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Calculus Volume 2

6.3 Taylor and Maclaurin Series

Calculus Volume 26.3 Taylor and Maclaurin 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

  • 6.3.1. Describe the procedure for finding a Taylor polynomial of a given order for a function.
  • 6.3.2. Explain the meaning and significance of Taylor’s theorem with remainder.
  • 6.3.3. Estimate the remainder for a Taylor series approximation of a given function.

In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function ff and the series converges on some interval, how do we prove that the series actually converges to f?f?

Overview of Taylor/Maclaurin Series

Consider a function ff that has a power series representation at x=a.x=a. Then the series has the form

n=0cn(xa)n=c0+c1(xa)+c2(xa)2+.n=0cn(xa)n=c0+c1(xa)+c2(xa)2+.
6.4

What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series Equation 6.4 is a representation for ff at x=a,x=a, we certainly want the series to equal f(a)f(a) at x=a.x=a. Evaluating the series at x=a,x=a, we see that

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

Thus, the series equals f(a)f(a) if the coefficient c0=f(a).c0=f(a). In addition, we would like the first derivative of the power series to equal f(a)f(a) at x=a.x=a. Differentiating Equation 6.4 term-by-term, we see that

ddx(n=0cn(xa)n)=c1+2c2(xa)+3c3(xa)2+.ddx(n=0cn(xa)n)=c1+2c2(xa)+3c3(xa)2+.

Therefore, at x=a,x=a, the derivative is

ddx(n=0cn(xa)n)=c1+2c2(aa)+3c3(aa)2+=c1.ddx(n=0cn(xa)n)=c1+2c2(aa)+3c3(aa)2+=c1.

Therefore, the derivative of the series equals f(a)f(a) if the coefficient c1=f(a).c1=f(a). Continuing in this way, we look for coefficients cn such that all the derivatives of the power series Equation 6.4 will agree with all the corresponding derivatives of ff at x=a.x=a. The second and third derivatives of Equation 6.4 are given by

d2dx2(n=0cn(xa)n)=2c2+3·2c3(xa)+4·3c4(xa)2+d2dx2(n=0cn(xa)n)=2c2+3·2c3(xa)+4·3c4(xa)2+

and

d3dx3(n=0cn(xa)n)=3·2c3+4·3·2c4(xa)+5·4·3c5(xa)2+.d3dx3(n=0cn(xa)n)=3·2c3+4·3·2c4(xa)+5·4·3c5(xa)2+.

Therefore, at x=a,x=a, the second and third derivatives

d2dx2(n=0cn(xa)n)=2c2+3·2c3(aa)+4·3c4(aa)2+=2c2d2dx2(n=0cn(xa)n)=2c2+3·2c3(aa)+4·3c4(aa)2+=2c2

and

d3dx3(n=0cn(xa)n)=3·2c3+4·3·2c4(aa)+5·4·3c5(aa)2+=3·2c3d3dx3(n=0cn(xa)n)=3·2c3+4·3·2c4(aa)+5·4·3c5(aa)2+=3·2c3

equal f(a)f(a) and f(a),f(a), respectively, if c2=f(a)2c2=f(a)2 and c3=f(a)3·2.c3=f(a)3·2. More generally, we see that if ff has a power series representation at x=a,x=a, then the coefficients should be given by cn=f(n)(a)n!.cn=f(n)(a)n!. That is, the series should be

n=0f(n)(a)n!(xa)n=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3+.n=0f(n)(a)n!(xa)n=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3+.

This power series for ff is known as the Taylor series for ff at a.a. If a=0,a=0, then this series is known as the Maclaurin series for f.f.

Definition

If ff has derivatives of all orders at x=a,x=a, then the Taylor series for the function ff at aa is

n=0f(n)(a)n!(xa)n=f(a)+f(a)(xa)+f(a)2!(xa)2++f(n)(a)n!(xa)n+.n=0f(n)(a)n!(xa)n=f(a)+f(a)(xa)+f(a)2!(xa)2++f(n)(a)n!(xa)n+.
6.5

The Taylor series for ff at 0 is known as the Maclaurin series for f.f.

Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from Uniqueness of Power Series that power series representations are unique. Therefore, if a function ff has a power series at a,a, then it must be the Taylor series for ff at a.a.

Theorem 6.6

Uniqueness of Taylor Series

If a function ff has a power series at a that converges to ff on some open interval containing a, then that power series is the Taylor series for ff at a.

The proof follows directly from Uniqueness of Power Series.

To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials.

Media

Visit the MacTutor History of Mathematics archive to read brief biographies of Brook Taylor and Colin Maclaurin and how they developed the concepts named after them.

Taylor Polynomials

The nth partial sum of the Taylor series for a function ff at aa is known as the nth Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by

p0(x)=f(a),p1(x)=f(a)+f(a)(xa),p2(x)=f(a)+f(a)(xa)+f(a)2!(xa)2,p3(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3,p0(x)=f(a),p1(x)=f(a)+f(a)(xa),p2(x)=f(a)+f(a)(xa)+f(a)2!(xa)2,p3(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3,

respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of ff at a,a, respectively. If a=0,a=0, then these polynomials are known as Maclaurin polynomials for f.f. We now provide a formal definition of Taylor and Maclaurin polynomials for a function f.f.

Definition

If ff has n derivatives at x=a,x=a, then the nth Taylor polynomial for ff at aa is

pn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3++f(n)(a)n!(xa)n.pn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3++f(n)(a)n!(xa)n.

