Section 11.11 Applications of Taylor Polynomials 33. An electric dipole consists of two electric charges of equal magnitude and opposite sign. If the charges are

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Section 11.11 Applications of Taylor Polynomials

33. An electric dipole consists of two electric charges of equal magnitude and opposite sign. If the charges are q and

−q and are located at a distance d from each other, then the electric field E at the point P in the figure is E = q

D² − q (D + d)²

By expanding this expression for E as a series in powers of d/D, show that E is approximately proportional to 1/D³when P is far away from the dipole.

SeCtION11.11 Applications of Taylor Polynomials 781

15. fsxd − x^2y3, a − 1, n − 3, 0.8 < x < 1.2 16. fsxd − sin x, a − y6, n − 4, 0 < x < y3 17. fsxd − sec x, a − 0, n − 2, 20.2 < x < 0.2 18. fsxd − lns1 1 2xd, a − 1, n − 3, 0.5 < x < 1.5 19. fsxd − e^x², a − 0, n − 3, 0 < x < 0.1

20. fsxd − x ln x, a − 1, n − 3, 0.5 < x < 1.5 21. fsxd − x sin x, a − 0, n − 4, 21 < x < 1 22. fsxd − sinh 2x, a − 0, n − 5, 21 < x < 1

23. Use the information from Exercise 5 to estimate cos 80°

correct to five decimal places.

24. Use the information from Exercise 16 to estimate sin 38°

correct to five decimal places.

25. Use Taylor’s Inequality to determine the number of terms of the Maclaurin series for e^x that should be used to esti- mate e^0.1 to within 0.00001.

26. How many terms of the Maclaurin series for lns1 1 xd do you need to use to estimate ln 1.4 to within 0.001?

27–29 Use the Alternating Series Estimation Theorem or Taylor’s Inequality to estimate the range of values of x for which the given approximation is accurate to within the stated error. Check your answer graphically.

27. sin x< x 2 x³

6

( |

^error

|

^,^0.01

⁾

28. cos x< 1 2 x² 2 1 x⁴

24

( |

^error

|

^,^0.005

⁾

29. arctan x< x 2 x³ 3 1 x⁵

5

( |

^error

|

^,^0.05

⁾

30. Suppose you know that

f^snds4d − s21dⁿn!

3ⁿsn 1 1d

and the Taylor series of f centered at 4 converges to fsxd for all x in the interval of convergence. Show that the fifth- degree Taylor polynomial approximates fs5d with error less than 0.0002.

31. A car is moving with speed 20 mys and acceleration 2 mys² at a given instant. Using a second-degree Taylor polyno mial, estimate how far the car moves in the next second. Would it be reasonable to use this polynomial to estimate the distance traveled during the next minute?

;

32. The resistivity of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters (V-m).

The resistivity of a given metal depends on the temperature according to the equation

std − 20e^st220d

where t is the temperature in °C. There are tables that list the values of (called the temperature coefficient) and 20 (the resistivity at 20°C) for various metals. Except at very low temperatures, the resis tivity varies almost linearly with temperature and so it is common to approximate the expression for std by its first- or second-degree Taylor polynomial at t − 20.

(a) Find expressions for these linear and quadratic approximations.

(b) For copper, the tables give − 0.0039y°C and

20− 1.7 3 10²⁸ V-m. Graph the resistivity of copper and the linear and quadratic approximations for 2250°C < t < 1000°C.

(c) For what values of t does the linear approximation agree with the exponential expression to within one percent?

33. An electric dipole consists of two electric charges of equal magnitude and opposite sign. If the charges are q and 2q and are located at a distance d from each other, then the electric field E at the point P in the figure is

E − q

D² 2 q sD 1 dd²

By expanding this expression for E as a series in powers of dyD, show that E is approximately proportional to 1yD³ when P is far away from the dipole.

P D d

q -q

34. (a) Derive Equation 3 for Gaussian optics from Equation 1 by approximating cos in Equation 2 by its first-degree Taylor polynomial.

(b) Show that if cos is replaced by its third-degree Taylor polynomial in Equation 2, then Equation 1 becomes Equation 4 for third-order optics. [Hint: Use the first two terms in the binomial series for ,o21 and ,i21. Also, use

< sin .]

35. If a water wave with length L moves with velocity v across a body of water with depth d, as in the figure on page 782, then

v²− tL

2 tanh 2d L

(a) If the water is deep, show that v<stLys2d.

(b) If the water is shallow, use the Maclaurin series for tanh to show that v<std. (Thus in shallow water the

;

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Solution:

206 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 32. (a)

 ^()() ^()(20) 0 ₂₀^(⁻²⁰⁾ ₂₀ 1 ₂₀^(⁻²⁰⁾ ₂₀ 2 ²₂₀^(⁻²⁰⁾ ²₂₀

The linear approximation is

1() = (20) + ⁰(20)( − 20) = 20[1 + ( − 20)]

The quadratic approximation is

2() = (20) + ⁰(20)( − 20) +⁰⁰(20)

2 ( − 20)²

= ₂₀

1 + ( − 20) + ¹2²( − 20)²

(b) (c)

From the graph, it seems that 1()is within 1% of (), that is, 099() ≤ ¹() ≤ 101(), for −14^◦C ≤  ≤ 58^◦C.

