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Calculus Volume 2

6.4 Working with Taylor Series

Calculus Volume 26.4 Working with Taylor 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

  • 6.4.1. Write the terms of the binomial series.
  • 6.4.2. Recognize the Taylor series expansions of common functions.
  • 6.4.3. Recognize and apply techniques to find the Taylor series for a function.
  • 6.4.4. Use Taylor series to solve differential equations.
  • 6.4.5. Use Taylor series to evaluate nonelementary integrals.

In the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. In this section we show how to use those Taylor series to derive Taylor series for other functions. We then present two common applications of power series. First, we show how power series can be used to solve differential equations. Second, we show how power series can be used to evaluate integrals when the antiderivative of the integrand cannot be expressed in terms of elementary functions. In one example, we consider ex2dx,ex2dx, an integral that arises frequently in probability theory.

The Binomial Series

Our first goal in this section is to determine the Maclaurin series for the function f(x)=(1+x)rf(x)=(1+x)r for all real numbers r.r. The Maclaurin series for this function is known as the binomial series. We begin by considering the simplest case: rr is a nonnegative integer. We recall that, for r=0,1,2,3,4,f(x)=(1+x)rr=0,1,2,3,4,f(x)=(1+x)r can be written as

f(x)=(1+x)0=1,f(x)=(1+x)1=1+x,f(x)=(1+x)2=1+2x+x2,f(x)=(1+x)3=1+3x+3x2+x3,f(x)=(1+x)4=1+4x+6x2+4x3+x4.f(x)=(1+x)0=1,f(x)=(1+x)1=1+x,f(x)=(1+x)2=1+2x+x2,f(x)=(1+x)3=1+3x+3x2+x3,f(x)=(1+x)4=1+4x+6x2+4x3+x4.

The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer r,r, the binomial coefficient of xnxn in the binomial expansion of (1+x)r(1+x)r is given by

(rn)=r!n!(rn)!(rn)=r!n!(rn)!
6.6

and

f(x)=(1+x)r=(r0)1+(r1)x+(r2)x2+(r3)x3++(rr1)xr1+(rr)xr=n=0r(rn)xn.f(x)=(1+x)r=(r0)1+(r1)x+(r2)x2+(r3)x3++(rr1)xr1+(rr)xr=n=0r(rn)xn.
6.7

For example, using this formula for r=5,r=5, we see that

f(x)=(1+x)5=(50)1+(51)x+(52)x2+(53)x3+(54)x4+(55)x5=5!0!5!1+5!1!4!x+5!2!3!x2+5!3!2!x3+5!4!1!x4+5!5!0!x5=1+5x+10x2+10x3+5x4+x5.f(x)=(1+x)5=(50)1+(51)x+(52)x2+(53)x3+(54)x4+(55)x5=5!0!5!1+5!1!4!x+5!2!3!x2+5!3!2!x3+5!4!1!x4+5!5!0!x5=1+5x+10x2+10x3+5x4+x5.

We now consider the case when the exponent rr is any real number, not necessarily a nonnegative integer. If rr is not a nonnegative integer, then f(x)=(1+x)rf(x)=(1+x)r cannot be written as a finite polynomial. However, we can find a power series for f.f. Specifically, we look for the Maclaurin series for f.f. To do this, we find the derivatives of ff and evaluate them at x=0.x=0.

f(x)=(1+x)rf(0)=1f(x)=r(1+x)r1f(0)=rf(x)=r(r1)(1+x)r2f(0)=r(r1)f(x)=r(r1)(r2)(1+x)r3f(0)=r(r1)(r2)f(n)(x)=r(r1)(r2)(rn+1)(1+x)rnf(n)(0)=r(r1)(r2)(rn+1)f(x)=(1+x)rf(0)=1f(x)=r(1+x)r1f(0)=rf(x)=r(r1)(1+x)r2f(0)=r(r1)f(x)=r(r1)(r2)(1+x)r3f(0)=r(r1)(r2)f(n)(x)=r(r1)(r2)(rn+1)(1+x)rnf(n)(0)=r(r1)(r2)(rn+1)

We conclude that the coefficients in the binomial series are given by

f(n)(0)n!=r(r1)(r2)(rn+1)n!.f(n)(0)n!=r(r1)(r2)(rn+1)n!.
6.8

We note that if rr is a nonnegative integer, then the (r+1)st(r+1)st derivative f(r+1)f(r+1) is the zero function, and the series terminates. In addition, if rr is a nonnegative integer, then Equation 6.8 for the coefficients agrees with Equation 6.6 for the coefficients, and the formula for the binomial series agrees with Equation 6.7 for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number r,r, we define

(rn)=r(r1)(r2)(rn+1)n!.(rn)=r(r1)(r2)(rn+1)n!.

