2)^n}{(n+1)3^n}=1+\dfrac{x-2}{2 \cdot 3}+\dfrac{(x-2)^2}{3 \cdot 3^2}+\dfrac{(x-2)^3}{4 \cdot 3^3}+\cdots\nonumber\]is a power series centered at \(x=2\).

Convergence of a Power Series

Since the terms in a power series involve a variable \(x\), the series may converge for certain values of \(x\) and diverge for other values of \(x\). For a power series centered at \(x=a\), the value of the series at \(x=a\) is given by \(c_0\). Therefore, a power series always converges at its center. Some power series converge only at that value of \(x\). Most power series, however, converge for more than one value of \(x\). In that case, the power series either converges for all real numbers \(x\) or converges for all \(x\) in a finite interval. For example, the geometric series \(\displaystyle \sum_{n=0}^ \infty x^n\) converges for all \(x\) in the interval \((-1,1)\), but diverges for all \(x\) outside that interval. We now summarize these three possibilities for a general power series.

Theorem: Convergence of a Power Series

The power series \(\displaystyle \sum_{n=0}^ \infty c_n(x-a)^n\) satisfies exactly one of the following properties:

1. The series converges at \(x=a\) and diverges for all \(x \neq a\).
2. The series converges for all real numbers \(x\).
3. There exists a real number \(R \gt 0\) such that the series converges if \(|x-a| \lt R\) and diverges if \(|x-a| \gt R\). At the values \(x\) where \(|x-a|=R\), the series could converge or diverge.

Proof

Suppose that the power series is centered at \(a=0\). We must first prove the following fact:

If there exists a real number \(d \neq 0\) such that \(\displaystyle \sum_{n=0}^\infty c_n d^n\) converges, then the series \(\displaystyle \sum_{n=0}^ \infty c_nx^n\) converges absolutely for all \(x\) such that \(|x| \lt |d|\).

Suppose, then, that \(\displaystyle \sum_{n=0}^ \infty c_nd^n\) converges. Then the \(n^{\text{th}}\) term \(c_nd^n \to 0\) as \(n \to \infty\). Therefore, there exists an integer \(N\) such that \(|c_nd^n| \leq 1\) for all \(n \geq N\). Writing\[|c_nx^n|=|c_nd^n| \left|\dfrac{x}{d}\right|^n,\nonumber\]we conclude that, for all \(n \geq N\),\[|c_nx^n| \leq \left|\dfrac{x}{d}\right|^n.\nonumber\]

The series\[\sum_{n=N}^ \infty \left|\dfrac{x}{d}\right|^n\nonumber\]is a geometric series that converges if \(\left|\frac{x}{d}\right| \lt 1\). Therefore, by the Series Comparison Test, we conclude that \(\displaystyle \sum_{n=N}^ \infty c_n x^n\) also converges for \(|x| \lt |d|\). Since we can add a finite number of terms to a convergent series, we conclude that \(\displaystyle \sum_{n=0}^ \infty c_n x^n\) converges for \(|x| \lt |d|\).

With this result, we can now prove the theorem. Consider the series\[\sum_{n=0}^ \infty a_nx^n\nonumber\]and let \(S\) be the set of real numbers for which the series converges. Suppose that the set contains only the center. That is, \(S=\{0\}\). Then the series falls under case i.

Suppose that the set \(S\) is the set of all real numbers. Then the series falls under case ii.

Suppose that \(S \neq {0}\) and \(S\) is not the set of real numbers. Then there exists a real number \(x^* \neq 0\) such that the series does not converge. Thus, the series cannot converge for any \(x\) such that \(|x| \gt |x^*|\). Therefore, the set \(S\) must be a bounded set, meaning it must have a smallest upper bound (this fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in Real Analysis). Call that smallest upper bound \(R\). Since \(S \neq \{0\}\), the number \(R \gt 0\). Therefore, the series converges for all \(x\) such that \(|x| \lt R\), and the series falls into case iii.

A more general proof

For a series centered at a value of \(a\) other than zero, the proof is adjusted by letting \(y=x-a\) and considering the series\[\sum_{n=0}^ \infty c_ny^n.\nonumber\]

If a series \(\displaystyle \sum_{n=0}^ \infty c_n(x-a)^n\) falls into case iii., then the series converges for all \(x\) such that \(|x-a| \lt R\) for some \(R \gt 0\), and diverges for all \(x\) such that \(|x-a| \gt R\). The series may converge or diverge at the values \(x\) where \(|x-a| = R\). The set of values \(x\) for which the series \(\displaystyle \sum_{n=0}^ \infty c_n(x-a)^n\) converges is known as the interval of convergence. Since the series diverges for all values \(x\) where \(|x-a| \gt R\), the length of the interval of convergence is \(2R\), and therefore, the radius of the interval is \(R\). The value \(R\) is called the radius of convergence.

