# 10.1: Power Series and Functions

Learning Objectives

• 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 type of 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

A series of the form

[sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+ldots ,]

where (x) is a variable and the coefficients (c_n) are constants, is known as a power series. The series

[1+x+x^2+ldots =sum_{n=0}^∞x^n]

is an example of a power series. Since this series is a geometric series with ratio (r=|x|), we know that it converges if (|x|<1) and diverges if (|x|≥1.)

Definition (PageIndex{1}): Power Series

A series of the form

[sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+ldots ]

is a power series centered at (x=0.) A series of the form

[sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+ldots ]

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.

The series

[sum_{n=0}^∞dfrac{x^n}{n!}=1+x+dfrac{x^2}{2!}+dfrac{x^3}{3!}+ldots ]

and

[sum_{n=0}^∞n!x^n=1+x+2!x^2+3!x^3+ldots ]

are both power series centered at (x=0.) The series

[sum_{n=0}^∞dfrac{(x−2)^n}{(n+1)3^n}=1+dfrac{x−2}{2⋅3}+dfrac{(x−2)^2}{3⋅3^2}+dfrac{(x−2)^3}{4⋅3^3}+ldots ]

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}^∞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.

Note (PageIndex{1}): Convergence of a Power Series

Consider the power series (displaystyle sum_{n=0}^∞c_n(x−a)^n.) The series satisfies exactly one of the following properties:

1. The series converges at (x=a) and diverges for all (x≠a.)
2. The series converges for all real numbers (x).
3. There exists a real number (R>0) such that the series converges if (|x−a|R). At the values (x) where |x−a|=R, the series may converge or diverge.

Proof

Suppose that the power series is centered at (a=0). (For a series centered at a value of a other than zero, the result follows by letting (y=x−a) and considering the series

[ sum_{n=1}^∞c_ny^n. onumber]

We must first prove the following fact:

If there exists a real number (d≠0) such that (displaystyle sum_{n=0}^∞c_nd^n) converges, then the series (displaystyle sum_{n=0}^∞c_nx^n) converges absolutely for all (x) such that (|x|<|d|.)

Since (displaystyle sum_{n=0}^∞c_nd^n) converges, the nth term (c_nd^n→0) as (n→∞). Therefore, there exists an integer (N) such that (|c_nd^n|≤1) for all (n≥N.) Writing

[|c_nx^n|=|c_nd^n| left|dfrac{x}{d} ight|^n, onumber]

we conclude that, for all n≥N,

[|c_nx^n|≤left|dfrac{x}{d} ight|^n. onumber]

The series

[sum_{n=N}^∞left|dfrac{x}{d} ight|^n onumber]

is a geometric series that converges if (|dfrac{x}{d}|<1.) Therefore, by the comparison test, we conclude that (displaystyle sum_{n=N}^∞c_nx^n) also converges for (|x|<|d|). Since we can add a finite number of terms to a convergent series, we conclude that (displaystyle sum_{n=0}^∞c_nx^n) converges for (|x|<|d|.)

With this result, we can now prove the theorem. Consider the series

[sum_{n=0}^∞a_nx^n onumber]

and let (S) be the set of real numbers for which the series converges. Suppose that the set (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≠{0}) and (S) is not the set of real numbers. Then there exists a real number (x*≠0) such that the series does not converge. Thus, the series cannot converge for any (x) such that (|x|>|x*|). Therefore, the set (S) must be a bounded set, which means that 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 courses.) Call that smallest upper bound (R). Since (S≠{0}), the number (R>0). Therefore, the series converges for all (x) such that (|x|

If a series (displaystyle sum_{n=0}^∞c_n(x−a)^n) falls into case iii. of Note, then the series converges for all (x) such that (|x−a|0), and diverges for all (x) such that (|x−a|>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}^∞c_n(x−a)^n) converges is known as the interval of convergence. Since the series diverges for all values (x) where (|x−a|>R), the length of the interval 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}^∞x^n) converges for all values (x) in the interval ((−1,1)) and diverges for all values (x) such that (|x|≥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).

Consider the power series (displaystyle sum_{n=0}^∞c_n(x−a)^n). The set of real numbers (x) where the series converges is the interval of convergence. If there exists a real number (R>0) such that the series converges for (|x−a|R,) then (R) is 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=∞) (Figure (PageIndex{1})).