33.  = 

² − 

( + )² = 

² − 

²(1 + )² = 

²

 1 −

 1 + 



−2 .

We use the Binomial Series to expand (1 + )⁻²:

 = 

²

 1 −

 1 − 2





 + 2 · 3

2!





2

−2 · 3 · 4 3!





3

+ · · ·



= 

²

 2







− 3





2

+ 4





3

− · · ·



≈ 

² · 2







= 2 · 1

³

when  is much larger than ; that is, when  is far away from the dipole.

34. (a) 1



+2



= 1



2

 − 1





[Equation 1] where

=

²+ (+ )²− 2(^+ ) cos  and =

²+ (− )²+ 2(− ) cos  (2) Using cos  ≈ 1 gives

=



²+ (+ )²− 2(^+ ) =

²+ ²+ 2+ ²− 2^− 2²=

²= 

and similarly, = . Thus, Equation 1 becomes 1



+2



= 1



2

 −1





⇒ 1



+2



= 2− ¹

 .

(b) Using cos  ≈ 1 − ¹₂²in (2) gives us

=

²+ (+ )²− 2(^+ )

1 −¹2²

=

²+ ²_+ 2+ ²− 2^+ ²− 2²+ ²²=

²_+ ²+ ²²

37. If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the earth.

(a) If R is the radius of the earth and L is the length of the highway, show that the correction is C = R sec(L/R) − R

(b) Use a Taylor polynomial to show that C ≈ _2R^L² +_24R^5L⁴3

(c) Compare the corrections given by the formulas in parts (a) and (b) for a highway that is 100 km long. (Take the radius of the earth to be 6370 km.)

782 Chapter 11 Infinite Sequences and Series

velocity of a wave tends to be independent of the length of the wave.)

(c) Use the Alternating Series Estimation Theorem to show that if L . 10d, then the estimate v²<td is accurate to within 0.014tL.

d L

36. A uniformly charged disk has radius R and surface charge den- sity as in the figure. The electric potential V at a point P at a distance d along the perpendicular central axis of the disk is

V − 2ke

(

_sd²1R²2d

)

where ke is a constant (called Coulomb’s constant). Show that

V< keR²

d for large d

R

d P

37. If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the earth.

(a) If R is the radius of the earth and L is the length of the highway, show that the correction is

C − R secsLyRd 2 R (b) Use a Taylor polynomial to show that

C< L²

2R 1 5L⁴ 24R³

(c) Compare the corrections given by the formulas in parts (a) and (b) for a highway that is 100 km long. (Take the radius of the earth to be 6370 km.)

R

L C

R

38. The period of a pendulum with length L that makes a maxi- mum angle 0 with the vertical is

T − 4

Î

^Lt

y

0^y2

dx s1 2 k² sin²x

where k − sin

s

2¹0

d

^andt is the acceleration due to gravity.

(In Exercise 7.7.42 we approximated this integral using Simpson’s Rule.)

(a) Expand the integrand as a binomial series and use the result of Exercise 7.1.50 to show that

T − 2

Î

^L_t

F

^{1 1} ¹²²²^k²¹ ¹²²²³⁴²²^k⁴¹ ¹²²²³⁴²²⁵⁶²²^k⁶^{1 ∙ ∙ ∙}

G

If 0 is not too large, the approximation T< 2sLyt, obtained by using only the first term in the series, is often used. A better approximation is obtained by using two terms:

T< 2

Î

^Lt

s

^{1 1}¹4k²

d

(b) Notice that all the terms in the series after the first one have coefficients that are at most ¹₄. Use this fact to compare this series with a geometric series and show that

2

Î

^L_t

^s

^{1 1}¹⁴^k²

^d

^<^{T <}²

Î

^L_t 4 2 4k^{4 2 3k}²² (c) Use the inequalities in part (b) to estimate the period of

a pendulum with L − 1 meter and 0− 10°. How does it compare with the estimate T< 2sLyt? What if

0− 42°?

39. In Section 4.8 we considered Newton’s method for approxi- mating a root r of the equation fsxd − 0, and from an initial approximation x1 we obtained successive approximations x2, x3, . . . , where

xn11− xn2 fsxnd f 9sxnd

Use Taylor’s Inequality with n − 1, a − xn, and x − r to show that if f 0sxd exists on an interval I containing r, xn, and xn11, and

|

^{f 0}^sxd

|

^<^M,

|

^{f 9}^sxd

|

^>K for all x [ I, then

|

^xⁿ¹¹²^r

|

^< _2K^M

|

^xⁿ²^r

|

²

[This means that if xn is accurate to d decimal places, then xn11

is accurate to about 2d decimal places. More precisely, if the error at stage n is at most 10^2m, then the error at stage n 1 1 is at most sMy2Kd10^22m.]

Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

1

(2)

208 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

(c) Since tanh  is an odd function, its Maclaurin series is alternating, so the error in the approximation

tanh2

 ≈ 2

 is less than the first neglected term, which is |⁰⁰⁰(0)|

3!

2



3

= 1 3

2



3

.

If   10, then 1 3

2



3

 1 3

 2 · 1

10

3

= ³

375, so the error in the approximation ²= is less than 

2 · ³

375 ≈ 00132.

36. First note that 2√

²+ ²− 

= 2

√

²

 1 +²

² − 



≈ 2





 1 + ²

² · 1 2+ · · ·



− 

 use the binomial series 1 +¹₂ + · · · for√ 1 + 

= 2



 +² 2 + · · ·



− 



≈ ²



since for large  the other terms are comparatively small. Now  = 2√

²+ ²− 

≈ ²

 by the preceding approximation.

37. (a)  is the length of the arc subtended by the angle , so  =  ⇒

 = . Now sec  = ( + ) ⇒  sec  =  +  ⇒

 =  sec  −  =  sec() − .

(b) First we’ll find a Taylor polynomial 4()for () = sec  at  = 0.

 ^()() ^()(0)

0 sec  1

1 sec  tan  0

2 sec (2 tan² + 1) 1

3 sec  tan (6 tan² + 5) 0 4 sec (24 tan⁴ + 28 tan² + 5) 5

Thus, () = sec  ≈ ⁴() = 1 +_2!¹( − 0)²+ _4!⁵( − 0)⁴ = 1 +¹₂²+₂₄⁵⁴. By part (a),

 ≈ 

 1 +1

2





2

+ 5 24





4

−  =  + 1 2 · ²

² + 5 24 · ⁴

⁴ −  = ²

2 + 5⁴ 24³. (c) Taking  = 100 km and  = 6370 km, the formula in part (a) says that

 =  sec() −  = 6370 sec(1006370) − 6370 ≈ 0785 009 965 44 km.

The formula in part (b) says that  ≈ ²

2+ 5⁴

24³ = 100²

2 · 6370 + 5 · 100⁴

24 · 6370³ ≈ 0785 009 957 36 km.

The difference between these two results is only 0000 000 008 08 km, or 0000 008 08 m!

60. (Review, p823) The force due to gravity on an object with mass m at a height h above the surface of the earth is F = mgR²

(R + h)²

where R is the radius of the earth and g is the acceleration due to gravity for an object on the surface of the earth.

(a) Express F as a series in powers of h/R.

(b) Observe that if we approximate F by the first term in the series, we get the expression F ≈ mg that is usually used when h is much smaller than R. Use the Alternating Series Estimation Theorem to estimate the range of values of h for which the approximation F ≈ mg is accurate to within one percent. (Use R = 6400 km.)

Solution:

2

(3)

CHAPTER 11 REVIEW ¤ 221 60. (a)  = ²

( + )² = 

(1 + )² =  ^∞

=0

−2









[binomial series]

(b) We expand  = 

1 − 2 () + 3 ()²− · · ·. This is an alternating series, so by the Alternating Series Estimation Theorem, the error in the approximation  = 

is less than 2, so for accuracy within 1% we want



 2

²( + )²



  001 ⇔ 2( + )²

³  001.

This inequality would be difficult to solve for , so we substitute  = 6,400 km and plot both sides of the inequality.

It appears that the approximation is accurate to within 1% for   31 km.

61. () = ^∞

=0

^ ⇒ (−) = ^∞

=0

(−)^= ^∞

=0(−1)^^ (a) If  is an odd function, then (−) = −() ⇒ ^∞

=0(−1)^^= ^∞

=0−^^. The coefficients of any power series are uniquely determined (by Theorem 11.10.5), so (−1)^= −^.

If  is even, then (−1)^= 1, so = −^ ⇒ 2^= 0 ⇒ ^= 0. Thus, all even coefficients are 0, that is,

0= 2= 4= · · · = 0.

(b) If  is even, then (−) = () ⇒ ^∞

=0(−1)^^= ^∞

=0

^ ⇒ (−1)^= .

If  is odd, then (−1)^= −1, so −^=  ⇒ 2^= 0 ⇒ ^= 0. Thus, all odd coefficients are 0, that is, 1= 3= 5= · · · = 0.

62. ^ = ^∞

=0

^

! ⇒ () = ^²= ^∞

=0

(²)^

! = ^∞

=0

^2

! = ^∞

=0

1

!^2. By Theorem 11.10.6 with  = 0, we also have

 () = ^∞

=0

^()(0)

! ^. Comparing coefficients for  = 2, we have ^(2)(0) (2)! = 1

! ⇒ ^(2)(0) =(2)!

! .

3