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Calculus Volume 2

6.1 Power Series and Functions

Calculus Volume 26.1 Power Series and Functions
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 6.1.1. Identify a power series and provide examples of them.
  • 6.1.2. Determine the radius of convergence and interval of convergence of a power series.
  • 6.1.3. 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

n=0cnxn=c0+c1x+c2x2+,n=0cnxn=c0+c1x+c2x2+,

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

1+x+x2+=n=0xn1+x+x2+=n=0xn

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

Definition

A series of the form

n=0cnxn=c0+c1x+c2x2+n=0cnxn=c0+c1x+c2x2+
6.1

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

n=0cn(xa)n=c0+c1(xa)+c2(xa)2+n=0cn(xa)n=c0+c1(xa)+c2(xa)2+
6.2

is a power series centered at x=a.x=a.

To make this definition precise, we stipulate that x0=1x0=1 and (xa)0=1(xa)0=1 even when x=0x=0 and x=a,x=a, respectively.

The series

n=0xnn!=1+x+x22!+x33!+n=0xnn!=1+x+x22!+x33!+

and

n=0n!xn=1+x+2!x2+3!x3+n=0n!xn=1+x+2!x2+3!x3+

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

n=0(x2)n(n+1)3n=1+x22·3+(x2)23·32+(x2)34·33+n=0(x2)n(n+1)3n=1+x22·3+(x2)23·32+(x2)34·33+

is a power series centered at x=2.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,x=a, the value of the series at x=ax=a is given by c0.c0. 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 n=0xnn=0xn converges for all x in the interval (−1,1),(−1,1), but diverges for all x outside that interval. We now summarize these three possibilities for a general power series.

Theorem 6.1

Convergence of a Power Series

Consider the power series n=0cn(xa)n.n=0cn(xa)n. The series satisfies exactly one of the following properties:

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

Proof

Suppose that the power series is centered at a=0.a=0. (For a series centered at a value of a other than zero, the result follows by letting y=xay=xa and considering the series n=1cnyn.)n=1cnyn.) We must first prove the following fact:

If there exists a real number d0d0 such that n=0cndnn=0cndn converges, then the series n=0cnxnn=0cnxn converges absolutely for all x such that |x|<|d|.|x|<|d|.

Since n=0cndnn=0cndn converges, the nth term cndn0cndn0 as n.n. Therefore, there exists an integer N such that |cndn|1|cndn|1 for all nN.nN. Writing

|cnxn|=|cndn||xd|n,|cnxn|=|cndn||xd|n,

we conclude that, for all nN,nN,

|cnxn||xd|n.|cnxn||xd|n.

The series

n=N|xd|nn=N|xd|n

is a geometric series that converges if |xd|<1.|xd|<1. Therefore, by the comparison test, we conclude that n=Ncnxnn=Ncnxn also converges for |x|<|d|.|x|<|d|. Since we can add a finite number of terms to a convergent series, we conclude that n=0cnxnn=0cnxn converges for |x|<|d|.|x|<|d|.

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

n=0anxnn=0anxn

and let S be the set of real numbers for which the series converges. Suppose that the set S={0}.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}S{0} and S is not the set of real numbers. Then there exists a real number x*0x*0 such that the series does not converge. Thus, the series cannot converge for any x such that |x|>|x*|.|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},S{0}, the number R>0.R>0. Therefore, the series converges for all x such that |x|<R,|x|<R, and the series falls into case iii.

If a series n=0cn(xa)nn=0cn(xa)n falls into case iii. of Convergence of a Power Series, then the series converges for all x such that |xa|<R|xa|<R for some R>0,R>0, and diverges for all x such that |xa|>R.|xa|>R. The series may converge or diverge at the values x where |xa|=R.|xa|=R. The set of values x for which the series n=0cn(xa)nn=0cn(xa)n converges is known as the interval of convergence. Since the series diverges for all values x where |xa|>R,|xa|>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 n=0xnn=0xn converges for all values x in the interval (−1,1)(−1,1) and diverges for all values x such that |x|1,|x|1, the interval of convergence of this series is (−1,1).(−1,1). Since the length of the interval is 2, the radius of convergence is 1.