With this notation, we can write the binomial series for (1+x)r(1+x)r as

n=0(rn)xn=1+rx+r(r1)2!x2++r(r1)(rn+1)n!xn+.n=0(rn)xn=1+rx+r(r1)2!x2++r(r1)(rn+1)n!xn+.
6.9

We now need to determine the interval of convergence for the binomial series Equation 6.9. We apply the ratio test. Consequently, we consider

|an+1||an|=|r(r1)(r2)(rn)|x||n+1(n+1)!·n!|r(r1)(r2)(rn+1)||x|n=|rn||x||n+1|.|an+1||an|=|r(r1)(r2)(rn)|x||n+1(n+1)!·n!|r(r1)(r2)(rn+1)||x|n=|rn||x||n+1|.

Since

limn|an+1||an|=|x|<1limn|an+1||an|=|x|<1

if and only if |x|<1,|x|<1, we conclude that the interval of convergence for the binomial series is (−1,1).(−1,1). The behavior at the endpoints depends on r.r. It can be shown that for r0r0 the series converges at both endpoints; for −1<r<0,−1<r<0, the series converges at x=1x=1 and diverges at x=−1;x=−1; and for r<−1,r<−1, the series diverges at both endpoints. The binomial series does converge to (1+x)r(1+x)r in (−1,1)(−1,1) for all real numbers r,r, but proving this fact by showing that the remainder Rn(x)0Rn(x)0 is difficult.

Definition

For any real number r,r, the Maclaurin series for f(x)=(1+x)rf(x)=(1+x)r is the binomial series. It converges to ff for |x|<1,|x|<1, and we write

(1+x)r=n=0(rn)xn=1+rx+r(r1)2!x2++r(r1)(rn+1)n!xn+(1+x)r=n=0(rn)xn=1+rx+r(r1)2!x2++r(r1)(rn+1)n!xn+

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

We can use this definition to find the binomial series for f(x)=1+xf(x)=1+x and use the series to approximate 1.5.1.5.

Example 6.17

Finding Binomial Series

  1. Find the binomial series for f(x)=1+x.f(x)=1+x.
  2. Use the third-order Maclaurin polynomial p3(x)p3(x) to estimate 1.5.1.5. Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of ff and p3.p3.

Solution

  1. Here r=12.r=12. Using the definition for the binomial series, we obtain
    1+x=1+12x+(1/2)(1/2)2!x2+(1/2)(1/2)(3/2)3!x3+=1+12x12!122x2+13!1·323x3+(−1)n+1n!1·3·5(2n3)2nxn+=1+n=1(−1)n+1n!1·3·5(2n3)2nxn.1+x=1+12x+(1/2)(1/2)2!x2+(1/2)(1/2)(3/2)3!x3+=1+12x12!122x2+13!1·323x3+(−1)n+1n!1·3·5(2n3)2nxn+=1+n=1(−1)n+1n!1·3·5(2n3)2nxn.
  2. From the result in part a. the third-order Maclaurin polynomial is
    p3(x)=1+12x18x2+116x3.p3(x)=1+12x18x2+116x3.

    Therefore,
    1.5=1+0.51+12(0.5)18(0.5)2+116(0.5)31.2266.1.5=1+0.51+12(0.5)18(0.5)2+116(0.5)31.2266.

    From Taylor’s theorem, the error satisfies
    R3(0.5)=f(4)(c)4!(0.5)4R3(0.5)=f(4)(c)4!(0.5)4

    for some cc between 00 and 0.5.0.5. Since f(4)(x)=1524(1+x)7/2,f(4)(x)=1524(1+x)7/2, and the maximum value of |f(4)(x)||f(4)(x)| on the interval (0,0.5)(0,0.5) occurs at x=0,x=0, we have
    |R3(0.5)|154!24(0.5)40.00244.|R3(0.5)|154!24(0.5)40.00244.

    The function and the Maclaurin polynomial p3p3 are graphed in Figure 6.10.
    This graph has two curves. The first one is f(x)= the square root of (1+x) and the second is psub3(x). The curves are very close at y = 1.
    Figure 6.10 The third-order Maclaurin polynomial p3(x)p3(x) provides a good approximation for f(x)=1+xf(x)=1+x for xx near zero.
Checkpoint 6.16

Find the binomial series for f(x)=1(1+x)2.f(x)=1(1+x)2.

Common Functions Expressed as Taylor Series

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form f(x)=(1+x)r.f(x)=(1+x)r. In Table 6.1, we summarize the results of these series. We remark that the convergence of the Maclaurin series for f(x)=ln(1+x)f(x)=ln(1+x) at the endpoint x=1x=1 and the Maclaurin series for f(x)=tan−1xf(x)=tan−1x at the endpoints x=1x=1 and x=−1x=−1 relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Function Maclaurin Series Interval of Convergence
f(x)=11xf(x)=11x n=0xnn=0xn −1<x<1−1<x<1
f(x)=exf(x)=ex n=0xnn!n=0xnn! <x<<x<
f(x)=sinxf(x)=sinx n=0(−1)nx2n+1(2n+1)!n=0(−1)nx2n+1(2n+1)! <x<<x<
f(x)=cosxf(x)=cosx n=0(−1)nx2n(2n)!n=0(−1)nx2n(2n)! <x<<x<
f(x)=ln(1+x)f(x)=ln(1+x) n=1(−1)n+1xnnn=1(−1)n+1xnn −1<x1−1<x1
f(x)=tan−1xf(x)=tan−1x n=0(−1)nx2n+12n+1n=0(−1)nx2n+12n+1 −1x1−1x1
f(x)=(1+x)rf(x)=(1+x)r n=0(rn)xnn=0(rn)xn −1<x<1−1<x<1
Table 6.1 Maclaurin Series for Common Functions

Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in Table 6.1, to create Maclaurin series for other functions.