The nth Taylor polynomial for ff at 0 is known as the nth Maclaurin polynomial for f.f.

We now show how to use this definition to find several Taylor polynomials for f(x)=lnxf(x)=lnx at x=1.x=1.

Example 6.11

Finding Taylor Polynomials

Find the Taylor polynomials p0,p1,p2p0,p1,p2 and p3p3 for f(x)=lnxf(x)=lnx at x=1.x=1. Use a graphing utility to compare the graph of ff with the graphs of p0,p1,p2p0,p1,p2 and p3.p3.

Solution

To find these Taylor polynomials, we need to evaluate ff and its first three derivatives at x=1.x=1.

f(x)=lnxf(1)=0f(x)=1xf(1)=1f(x)=1x2f(1)=−1f(x)=2x3f(1)=2f(x)=lnxf(1)=0f(x)=1xf(1)=1f(x)=1x2f(1)=−1f(x)=2x3f(1)=2

Therefore,

p0(x)=f(1)=0,p1(x)=f(1)+f(1)(x1)=x1,p2(x)=f(1)+f(1)(x1)+f(1)2(x1)2=(x1)12(x1)2,p3(x)=f(1)+f(1)(x1)+f(1)2(x1)2+f(1)3!(x1)3=(x1)12(x1)2+13(x1)3.p0(x)=f(1)=0,p1(x)=f(1)+f(1)(x1)=x1,p2(x)=f(1)+f(1)(x1)+f(1)2(x1)2=(x1)12(x1)2,p3(x)=f(1)+f(1)(x1)+f(1)2(x1)2+f(1)3!(x1)3=(x1)12(x1)2+13(x1)3.

The graphs of y=f(x)y=f(x) and the first three Taylor polynomials are shown in Figure 6.5.

This graph has four curves. The first is the function f(x)=ln(x). The second function is psub1(x)=x-1. The third is psub2(x)=(x-1)-1/2(x-1)^2. The fourth is psub3(x)=(x-1)-1/2(x-1)^2 +1/3(x-1)^3. The curves are very close around x = 1.
Figure 6.5 The function y=lnxy=lnx and the Taylor polynomials p0,p1,p2p0,p1,p2 and p3p3 at x=1x=1 are plotted on this graph.
Checkpoint 6.10

Find the Taylor polynomials p0,p1,p2p0,p1,p2 and p3p3 for f(x)=1x2f(x)=1x2 at x=1.x=1.

We now show how to find Maclaurin polynomials for ex, sinx,sinx, and cosx.cosx. As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.

Example 6.12

Finding Maclaurin Polynomials

For each of the following functions, find formulas for the Maclaurin polynomials p0,p1,p2p0,p1,p2 and p3.p3. Find a formula for the nth Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of p0,p1,p2p0,p1,p2 and p3p3 with f.f.

  1. f(x)=exf(x)=ex
  2. f(x)=sinxf(x)=sinx
  3. f(x)=cosxf(x)=cosx

Solution

  1. Since f(x)=ex,f(x)=ex, we know that f(x)=f(x)=f(x)==f(n)(x)=exf(x)=f(x)=f(x)==f(n)(x)=ex for all positive integers n. Therefore,
    f(0)=f(0)=f(0)==f(n)(0)=1f(0)=f(0)=f(0)==f(n)(0)=1

    for all positive integers n. Therefore, we have
    p0(x)=f(0)=1,p1(x)=f(0)+f(0)x=1+x,p2(x)=f(0)+f(0)x+f(0)2!x2=1+x+12x2,p3(x)=f(0)+f(0)x+f(0)2x2+f(0)3!x3=1+x+12x2+13!x3,pn(x)=f(0)+f(0)x+f(0)2x2+f(0)3!x3++f(n)(0)n!xn=1+x+x22!+x33!++xnn!=k=0nxkk!.p0(x)=f(0)=1,p1(x)=f(0)+f(0)x=1+x,p2(x)=f(0)+f(0)x+f(0)2!x2=1+x+12x2,p3(x)=f(0)+f(0)x+f(0)2x2+f(0)3!x3=1+x+12x2+13!x3,pn(x)=f(0)+f(0)x+f(0)2x2+f(0)3!x3++f(n)(0)n!xn=1+x+x22!+x33!++xnn!=k=0nxkk!.

    The function and the first three Maclaurin polynomials are shown in Figure 6.6.
    This graph has four curves. The first is the function f(x)=e^x. The second function is psub0(x)=1. The third is psub1(x) which is an increasing line passing through y=1. The fourth function is psub3(x) which is a curve passing through y=1. The curves are very close around y= 1.
    Figure 6.6 The graph shows the function y=exy=ex and the Maclaurin polynomials p0,p1,p2p0,p1,p2 and p3.p3.
  2. For f(x)=sinx,f(x)=sinx, the values of the function and its first four derivatives at x=0x=0 are given as follows:
    f(x)=sinxf(0)=0f(x)=cosxf(0)=1f(x)=sinxf(0)=0f(x)=cosxf(0)=−1f(4)(x)=sinxf(4)(0)=0.f(x)=sinxf(0)=0f(x)=cosxf(0)=1f(x)=sinxf(0)=0f(x)=cosxf(0)=−1f(4)(x)=sinxf(4)(0)=0.