Definition

Consider the power series n=0cn(xa)n.n=0cn(xa)n. The set of real numbers x where the series converges is the interval of convergence. If there exists a real number R>0R>0 such that the series converges for |xa|<R|xa|<R and diverges for |xa|>R,|xa|>R, then R is the radius of convergence. If the series converges only at x=a,x=a, we say the radius of convergence is R=0.R=0. If the series converges for all real numbers x, we say the radius of convergence is R=R= (Figure 6.2).

This figure has three number lines, each labeled with x. In the middle of each number line is a point labeled a. The first number line has “diverges” over all of the line to the left of a and “diverges” over the line to the right of a. At the point a itself, it is labeled as “converges”. The second number line has “converges” labeled for the entire line. The third number line has points labeled at a-R, a, and a+R. To the left of a-R, the number line is labeled “diverges”. Between a-R and a+R the number line is labeled “converges” and to the right of a+R the number line is labeled “diverges”.
Figure 6.2 For a series n=0cn(xa)nn=0cn(xa)n graph (a) shows a radius of convergence at R=0,R=0, graph (b) shows a radius of convergence at R=,R=, and graph (c) shows a radius of convergence at R. For graph (c) we note that the series may or may not converge at the endpoints x=a+Rx=a+R and x=aR.x=aR.

To determine the interval of convergence for a power series, we typically apply the ratio test. In Example 6.1, we show the three different possibilities illustrated in Figure 6.2.

Example 6.1

Finding the Interval and Radius of Convergence

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

  1. n=0xnn!n=0xnn!
  2. n=0n!xnn=0n!xn
  3. n=0(x2)n(n+1)3nn=0(x2)n(n+1)3n

Solution

  1. To check for convergence, apply the ratio test. We have
    ρ=limn|xn+1(n+1)!xnn!|=limn|xn+1(n+1)!·n!xn|=limn|xn+1(n+1)·n!·n!xn|=limn|xn+1|=|x|limn1n+1=0<1ρ=limn|xn+1(n+1)!xnn!|=limn|xn+1(n+1)!·n!xn|=limn|xn+1(n+1)·n!·n!xn|=limn|xn+1|=|x|limn1n+1=0<1

    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=.R=.
  2. Apply the ratio test. For x0,x0, we see that
    ρ=limn|(n+1)!xn+1n!xn|=limn|(n+1)x|=|x|limn(n+1)=.ρ=limn|(n+1)!xn+1n!xn|=limn|(n+1)x|=|x|limn(n+1)=.

    Therefore, the series diverges for all x0.x0. Since the series is centered at x=0,x=0, it must converge there, so the series converges only for x0.x0. The interval of convergence is the single value x=0x=0 and the radius of convergence is R=0.R=0.
  3. In order to apply the ratio test, consider
    ρ=limn|(x2)n+1(n+2)3n+1(x2)n(n+1)3n|=limn|(x2)n+1(n+2)3n+1·(n+1)3n(x2)n|=limn|(x2)(n+1)3(n+2)|=|x2|3.ρ=limn|(x2)n+1(n+2)3n+1(x2)n(n+1)3n|=limn|(x2)n+1(n+2)3n+1·(n+1)3n(x2)n|=limn|(x2)(n+1)3(n+2)|=|x2|3.

    The ratio ρ<1ρ<1 if |x2|<3.|x2|<3. Since |x2|<3|x2|<3 implies that −3<x2<3,−3<x2<3, the series converges absolutely if −1<x<5.−1<x<5. The ratio ρ>1ρ>1 if |x2|>3.|x2|>3. Therefore, the series diverges if x<−1x<−1 or x>5.x>5. The ratio test is inconclusive if ρ=1.ρ=1. The ratio ρ=1ρ=1 if and only if x=−1x=−1 or x=5.x=5. We need to test these values of x separately. For x=−1,x=−1, the series is given by
    n=0(−1)nn+1=112+1314+.n=0(−1)nn+1=112+1314+.

    Since this is the alternating harmonic series, it converges. Thus, the series converges at x=−1.x=−1. For x=5,x=5, the series is given by
    n=01n+1=1+12+13+14+.n=01n+1=1+12+13+14+.