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}^∞dfrac{x^n}{n!})
2. (displaystyle sum_{n=0}^∞n!x^n)
3. (displaystyle sum_{n=0}^∞dfrac{(x−2)^n}{(n+1)3^n})

Solution

a. To check for convergence, apply the ratio test. We have

[ egin{align*} ρ &=lim_{n→∞} left|dfrac{dfrac{x^{n+1}}{(n+1)!}}{dfrac{x^n}{n!}} ight| [4pt]
&=lim_{n→∞} left|dfrac{x^{n+1}}{(n+1)!}⋅dfrac{n!}{x^n} ight| [4pt]
&=lim_{n→∞}left|dfrac{x^{n+1}}{(n+1)⋅n!}⋅dfrac{n!}{x^n} ight| [4pt]
&=lim_{n→∞}left|dfrac{x}{n+1} ight| [4pt]
&=|x|lim_{n→∞}dfrac{1}{n+1} [4pt]
&=0<1end{align*}]

for all values of (x). Therefore, the series converges for all real numbers (x). The interval of convergence is ((−∞,∞)) and the radius of convergence is (R=∞.)

b. Apply the ratio test. For (x≠0), we see that

[ egin{align*} ρ &=lim_{n→∞}left|dfrac{(n+1)!x^{n+1}}{n!x^n} ight| [4pt]
&=lim_{n→∞}|(n+1)x| [4pt]
&=|x|lim_{n→∞}(n+1) [4pt]
&=∞. end{align*}]

Therefore, the series diverges for all (x≠0). Since the series is centered at (x=0), it must converge there, so the series converges only for (x≠0). The interval of convergence is the single value (x=0) and the radius of convergence is (R=0).

c. In order to apply the ratio test, consider

[ egin{align*} ρ &=lim_{n→∞}left|dfrac{dfrac{(x−2)^{n+1}}{(n+2)3^{n+1}}}{dfrac{(x−2)^n}{(n+1)3^n}} ight| [4pt]
&=lim_{n→∞} left|dfrac{(x−2)^{n+1}}{(n+2)3^{n+1}}⋅dfrac{(n+1)3^n}{(x−2)^n} ight| [4pt]
&=lim_{n→∞} left|dfrac{(x−2)(n+1)}{3(n+2)} ight|[4pt]
&=dfrac{|x−2|}{3}.end{align*}]

The ratio (ρ<1) if (|x−2|<3). Since (|x−2|<3) implies that (−31) if (|x−2|>3). Therefore, the series diverges if (x<−1) or (x>5). The ratio test is inconclusive if (ρ=1). The ratio (ρ=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}^∞dfrac{(−1)^n}{n+1}=1−dfrac{1}{2}+dfrac{1}{3}−dfrac{1}{4}+ldots . onumber]

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}^∞dfrac{1}{n+1}=1+dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+ldots . onumber]

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).

Exercise (PageIndex{1})

Find the interval and radius of convergence for the series

[ sum_{n=1}^∞dfrac{x^n}{sqrt{n}}. onumber]

Hint

Apply the ratio test to check for absolute convergence.

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 easiest 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+ldots =sum_{n=0}^∞x^n.]

Recall that the geometric series

[a+ar+ar^2+ar^3+ldots ]

converges if and only if (|r|<1.) In that case, it converges to (dfrac{a}{1−r}). Therefore, if (|x|<1), the series in Example (PageIndex{1}) converges to (dfrac{1}{1−x}) and we write

[1+x+x^2+x^3+ldots =dfrac{1}{1−x} for|x|<1.]

As a result, we are able to represent the function (f(x)=dfrac{1}{1−x}) by the power series

[1+x+x^2+x^3+ldots when|x|<1.]

We now show graphically how this series provides a representation for the function (f(x)=dfrac{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)=dfrac{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 you see that as (N) increases, (S_N) becomes a better approximation for (f(x)=dfrac{1}{1−x}) for (x) in the interval ((−1,1)).

Exercise (PageIndex{2})

Sketch a graph of (f(x)=dfrac{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).)

Hint
(S_N(x)=1+x^2+ldots +x^{2N}=dfrac{1−x^{2(N+1)}}{1−x^2})

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)=dfrac{1}{1+x^3})
2. (f(x)=dfrac{x^2}{4−x^2})

Solution

a. You should recognize this function (f) as the sum of a geometric series, because

[ dfrac{1}{1+x^3}=dfrac{1}{1−(−x^3)}. onumber]

Using the fact that, for (|r|<1,dfrac{a}{1−r}) is the sum of the geometric series

[ sum_{n=0}^∞ar^n=a+ar+ar^2+ldots , onumber]

we see that, for (|−x^3|<1,)

[ egin{align*} dfrac{1}{1+x^3} =dfrac{1}{1−(−x^3)} [4pt] =sum_{n=0}^∞(−x^3)^n [4pt] =1−x^3+x^6−x^9+ldots . end{align*}]

Since this series converges if and only if (|−x^3|<1), the interval of convergence is ((−1,1)), and we have