For example, since the series \(\displaystyle \sum_{n=0}^ \infty x^n\) converges for all values \(x\) in the interval \((-1,1)\) and diverges for all values \(x\) such that \(|x| \geq 1\), the interval of convergence of this series is \((-1,1)\). Since the length of the interval is \(2\), the radius of convergence is \(1\).

We formalize these statements in the following definition.

Definition: Radius of Convergence

Consider the power series \(\displaystyle \sum_{n=0}^ \infty c_n(x-a)^n\). The set of real numbers \(x\) where the series converges is called the interval of convergence. If there exists a real number \(R \gt 0\) such that the series converges for \(|x-a| \lt R\) and diverges for \(|x-a| \gt R\), then \(R\) is called the radius of convergence. If the series converges only at \(x=a\), we say the radius of convergence is \(R=0\). If the series converges for all real numbers \(x\), we say the radius of convergence is \(R= \infty\) (see Figure \(\PageIndex{1}\)).

For a series \(\displaystyle \sum_{n=0}^ \infty c_n(x-a)^n\)
Figure \(\PageIndex{1A}\): A radius of convergence of \(R=0\)
Figure \(\PageIndex{1B}\): A radius of convergence of \(R= \infty\)
Figure \(\PageIndex{1C}\): A radius of convergence of \(R\), where the series may or may not converge at the endpoints \(x=a+R\) and \(x=a-R\).

To determine the interval of convergence for a power series, we typically apply the Ratio Test. In Example \(\PageIndex{1}\), we show the three different possibilities illustrated in Figure \(\PageIndex{1}\).

Example \(\PageIndex{1}\): Finding the Interval and Radius of Convergence

For each of the following series, find the interval and radius of convergence.

1. \(\displaystyle \sum_{n=0}^ \infty \frac{x^n}{n!}\)
2. \(\displaystyle \sum_{n=0}^ \infty n!x^n\)
3. \(\displaystyle \sum_{n=0}^ \infty \frac{(x-2)^n}{(n+1)3^n}\)

Solutions

1. To check for convergence, apply the Ratio Test. We have\[\begin{array}{rcl} \rho & = & \displaystyle \lim_{n \to \infty} \left|\dfrac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}\right| \\[6pt] & = & \displaystyle \lim_{n \to \infty} \left|\dfrac{x^{n+1}}{(n+1)!} \cdot \dfrac{n!}{x^n}\right| \\[6pt] & = & \displaystyle \lim_{n \to \infty}\left|\dfrac{x^{n+1}}{(n+1) \cdot n!} \cdot \dfrac{n!}{x^n}\right| \\[6pt] & = & \displaystyle \lim_{n \to \infty}\left|\dfrac{x}{n+1}\right| \\[6pt] & = & |x| \displaystyle \lim_{n \to \infty}\dfrac{1}{n+1} \\[6pt] & = & 0 \\[6pt] & \lt & 1 \\[6pt] \end{array}\nonumber\]for all values of \(x\). Therefore, the series converges for all real numbers \(x\). The interval of convergence is \((- \infty , \infty )\) and the radius of convergence is \(R= \infty\).
2. Apply the Ratio Test. For \(x \neq 0\), we see that\[\begin{array}{rcl} \rho & = & \displaystyle \lim_{n \to \infty}\left|\dfrac{(n+1)!x^{n+1}}{n!x^n}\right| \\[6pt] & = & \displaystyle \lim_{n \to \infty}|(n+1)x| \\[6pt] & = & |x| \displaystyle \lim_{n \to \infty}(n+1) \\[6pt] & = & \infty . \\[6pt] \end{array}\nonumber\]Therefore, the series diverges for all \(x \neq 0\). Since the series is centered at \(x=0\), it must converge there, so it converges only for \(x = 0\). The interval of convergence is the single value \(x=0\) and the radius of convergence is \(R=0\).
3. In order to apply the Ratio Test, consider\[\begin{array}{rcl} \rho & = & \displaystyle \lim_{n \to \infty}\left|\dfrac{\frac{(x-2)^{n+1}}{(n+2)3^{n+1}}}{\frac{(x-2)^n}{(n+1)3^n}}\right| \\[6pt] & = & \displaystyle \lim_{n \to \infty} \left|\dfrac{(x-2)^{n+1}}{(n+2)3^{n+1}} \cdot \dfrac{(n+1)3^n}{(x-2)^n}\right| \\[6pt] & = & \displaystyle \lim_{n \to \infty} \left|\dfrac{(x-2)(n+1)}{3(n+2)}\right|\\[6pt] & = & \dfrac{|x-2|}{3}. \\[6pt] \end{array}\nonumber\]The ratio \(\rho \lt 1\) if \(|x-2| \lt 3\). Since \(|x-2| \lt 3\) implies that \(-3 \lt x-2 \lt 3\), the series converges absolutely if \(-1 \lt x \lt 5\). The ratio \(\rho \gt 1\) if \(|x-2| \gt 3\). Therefore, the series diverges if \(x \lt -1\) or \(x \gt 5\). The Ratio Test is inconclusive if \(\rho =1\). The ratio \(\rho =1\) if and only if \(x=-1\) or \(x=5\). We need to test these values of \(x\) separately. For \(x=-1\), the series is given by\[\sum_{n=0}^ \infty \dfrac{(-1)^n}{n+1}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots .\nonumber\]Since this is the Alternating Harmonic Series, it converges. Thus, the series converges at \(x=-1\). For \(x=5\), the series is given by\[\sum_{n=0}^ \infty \dfrac{1}{n+1}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ldots .\nonumber\]This is the Harmonic Series, which is divergent. Therefore, the power series diverges at \(x=5\). We conclude that the interval of convergence is \([-1,5)\) and the radius of convergence is \(R=3\).