    This is the harmonic series, which is divergent. Therefore, the power series diverges at x=5.x=5. We conclude that the interval of convergence is [−1,5)[−1,5) and the radius of convergence is R=3.R=3.

Checkpoint 6.1

Find the interval and radius of convergence for the series n=1xnn.n=1xnn.

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+x2+x3+=n=0xn.1+x+x2+x3+=n=0xn.
6.3

Recall that the geometric series

a+ar+ar2+ar3+a+ar+ar2+ar3+

converges if and only if |r|<1.|r|<1. In that case, it converges to a1r.a1r. Therefore, if |x|<1,|x|<1, the series in Example 6.3 converges to 11x11x and we write

1+x+x2+x3+=11xfor|x|<1.1+x+x2+x3+=11xfor|x|<1.

As a result, we are able to represent the function f(x)=11xf(x)=11x by the power series

1+x+x2+x3+when|x|<1.1+x+x2+x3+when|x|<1.

We now show graphically how this series provides a representation for the function f(x)=11xf(x)=11x by comparing the graph of f with the graphs of several of the partial sums of this infinite series.

Example 6.2

Graphing a Function and Partial Sums of its Power Series

Sketch a graph of f(x)=11xf(x)=11x and the graphs of the corresponding partial sums SN(x)=n=0NxnSN(x)=n=0Nxn for N=2,4,6N=2,4,6 on the interval (−1,1).(−1,1). Comment on the approximation SNSN as N increases.

Solution

From the graph in Figure 6.3 you see that as N increases, SNSN becomes a better approximation for f(x)=11xf(x)=11x for x in the interval (−1,1).(−1,1).

This figure is the graph of y = 1/(1-x), which is an increasing curve with vertical asymptote at 1. Also on this graph are three partial sums of the function, S sub 6, S sub 4, and S sub 2. These curves, in order, gradually become flatter.
Figure 6.3 The graph shows a function and three approximations of it by partial sums of a power series.
Checkpoint 6.2

Sketch a graph of f(x)=11x2f(x)=11x2 and the corresponding partial sums SN(x)=n=0Nx2nSN(x)=n=0Nx2n for N=2,4,6N=2,4,6 on the interval (−1,1).(−1,1).

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 6.3

Representing a Function with a Power Series

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

  1. f(x)=11+x3f(x)=11+x3
  2. f(x)=x24x2f(x)=x24x2

Solution

  1. You should recognize this function f as the sum of a geometric series, because
    11+x3=11(x3).11+x3=11(x3).

    Using the fact that, for |r|<1,a1r|r|<1,a1r is the sum of the geometric series
    n=0arn=a+ar+ar2+,n=0arn=a+ar+ar2+,

    we see that, for |x3|<1,|x3|<1,
    11+x3=11(x3)=n=0(x3)n=1x3+x6x9+.11+x3=11(x3)=n=0(x3)n=1x3+x6x9+.

    Since this series converges if and only if |x3|<1,|x3|<1, the interval of convergence is (−1,1),(−1,1), and we have
    11+x3=1x3+x6x9+for|x|<1.11+x3=1x3+x6x9+for|x|<1.
  2. 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
    x24x2=x24(1x24)=x24(1(x2)2).x24x2=x24(1x24)=x24(1(x2)2).

    Therefore, we have
    x24x2=x24(1(x2)2)=x241(x2)2=n=0x24(x2)2n.x24x2=x24(1(x2)2)=x241(x2)2=n=0x24(x2)2n.

    The series converges as long as |(x2)2|<1|(x2)2|<1 (note that when |(x2)2|=1|(x2)2|=1 the series does not converge). Solving this inequality, we conclude that the interval of convergence is (−2,2)(−2,2) and
    x24x2=n=0x2n+24n+1=x24+x442+x643+x24x2=n=0x2n+24n+1=x24+x442+x643+

    for |x|<2.|x|<2.
Checkpoint 6.3

Represent the function f(x)=x32xf(x)=x32x using a power series and find the interval of convergence.

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.

Section 6.1 Exercises

In the following exercises, state whether each statement is true, or give an example to show that it is false.

1.