[ dfrac{1}{1+x^3}=1−x^3+x^6−x^9+ldots for|x|<1. onumber]

b. This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate f to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain

[ egin{align*} dfrac{x^2}{4−x^2} =dfrac{x^2}{4(dfrac{1−x^2}{4})}[4pt] =dfrac{x^2}{4(1−(dfrac{x}{2})^2)}.end{align*}]

Therefore, we have

[ egin{align*} dfrac{x^2}{4−x^2} &=dfrac{x^2}{4(1−(dfrac{x}{2})^2)} [4pt]
&= dfrac{dfrac{x^2}{4}}{1−(dfrac{x}{2})^2} [4pt]
&= sum_{n=0}^∞dfrac{x^2}{4}(dfrac{x}{2})^{2n}. end{align*}]

The series converges as long as (|(dfrac{x}{2})^2|<1) (note that when (|(dfrac{x}{2})^2|=1) the series does not converge). Solving this inequality, we conclude that the interval of convergence is ((−2,2)) and

[ egin{align*} dfrac{x^2}{4−x^2} &=sum_{n=0}^∞dfrac{x^{2n+2}}{4^{n+1}}[4pt]
&=dfrac{x^2}{4}+dfrac{x^4}{4^2}+dfrac{x^6}{4^3}+ ldots end{align*}]

for (|x|<2.)

Exercise (PageIndex{3})

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

Hint

Rewrite f in the form (f(x)=dfrac{g(x)}{1−h(x)}) for some functions (g) and (h).

(displaystyle sum_{n=0}^∞dfrac{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, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.

## Key Concepts

• For a power series centered at (x=a), one of the following three properties hold:
• i. The power series converges only at (x=a). In this case, we say that the radius of convergence is (R=0).
• ii. The power series converges for all real numbers (x). In this case, we say that the radius of convergence is (R=∞).
• iii. There is a real number R such that the series converges for (|x−a|R). In this case, the radius of convergence is (R.)
• If a power series converges on a finite interval, the series may or may not converge at the endpoints.
• The ratio test may often be used to determine the radius of convergence.
• The geometric series (displaystyle sum_{n=0}^∞x^n=dfrac{1}{1−x}) for (|x|<1) allows us to represent certain functions using geometric series.

## Key Equations

• Power series centered at (x=0)

[ sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+ldots n onumber]

• Power series centered at (x=a)

[ sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+ldots onumber ]

## Glossary

interval of convergence
the set of real numbers (x) for which a power series converges
power series
a series of the form (displaystyle sum_{n=0}^∞c_nx^n) is a power series centered at (x=0); a series of the form (displaystyle sum_{n=0}^∞c_n(x−a)^n) is a power series centered at (x=a)
if there exists a real number (R>0) such that a power series centered at (x=a) converges for (|x−a|R), then (R) is the radius of convergence; if the power series only converges at (x=a), the radius of convergence is (R=0); if the power series converges for all real numbers (x), the radius of convergence is (R=∞)

## Power series

Beyond their role in mathematical analysis, power series also occur in combinatorics as generating functions (a kind of formal power series) and in electronic engineering (under the name of the Z-transform). The familiar decimal notation for real numbers can also be viewed as an example of a power series, with integer coefficients, but with the argument x fixed at 1 ⁄ 10 . In number theory, the concept of p-adic numbers is also closely related to that of a power series.

## 11.9 Representations of Functions as Power Series (# 1)

Introduction: In this lesson we will use the formula for the sum of a geometric series to find power series representations of a variety of functions.

Objectives: After this lesson you should be able to:

• Find a geometric power series that represents a function.
• Construct a power series using series operations.

Video & Notes: Fill out the note sheet for this lesson (11-9-Representations-of-Functions-by-Power-Series-1) as you watch the video. If you prefer, you could read Section 11.9 of your textbook and work out the problems on the notes on your own as practice. Remember, notes must be uploaded to Blackboard weekly for a grade! If for some reason the video below does not load you can access it on YouTube here.

Homework: Go to WebAssign and complete the 󈫻.9 Representations of Functions by Power Series” assignment. There will be only one assignment for both parts of this lesson.

## How do you find the power series for #f(x)=xln(1+2x)# and determine its radius of convergence?

The power series is #sum_(n=0)^oo((-1)^n2^(n+1))/(n+1)x^(n+2)# , its radius of convergence is #R=1/2# .

#### Explanation:

Integrate both sides. In the series, we integrate just the part with #x# .

Don't forget the #1//2# in the integral. Absolute value bars aren't needed for #ln(1+2x)# since we're only working in #x>=0# .

