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Calculus Volume 2

6.2 Properties of Power Series

Calculus Volume 26.2 Properties of Power Series
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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

  • 6.2.1. Combine power series by addition or subtraction.
  • 6.2.2. Create a new power series by multiplication by a power of the variable or a constant, or by substitution.
  • 6.2.3. Multiply two power series together.
  • 6.2.4. Differentiate and integrate power series term-by-term.

In the preceding section on power series and functions we showed how to represent certain functions using power series. In this section we discuss how power series can be combined, differentiated, or integrated to create new power series. This capability is particularly useful for a couple of reasons. First, it allows us to find power series representations for certain elementary functions, by writing those functions in terms of functions with known power series. For example, given the power series representation for f(x)=11x,f(x)=11x, we can find a power series representation for f(x)=1(1x)2.f(x)=1(1x)2. Second, being able to create power series allows us to define new functions that cannot be written in terms of elementary functions. This capability is particularly useful for solving differential equations for which there is no solution in terms of elementary functions.

Combining Power Series

If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of x or evaluate a power series at xmxm for a positive integer m to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for f(x)=11x,f(x)=11x, we can find power series representations for related functions, such as

y=3x1x2andy=1(x1)(x3).y=3x1x2andy=1(x1)(x3).

In Combining Power Series we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at x=0.x=0. Similar results hold for power series centered at x=a.x=a.

Theorem 6.2

Combining Power Series

Suppose that the two power series n=0cnxnn=0cnxn and n=0dnxnn=0dnxn converge to the functions f and g, respectively, on a common interval I.

  1. The power series n=0(cnxn±dnxn)n=0(cnxn±dnxn) converges to f±gf±g on I.
  2. For any integer m0m0 and any real number b, the power series n=0bxmcnxnn=0bxmcnxn converges to bxmf(x)bxmf(x) on I.
  3. For any integer m0m0 and any real number b, the series n=0cn(bxm)nn=0cn(bxm)n converges to f(bxm)f(bxm) for all x such that bxmbxm is in I.

Proof

We prove i. in the case of the series n=0(cnxn+dnxn).n=0(cnxn+dnxn). Suppose that n=0cnxnn=0cnxn and n=0dnxnn=0dnxn converge to the functions f and g, respectively, on the interval I. Let x be a point in I and let SN(x)SN(x) and TN(x)TN(x) denote the Nth partial sums of the series n=0cnxnn=0cnxn and n=0dnxn,n=0dnxn, respectively. Then the sequence {SN(x)}{SN(x)} converges to f(x)f(x) and the sequence {TN(x)}{TN(x)} converges to g(x).g(x). Furthermore, the Nth partial sum of n=0(cnxn+dnxn)n=0(cnxn+dnxn) is

n=0N(cnxn+dnxn)=n=0Ncnxn+n=0Ndnxn=SN(x)+TN(x).n=0N(cnxn+dnxn)=n=0Ncnxn+n=0Ndnxn=SN(x)+TN(x).

Because

limN(SN(x)+TN(x))=limNSN(x)+limNTN(x)=f(x)+g(x),limN(SN(x)+TN(x))=limNSN(x)+limNTN(x)=f(x)+g(x),

we conclude that the series n=0(cnxn+dnxn)n=0(cnxn+dnxn) converges to f(x)+g(x).f(x)+g(x).

We examine products of power series in a later theorem. First, we show several applications of Combining Power Series and how to find the interval of convergence of a power series given the interval of convergence of a related power series.

Example 6.4

Combining Power Series

Suppose that n=0anxnn=0anxn is a power series whose interval of convergence is (−1,1),(−1,1), and suppose that n=0bnxnn=0bnxn is a power series whose interval of convergence is (−2,2).(−2,2).

  1. Find the interval of convergence of the series n=0(anxn+bnxn).n=0(anxn+bnxn).
  2. Find the interval of convergence of the series n=0an3nxn.n=0an3nxn.

Solution

  1. Since the interval (−1,1)(−1,1) is a common interval of convergence of the series n=0anxnn=0anxn and n=0bnxn,n=0bnxn, the interval of convergence of the series n=0(anxn+bnxn)n=0(anxn+bnxn) is (−1,1).(−1,1).
  2. Since n=0anxnn=0anxn is a power series centered at zero with radius of convergence 1, it converges for all x in the interval (−1,1).(−1,1). By Combining Power Series, the series
    n=0an3nxn=n=0an(3x)nn=0an3nxn=n=0an(3x)n

    converges if 3x is in the interval (−1,1).(−1,1). Therefore, the series converges for all x in the interval (13,13).(13,13).