If n=1anxnn=1anxn converges, then anxn0anxn0 as n.n.

2.

n=1anxnn=1anxn converges at x=0x=0 for any real numbers an.an.

3.

Given any sequence an,an, there is always some R>0,R>0, possibly very small, such that n=1anxnn=1anxn converges on (R,R).(R,R).

4.

If n=1anxnn=1anxn has radius of convergence R>0R>0 and if |bn||an||bn||an| for all n, then the radius of convergence of n=1bnxnn=1bnxn is greater than or equal to R.

5.

Suppose that n=0an(x3)nn=0an(x3)n converges at x=6.x=6. At which of the following points must the series also converge? Use the fact that if an(xc)nan(xc)n converges at x, then it converges at any point closer to c than x.

  1. x=1x=1
  2. x=2x=2
  3. x=3x=3
  4. x=0x=0
  5. x=5.99x=5.99
  6. x=0.000001x=0.000001
6.

Suppose that n=0an(x+1)nn=0an(x+1)n converges at x=−2.x=−2. At which of the following points must the series also converge? Use the fact that if an(xc)nan(xc)n converges at x, then it converges at any point closer to c than x.

  1. x=2x=2
  2. x=−1x=−1
  3. x=−3x=−3
  4. x=0x=0
  5. x=0.99x=0.99
  6. x=0.000001x=0.000001

In the following exercises, suppose that |an+1an|1|an+1an|1 as n.n. Find the radius of convergence for each series.

7.

n=0an2nxnn=0an2nxn

8.

n=0anxn2nn=0anxn2n

9.

n=0anπnxnenn=0anπnxnen

10.

n=0an(−1)nxn10nn=0an(−1)nxn10n

11.

n=0an(−1)nx2nn=0an(−1)nx2n

12.

n=0an(−4)nx2nn=0an(−4)nx2n

In the following exercises, find the radius of convergence R and interval of convergence for anxnanxn with the given coefficients an.an.

13.

n=1(2x)nnn=1(2x)nn

14.

n=1(−1)nxnnn=1(−1)nxnn

15.

n=1nxn2nn=1nxn2n

16.

n=1nxnenn=1nxnen

17.

n=1n2xn2nn=1n2xn2n

18.

k=1kexkekk=1kexkek

19.

k=1πkxkkπk=1πkxkkπ

20.

n=1xnn!n=1xnn!

21.

n=110nxnn!n=110nxnn!

22.

n=1(−1)nxnln(2n)n=1(−1)nxnln(2n)

In the following exercises, find the radius of convergence of each series.

23.

k=1(k!)2xk(2k)!k=1(k!)2xk(2k)!

24.

n=1(2n)!xnn2nn=1(2n)!xnn2n

25.

k=1k!1·3·5(2k1)xkk=1k!1·3·5(2k1)xk

26.

k=12·4·62k(2k)!xkk=12·4·62k(2k)!xk

27.

n=1xn(2nn)n=1xn(2nn) where (nk)=n!k!(nk)!(nk)=n!k!(nk)!

28.

n=1sin2nxnn=1sin2nxn

In the following exercises, use the ratio test to determine the radius of convergence of each series.

29.

n=1(n!)3(3n)!xnn=1(n!)3(3n)!xn

30.

n=123n(n!)3(3n)!xnn=123n(n!)3(3n)!xn

31.

n=1n!nnxnn=1n!nnxn

32.

n=1(2n)!n2nxnn=1(2n)!n2nxn

In the following exercises, given that 11x=n=0xn11x=n=0xn with convergence in (−1,1),(−1,1), find the power series for each function with the given center a, and identify its interval of convergence.

33.

f(x)=1x;a=1f(x)=1x;a=1 (Hint: 1x=11(1x))1x=11(1x))

34.

f(x)=11x2;a=0f(x)=11x2;a=0

35.

f(x)=x1x2;a=0f(x)=x1x2;a=0

36.

f(x)=11+x2;a=0f(x)=11+x2;a=0

37.

f(x)=x21+x2;a=0f(x)=x21+x2;a=0

38.

f(x)=12x;a=1f(x)=12x;a=1

39.

f(x)=112x;a=0.f(x)=112x;a=0.