    Since the fourth derivative is sinx,sinx, the pattern repeats. That is, f(2m)(0)=0f(2m)(0)=0 and f(2m+1)(0)=(−1)mf(2m+1)(0)=(−1)m for m0.m0. Thus, we have
    p0(x)=0,p1(x)=0+x=x,p2(x)=0+x+0=x,p3(x)=0+x+013!x3=xx33!,p4(x)=0+x+013!x3+0=xx33!,p5(x)=0+x+013!x3+0+15!x5=xx33!+x55!,p0(x)=0,p1(x)=0+x=x,p2(x)=0+x+0=x,p3(x)=0+x+013!x3=xx33!,p4(x)=0+x+013!x3+0=xx33!,p5(x)=0+x+013!x3+0+15!x5=xx33!+x55!,

    and for m0,m0,
    p2m+1(x)=p2m+2(x)=xx33!+x55!+(−1)mx2m+1(2m+1)!=k=0m(−1)kx2k+1(2k+1)!.p2m+1(x)=p2m+2(x)=xx33!+x55!+(−1)mx2m+1(2m+1)!=k=0m(−1)kx2k+1(2k+1)!.

    Graphs of the function and its Maclaurin polynomials are shown in Figure 6.7.
    This graph has four curves. The first is the function f(x)=sin(x). The second function is psub1(x). The third is psub3(x). The fourth function is psub5(x). The curves are very close around x=0.
    Figure 6.7 The graph shows the function y=sinxy=sinx and the Maclaurin polynomials p1,p3p1,p3 and p5.p5.
  3. For f(x)=cosx,f(x)=cosx, the values of the function and its first four derivatives at x=0x=0 are given as follows:
    f(x)=cosxf(0)=1f(x)=sinxf(0)=0f(x)=cosxf(0)=−1f(x)=sinxf(0)=0f(4)(x)=cosxf(4)(0)=1.f(x)=cosxf(0)=1f(x)=sinxf(0)=0f(x)=cosxf(0)=−1f(x)=sinxf(0)=0f(4)(x)=cosxf(4)(0)=1.

    Since the fourth derivative is sinx,sinx, the pattern repeats. In other words, f(2m)(0)=(−1)mf(2m)(0)=(−1)m and f(2m+1)=0f(2m+1)=0 for m0.m0. Therefore,
    p0(x)=1,p1(x)=1+0=1,p2(x)=1+012!x2=1x22!,p3(x)=1+012!x2+0=1x22!,p4(x)=1+012!x2+0+14!x4=1x22!+x44!,p5(x)=1+012!x2+0+14!x4+0=1x22!+x44!,p0(x)=1,p1(x)=1+0=1,p2(x)=1+012!x2=1x22!,p3(x)=1+012!x2+0=1x22!,p4(x)=1+012!x2+0+14!x4=1x22!+x44!,p5(x)=1+012!x2+0+14!x4+0=1x22!+x44!,

    and for n0,n0,
    p2m(x)=p2m+1(x)=1x22!+x44!+(−1)mx2m(2m)!=k=0m(−1)kx2k(2k)!.p2m(x)=p2m+1(x)=1x22!+x44!+(−1)mx2m(2m)!=k=0m(−1)kx2k(2k)!.

    Graphs of the function and the Maclaurin polynomials appear in Figure 6.8.
    This graph has four curves. The first is the function f(x)=cos(x). The second function is psub0(x). The third is psub2(x). The fourth function is psub4(x). The curves are very close around y=1
    Figure 6.8 The function y=cosxy=cosx and the Maclaurin polynomials p0,p2p0,p2 and p4p4 are plotted on this graph.
Checkpoint 6.11

Find formulas for the Maclaurin polynomials p0,p1,p2p0,p1,p2 and p3p3 for f(x)=11+x.f(x)=11+x. Find a formula for the nth Maclaurin polynomial. Write your anwer using sigma notation.

Taylor’s Theorem with Remainder

Recall that the nth Taylor polynomial for a function ff at a is the nth partial sum of the Taylor series for ff at a. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials {pn}{pn} converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to f.f. To answer this question, we define the remainder Rn(x)Rn(x) as

Rn(x)=f(x)pn(x).Rn(x)=f(x)pn(x).

For the sequence of Taylor polynomials to converge to f,f, we need the remainder Rn to converge to zero. To determine if Rn converges to zero, we introduce Taylor’s theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the nth Taylor polynomial approximates the function.

Here we look for a bound on |Rn|.|Rn|. Consider the simplest case: n=0.n=0. Let p0 be the 0th Taylor polynomial at a for a function f.f. The remainder R0 satisfies

R0(x)=f(x)p0(x)=f(x)f(a).R0(x)=f(x)p0(x)=f(x)f(a).

If ff is differentiable on an interval I containing a and x, then by the Mean Value Theorem there exists a real number c between a and x such that f(x)f(a)=f(c)(xa).f(x)f(a)=f(c)(xa). Therefore,

R0(x)=f(c)(xa).R0(x)=f(c)(xa).

Using the Mean Value Theorem in a similar argument, we can show that if ff is n times differentiable on an interval I containing a and x, then the nth remainder Rn satisfies

Rn(x)=f(n+1)(c)(n+1)!(xa)n+1Rn(x)=f(n+1)(c)(n+1)!(xa)n+1

for some real number c between a and x. It is important to note that the value c in the numerator above is not the center a, but rather an unknown value c between a and x. This formula allows us to get a bound on the remainder Rn. If we happen to know that |f(n+1)(x)||f(n+1)(x)| is bounded by some real number M on this interval I, then

|Rn(x)|M(n+1)!|xa|n+1|Rn(x)|M(n+1)!|xa|n+1

for all x in the interval I.