Checkpoint 6.4

Suppose that n=0anxnn=0anxn has an interval of convergence of (−1,1).(−1,1). Find the interval of convergence of n=0an(x2)n.n=0an(x2)n.

In the next example, we show how to use Combining Power Series and the power series for a function f to construct power series for functions related to f. Specifically, we consider functions related to the function f(x)=11xf(x)=11x and we use the fact that

11x=n=0xn=1+x+x2+x3+11x=n=0xn=1+x+x2+x3+

for |x|<1.|x|<1.

Example 6.5

Constructing Power Series from Known Power Series

Use the power series representation for f(x)=11xf(x)=11x combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series.

  1. f(x)=3x1+x2f(x)=3x1+x2
  2. f(x)=1(x1)(x3)f(x)=1(x1)(x3)

Solution

  1. First write f(x)f(x) as
    f(x)=3x(11(x2)).f(x)=3x(11(x2)).

    Using the power series representation for f(x)=11xf(x)=11x and parts ii. and iii. of Combining Power Series, we find that a power series representation for f is given by
    n=03x(x2)n=n=03(−1)nx2n+1.n=03x(x2)n=n=03(−1)nx2n+1.

    Since the interval of convergence of the series for 11x11x is (−1,1),(−1,1), the interval of convergence for this new series is the set of real numbers x such that |x2|<1.|x2|<1. Therefore, the interval of convergence is (−1,1).(−1,1).
  2. To find the power series representation, use partial fractions to write f(x)=1(1x)(x3)f(x)=1(1x)(x3) as the sum of two fractions. We have
    1(x1)(x3)=1/2x1+1/2x3=1/21x1/23x=1/21x1/61x3.1(x1)(x3)=1/2x1+1/2x3=1/21x1/23x=1/21x1/61x3.

    First, using part ii. of Combining Power Series, we obtain
    1/21x=n=012xnfor|x|<1.1/21x=n=012xnfor|x|<1.

    Then, using parts ii. and iii. of Combining Power Series, we have
    1/61x/3=n=016(x3)nfor|x|<3.1/61x/3=n=016(x3)nfor|x|<3.

    Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we have
    1(x1)(x3)=n=0(1216·3n)xn1(x1)(x3)=n=0(1216·3n)xn

    where the interval of convergence is (−1,1).(−1,1).

Checkpoint 6.5

Use the series for f(x)=11xf(x)=11x on |x|<1|x|<1 to construct a series for 1(1x)(x2).1(1x)(x2). Determine the interval of convergence.

In Example 6.5, we showed how to find power series for certain functions. In Example 6.6 we show how to do the opposite: given a power series, determine which function it represents.

Example 6.6

Finding the Function Represented by a Given Power Series

Consider the power series n=02nxn.n=02nxn. Find the function f represented by this series. Determine the interval of convergence of the series.

Solution

Writing the given series as

n=02nxn=n=0(2x)n,n=02nxn=n=0(2x)n,

we can recognize this series as the power series for

f(x)=112x.f(x)=112x.

Since this is a geometric series, the series converges if and only if |2x|<1.|2x|<1. Therefore, the interval of convergence is (12,12).(12,12).

Checkpoint 6.6

Find the function represented by the power series n=013nxn.n=013nxn. Determine its interval of convergence.

Recall the questions posed in the chapter opener about which is the better way of receiving payouts from lottery winnings. We now revisit those questions and show how to use series to compare values of payments over time with a lump sum payment today. We will compute how much future payments are worth in terms of today’s dollars, assuming we have the ability to invest winnings and earn interest. The value of future payments in terms of today’s dollars is known as the present value of those payments.

Example 6.7

Chapter Opener: Present Value of Future Winnings

This is a picture of a stack of money. There are 100 dollar bills wrapped in groups of $10,000.
Figure 6.4 (credit: modification of work by Robert Huffstutter, Flickr)

Suppose you win the lottery and are given the following three options: (1) Receive 20 million dollars today; (2) receive 1.5 million dollars per year over the next 20 years; or (3) receive 1 million dollars per year indefinitely (being passed on to your heirs). Which is the best deal, assuming that the annual interest rate is 5%? We answer this by working through the following sequence of questions.