40.

f(x)=114x2;a=0f(x)=114x2;a=0

41.

f(x)=x214x2;a=0f(x)=x214x2;a=0

42.

f(x)=x254x+x2;a=2f(x)=x254x+x2;a=2

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

43.

Explain why, if |an|1/nr>0,|an|1/nr>0, then |anxn|1/n|x|r<1|anxn|1/n|x|r<1 whenever |x|<1r|x|<1r and, therefore, the radius of convergence of n=1anxnn=1anxn is R=1r.R=1r.

44.

n=1xnnnn=1xnnn

45.

k=1(k12k+3)kxkk=1(k12k+3)kxk

46.

k=1(2k21k2+3)kxkk=1(2k21k2+3)kxk

47.

n=1an=(n1/n1)nxnn=1an=(n1/n1)nxn

48.

Suppose that p(x)=n=0anxnp(x)=n=0anxn such that an=0an=0 if n is even. Explain why p(x)=p(x).p(x)=p(x).

49.

Suppose that p(x)=n=0anxnp(x)=n=0anxn such that an=0an=0 if n is odd. Explain why p(x)=p(x).p(x)=p(x).

50.

Suppose that p(x)=n=0anxnp(x)=n=0anxn converges on (−1,1].(−1,1]. Find the interval of convergence of p(Ax).p(Ax).

51.

Suppose that p(x)=n=0anxnp(x)=n=0anxn converges on (−1,1].(−1,1]. Find the interval of convergence of p(2x1).p(2x1).

In the following exercises, suppose that p(x)=n=0anxnp(x)=n=0anxn satisfies limnan+1an=1limnan+1an=1 where an0an0 for each n. State whether each series converges on the full interval (−1,1),(−1,1), or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

52.

n=0anx2nn=0anx2n

53.

n=0a2nx2nn=0a2nx2n

54.

n=0a2nxn(Hint:x=±x2)n=0a2nxn(Hint:x=±x2)

55.

n=0an2xn2n=0an2xn2 (Hint: Let bk=akbk=ak if k=n2k=n2 for some n, otherwise bk=0.)bk=0.)

56.

Suppose that p(x)p(x) is a polynomial of degree N. Find the radius and interval of convergence of n=1p(n)xn.n=1p(n)xn.

57.

[T] Plot the graphs of 11x11x and of the partial sums SN=n=0NxnSN=n=0Nxn for n=10,20,30n=10,20,30 on the interval [−0.99,0.99].[−0.99,0.99]. Comment on the approximation of 11x11x by SNSN near x=−1x=−1 and near x=1x=1 as N increases.

58.

[T] Plot the graphs of ln(1x)ln(1x) and of the partial sums SN=n=1NxnnSN=n=1Nxnn for n=10,50,100n=10,50,100 on the interval [−0.99,0.99].[−0.99,0.99]. Comment on the behavior of the sums near x=−1x=−1 and near x=1x=1 as N increases.

59.

[T] Plot the graphs of the partial sums Sn=n=1Nxnn2Sn=n=1Nxnn2 for n=10,50,100n=10,50,100 on the interval [−0.99,0.99].[−0.99,0.99]. Comment on the behavior of the sums near x=−1x=−1 and near x=1x=1 as N increases.

60.

[T] Plot the graphs of the partial sums SN=n=1NsinnxnSN=n=1Nsinnxn for n=10,50,100n=10,50,100 on the interval [−0.99,0.99].[−0.99,0.99]. Comment on the behavior of the sums near x=−1x=−1 and near x=1x=1 as N increases.

61.

[T] Plot the graphs of the partial sums SN=n=0N(−1)nx2n+1(2n+1)!SN=n=0N(−1)nx2n+1(2n+1)! for n=3,5,10n=3,5,10 on the interval [−2π,2π].[−2π,2π]. Comment on how these plots approximate sinxsinx as N increases.

62.

[T] Plot the graphs of the partial sums SN=n=0N(−1)nx2n(2n)!SN=n=0N(−1)nx2n(2n)! for n=3,5,10n=3,5,10 on the interval [−2π,2π].[−2π,2π]. Comment on how these plots approximate cosxcosx as N increases.

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