We now state Taylor’s theorem, which provides the formal relationship between a function ff and its nth degree Taylor polynomial pn(x).pn(x). This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for ff converges to f.f.

Theorem 6.7

Taylor’s Theorem with Remainder

Let ff be a function that can be differentiated n+1n+1 times on an interval I containing the real number a. Let pn be the nth Taylor polynomial of ff at a and let

Rn(x)=f(x)pn(x)Rn(x)=f(x)pn(x)

be the nth remainder. Then for each x in the interval I, there exists a real number c between a and x such that

Rn(x)=f(n+1)(c)(n+1)!(xa)n+1.Rn(x)=f(n+1)(c)(n+1)!(xa)n+1.

If there exists a real number M such that |f(n+1)(x)|M|f(n+1)(x)|M for all xI,xI, then

|Rn(x)|M(n+1)!|xa|n+1|Rn(x)|M(n+1)!|xa|n+1

for all x in I.

Proof

Fix a point xIxI and introduce the function g such that

g(t)=f(x)f(t)f(t)(xt)f(t)2!(xt)2f(n)(t)n!(xt)nRn(x)(xt)n+1(xa)n+1.g(t)=f(x)f(t)f(t)(xt)f(t)2!(xt)2f(n)(t)n!(xt)nRn(x)(xt)n+1(xa)n+1.

We claim that g satisfies the criteria of Rolle’s theorem. Since g is a polynomial function (in t), it is a differentiable function. Also, g is zero at t=at=a and t=xt=x because

g(a)=f(x)f(a)f(a)(xa)f(a)2!(xa)2++f(n)(a)n!(xa)nRn(x)=f(x)pn(x)Rn(x)=0,g(x)=f(x)f(x)00=0.g(a)=f(x)f(a)f(a)(xa)f(a)2!(xa)2++f(n)(a)n!(xa)nRn(x)=f(x)pn(x)Rn(x)=0,g(x)=f(x)f(x)00=0.

Therefore, g satisfies Rolle’s theorem, and consequently, there exists c between a and x such that g(c)=0.g(c)=0. We now calculate g.g. Using the product rule, we note that

ddt[f(n)(t)n!(xt)n]=f(n)(t)(n1)!(xt)n1+f(n+1)(t)n!(xt)n.ddt[f(n)(t)n!(xt)n]=f(n)(t)(n1)!(xt)n1+f(n+1)(t)n!(xt)n.

Consequently,

g(t)=f(t)+[f(t)f(t)(xt)]+[f(t)(xt)f(t)2!(xt)2]++[f(n)(t)(n1)!(xt)n1f(n+1)(t)n!(xt)n]+(n+1)Rn(x)(xt)n(xa)n+1.g(t)=f(t)+[f(t)f(t)(xt)]+[f(t)(xt)f(t)2!(xt)2]++[f(n)(t)(n1)!(xt)n1f(n+1)(t)n!(xt)n]+(n+1)Rn(x)(xt)n(xa)n+1.

Notice that there is a telescoping effect. Therefore,

g(t)=f(n+1)(t)n!(xt)n+(n+1)Rn(x)(xt)n(xa)n+1.g(t)=f(n+1)(t)n!(xt)n+(n+1)Rn(x)(xt)n(xa)n+1.

By Rolle’s theorem, we conclude that there exists a number c between a and x such that g(c)=0.g(c)=0. Since

g(c)=f(n+1)(c)n!(xc)n+(n+1)Rn(x)(xc)n(xa)n+1g(c)=f(n+1)(c)n!(xc)n+(n+1)Rn(x)(xc)n(xa)n+1

we conclude that

f(n+1)(c)n!(xc)n+(n+1)Rn(x)(xc)n(xa)n+1=0.f(n+1)(c)n!(xc)n+(n+1)Rn(x)(xc)n(xa)n+1=0.

Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by n+1,n+1, we conclude that

Rn(x)=f(n+1)(c)(n+1)!(xa)n+1Rn(x)=f(n+1)(c)(n+1)!(xa)n+1

as desired. From this fact, it follows that if there exists M such that |f(n+1)(x)|M|f(n+1)(x)|M for all x in I, then

|Rn(x)|M(n+1)!|xa|n+1.|Rn(x)|M(n+1)!|xa|n+1.

Not only does Taylor’s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of f(x)=x3f(x)=x3 at x=8x=8 and determine how accurate these approximations are at estimating 113.113.

Example 6.13

Using Linear and Quadratic Approximations to Estimate Function Values

Consider the function f(x)=x3.f(x)=x3.

  1. Find the first and second Taylor polynomials for ff at x=8.x=8. Use a graphing utility to compare these polynomials with ff near x=8.x=8.
  2. Use these two polynomials to estimate 113.113.
  3. Use Taylor’s theorem to bound the error.

Solution

  1. For f(x)=x3,f(x)=x3, the values of the function and its first two derivatives at x=8x=8 are as follows:
    f(x)=x3f(8)=2f(x)=13x2/3f(8)=112f(x)=−29x5/3f(8)=1144.f(x)=x3f(8)=2f(x)=13x2/3f(8)=112f(x)=−29x5/3f(8)=1144.