  1. How much is the 1.5 million dollars received annually over the course of 20 years worth in terms of today’s dollars, assuming an annual interest rate of 5%?
  2. Use the answer to part a. to find a general formula for the present value of payments of C dollars received each year over the next n years, assuming an average annual interest rate r.
  3. Find a formula for the present value if annual payments of C dollars continue indefinitely, assuming an average annual interest rate r.
  4. Use the answer to part c. to determine the present value of 1 million dollars paid annually indefinitely.
  5. Use your answers to parts a. and d. to determine which of the three options is best.

Solution

  1. Consider the payment of 1.5 million dollars made at the end of the first year. If you were able to receive that payment today instead of one year from now, you could invest that money and earn 5% interest. Therefore, the present value of that money P1 satisfies P1(1+0.05)=1.5million dollars.P1(1+0.05)=1.5million dollars. We conclude that
    P1=1.51.05=$1.429million dollars.P1=1.51.05=$1.429million dollars.

    Similarly, consider the payment of 1.5 million dollars made at the end of the second year. If you were able to receive that payment today, you could invest that money for two years, earning 5% interest, compounded annually. Therefore, the present value of that money P2 satisfies P2(1+0.05)2=1.5million dollars.P2(1+0.05)2=1.5million dollars. We conclude that
    P2=1.5(1.05)2=$1.361million dollars.P2=1.5(1.05)2=$1.361million dollars.

    The value of the future payments today is the sum of the present values P1,P2,,P20P1,P2,,P20 of each of those annual payments. The present value Pk satisfies
    Pk=1.5(1.05)k.Pk=1.5(1.05)k.

    Therefore,
    P=1.51.05+1.5(1.05)2++1.5(1.05)20=$18.693million dollars.P=1.51.05+1.5(1.05)2++1.5(1.05)20=$18.693million dollars.
  2. Using the result from part a. we see that the present value P of C dollars paid annually over the course of n years, assuming an annual interest rate r, is given by
    P=C1+r+C(1+r)2++C(1+r)ndollars.P=C1+r+C(1+r)2++C(1+r)ndollars.
  3. Using the result from part b. we see that the present value of an annuity that continues indefinitely is given by the infinite series
    P=n=0C(1+r)n+1.P=n=0C(1+r)n+1.

    We can view the present value as a power series in r, which converges as long as |11+r|<1.|11+r|<1. Since r>0,r>0, this series converges. Rewriting the series as
    P=C(1+r)n=0(11+r)n,P=C(1+r)n=0(11+r)n,

    we recognize this series as the power series for
    f(r)=11(11+r)=1(r1+r)=1+rr.f(r)=11(11+r)=1(r1+r)=1+rr.

    We conclude that the present value of this annuity is
    P=C1+r·1+rr=Cr.P=C1+r·1+rr=Cr.
  4. From the result to part c. we conclude that the present value P of C=1million dollarsC=1million dollars paid out every year indefinitely, assuming an annual interest rate r=0.05,r=0.05, is given by
    P=10.05=20million dollars.P=10.05=20million dollars.
  5. From part a. we see that receiving $1.5 million dollars over the course of 20 years is worth $18.693 million dollars in today’s dollars. From part d. we see that receiving $1 million dollars per year indefinitely is worth $20 million dollars in today’s dollars. Therefore, either receiving a lump-sum payment of $20 million dollars today or receiving $1 million dollars indefinitely have the same present value.

Multiplication of Power Series

We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.

The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply

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

and

n=0dnxn=d0+d1x+d2x2+.n=0dnxn=d0+d1x+d2x2+.

It appears that the product should satisfy

(n=0cnxn)(n=−0dnxn)=(c0+c1x+c2x2+)·(d0+d1x+d2x2+)=c0d0+(c1d0+c0d1)x+(c2d0+c1d1+c0d2)x2+.(n=0cnxn)(n=−0dnxn)=(c0+c1x+c2x2+)·(d0+d1x+d2x2+)=c0d0+(c1d0+c0d1)x+(c2d0+c1d1+c0d2)x2+.