Checkpoint \(\PageIndex{1}\)

Find the interval and radius of convergence for the series\[\sum_{n=1}^ \infty \dfrac{x^n}{\sqrt{n}}.\nonumber\]

Answer

The interval of convergence is \([-1,1)\). The radius of convergence is \(R=1\).

Representing Functions as Power Series

Being able to represent a function by an "infinite polynomial" is a powerful tool. Polynomial functions are the most manageable functions to analyze since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is:

When can we represent a function by a power series?

Consider again the geometric series\[1+x+x^2+x^3+\cdots =\sum_{n=0}^ \infty x^n. \label{geoseries}\]Recall that the geometric series\[a+ar+ar^2+ar^3+\cdots\nonumber\]converges if and only if \(|r| \lt 1\). In that case, it converges to \(\frac{a}{1-r}\). Therefore, if \(|x| \lt 1\), the series in Equation \(\ref{geoseries}\) converges to \(\frac{1}{1-x}\) and we write\[1+x+x^2+x^3+\cdots =\dfrac{1}{1-x} \text{ for } |x| \lt 1.\nonumber\]As a result, we can represent the function \(f(x)=\frac{1}{1-x}\) by the power series\[1+x+x^2+x^3+\cdots \text{ when } |x| \lt 1.\nonumber\]We now show graphically how this series provides a representation for the function \(f(x)=\frac{1}{1-x}\) by comparing the graph of \(f\) with the graphs of several of the partial sums of this infinite series.

Example \(\PageIndex{2}\): Graphing a Function and Partial Sums of its Power Series

Sketch a graph of \(f(x)=\frac{1}{1-x}\) and the graphs of the corresponding partial sums \(\displaystyle S_N(x)=\sum_{n=0}^Nx^n\) for \(N=2,4,6\) on the interval \((-1,1)\). Comment on the approximation \(S_N\) as \(N\) increases.

Solution

From the graph in Figure \(\PageIndex{2}\), you see that as \(N\) increases, \(S_N\) becomes a better approximation for \(f(x)=\frac{1}{1-x}\) for \(x\) in the interval \((-1,1)\).

Figure \(\PageIndex{2}\): The graph shows a function and three approximations by partial sums of a power series.

Interactive Element: Partial Sums Closing In on \( \frac{1}{1-x} \)

The following Interactive Element might help you visualize how the partial sums \( S_N(x)=\displaystyle\sum_{n=0}^{N}x^n \) approximate \( f(x)=\frac{1}{1-x} \), and where they fail.

Interact: Drag the green point along the track at the bottom to change \( N \), the number of terms in the partial sum \( S_N \).

Observation: Inside the interval of convergence \( (-1,1) \), raising \( N \) makes \( S_N \) hug \( f(x)=\frac{1}{1-x} \) ever more closely; outside it (past \( x=\pm 1 \)), adding terms drives \( S_N \) away from \( f \) — the series diverges.