Example 6.18

Deriving Maclaurin Series from Known Series

Find the Maclaurin series of each of the following functions by using one of the series listed in Table 6.1.

  1. f(x)=cosxf(x)=cosx
  2. f(x)=sinhxf(x)=sinhx

Solution

  1. Using the Maclaurin series for cosxcosx we find that the Maclaurin series for cosxcosx is given by
    n=0(−1)n(x)2n(2n)!=n=0(−1)nxn(2n)!=1x2!+x24!x36!+x48!.n=0(−1)n(x)2n(2n)!=n=0(−1)nxn(2n)!=1x2!+x24!x36!+x48!.

    This series converges to cosxcosx for all xx in the domain of cosx;cosx; that is, for all x0.x0.
  2. To find the Maclaurin series for sinhx,sinhx, we use the fact that
    sinhx=exex2.sinhx=exex2.

    Using the Maclaurin series for ex,ex, we see that the nthnth term in the Maclaurin series for sinhxsinhx is given by
    xnn!(x)nn!.xnn!(x)nn!.

    For nn even, this term is zero. For nn odd, this term is 2xnn!.2xnn!. Therefore, the Maclaurin series for sinhxsinhx has only odd-order terms and is given by
    n=0x2n+1(2n+1)!=x+x33!+x55!+.n=0x2n+1(2n+1)!=x+x33!+x55!+.
Checkpoint 6.17

Find the Maclaurin series for sin(x2).sin(x2).

We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In Example 6.19, we differentiate the binomial series for 1+x1+x term by term to find the binomial series for 11+x.11+x. Note that we could construct the binomial series for 11+x11+x directly from the definition, but differentiating the binomial series for 1+x1+x is an easier calculation.

Example 6.19

Differentiating a Series to Find a New Series

Use the binomial series for 1+x1+x to find the binomial series for 11+x.11+x.

Solution

The two functions are related by

ddx1+x=121+x,ddx1+x=121+x,

so the binomial series for 11+x11+x is given by

11+x=2ddx1+x=1+n=1(−1)nn!1·3·5(2n1)2nxn.11+x=2ddx1+x=1+n=1(−1)nn!1·3·5(2n1)2nxn.
Checkpoint 6.18

Find the binomial series for f(x)=1(1+x)3/2f(x)=1(1+x)3/2

In this example, we differentiated a known Taylor series to construct a Taylor series for another function. The ability to differentiate power series term by term makes them a powerful tool for solving differential equations. We now show how this is accomplished.

Solving Differential Equations with Power Series

Consider the differential equation

y(x)=y.y(x)=y.

Recall that this is a first-order separable equation and its solution is y=Cex.y=Cex. This equation is easily solved using techniques discussed earlier in the text. For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form y=n=0cnxny=n=0cnxn and determine what the coefficients would need to be. In the next example, we consider an initial-value problem involving y=yy=y to illustrate the technique.

Example 6.20

Power Series Solution of a Differential Equation

Use power series to solve the initial-value problem

y=y,y(0)=3.y=y,y(0)=3.

Solution

Suppose that there exists a power series solution

y(x)=n=0cnxn=c0+c1x+c2x2+c3x3+c4x4+.y(x)=n=0cnxn=c0+c1x+c2x2+c3x3+c4x4+.

Differentiating this series term by term, we obtain

y=c1+2c2x+3c3x2+4c4x3+.y=c1+2c2x+3c3x2+4c4x3+.

If y satisfies the differential equation, then

c0+c1x+c2x2+c3x3+=c1+2c2x+3c3x2+4c3x3+.c0+c1x+c2x2+c3x3+=c1+2c2x+3c3x2+4c3x3+.

Using Uniqueness of Power Series on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,

c0=c1,c1=2c2,c2=3c3,c3=4c4,.c0=c1,c1=2c2,c2=3c3,c3=4c4,.

Using the initial condition y(0)=3y(0)=3 combined with the power series representation

y(x)=c0+c1x+c2x2+c3x3+,y(x)=c0+c1x+c2x2+c3x3+,

we find that c0=3.c0=3. We are now ready to solve for the rest of the coefficients. Using the fact that c0=3,c0=3, we have

c1=c0=3=31!,c2=c12=32=32!,c3=c23=33·2=33!,c4=c34=34·3·2=34!.c1=c0=3=31!,c2=c12=32=32!,c3=c23=33·2=33!,c4=c34=34·3·2=34!.

Therefore,

y=3[1+11!x+12!x2+13!x314!x4+]=3n=0xnn!.y=3[1+11!x+12!x2+13!x314!x4+]=3n=0xnn!.