In Multiplying Power Series, we state the main result regarding multiplying power series, showing that if n=0cnxnn=0cnxn and n=0dnxnn=0dnxn converge on a common interval I, then we can multiply the series in this way, and the resulting series also converges on the interval I.

Theorem 6.3

Multiplying Power Series

Suppose that the power series n=0cnxnn=0cnxn and n=0dnxnn=0dnxn converge to f and g, respectively, on a common interval I. Let

en=c0dn+c1dn1+c2dn2++cn1d1+cnd0=k=0nckdnk.en=c0dn+c1dn1+c2dn2++cn1d1+cnd0=k=0nckdnk.

Then

(n=0cnxn)(n=0dnxn)=n=0enxn(n=0cnxn)(n=0dnxn)=n=0enxn

and

n=0enxnconverges tof(x)·g(x)onI.n=0enxnconverges tof(x)·g(x)onI.

The series n=0enxnn=0enxn is known as the Cauchy product of the series n=0cnxnn=0cnxn and n=0dnxn.n=0dnxn.

We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for

f(x)=1(1x)(1x2)f(x)=1(1x)(1x2)

using the power series representations for

y=11xandy=11x2.y=11xandy=11x2.

Example 6.8

Multiplying Power Series

Multiply the power series representation

11x=n=0xn=1+x+x2+x3+11x=n=0xn=1+x+x2+x3+

for |x|<1|x|<1 with the power series representation

11x2=n=0(x2)n=1+x2+x4+x6+11x2=n=0(x2)n=1+x2+x4+x6+

for |x|<1|x|<1 to construct a power series for f(x)=1(1x)(1x2)f(x)=1(1x)(1x2) on the interval (−1,1).(−1,1).

Solution

We need to multiply

(1+x+x2+x3+)(1+x2+x4+x6+).(1+x+x2+x3+)(1+x2+x4+x6+).

Writing out the first several terms, we see that the product is given by

(1+x2+x4+x6+)+(x+x3+x5+x7+)+(x2+x4+x6+x8+)+(x3+x5+x7+x9+)=1+x+(1+1)x2+(1+1)x3+(1+1+1)x4+(1+1+1)x5+=1+x+2x2+2x3+3x4+3x5+.(1+x2+x4+x6+)+(x+x3+x5+x7+)+(x2+x4+x6+x8+)+(x3+x5+x7+x9+)=1+x+(1+1)x2+(1+1)x3+(1+1+1)x4+(1+1+1)x5+=1+x+2x2+2x3+3x4+3x5+.

Since the series for y=11xy=11x and y=11x2y=11x2 both converge on the interval (−1,1),(−1,1), the series for the product also converges on the interval (−1,1).(−1,1).

Checkpoint 6.7

Multiply the series 11x=n=0xn11x=n=0xn by itself to construct a series for 1(1x)(1x).1(1x)(1x).

Differentiating and Integrating Power Series

Consider a power series n=0cnxn=c0+c1x+c2x2+n=0cnxn=c0+c1x+c2x2+ that converges on some interval I, and let ff be the function defined by this series. Here we address two questions about f.f.

  • Is ff differentiable, and if so, how do we determine the derivative f?f?
  • How do we evaluate the indefinite integral f(x)dx?f(x)dx?

We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Here we show that we can do the same thing for convergent power series. That is, if

f(x)=cnxn=c0+c1x+c2x2+f(x)=cnxn=c0+c1x+c2x2+

converges on some interval I, then

f(x)=c1+2c2x+3c3x2+f(x)=c1+2c2x+3c3x2+

and

f(x)dx=C+c0x+c1x22+c2x33+.f(x)dx=C+c0x+c1x22+c2x33+.

Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for f(x)=11x,f(x)=11x, we can differentiate term-by-term to find the power series for f(x)=1(1x)2.f(x)=1(1x)2. Similarly, using the power series for g(x)=11+x,g(x)=11+x, we can integrate term-by-term to find the power series for G(x)=ln(1+x),G(x)=ln(1+x), an antiderivative of g. We show how to do this in Example 6.9 and Example 6.10. First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series.