    Thus, the first and second Taylor polynomials at x=8x=8 are given by
    p1(x)=f(8)+f(8)(x8)=2+112(x8)p2(x)=f(8)+f(8)(x8)+f(8)2!(x8)2=2+112(x8)1288(x8)2.p1(x)=f(8)+f(8)(x8)=2+112(x8)p2(x)=f(8)+f(8)(x8)+f(8)2!(x8)2=2+112(x8)1288(x8)2.

    The function and the Taylor polynomials are shown in Figure 6.9.
    This graph has four curves. The first is the function f(x)=cube root of x. The second function is psub1(x). The third is psub2(x). The curves are very close around x=8.
    Figure 6.9 The graphs of f(x)=x3f(x)=x3 and the linear and quadratic approximations p1(x)p1(x) and p2(x).p2(x).
  2. Using the first Taylor polynomial at x=8,x=8, we can estimate
    113p1(11)=2+112(118)=2.25.113p1(11)=2+112(118)=2.25.

    Using the second Taylor polynomial at x=8,x=8, we obtain
    113p2(11)=2+112(118)1288(118)2=2.21875.113p2(11)=2+112(118)1288(118)2=2.21875.
  3. By Uniqueness of Taylor Series, there exists a c in the interval (8,11)(8,11) such that the remainder when approximating 113113 by the first Taylor polynomial satisfies
    R1(11)=f(c)2!(118)2.R1(11)=f(c)2!(118)2.

    We do not know the exact value of c, so we find an upper bound on R1(11)R1(11) by determining the maximum value of ff on the interval (8,11).(8,11). Since f(x)=29x5/3,f(x)=29x5/3, the largest value for |f(x)||f(x)| on that interval occurs at x=8.x=8. Using the fact that f(8)=1144,f(8)=1144, we obtain
    |R1(11)|1144·2!(118)2=0.03125.|R1(11)|1144·2!(118)2=0.03125.

    Similarly, to estimate R2(11),R2(11), we use the fact that
    R2(11)=f(c)3!(118)3.R2(11)=f(c)3!(118)3.

    Since f(x)=1027x8/3,f(x)=1027x8/3, the maximum value of ff on the interval (8,11)(8,11) is f(8)0.0014468.f(8)0.0014468. Therefore, we have
    |R2(11)|0.00114683!(118)30.0065104.|R2(11)|0.00114683!(118)30.0065104.

Checkpoint 6.12

Find the first and second Taylor polynomials for f(x)=xf(x)=x at x=4.x=4. Use these polynomials to estimate 6.6. Use Taylor’s theorem to bound the error.

Example 6.14

Approximating sin x Using Maclaurin Polynomials

From Example 6.12b., the Maclaurin polynomials for sinxsinx are given by

p2m+1(x)=p2m+2(x)=xx33!+x55!x77!++(−1)mx2m+1(2m+1)!p2m+1(x)=p2m+2(x)=xx33!+x55!x77!++(−1)mx2m+1(2m+1)!

for m=0,1,2,.m=0,1,2,.

  1. Use the fifth Maclaurin polynomial for sinxsinx to approximate sin(π18)sin(π18) and bound the error.
  2. For what values of x does the fifth Maclaurin polynomial approximate sinxsinx to within 0.0001?

Solution

  1. The fifth Maclaurin polynomial is
    p5(x)=xx33!+x55!.p5(x)=xx33!+x55!.

    Using this polynomial, we can estimate as follows:
    sin(π18)p5(π18)=π1813!(π18)3+15!(π18)50.173648.sin(π18)p5(π18)=π1813!(π18)3+15!(π18)50.173648.

    To estimate the error, use the fact that the sixth Maclaurin polynomial is p6(x)=p5(x)p6(x)=p5(x) and calculate a bound on R6(π18).R6(π18). By Uniqueness of Taylor Series, the remainder is
    R6(π18)=f(7)(c)7!(π18)7R6(π18)=f(7)(c)7!(π18)7

    for some c between 0 and π18.π18. Using the fact that |f(7)(x)|1|f(7)(x)|1 for all x, we find that the magnitude of the error is at most
    17!·(π18)79.8×10−10.17!·(π18)79.8×10−10.
  2. We need to find the values of x such that
    17!|x|70.0001.17!|x|70.0001.

    Solving this inequality for x, we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as |x|<0.907.|x|<0.907.

Checkpoint 6.13

Use the fourth Maclaurin polynomial for cosxcosx to approximate cos(π12).cos(π12).

Now that we are able to bound the remainder Rn(x),Rn(x), we can use this bound to prove that a Taylor series for ff at a converges to f.f.

Representing Functions with Taylor and Maclaurin Series

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

Example 6.15

Finding a Taylor Series

Find the Taylor series for f(x)=1xf(x)=1x at x=1.x=1. Determine the interval of convergence.

Solution

For f(x)=1x,f(x)=1x, the values of the function and its first four derivatives at x=1x=1 are

f(x)=1xf(1)=1f(x)=1x2f(1)=−1f(x)=2x3f(1)=2!f(x)=3·2x4f(1)=−3!f(4)(x)=4·3·2x5f(4)(1)=4!.f(x)=1xf(1)=1f(x)=1x2f(1)=−1f(x)=2x3f(1)=2!f(x)=3·2x4f(1)=−3!f(4)(x)=4·3·2x5f(4)(1)=4!.