You might recognize

n=0xnn!n=0xnn!

as the Taylor series for ex.ex. Therefore, the solution is y=3ex.y=3ex.

Checkpoint 6.19

Use power series to solve y=2y,y(0)=5.y=2y,y(0)=5.

We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation

yxy=0yxy=0

is known as Airy’s equation. It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.

Example 6.21

Power Series Solution of Airy’s Equation

Use power series to solve

yxy=0yxy=0

with the initial conditions y(0)=ay(0)=a and y(0)=b.y(0)=b.

Solution

We look for a solution of the form

y=n=0cnxn=c0+c1x+c2x2+c3x3+c4x4+.y=n=0cnxn=c0+c1x+c2x2+c3x3+c4x4+.

Differentiating this function term by term, we obtain

y=c1+2c2x+3c3x2+4c4x3+,y=2·1c2+3·2c3x+4·3c4x2+.y=c1+2c2x+3c3x2+4c4x3+,y=2·1c2+3·2c3x+4·3c4x2+.

If y satisfies the equation y=xy,y=xy, then

2·1c2+3·2c3x+4·3c4x2+=x(c0+c1x+c2x2+c3x3+).2·1c2+3·2c3x+4·3c4x2+=x(c0+c1x+c2x2+c3x3+).

Using Uniqueness of Power Series on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

2·1c2=0,3·2c3=c0,4·3c4=c1,5·4c5=c2,.2·1c2=0,3·2c3=c0,4·3c4=c1,5·4c5=c2,.

More generally, for n3,n3, we have n·(n1)cn=cn3.n·(n1)cn=cn3. In fact, all coefficients can be written in terms of c0c0 and c1.c1. To see this, first note that c2=0.c2=0. Then

c3=c03·2,c4=c14·3.c3=c03·2,c4=c14·3.

For c5,c6,c7,c5,c6,c7, we see that

c5=c25·4=0,c6=c36·5=c06·5·3·2,c7=c47·6=c17·6·4·3.c5=c25·4=0,c6=c36·5=c06·5·3·2,c7=c47·6=c17·6·4·3.

Therefore, the series solution of the differential equation is given by

y=c0+c1x+0·x2+c03·2x3+c14·3x4+0·x5+c06·5·3·2x6+c17·6·4·3x7+.y=c0+c1x+0·x2+c03·2x3+c14·3x4+0·x5+c06·5·3·2x6+c17·6·4·3x7+.

The initial condition y(0)=ay(0)=a implies c0=a.c0=a. Differentiating this series term by term and using the fact that y(0)=b,y(0)=b, we conclude that c1=b.c1=b. Therefore, the solution of this initial-value problem is

y=a(1+x33·2+x66·5·3·2+)+b(x+x44·3+x77·6·4·3+).y=a(1+x33·2+x66·5·3·2+)+b(x+x44·3+x77·6·4·3+).
Checkpoint 6.20

Use power series to solve y+x2y=0y+x2y=0 with the initial condition y(0)=ay(0)=a and y(0)=b.y(0)=b.

Evaluating Nonelementary Integrals

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

One integral that arises often in applications in probability theory is ex2dx.ex2dx. Unfortunately, the antiderivative of the integrand ex2ex2 is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with noncomplicated function. For example, the function f(x)=x23x+ex3sin(5x+4)f(x)=x23x+ex3sin(5x+4) is an elementary function, although not a particularly simple-looking function. Any integral of the form f(x)dxf(x)dx where the antiderivative of ff cannot be written as an elementary function is considered a nonelementary integral.

Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering ex2dx.ex2dx.

Example 6.22

Using Taylor Series to Evaluate a Definite Integral

  1. Express ex2dxex2dx as an infinite series.
  2. Evaluate 01ex2dx01ex2dx to within an error of 0.01.0.01.

Solution

  1. The Maclaurin series for ex2ex2 is given by
    ex2=n=0(x2)nn!=1x2+x42!x63!++(−1)nx2nn!+=n=0(−1)nx2nn!.ex2=n=0(x2)nn!=1x2+x42!x63!++(−1)nx2nn!+=n=0(−1)nx2nn!.

    Therefore,
    ex2dx=(1x2+x42!x63!++(−1)nx2nn!+)dx=C+xx33+x55.2!x77.3!++(−1)nx2n+1(2n+1)n!+.ex2dx=(1x2+x42!x63!++(−1)nx2nn!+)dx=C+xx33+x55.2!x77.3!++(−1)nx2n+1(2n+1)n!+.
  2. Using the result from part a. we have
    01ex2dx=113+110142+1216.01ex2dx=113+110142+1216.

    The sum of the first four terms is approximately 0.74.0.74. By the alternating series test, this estimate is accurate to within an error of less than 12160.0046296<0.01.12160.0046296<0.01.
Checkpoint 6.21

Express cosxdxcosxdx as an infinite series. Evaluate 01cosxdx01cosxdx to within an error of 0.01.0.01.