Theorem 6.4

Term-by-Term Differentiation and Integration for Power Series

Suppose that the power series n=0cn(xa)nn=0cn(xa)n converges on the interval (aR,a+R)(aR,a+R) for some R>0.R>0. Let f be the function defined by the series

f(x)=n=0cn(xa)n=c0+c1(xa)+c2(xa)2+c3(xa)3+f(x)=n=0cn(xa)n=c0+c1(xa)+c2(xa)2+c3(xa)3+

for |xa|<R.|xa|<R. Then f is differentiable on the interval (aR,a+R)(aR,a+R) and we can find ff by differentiating the series term-by-term:

f(x)=n=1ncn(xa)n1=c1+2c2(xa)+3c3(xa)2+f(x)=n=1ncn(xa)n1=c1+2c2(xa)+3c3(xa)2+

for |xa|<R.|xa|<R. Also, to find f(x)dx,f(x)dx, we can integrate the series term-by-term. The resulting series converges on (aR,a+R),(aR,a+R), and we have

f(x)dx=C+n=0cn(xa)n+1n+1=C+c0(xa)+c1(xa)22+c2(xa)33+f(x)dx=C+n=0cn(xa)n+1n+1=C+c0(xa)+c1(xa)22+c2(xa)33+

for |xa|<R.|xa|<R.

The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples.

Example 6.9

Differentiating Power Series

  1. Use the power series representation
    f(x)=11x=n=0xn=1+x+x2+x3+f(x)=11x=n=0xn=1+x+x2+x3+

    for |x|<1|x|<1 to find a power series representation for
    g(x)=1(1x)2g(x)=1(1x)2

    on the interval (−1,1).(−1,1). Determine whether the resulting series converges at the endpoints.
  2. Use the result of part a. to evaluate the sum of the series n=0n+14n.n=0n+14n.

Solution

  1. Since g(x)=1(1x)2g(x)=1(1x)2 is the derivative of f(x)=11x,f(x)=11x, we can find a power series representation for g by differentiating the power series for f term-by-term. The result is
    g(x)=1(1x)2=ddx(11x)=n=0ddx(xn)=ddx(1+x+x2+x3+)=0+1+2x+3x2+4x3+=n=0(n+1)xng(x)=1(1x)2=ddx(11x)=n=0ddx(xn)=ddx(1+x+x2+x3+)=0+1+2x+3x2+4x3+=n=0(n+1)xn

    for |x|<1.|x|<1. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints x=±1.x=±1. Note that this is the same result found in Example 6.8.
  2. From part a. we know that
    n=0(n+1)xn=1(1x)2.n=0(n+1)xn=1(1x)2.

    Therefore,
    n=0n+14n=n=0(n+1)(14)n=1(114)2=1(34)2=169.n=0n+14n=n=0(n+1)(14)n=1(114)2=1(34)2=169.
Checkpoint 6.8

Differentiate the series 1(1x)2=n=0(n+1)xn1(1x)2=n=0(n+1)xn term-by-term to find a power series representation for 2(1x)32(1x)3 on the interval (−1,1).(−1,1).

Example 6.10

Integrating Power Series

For each of the following functions f, find a power series representation for f by integrating the power series for ff and find its interval of convergence.

  1. f(x)=ln(1+x)f(x)=ln(1+x)
  2. f(x)=tan−1xf(x)=tan−1x

Solution

  1. For f(x)=ln(1+x),f(x)=ln(1+x), the derivative is f(x)=11+x.f(x)=11+x. We know that
    11+x=11(x)=n=0(x)n=1x+x2x3+11+x=11(x)=n=0(x)n=1x+x2x3+

    for |x|<1.|x|<1. To find a power series for f(x)=ln(1+x),f(x)=ln(1+x), we integrate the series term-by-term.
    f(x)dx=(1x+x2x3+)dx=C+xx22+x33x44+f(x)dx=(1x+x2x3+)dx=C+xx22+x33x44+

    Since f(x)=ln(1+x)f(x)=ln(1+x) is an antiderivative of 11+x,11+x, it remains to solve for the constant C. Since ln(1+0)=0,ln(1+0)=0, we have C=0.C=0. Therefore, a power series representation for f(x)=ln(1+x)f(x)=ln(1+x) is
    ln(1+x)=xx22+x33x44+=n=1(−1)n+1xnnln(1+x)=xx22+x33x44+=n=1(−1)n+1xnn