That is, we have f(n)(1)=(−1)nn!f(n)(1)=(−1)nn! for all n0.n0. Therefore, the Taylor series for ff at x=1x=1 is given by

n=0f(n)(1)n!(x1)n=n=0(−1)n(x1)n.n=0f(n)(1)n!(x1)n=n=0(−1)n(x1)n.

To find the interval of convergence, we use the ratio test. We find that

|an+1||an|=|(−1)n+1(x1)n+1||(−1)n(x1)n|=|x1|.|an+1||an|=|(−1)n+1(x1)n+1||(−1)n(x1)n|=|x1|.

Thus, the series converges if |x1|<1.|x1|<1. That is, the series converges for 0<x<2.0<x<2. Next, we need to check the endpoints. At x=2,x=2, we see that

n=0(−1)n(21)n=n=0(−1)nn=0(−1)n(21)n=n=0(−1)n

diverges by the divergence test. Similarly, at x=0,x=0,

n=0(−1)n(01)n=n=0(−1)2n=n=01n=0(−1)n(01)n=n=0(−1)2n=n=01

diverges. Therefore, the interval of convergence is (0,2).(0,2).

Checkpoint 6.14

Find the Taylor series for f(x)=12xf(x)=12x at x=2x=2 and determine its interval of convergence.

We know that the Taylor series found in this example converges on the interval (0,2),(0,2), but how do we know it actually converges to f?f? We consider this question in more generality in a moment, but for this example, we can answer this question by writing

f(x)=1x=11(1x).f(x)=1x=11(1x).

That is, ff can be represented by the geometric series n=0(1x)n.n=0(1x)n. Since this is a geometric series, it converges to 1x1x as long as |1x|<1.|1x|<1. Therefore, the Taylor series found in Example 6.15 does converge to f(x)=1xf(x)=1x on (0,2).(0,2).

We now consider the more general question: if a Taylor series for a function ff converges on some interval, how can we determine if it actually converges to f?f? To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for ff at a, the nth partial sum is given by the nth Taylor polynomial pn. Therefore, to determine if the Taylor series converges to f,f, we need to determine whether

limnpn(x)=f(x).limnpn(x)=f(x).

Since the remainder Rn(x)=f(x)pn(x),Rn(x)=f(x)pn(x), the Taylor series converges to ff if and only if

limnRn(x)=0.limnRn(x)=0.

We now state this theorem formally.

Theorem 6.8

Convergence of Taylor Series

Suppose that ff has derivatives of all orders on an interval I containing a. Then the Taylor series

n=0f(n)(a)n!(xa)nn=0f(n)(a)n!(xa)n

converges to f(x)f(x) for all x in I if and only if

limnRn(x)=0limnRn(x)=0

for all x in I.

With this theorem, we can prove that a Taylor series for ff at a converges to ff if we can prove that the remainder Rn(x)0.Rn(x)0. To prove that Rn(x)0,Rn(x)0, we typically use the bound

|Rn(x)|M(n+1)!|xa|n+1|Rn(x)|M(n+1)!|xa|n+1

from Taylor’s theorem with remainder.

In the next example, we find the Maclaurin series for ex and sinxsinx and show that these series converge to the corresponding functions for all real numbers by proving that the remainders Rn(x)0Rn(x)0 for all real numbers x.

Example 6.16

Finding Maclaurin Series

For each of the following functions, find the Maclaurin series and its interval of convergence. Use Taylor’s Theorem with Remainder to prove that the Maclaurin series for ff converges to ff on that interval.

  1. ex
  2. sinxsinx

Solution

  1. Using the nth Maclaurin polynomial for ex found in Example 6.12a., we find that the Maclaurin series for ex is given by
    n=0xnn!.n=0xnn!.

    To determine the interval of convergence, we use the ratio test. Since
    |an+1||an|=|x|n+1(n+1)!·n!|x|n=|x|n+1,|an+1||an|=|x|n+1(n+1)!·n!|x|n=|x|n+1,

    we have
    limn|an+1||an|=limn|x|n+1=0limn|an+1||an|=limn|x|n+1=0

    for all x. Therefore, the series converges absolutely for all x, and thus, the interval of convergence is (,).(,). To show that the series converges to ex for all x, we use the fact that f(n)(x)=exf(n)(x)=ex for all n0n0 and ex is an increasing function on (,).(,). Therefore, for any real number b, the maximum value of ex for all |x|b|x|b is eb. Thus,
    |Rn(x)|eb(n+1)!|x|n+1.|Rn(x)|eb(n+1)!|x|n+1.

    Since we just showed that
    n=0|x|nn!n=0|x|nn!

    converges for all x, by the divergence test, we know that
    limn|x|n+1(n+1)!=0limn|x|n+1(n+1)!=0

    for any real number x. By combining this fact with the squeeze theorem, the result is limnRn(x)=0.limnRn(x)=0.
  2. Using the nth Maclaurin polynomial for sinxsinx found in Example 6.12b., we find that the Maclaurin series for sinxsinx is given by
    n=0(−1)nx2n+1(2n+1)!.n=0(−1)nx2n+1(2n+1)!.

    In order to apply the ratio test, consider
    |an+1||an|=|x|2n+3(2n+3)!·(2n+1)!|x|2n+1=|x|2(2n+3)(2n+2).|an+1||an|=|x|2n+3(2n+3)!·(2n+1)!|x|2n+1=|x|2(2n+3)(2n+2).