As mentioned above, the integral ex2dxex2dx arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean μμ and standard deviation σ,σ, then the probability that a randomly chosen value lies between x=ax=a and x=bx=b is given by

1σ2πabe(xμ)2/(2σ2)dx.1σ2πabe(xμ)2/(2σ2)dx.
6.10

(See Figure 6.11.)

This graph is the normal distribution. It is a bell-shaped curve with the highest point above mu on the x-axis. Also, there is a shaded area under the curve above the x-axis. The shaded area is bounded by alpha on the left and b on the right.
Figure 6.11 If data values are normally distributed with mean μμ and standard deviation σ,σ, the probability that a randomly selected data value is between aa and bb is the area under the curve y=1σ2πe(xμ)2/(2σ2)y=1σ2πe(xμ)2/(2σ2) between x=ax=a and x=b.x=b.

To simplify this integral, we typically let z=xμσ.z=xμσ. This quantity zz is known as the zz score of a data value. With this simplification, integral Equation 6.10 becomes

12π(aμ)/σ(bμ)/σez2/2dz.12π(aμ)/σ(bμ)/σez2/2dz.
6.11

In Example 6.23, we show how we can use this integral in calculating probabilities.

Example 6.23

Using Maclaurin Series to Approximate a Probability

Suppose a set of standardized test scores are normally distributed with mean μ=100μ=100 and standard deviation σ=50.σ=50. Use Equation 6.11 and the first six terms in the Maclaurin series for ex2/2ex2/2 to approximate the probability that a randomly selected test score is between x=100x=100 and x=200.x=200. Use the alternating series test to determine how accurate your approximation is.

Solution

Since μ=100,σ=50,μ=100,σ=50, and we are trying to determine the area under the curve from a=100a=100 to b=200,b=200, integral Equation 6.11 becomes

12π02ez2/2dz.12π02ez2/2dz.

The Maclaurin series for ex2/2ex2/2 is given by

ex2/2=n=0(x22)nn!=1x221·1!+x422·2!x623·3!++(−1)nx2n2n·n!+=n=0(−1)nx2n2n·n!.ex2/2=n=0(x22)nn!=1x221·1!+x422·2!x623·3!++(−1)nx2n2n·n!+=n=0(−1)nx2n2n·n!.

Therefore,

12πez2/2dz=12π(1z221·1!+z422·2!z623·3!++(−1)nz2n2n·n!+)dz=12π(C+zz33·21·1!+z55·22·2!z77·23·3!++(−1)nz2n+1(2n+1)2n·n!+)12π02ez2/2dz=12π(286+3240128336+512345621111·25·5!+).12πez2/2dz=12π(1z221·1!+z422·2!z623·3!++(−1)nz2n2n·n!+)dz=12π(C+zz33·21·1!+z55·22·2!z77·23·3!++(−1)nz2n+1(2n+1)2n·n!+)12π02ez2/2dz=12π(286+3240128336+512345621111·25·5!+).

Using the first five terms, we estimate that the probability is approximately 0.4922.0.4922. By the alternating series test, we see that this estimate is accurate to within

12π21313·26·6!0.00546.12π21313·26·6!0.00546.

Analysis

If you are familiar with probability theory, you may know that the probability that a data value is within two standard deviations of the mean is approximately 95%.95%. Here we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around 47.5%.47.5%. The estimate, combined with the bound on the accuracy, falls within this range.

Checkpoint 6.22

Use the first five terms of the Maclaurin series for ex2/2ex2/2 to estimate the probability that a randomly selected test score is between 100100 and 150.150. Use the alternating series test to determine the accuracy of this estimate.

Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is

0π/2dθ1k2sin2θ.0π/2dθ1k2sin2θ.

An integral of this form is known as an elliptic integral of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.

Example 6.24

Period of a Pendulum

The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length LL that makes a maximum angle θmaxθmax with the vertical, its period TT is given by

T=4Lg0π/2dθ1k2sin2θT=4Lg0π/2dθ1k2sin2θ

where gg is the acceleration due to gravity and k=sin(θmax2)k=sin(θmax2) (see Figure 6.12). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and sinθsinθ is approximated by θ.)θ.) Use the binomial series

11+x=1+n=1(−1)nn!1·3·5(2n1)2nxn11+x=1+n=1(−1)nn!1·3·5(2n1)2nxn

to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if

  1. you use only the first term in the binomial series, and
  2. you use the first two terms in the binomial series.
    This figure is a pendulum. There are three positions of the pendulum shown. When the pendulum is to the far left, it is labeled negative theta max. When the pendulum is in the middle and vertical, it is labeled equilibrium position. When the pendulum is to the far right it is labeled theta max. Also, theta is the angle from equilibrium to the far right position. The length of the pendulum is labeled L.
    Figure 6.12 This pendulum has length LL and makes a maximum angle θmaxθmax with the vertical.

Solution

We use the binomial series, replacing xx with k2sin2θ.k2sin2θ. Then we can write the period as

T=4Lg0π/2(1+12k2sin2θ+1·32!22k4sin4θ+)dθ.T=4Lg0π/2(1+12k2sin2θ+1·32!22k4sin4θ+)dθ.
  1. Using just the first term in the integrand, the first-order estimate is
    T4Lg0π/2dθ=2πLg.T4Lg0π/2dθ=2πLg.