    for |x|<1.|x|<1. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at x=1x=1 the series is the alternating harmonic series, which converges. Also, at x=−1,x=−1, the series is the harmonic series, which diverges. It is important to note that, even though this series converges at x=1,x=1, Term-by-Term Differentiation and Integration for Power Series does not guarantee that the series actually converges to ln(2).ln(2). In fact, the series does converge to ln(2),ln(2), but showing this fact requires more advanced techniques. (Abel’s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is (−1,1].(−1,1].
  2. The derivative of f(x)=tan−1xf(x)=tan−1x is f(x)=11+x2.f(x)=11+x2. We know that
    11+x2=11(x2)=n=0(x2)n=1x2+x4x6+11+x2=11(x2)=n=0(x2)n=1x2+x4x6+

    for |x|<1.|x|<1. To find a power series for f(x)=tan−1x,f(x)=tan−1x, we integrate this series term-by-term.
    f(x)dx=(1x2+x4x6+)dx=C+xx33+x55x77+f(x)dx=(1x2+x4x6+)dx=C+xx33+x55x77+

    Since tan−1(0)=0,tan−1(0)=0, we have C=0.C=0. Therefore, a power series representation for f(x)=tan−1xf(x)=tan−1x is
    tan−1x=xx33+x55x77+=n=0(−1)nx2n+12n+1tan−1x=xx33+x55x77+=n=0(−1)nx2n+12n+1

    for |x|<1.|x|<1. Again, Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at x=1x=1 and x=−1.x=−1. As discussed in part a., using Abel’s theorem, it can be shown that the series actually converges to tan−1(1)tan−1(1) and tan−1(−1)tan−1(−1) at x=1x=1 and x=−1,x=−1, respectively. Thus, the interval of convergence is [−1,1].[−1,1].
Checkpoint 6.9

Integrate the power series ln(1+x)=n=1(−1)n+1xnnln(1+x)=n=1(−1)n+1xnn term-by-term to evaluate ln(1+x)dx.ln(1+x)dx.

Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function f and a power series for f at a, is it possible that there is a different power series for f at a that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if

c0+c1x+c2x2+=d0+d1x+d2x2+c0+c1x+c2x2+=d0+d1x+d2x2+

for all values x in some open interval I about zero, then the coefficients cn should equal dn for n0.n0. We now state this result formally in Uniqueness of Power Series.

Theorem 6.5

Uniqueness of Power Series

Let n=0cn(xa)nn=0cn(xa)n and n=0dn(xa)nn=0dn(xa)n be two convergent power series such that

n=0cn(xa)n=n=0dn(xa)nn=0cn(xa)n=n=0dn(xa)n

for all x in an open interval containing a. Then cn=dncn=dn for all n0.n0.

Proof

Let

f(x)=c0+c1(xa)+c2(xa)2+c3(xa)3+=d0+d1(xa)+d2(xa)2+d3(xa)3+.f(x)=c0+c1(xa)+c2(xa)2+c3(xa)3+=d0+d1(xa)+d2(xa)2+d3(xa)3+.

Then f(a)=c0=d0.f(a)=c0=d0. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,

f(x)=c1+2c2(xa)+3c3(xa)2+=d1+2d2(xa)+3d3(xa)2+,f(x)=c1+2c2(xa)+3c3(xa)2+=d1+2d2(xa)+3d3(xa)2+,

and thus, f(a)=c1=d1.f(a)=c1=d1. Similarly,

f(x)=2c2+3·2c3(xa)+=2d2+3·2d3(xa)+f(x)=2c2+3·2c3(xa)+=2d2+3·2d3(xa)+

implies that f(a)=2c2=2d2,f(a)=2c2=2d2, and therefore, c2=d2.c2=d2. More generally, for any integer n0,f(n)(a)=n!cn=n!dn,n0,f(n)(a)=n!cn=n!dn, and consequently, cn=dncn=dn for all n0.n0.

In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series.

Section 6.2 Exercises

63.

If f(x)=n=0xnn!f(x)=n=0xnn! and g(x)=n=0(−1)nxnn!,g(x)=n=0(−1)nxnn!, find the power series of 12(f(x)+g(x))12(f(x)+g(x)) and of 12(f(x)g(x)).12(f(x)g(x)).

64.

If C(x)=n=0x2n(2n)!C(x)=n=0x2n(2n)! and S(x)=n=0x2n+1(2n+1)!,S(x)=n=0x2n+1(2n+1)!, find the power series of C(x)+S(x)C(x)+S(x) and of C(x)S(x).C(x)S(x).