Add a constant of integration. Fortunately, letting #x=0# shows that #C=0# , so this isn't really an issue.

The next step to reaching the "goal function" of #xln(1+2x)# . To do this, multiply both sides by #2x# .

For the radius of convergence of the series #suma_n# , find the ratio #abs(a_(n+1)/a_n)# . Here, #a_n=((-1)^n2^(n+1))/(n+1)x^(n+2)# .

Find its limit as #nrarroo# :

The limit only concerns how #n# changes, so #abs(2x)# can be extracted from the limit (we can also disregard the #-1# as we're working within absolute values):

The ratio test states that #suma_n# converges when #lim_(nrarroo)abs(a_(n+1)/a_n)<1# . So, our interval of convergence will occur when:

## Finding a power series representation for the function $f(x) = frac$ with the center of the series at a = 1

I'm slightly stuck on a homework question. I'm supposed to:

(a) Find a power series representation for the function $f(x) = frac<1>$ with the center of the series at $a = 1$ by realizing that $f$ is the sum of a convergent geometric series. Then, find the series interval and radius of convergence using the established fact that a geometric series, $sum_^infty ar^$ , converges when $|r| < 1$ .

(b) Find the Taylor series expansion of $f(x) = frac<1>$ about $a = 1$ , making sure to show steps involved in finding the coefficients $c_n = frac$ . Then, using the ratio test, find the series interval and radius of convergence.

So far, this is what I've figured out so far:

(b) I found the Taylor series for $f(x) = frac <1>$ about $a=1$ and ended up with $frac<1> = 1+(-1)+frac<2><2!>(x-1)^2+frac<-6><3!>(x-1)^3+frac<24><4!>(x-1)^4+.$ $= 1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4+.$

I know I still have to figure out a very large chunk of these problems. Any help is greatly appreciated!

## Moment Problem

Using the moment generating function, we can now show, at least in the case of a discrete random variable with finite range, that its distribution function is completely determined by its moments.

Add exercises text here. For the automatic number to work, you need to Let (X) be a discrete random variable with finite range (), distribution function (p), and moment generating function (g). Then (g) is uniquely determined by (p), and conversely.

We know that (p) determines (g), since [g(t) = sum_^n e^ p(x_j) .] Conversely, assume that (g(t)) is known. We wish to determine the values of (x_j) and (p(x_j)), for (1 le j le n). We assume, without loss of generality, that (p(x_j) > 0) for (1 le j le n), and that [x_1 < x_2 < ldots < x_n .] We note that (g(t)) is differentiable for all (t), since it is a finite linear combination of exponential functions. If we compute (g'(t)/g(t)), we obtain [

If we delete the hypothesis that (X) have finite range in the above theorem, then the conclusion is no longer necessarily true.

## Power Functions - PowerPoint PPT Presentation

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## Power Series Expansions

1. Taylor series
If a function (fleft( x ight)) has continuous derivatives up to (left( ight))th order inclusive, then this function can be expanded in a power series about the point (x = a) by the Taylor formula :
(fleft( x ight) =) (sumlimits_^infty <>left( a ight) ight)>^n>>><> ormalsize>> =) ( ight)> ) (+<left( a ight) < ight)>^2>>><<2!>> ormalsize> + ldots> ) (+<>left( a ight) < ight)>^n>>><> ormalsize> + >,)
where the remainder term () in the Lagrange term is given by the expression
( =
ight)>>left( xi ight) < ight)>^>>>< ight)!>> ormalsize>,) (a lt xi lt x.)
If this expansion converges over a certain range of (x) centered at (a), that is, (limlimits_ = 0,) then the expansion is called Taylor series of the function (fleft( x ight)) expanded about the point (a).
2. A Maclaurin series is a special case of a Taylor series when the power series expansion is performed at the point (a = 0:)
(fleft( x ight) = sumlimits_^infty <>left( 0 ight)>><>> ormalsize> =) (fleft( 0 ight) + f’left( 0 ight)x ) (+left( 0 ight)>><<2!>> ormalsize> + ldots) (+>left( 0 ight)>><> ormalsize> + .)

## 10.1: Power Series and Functions

The links on the right side of this page are for video recordings of the PowerPoint lectures given in AB and BC Calculus class. You may click on either the VIDEO link or the YouTube link, whichever works better for you.

Determine a function $f(x)$ such that $f(x) = sum_^ nx^$ .
In solving problems such as this one, our goal is essentially to use any power series operations we have under our belt to modify the series into a more recognizable series that we can substitute in to find $f$ .
We note that since we used the derivative of the geometric series in our substitution we have that $frac <(1 - x)^2>= sum_^ nx^$ for $mid x mid < 1$ .