    If θmaxθmax is small, then k=sin(θmax2)k=sin(θmax2) is small. We claim that when kk is small, this is a good estimate. To justify this claim, consider
    0π/2(1+12k2sin2θ+1·32!22k4sin4θ+)dθ.0π/2(1+12k2sin2θ+1·32!22k4sin4θ+)dθ.

    Since |sinx|1,|sinx|1, this integral is bounded by
    0π/2(12k2+1.32!22k4+)dθ<π2(12k2+1·32!22k4+).0π/2(12k2+1.32!22k4+)dθ<π2(12k2+1·32!22k4+).

    Furthermore, it can be shown that each coefficient on the right-hand side is less than 11 and, therefore, that this expression is bounded by
    πk22(1+k2+k4+)=πk22·11k2,πk22(1+k2+k4+)=πk22·11k2,

    which is small for kk small.
  2. For larger values of θmax,θmax, we can approximate TT by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate
    T4Lg0π/2(1+12k2sin2θ)dθ=2πLg(1+k24).T4Lg0π/2(1+12k2sin2θ)dθ=2πLg(1+k24).

The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.

Section 6.4 Exercises

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

174.

(1x)1/3(1x)1/3

175.

(1+x2)−1/3(1+x2)−1/3

176.

(1x)1.01(1x)1.01

177.

(12x)2/3(12x)2/3

In the following exercises, use the substitution (b+x)r=(b+a)r(1+xab+a)r(b+x)r=(b+a)r(1+xab+a)r in the binomial expansion to find the Taylor series of each function with the given center.

178.

x+2x+2 at a=0a=0

179.

x2+2x2+2 at a=0a=0

180.

x+2x+2 at a=1a=1

181.

2xx22xx2 at a=1a=1 (Hint: 2xx2=1(x1)2)2xx2=1(x1)2)

182.

(x8)1/3(x8)1/3 at a=9a=9

183.

xx at a=4a=4

184.

x1/3x1/3 at a=27a=27

185.

xx at x=9x=9

In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most 1/1000.1/1000.

186.

[T] (15)1/4(15)1/4 using (16x)1/4(16x)1/4

187.

[T] (1001)1/3(1001)1/3 using (1000+x)1/3(1000+x)1/3

In the following exercises, use the binomial approximation 1x1x2x28x3165x41287x52561x1x2x28x3165x41287x5256 for |x|<1|x|<1 to approximate each number. Compare this value to the value given by a scientific calculator.

188.

[T] 1212 using x=12x=12 in (1x)1/2(1x)1/2

189.

[T] 5=5×155=5×15 using x=45x=45 in (1x)1/2(1x)1/2

190.

[T] 3=333=33 using x=23x=23 in (1x)1/2(1x)1/2

191.

[T] 66 using x=56x=56 in (1x)1/2(1x)1/2

192.

Integrate the binomial approximation of 1x1x to find an approximation of 0x1tdt.0x1tdt.

193.

[T] Recall that the graph of 1x21x2 is an upper semicircle of radius 1.1. Integrate the binomial approximation of 1x21x2 up to order 88 from x=−1x=−1 to x=1x=1 to estimate π2.π2.

In the following exercises, use the expansion (1+x)1/3=1+13x19x2+581x310243x4+(1+x)1/3=1+13x19x2+581x310243x4+ to write the first five terms (not necessarily a quartic polynomial) of each expression.

194.

(1+4x)1/3;a=0(1+4x)1/3;a=0

195.

(1+4x)4/3;a=0(1+4x)4/3;a=0

196.

(3+2x)1/3;a=−1(3+2x)1/3;a=−1

197.

(x2+6x+10)1/3;a=−3(x2+6x+10)1/3;a=−3

198.

Use (1+x)1/3=1+13x19x2+581x310243x4+(1+x)1/3=1+13x19x2+581x310243x4+ with x=1x=1 to approximate 21/3.21/3.

199.

Use the approximation (1x)2/3=12x3x294x3817x424314x5729+(1x)2/3=12x3x294x3817x424314x5729+ for |x|<1|x|<1 to approximate 21/3=2.2−2/3.21/3=2.2−2/3.

200.

Find the 25th25th derivative of f(x)=(1+x2)13f(x)=(1+x2)13 at x=0.x=0.

201.

Find the 9999 th derivative of f(x)=(1+x4)25.f(x)=(1+x4)25.

In the following exercises, find the Maclaurin series of each function.

202.

f(x)=xe2xf(x)=xe2x

203.

f(x)=2xf(x)=2x

204.

f(x)=sinxxf(x)=sinxx

205.

f(x)=sin(x)x,(x>0),f(x)=sin(x)x,(x>0),

206.

f(x)=sin(x2)f(x)=sin(x2)

207.

f(x)=ex3f(x)=ex3

208.

f(x)=cos2xf(x)=cos2x using the identity cos2x=12+12cos(2x)cos2x=12+12cos(2x)

209.

f(x)=sin2xf(x)=sin2x using the identity sin2x=1212cos(2x)sin2x=1212cos(2x)

In the following exercises, find the Maclaurin series of F(x)=0xf(t)dtF(x)=0xf(t)dt by integrating the Maclaurin series of ff term by term. If ff is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

210.