In the following exercises, use partial fractions to find the power series of each function.

65.

4(x3)(x+1)4(x3)(x+1)

66.

3(x+2)(x1)3(x+2)(x1)

67.

5(x2+4)(x21)5(x2+4)(x21)

68.

30(x2+1)(x29)30(x2+1)(x29)

In the following exercises, express each series as a rational function.

69.

n=11xnn=11xn

70.

n=11x2nn=11x2n

71.

n=11(x3)2n1n=11(x3)2n1

72.

n=1(1(x3)2n11(x2)2n1)n=1(1(x3)2n11(x2)2n1)

The following exercises explore applications of annuities.

73.

Calculate the present values P of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of r=0.03,r=0.05,r=0.03,r=0.05, and r=0.07.r=0.07.

74.

Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of r=0.03,r=0.05r=0.03,r=0.05 and r=0.07.r=0.07.

75.

Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of r=0.03,r=0.05,r=0.03,r=0.05, and r=0.07.r=0.07.

76.

Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interest rates of r=0.03,r=0.05,r=0.03,r=0.05, and r=0.07.r=0.07.

77.

Suppose that an annuity has a present value P=1million dollars.P=1million dollars. What interest rate r would allow for perpetual annual payouts of $50,000?

78.

Suppose that an annuity has a present value P=10million dollars.P=10million dollars. What interest rate r would allow for perpetual annual payouts of $100,000?

In the following exercises, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.

79.

x+x2x3+x4+x5x6+x+x2x3+x4+x5x6+ (Hint: Group powers x3k, x3k1,x3k1, and x3k2.)x3k2.)

80.

x+x2x3x4+x5+x6x7x8+x+x2x3x4+x5+x6x7x8+ (Hint: Group powers x4k, x4k1,x4k1, etc.)

81.

xx2x3+x4x5x6+x7xx2x3+x4x5x6+x7 (Hint: Group powers x3k, x3k1,x3k1, and x3k2.)x3k2.)

82.

x2+x24x38+x416+x532x664+x2+x24x38+x416+x532x664+ (Hint: Group powers (x2)3k,(x2)3k1,(x2)3k,(x2)3k1, and (x2)3k2.)(x2)3k2.)

In the following exercises, find the power series of f(x)g(x)f(x)g(x) given f and g as defined.

83.

f(x)=2n=0xn,g(x)=n=0nxnf(x)=2n=0xn,g(x)=n=0nxn

84.

f(x)=n=1xn,g(x)=n=11nxn.f(x)=n=1xn,g(x)=n=11nxn. Express the coefficients of f(x)g(x)f(x)g(x) in terms of Hn=k=1n1k.Hn=k=1n1k.

85.

f(x)=g(x)=n=1(x2)nf(x)=g(x)=n=1(x2)n

86.

f(x)=g(x)=n=1nxnf(x)=g(x)=n=1nxn

In the following exercises, differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion for the derivative of f.

87.

f(x)=11+x=n=0(−1)nxnf(x)=11+x=n=0(−1)nxn

88.

f(x)=11x2=n=0x2nf(x)=11x2=n=0x2n

In the following exercises, integrate the given series expansion of ff term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of f.f.

89.

f(x)=2x(1+x2)2=n=1(−1)n(2n)x2n1f(x)=2x(1+x2)2=n=1(−1)n(2n)x2n1

90.

f(x)=2x1+x2=2n=0(−1)nx2n+1f(x)=2x1+x2=2n=0(−1)nx2n+1

In the following exercises, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.

91.

Evaluate n=1n2nn=1n2n as f(12)f(12) where f(x)=n=0xn.f(x)=n=0xn.

92.

Evaluate n=1n3nn=1n3n as f(13)f(13) where f(x)=n=0xn.f(x)=n=0xn.

93.

Evaluate n=2n(n1)2nn=2n(n1)2n as f(12)f(12) where f(x)=n=0xn.f(x)=n=0xn.

94.

Evaluate n=0(−1)nn+1n=0(−1)nn+1 as 01f(t)dt01f(t)dt where f(x)=n=0(−1)nx2n=11+x2.f(x)=n=0(−1)nx2n=11+x2.

In the following exercises, given that 11x=n=0xn,11x=n=0xn, use term-by-term differentiation or integration to find power series for each function centered at the given point.