    Since
    limn|x|2(2n+3)(2n+2)=0limn|x|2(2n+3)(2n+2)=0

    for all x, we obtain the interval of convergence as (,).(,). To show that the Maclaurin series converges to sinx,sinx, look at Rn(x).Rn(x). For each x there exists a real number c between 0 and x such that
    Rn(x)=f(n+1)(c)(n+1)!xn+1.Rn(x)=f(n+1)(c)(n+1)!xn+1.

    Since |f(n+1)(c)|1|f(n+1)(c)|1 for all integers n and all real numbers c, we have
    |Rn(x)||x|n+1(n+1)!|Rn(x)||x|n+1(n+1)!

    for all real numbers x. Using the same idea as in part a., the result is limnRn(x)=0limnRn(x)=0 for all x, and therefore, the Maclaurin series for sinxsinx converges to sinxsinx for all real x.
Checkpoint 6.15

Find the Maclaurin series for f(x)=cosx.f(x)=cosx. Use the ratio test to show that the interval of convergence is (,).(,). Show that the Maclaurin series converges to cosxcosx for all real numbers x.

Student Project

Proving that e is Irrational

In this project, we use the Maclaurin polynomials for ex to prove that e is irrational. The proof relies on supposing that e is rational and arriving at a contradiction. Therefore, in the following steps, we suppose e=r/se=r/s for some integers r and s where s0.s0.

  1. Write the Maclaurin polynomials p0(x),p1(x),p2(x),p3(x),p4(x)p0(x),p1(x),p2(x),p3(x),p4(x) for ex. Evaluate p0(1),p1(1),p2(1),p3(1),p4(1)p0(1),p1(1),p2(1),p3(1),p4(1) to estimate e.
  2. Let Rn(x)Rn(x) denote the remainder when using pn(x)pn(x) to estimate ex. Therefore, Rn(x)=expn(x),Rn(x)=expn(x), and Rn(1)=epn(1).Rn(1)=epn(1). Assuming that e=rse=rs for integers r and s, evaluate R0(1),R1(1),R2(1),R3(1),R4(1).R0(1),R1(1),R2(1),R3(1),R4(1).
  3. Using the results from part 2, show that for each remainder R0(1),R1(1),R2(1),R3(1),R4(1),R0(1),R1(1),R2(1),R3(1),R4(1), we can find an integer k such that kRn(1)kRn(1) is an integer for n=0,1,2,3,4.n=0,1,2,3,4.
  4. Write down the formula for the nth Maclaurin polynomial pn(x)pn(x) for ex and the corresponding remainder Rn(x).Rn(x). Show that sn!Rn(1)sn!Rn(1) is an integer.
  5. Use Taylor’s theorem to write down an explicit formula for Rn(1).Rn(1). Conclude that Rn(1)0,Rn(1)0, and therefore, sn!Rn(1)0.sn!Rn(1)0.
  6. Use Taylor’s theorem to find an estimate on Rn(1).Rn(1). Use this estimate combined with the result from part 5 to show that |sn!Rn(1)|<sen+1.|sn!Rn(1)|<sen+1. Conclude that if n is large enough, then |sn!Rn(1)|<1.|sn!Rn(1)|<1. Therefore, sn!Rn(1)sn!Rn(1) is an integer with magnitude less than 1. Thus, sn!Rn(1)=0.sn!Rn(1)=0. But from part 5, we know that sn!Rn(1)0.sn!Rn(1)0. We have arrived at a contradiction, and consequently, the original supposition that e is rational must be false.

Section 6.3 Exercises

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

116.

f(x)=1+x+x2f(x)=1+x+x2 at a=1a=1

117.

f(x)=1+x+x2f(x)=1+x+x2 at a=−1a=−1

118.

f(x)=cos(2x)f(x)=cos(2x) at a=πa=π

119.

f(x)=sin(2x)f(x)=sin(2x) at a=π2a=π2

120.

f(x)=xf(x)=x at a=4a=4

121.

f(x)=lnxf(x)=lnx at a=1a=1

122.

f(x)=1xf(x)=1x at a=1a=1

123.

f(x)=exf(x)=ex at a=1a=1

In the following exercises, verify that the given choice of n in the remainder estimate |Rn|M(n+1)!(xa)n+1,|Rn|M(n+1)!(xa)n+1, where M is the maximum value of |f(n+1)(z)||f(n+1)(z)| on the interval between a and the indicated point, yields |Rn|11000.|Rn|11000. Find the value of the Taylor polynomial pn of ff at the indicated point.

124.

[T] 10;a=9,n=310;a=9,n=3

125.

[T] (28)1/3;a=27,n=1(28)1/3;a=27,n=1

126.

[T] sin(6);a=2π,n=5sin(6);a=2π,n=5

127.

[T] e2; a=0,n=9a=0,n=9

128.

[T] cos(π5);a=0,n=4cos(π5);a=0,n=4

129.

[T] ln(2);a=1,n=1000ln(2);a=1,n=1000

130.

Integrate the approximation sinttt36+t5120t75040sinttt36+t5120t75040 evaluated at πt to approximate 01sinπtπtdt.01sinπtπtdt.

131.

Integrate the approximation ex1+x+x22++x6720ex1+x+x22++x6720 evaluated at −x2 to approximate 01ex2dx.01ex2dx.

In the following exercises, find the smallest value of n such that the remainder estimate |Rn|M(n+1)!(xa)n+1,|Rn|M(n+1)!(xa)n+1, where M is the maximum value of |f(n+1)(z)||f(n+1)(z)| on the interval between a and the indicated point, yields |Rn|11000|Rn|11000 on the indicated interval.