F(x)=0xet2dt;f(t)=et2=n=0(−1)nt2nn!F(x)=0xet2dt;f(t)=et2=n=0(−1)nt2nn!

211.

F(x)=tan−1x;f(t)=11+t2=n=0(−1)nt2nF(x)=tan−1x;f(t)=11+t2=n=0(−1)nt2n

212.

F(x)=tanh−1x;f(t)=11t2=n=0t2nF(x)=tanh−1x;f(t)=11t2=n=0t2n

213.

F(x)=sin−1x;f(t)=11t2=k=0(12k)t2kk!F(x)=sin−1x;f(t)=11t2=k=0(12k)t2kk!

214.

F(x)=0xsinttdt;f(t)=sintt=n=0(−1)nt2n(2n+1)!F(x)=0xsinttdt;f(t)=sintt=n=0(−1)nt2n(2n+1)!

215.

F(x)=0xcos(t)dt;f(t)=n=0(−1)nxn(2n)!F(x)=0xcos(t)dt;f(t)=n=0(−1)nxn(2n)!

216.

F(x)=0x1costt2dt;f(t)=1costt2=n=0(−1)nt2n(2n+2)!F(x)=0x1costt2dt;f(t)=1costt2=n=0(−1)nt2n(2n+2)!

217.

F(x)=0xln(1+t)tdt;f(t)=n=0(−1)ntnn+1F(x)=0xln(1+t)tdt;f(t)=n=0(−1)ntnn+1

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of f.f.

218.

f(x)=sin(x+π4)=sinxcos(π4)+cosxsin(π4)f(x)=sin(x+π4)=sinxcos(π4)+cosxsin(π4)

219.

f(x)=tanxf(x)=tanx

220.

f(x)=ln(cosx)f(x)=ln(cosx)

221.

f(x)=excosxf(x)=excosx

222.

f(x)=esinxf(x)=esinx

223.

f(x)=sec2xf(x)=sec2x

224.

f(x)=tanhxf(x)=tanhx

225.

f(x)=tanxxf(x)=tanxx (see expansion for tanx)tanx)

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

226.

ln(1+x)ln(1+x)

227.

11+x211+x2

228.

tan−1xtan−1x

229.

ln(1+x2)ln(1+x2)

230.

Find the Maclaurin series of sinhx=exex2.sinhx=exex2.

231.

Find the Maclaurin series of coshx=ex+ex2.coshx=ex+ex2.

232.

Differentiate term by term the Maclaurin series of sinhxsinhx and compare the result with the Maclaurin series of coshx.coshx.

233.

[T] Let Sn(x)=k=0n(−1)kx2k+1(2k+1)!Sn(x)=k=0n(−1)kx2k+1(2k+1)! and Cn(x)=n=0n(−1)kx2k(2k)!Cn(x)=n=0n(−1)kx2k(2k)! denote the respective Maclaurin polynomials of degree 2n+12n+1 of sinxsinx and degree 2n2n of cosx.cosx. Plot the errors Sn(x)Cn(x)tanxSn(x)Cn(x)tanx for n=1,..,5n=1,..,5 and compare them to x+x33+2x515+17x7315tanxx+x33+2x515+17x7315tanx on (π4,π4).(π4,π4).

234.

Use the identity 2sinxcosx=sin(2x)2sinxcosx=sin(2x) to find the power series expansion of sin2xsin2x at x=0.x=0. (Hint: Integrate the Maclaurin series of sin(2x)sin(2x) term by term.)

235.

If y=n=0anxn,y=n=0anxn, find the power series expansions of xyxy and x2y.x2y.

236.

[T] Suppose that y=k=0akxky=k=0akxk satisfies y=−2xyy=−2xy and y(0)=0.y(0)=0. Show that a2k+1=0a2k+1=0 for all kk and that a2k+2=a2kk+1.a2k+2=a2kk+1. Plot the partial sum S20S20 of yy on the interval [−4,4].[−4,4].

237.

[T] Suppose that a set of standardized test scores is normally distributed with mean μ=100μ=100 and standard deviation σ=10.σ=10. Set up an integral that represents the probability that a test score will be between 9090 and 110110 and use the integral of the degree 1010 Maclaurin polynomial of 12πex2/212πex2/2 to estimate this probability.

238.

[T] Suppose that a set of standardized test scores is normally distributed with mean μ=100μ=100 and standard deviation σ=10.σ=10. Set up an integral that represents the probability that a test score will be between 7070 and 130130 and use the integral of the degree 5050 Maclaurin polynomial of 12πex2/212πex2/2 to estimate this probability.

239.

[T] Suppose that n=0anxnn=0anxn converges to a function f(x)f(x) such that f(0)=1,f(0)=0,f(0)=1,f(0)=0, and f(x)=f(x).f(x)=f(x). Find a formula for anan and plot the partial sum SNSN for N=20N=20 on [−5,5].[−5,5].