95.

f(x)=lnxf(x)=lnx centered at x=1x=1 (Hint: x=1(1x))x=1(1x))

96.

ln(1x)ln(1x) at x=0x=0

97.

ln(1x2)ln(1x2) at x=0x=0

98.

f(x)=2x(1x2)2f(x)=2x(1x2)2 at x=0x=0

99.

f(x)=tan−1(x2)f(x)=tan−1(x2) at x=0x=0

100.

f(x)=ln(1+x2)f(x)=ln(1+x2) at x=0x=0

101.

f(x)=0xlntdtf(x)=0xlntdt where ln(x)=n=1(−1)n1(x1)nnln(x)=n=1(−1)n1(x1)nn

102.

[T] Evaluate the power series expansion ln(1+x)=n=1(−1)n1xnnln(1+x)=n=1(−1)n1xnn at x=1x=1 to show that ln(2)ln(2) is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate ln(2)ln(2) accurate to within 0.001, and find such an approximation.

103.

[T] Subtract the infinite series of ln(1x)ln(1x) from ln(1+x)ln(1+x) to get a power series for ln(1+x1x).ln(1+x1x). Evaluate at x=13.x=13. What is the smallest N such that the Nth partial sum of this series approximates ln(2)ln(2) with an error less than 0.001?

In the following exercises, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.

104.

k=0(xkx2k+1)k=0(xkx2k+1)

105.

k=1x3k6kk=1x3k6k

106.

k=1(1+x2)kk=1(1+x2)k using y=11+x2y=11+x2

107.

k=12kxk=12kx using y=2xy=2x

108.

Show that, up to powers x3 and y3, E(x)=n=0xnn!E(x)=n=0xnn! satisfies E(x+y)=E(x)E(y).E(x+y)=E(x)E(y).

109.

Differentiate the series E(x)=n=0xnn!E(x)=n=0xnn! term-by-term to show that E(x)E(x) is equal to its derivative.

110.

Show that if f(x)=n=0anxnf(x)=n=0anxn is a sum of even powers, that is, an=0an=0 if n is odd, then F=0xf(t)dtF=0xf(t)dt is a sum of odd powers, while if f is a sum of odd powers, then F is a sum of even powers.

111.

[T] Suppose that the coefficients an of the series n=0anxnn=0anxn are defined by the recurrence relation an=an1n+an2n(n1).an=an1n+an2n(n1). For a0=0a0=0 and a1=1,a1=1, compute and plot the sums SN=n=0NanxnSN=n=0Nanxn for N=2,3,4,5N=2,3,4,5 on [−1,1].[−1,1].

112.

[T] Suppose that the coefficients an of the series n=0anxnn=0anxn are defined by the recurrence relation an=an1nan2n(n1).an=an1nan2n(n1). For a0=1a0=1 and a1=0,a1=0, compute and plot the sums SN=n=0NanxnSN=n=0Nanxn for N=2,3,4,5N=2,3,4,5 on [−1,1].[−1,1].

113.

[T] Given the power series expansion ln(1+x)=n=1(−1)n1xnn,ln(1+x)=n=1(−1)n1xnn, determine how many terms N of the sum evaluated at x=−1/2x=−1/2 are needed to approximate ln(2)ln(2) accurate to within 1/1000. Evaluate the corresponding partial sum n=1N(−1)n1xnn.n=1N(−1)n1xnn.

114.

[T] Given the power series expansion tan−1(x)=k=0(−1)kx2k+12k+1,tan−1(x)=k=0(−1)kx2k+12k+1, use the alternating series test to determine how many terms N of the sum evaluated at x=1x=1 are needed to approximate tan−1(1)=π4tan−1(1)=π4 accurate to within 1/1000. Evaluate the corresponding partial sum k=0N(−1)kx2k+12k+1.k=0N(−1)kx2k+12k+1.

115.

[T] Recall that tan−1(13)=π6.tan−1(13)=π6. Assuming an exact value of (13),(13), estimate π6π6 by evaluating partial sums SN(13)SN(13) of the power series expansion tan−1(x)=k=0(−1)kx2k+12k+1tan−1(x)=k=0(−1)kx2k+12k+1 at x=13.x=13. What is the smallest number N such that 6SN(13)6SN(13) approximates π accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?

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