Checkpoint \(\PageIndex{2}\)

Sketch a graph of \(f(x)=\frac{1}{1-x^2}\) and the corresponding partial sums \(\displaystyle S_N(x)=\sum_{n=0}^Nx^{2n}\) for \(N=2,4,6\) on the interval \((-1,1)\).

Answer

Figure \(\PageIndex{3}\)

Next, we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.

Example \(\PageIndex{3}\): Representing a Function with a Power Series

Use a power series to represent each of the following functions \(f\). Find the interval of convergence.

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

Solutions

1. You should recognize this function \(f\) as the sum of a geometric series, because\[\dfrac{1}{1+x^3}=\dfrac{1}{1-(-x^3)}.\nonumber\]Using the fact that, for \(|r| \lt 1\), \(\frac{a}{1-r}\) is the sum of the geometric series\[\sum_{n=0}^ \infty a r^n = a+ar+ar^2+\cdots ,\nonumber\]we see that, for \(|-x^3| \lt 1\),\[\begin{array}{rcl} \dfrac{1}{1+x^3} & = & \dfrac{1}{1-(-x^3)} \\[6pt] & = & \sum_{n=0}^ \infty (-x^3)^n \\[6pt] & = & 1-x^3+x^6-x^9+\cdots . \\[6pt] \end{array}\nonumber\]Since this series converges if and only if \(|-x^3| \lt 1\), the interval of convergence is \((-1,1)\), and we have\[\dfrac{1}{1+x^3}=1-x^3+x^6-x^9+\cdots \text{ for } |x| \lt 1.\nonumber\]
2. This function is not in the exact form of a sum of a geometric series. However, we can relate \(f\) to a geometric series with a little algebraic manipulation. By factoring \(4\) out of the two terms in the denominator, we obtain\[\begin{array}{rcl} \dfrac{x^2}{4-x^2} & = & \dfrac{x^2}{4(\frac{1-x^2}{4})} \\[6pt] & = & \dfrac{x^2}{4(1-(\frac{x}{2})^2)}. \\[6pt] \end{array}\nonumber\]Therefore, we have\[\begin{array}{rcl} \dfrac{x^2}{4-x^2} & = & \dfrac{x^2}{4(1-(\frac{x}{2})^2)} \\[6pt] & = & \dfrac{\frac{x^2}{4}}{1-(\frac{x}{2})^2} \\[6pt] & = & \displaystyle \sum_{n=0}^ \infty \dfrac{x^2}{4} \left(\dfrac{x}{2}\right)^{2n}. \\[6pt] \end{array}\nonumber\]The series converges as long as \(\left|(\frac{x}{2})^2\right| \lt 1\) (note that when \(\left|(\frac{x}{2})^2\right|=1\) the series does not converge). Solving this inequality, we conclude that the interval of convergence is \((-2,2)\) and\[\begin{array}{rcl} \dfrac{x^2}{4-x^2} & = & \displaystyle \sum_{n=0}^ \infty \dfrac{x^{2n+2}}{4^{n+1}}\\[6pt] & = & \dfrac{x^2}{4}+\dfrac{x^4}{4^2}+\dfrac{x^6}{4^3}+ \cdots \\[6pt] \end{array}\nonumber\]for \(|x| \lt 2\).

Checkpoint \(\PageIndex{3}\)

Represent the function \(f(x)=\frac{x^3}{2-x}\) using a power series and find the interval of convergence.

Answer

\(\displaystyle \sum_{n=0}^ \infty \frac{x^{n+3}}{2^{n+1}}\) with interval of convergence \((-2,2)\)

In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions. We also show how we can use these representations to evaluate, differentiate, and integrate various functions.

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Miscellaneous Topics: Thinking About Mathematics (Part 3)

When I would get frustrated in school, I often found it good to reflect upon some of my favorite teachers and think about how they would approach the subject.

- Professor Simpson, Cosumnes River College

1. Who was your favorite teacher, and what subject did they teach?
2. Can you make any suggestions I can incorporate into the classroom that worked for that teacher?

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Success in College: Self-Motivation Learning Preferences

Choose the statements that best describe you.

I come to class because I don’t want to miss anything.	or	I come to class because it is required.
I would probably do less work in this class if I were graded pass/fail instead of a letter grade.	or	I would probably do the same amount of work in this class if I was graded pass/fail instead of with a letter grade.
I need frequent homework assignments to keep me working.	or	I do homework frequently to know what is going on in class.

1. Do you consider the choices above to be good or bad for your education?
2. Some professors provide detailed schedules with assignments and test dates at the beginning of the term. Others announce the homework assignments at each class meeting and test days during the term. Which do you prefer?