240.

[T] Suppose that n=0anxnn=0anxn converges to a function f(x)f(x) such that f(0)=0,f(0)=1,f(0)=0,f(0)=1, and f(x)=f(x).f(x)=f(x). Find a formula for anan and plot the partial sum SNSN for N=10N=10 on [−5,5].[−5,5].

241.

Suppose that n=0anxnn=0anxn converges to a function yy such that yy+y=0yy+y=0 where y(0)=1y(0)=1 and y(0)=0.y(0)=0. Find a formula that relates an+2,an+1,an+2,an+1, and anan and compute a0,...,a5.a0,...,a5.

242.

Suppose that n=0anxnn=0anxn converges to a function yy such that yy+y=0yy+y=0 where y(0)=0y(0)=0 and y(0)=1.y(0)=1. Find a formula that relates an+2,an+1,an+2,an+1, and anan and compute a1,...,a5.a1,...,a5.

The error in approximating the integral abf(t)dtabf(t)dt by that of a Taylor approximation abPn(t)dtabPn(t)dt is at most abRn(t)dt.abRn(t)dt. In the following exercises, the Taylor remainder estimate RnM(n+1)!|xa|n+1RnM(n+1)!|xa|n+1 guarantees that the integral of the Taylor polynomial of the given order approximates the integral of ff with an error less than 110.110.

  1. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than 1100.1100.
  2. Compare the accuracy of the polynomial integral estimate with the remainder estimate.
243.

[T] 0πsinttdt;Ps=1x23!+x45!x67!+x89!0πsinttdt;Ps=1x23!+x45!x67!+x89! (You may assume that the absolute value of the ninth derivative of sinttsintt is bounded by 0.1.)0.1.)

244.

[T] 02ex2dx;p11=1x2+x42x63!+x2211!02ex2dx;p11=1x2+x42x63!+x2211! (You may assume that the absolute value of the 23rd23rd derivative of ex2ex2 is less than 2×1014.)2×1014.)

The following exercises deal with Fresnel integrals.

245.

The Fresnel integrals are defined by C(x)=0xcos(t2)dtC(x)=0xcos(t2)dt and S(x)=0xsin(t2)dt.S(x)=0xsin(t2)dt. Compute the power series of C(x)C(x) and S(x)S(x) and plot the sums CN(x)CN(x) and SN(x)SN(x) of the first N=50N=50 nonzero terms on [0,2π].[0,2π].

246.

[T] The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates (C(t),S(t)).(C(t),S(t)). Plot the curve (C50,S50)(C50,S50) for 0t2π,0t2π, the coordinates of which were computed in the previous exercise.

247.

Estimate 01/4xx2dx01/4xx2dx by approximating 1x1x using the binomial approximation 1x2x28x3165x421287x5256.1x2x28x3165x421287x5256.

248.

[T] Use Newton’s approximation of the binomial 1x21x2 to approximate ππ as follows. The circle centered at (12,0)(12,0) with radius 1212 has upper semicircle y=x1x.y=x1x. The sector of this circle bounded by the xx-axis between x=0x=0 and x=12x=12 and by the line joining (14,34)(14,34) corresponds to 1616 of the circle and has area π24.π24. This sector is the union of a right triangle with height 3434 and base 1414 and the region below the graph between x=0x=0 and x=14.x=14. To find the area of this region you can write y=x1x=x×(binomial expansion of1x)y=x1x=x×(binomial expansion of1x) and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate π.π.

249.

Use the approximation T2πLg(1+k24)T2πLg(1+k24) to approximate the period of a pendulum having length 1010 meters and maximum angle θmax=π6θmax=π6 where k=sin(θmax2).k=sin(θmax2). Compare this with the small angle estimate T2πLg.T2πLg.

250.

Suppose that a pendulum is to have a period of 22 seconds and a maximum angle of θmax=π6.θmax=π6. Use T2πLg(1+k24)T2πLg(1+k24) to approximate the desired length of the pendulum. What length is predicted by the small angle estimate T2πLg?T2πLg?

251.

Evaluate 0π/2sin4θdθ0π/2sin4θdθ in the approximation T=4Lg0π/2(1+12k2sin2θ+38k4sin4θ+)dθT=4Lg0π/2(1+12k2sin2θ+38k4sin4θ+)dθ to obtain an improved estimate for T.T.

252.

[T] An equivalent formula for the period of a pendulum with amplitude θmaxθmax is T(θmax)=22Lg0θmaxdθcosθcos(θmax)T(θmax)=22Lg0θmaxdθcosθcos(θmax) where LL is the pendulum length and gg is the gravitational acceleration constant. When θmax=π3θmax=π3 we get 1cost1/22(1+t22+t43+181t6720).1cost1/22(1+t22+t43+181t6720). Integrate this approximation to estimate T(π3)T(π3) in terms of LL and g.g. Assuming g=9.806g=9.806 meters per second squared, find an approximate length LL such that T(π3)=2T(π3)=2 seconds.

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