132.

f(x)=sinxf(x)=sinx on [π,π],a=0[π,π],a=0

133.

f(x)=cosxf(x)=cosx on [π2,π2],a=0[π2,π2],a=0

134.

f(x)=e−2xf(x)=e−2x on [−1,1],a=0[−1,1],a=0

135.

f(x)=exf(x)=ex on [−3,3],a=0[−3,3],a=0

In the following exercises, the maximum of the right-hand side of the remainder estimate |R1|max|f(z)|2R2|R1|max|f(z)|2R2 on [aR,a+R][aR,a+R] occurs at a or a±R.a±R. Estimate the maximum value of R such that max|f(z)|2R20.1max|f(z)|2R20.1 on [aR,a+R][aR,a+R] by plotting this maximum as a function of R.

136.

[T] ex approximated by 1+x,a=01+x,a=0

137.

[T] sinxsinx approximated by x, a=0a=0

138.

[T] lnxlnx approximated by x1,a=1x1,a=1

139.

[T] cosxcosx approximated by 1,a=01,a=0

In the following exercises, find the Taylor series of the given function centered at the indicated point.

140.

x4x4 at a=−1a=−1

141.

1+x+x2+x31+x+x2+x3 at a=−1a=−1

142.

sinxsinx at a=πa=π

143.

cosxcosx at a=2πa=2π

144.

sinxsinx at x=π2x=π2

145.

cosxcosx at x=π2x=π2

146.

exex at a=−1a=−1

147.

exex at a=1a=1

148.

1(x1)21(x1)2 at a=0a=0 (Hint: Differentiate 11x.)11x.)

149.

1(x1)31(x1)3 at a=0a=0

150.

F(x)=0xcos(t)dt;f(t)=n=0(−1)ntn(2n)!F(x)=0xcos(t)dt;f(t)=n=0(−1)ntn(2n)! at a=0a=0 (Note: ff is the Taylor series of cos(t).)cos(t).)

In the following exercises, compute the Taylor series of each function around x=1.x=1.

151.

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

152.

f(x)=x3f(x)=x3

153.

f(x)=(x2)2f(x)=(x2)2

154.

f(x)=lnxf(x)=lnx

155.

f(x)=1xf(x)=1x

156.

f(x)=12xx2f(x)=12xx2

157.

f(x)=x4x2x21f(x)=x4x2x21

158.

f(x)=exf(x)=ex

159.

f(x)=e2xf(x)=e2x

[T] In the following exercises, identify the value of x such that the given series n=0ann=0an is the value of the Maclaurin series of f(x)f(x) at x.x. Approximate the value of f(x)f(x) using S10=n=010an.S10=n=010an.

160.

n=01n!n=01n!

161.

n=02nn!n=02nn!

162.

n=0(−1)n(2π)2n(2n)!n=0(−1)n(2π)2n(2n)!

163.

n=0(−1)n(2π)2n+1(2n+1)!n=0(−1)n(2π)2n+1(2n+1)!

The following exercises make use of the functions S5(x)=xx36+x5120S5(x)=xx36+x5120 and C4(x)=1x22+x424C4(x)=1x22+x424 on [π,π].[π,π].

164.

[T] Plot sin2x(S5(x))2sin2x(S5(x))2 on [π,π].[π,π]. Compare the maximum difference with the square of the Taylor remainder estimate for sinx.sinx.

165.

[T] Plot cos2x(C4(x))2cos2x(C4(x))2 on [π,π].[π,π]. Compare the maximum difference with the square of the Taylor remainder estimate for cosx.cosx.

166.

[T] Plot |2S5(x)C4(x)sin(2x)||2S5(x)C4(x)sin(2x)| on [π,π].[π,π].

167.

[T] Compare S5(x)C4(x)S5(x)C4(x) on [−1,1][−1,1] to tanx.tanx. Compare this with the Taylor remainder estimate for the approximation of tanxtanx by x+x33+2x515.x+x33+2x515.

168.

[T] Plot exe4(x)exe4(x) where e4(x)=1+x+x22+x36+x424e4(x)=1+x+x22+x36+x424 on [0,2].[0,2]. Compare the maximum error with the Taylor remainder estimate.

169.

(Taylor approximations and root finding.) Recall that Newton’s method xn+1=xnf(xn)f(xn)xn+1=xnf(xn)f(xn) approximates solutions of f(x)=0f(x)=0 near the input x0.x0.

  1. If ff and gg are inverse functions, explain why a solution of g(x)=ag(x)=a is the value f(a)off.f(a)off.
  2. Let pN(x)pN(x) be the NthNth degree Maclaurin polynomial of ex.ex. Use Newton’s method to approximate solutions of pN(x)2=0pN(x)2=0 for N=4,5,6.N=4,5,6.
  3. Explain why the approximate roots of pN(x)2=0pN(x)2=0 are approximate values of ln(2).ln(2).

In the following exercises, use the fact that if q(x)=n=1an(xc)nq(x)=n=1an(xc)n converges in an interval containing c,c, then limxcq(x)=a0limxcq(x)=a0 to evaluate each limit using Taylor series.

170.

limx0cosx1x2limx0cosx1x2

171.

limx0ln(1x2)x2limx0ln(1x2)x2

172.

limx0ex2x21x4limx0ex2x21x4

173.

limx0+cos(x)12xlimx0+cos(x)12x

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