Approximating Functions 10.1 Taylor Polynomials

Chapter Ten ¢¡£ ¥¤§¦ In Section ??, for , we found the sum

APPROXIMATING of a geometric series as a function of the ¡ common ratio :

FUNCTIONS ¦

¦©¨ ¡ ¨ ¡ ¨¨¡¨ ¦¡

Viewed from another perspective, this formula

gives us a series whose sum is the function

¡$#%§¦'& ¦( ¡$# ! "! . We now work in the opposite direction, starting with a function and looking for a series of simpler functions whose sum is that function. We ﬁrst use polynomials, which lead us to Taylor series, and then trigonometric functions, which lead us to Fourier series.

Taylor approximations use polynomials, which may be considered the simplest functions. They are easy to use because they can be evaluated

by simple arithmetic, unlike transcendental

¡ +-, functions, such as )'* and .

Fourier approximations use sines and cosines, the simplest periodic functions, instead of polynomials. Taylor approximations are generally good approximations to the function locally (that is, near a speciﬁc point), whereas Fourier approximations are generally good approximations over an interval. 446 Chapter Ten APPROXIMATING FUNCTIONS 10.1 TAYLOR POLYNOMIALS

In this section, we see how to approximate a function by polynomials.

Linear Approximations We already know how to approximate a function using a degree 1 polynomial, namely the tangent

line approximation given in Section 4.8 : .0/2143657.0/2893:;.=<>/>893?/@1AB893?C

The tangent line and the curve share the same slope at 1£DE8 . As Figure 10.1 suggests, the tangent 8

line approximation to the function is generally more accurate for values of 1 close to . J4K

True value GVL J4K Approximate value of GVL

Tangent line

JNSTK K

FML GUQFNL

GQRF

P O

J4K J4K

FML FNL

I I

F G

J4K

G F

Figure 10.1: Tangent line approximation of GVL for near 1DZX We ﬁrst focus on 8(DYX . The tangent line approximation at is referred to as the ﬁrst

Taylor approximation at 1RD[X , or as follows:

¡$# ¡

Taylor Polynomial of Degree 1 Approximating \! for near 0 .0/@1=35^]`_a/21436D7.0/2XN3:(.=

We now look at approximations by polynomials of higher degree. Let us compare approxima-

/@1=36D[d?eMfV1 1D[X

tions c near by linear and quadratic functions.

/21436D^d?eNf91 1 1D[X

Example 1 Approximate c , with in radians, by its tangent line at .

1RD^X Dih

Solution The tangent line at is just the horizontal line g , as shown in Figure 10.2, so

/21436D[d?eNf915jhMk 1

c for near 0. XMl

If we take 1%DXVC , then

/2XmC XNln3`DdoeMfp/2XVC XMlM3`DXVC qnqMrsCoCtCuk

c 1DjAvXmCwh

which is quite close to the approximation doeMf915ih . Similarly, if , then

/xAvXmCwhy30D[d?eNfo/zAvXVC¢hp3`D^XmC qMqMl{CtCoC c

10.1 TAYLOR POLYNOMIALS 447

1RD^XVC |

is close to the approximation d?eNf91R5jh . However, if , then

/2XmC |}3DdoeMfp/2XVC |N3`D[XVC qM~VhCoCtCuk

c 1 so the approximation d?eMfV15h is less accurate. The graph suggests that the farther a point is

away from X , the worse the approximation is likely to be.

P p

G G© Figure 10.2: Graph of p and its tangent line at

Quadratic Approximations

To get a more accurate approximation, we use a quadratic function instead of a linear function.

/@143D^doeMf91 1 X

Example 2 Find the quadratic approximation to c for near .

/@1=3¥DdoeMf91 ]'M/2143 1DX

Solution To ensure that the quadratic, , is a good approximation to c at , we require

that doeMf91 and the quadratic have the same value, the same slope, and the same second derivative at

1RD^X ]'n/>XM3D />XM3 ] />XM3D /2XN3 ] /2XN3D />XM3

< < < < < <

c c . That is, we require c , , and . We take the quadratic

polynomial

] /@1=36D[¥:;s_1$:; 1 k

¥ _

and determine , , and . Since

] /@1=36D76{:(s_?1$:( 1 /@1=36DdoeMf91

and c

] /@1=36D7{_6:(~M 1 /@143¥DA£f-ws1

< <

] /@1=36D7~n /@143¥DA£d?eNf91k

< < < <

we have

UD[] /2XN3D /2XN3DdoeMfVXDjh UDjh

c so

s_vD[] /2XN3D /2XM3DjA£fz¢vXD^X s_vD^X

< <

C ~n D[] />XM3D /2XN3DA£d?eNfVXDjAhMk DjA

< < < <

Consequently, the quadratic approximation is

1 h

1 DihsA k doeMf91R5^] /@1=36Dih:X?1"A 1 ~ ~ for near 0.

Figure 10.3 suggests that the quadratic approximation doeMf915 ]M/@1=3 is better than the linear

doeMf915E] /@1=3 1

approximation _ for near 0. Let’s compare the accuracy of the two approximations.

1 1BDXVC | doeMft/2XVC |N3DXmC qN~9hCtCoC ] /2XmC |}3vDXVC qM~X Recall that ]`_a/@1=3Dh for all . At , we have and , so the quadratic approximation is a signiﬁcant improvement over the linear approximation. The magnitude of the error is about 0.001 instead of 0.08.

448 Chapter Ten APPROXIMATING FUNCTIONS

¡¥

G9L

Q¥

P ¡

GVLQ£¢u¤

K K

¡¥ ¡

G9L GVL G G Figure 10.3: Graph of p and its linear, , and quadratic, , approximations for near 0

Generalizing the computations in Example 2, we deﬁne the second Taylor approximation at

1RD^X .

¡$# ¡

Taylor Polynomial of Degree 2 Approximating \! for near 0

. /2XN3

< <

.0/214365[] /@1=36D7.0/2XM3:;. /2XM3x1$: 1

< ~

Higher-Degree Polynomials In a small interval around 1D¦X , the quadratic approximation to a function is usually a better approximation than the linear (tangent line) approximation. However, Figure 10.3 shows that the quadratic can still bend away from the original function for large 1 . We can attempt to ﬁx this by

using an approximating polynomial of higher degree. Suppose that we approximate a function .0/2143 1

for near 0 by a polynomial of degree § :

¨}© ¨

.0/21435[]¨/21436D7 :( 1$:;o1 :^oty:;¥¨}© 1 :(¥¨91 C

_ _

¨ ktCoCtC?k

We need to ﬁnd the values of the constants: nk¥{_k6 . To do this, we require that

.0/@1=3 ] /@143

the function and each of its ﬁrst § derivatives agree with those of the polynomial at the 1RD[X point 1D[X . In general, the more derivatives there are that agree at , the larger the interval on

which the function and the polynomial remain close to each other. Dª

To see how to ﬁnd the constants, let’s take § as an example

.0/214365^]«N/@1=36D[¥:;s_1$:; 1 :;«o1 C

Substituting 1%D^X gives

.0/>XM3D[] /2XN3D[ C

« ]'«N/@143

Differentiating yields

]< /@143D^s_:;~n 1$:(ªM¥«o1 k «

so substituting 1RDX shows that

. /2XN3D] />XM3D7{_aC

< < «

Differentiating and substituting again, we get

] /@1=36D7~Uhp :(ªt~Unhy¥«?1k

< < «

which gives

. /2XM3D[] />XM36D[~Unhynk

< < < < «

10.1 TAYLOR POLYNOMIALS 449

. /2XN3

< so that <

D C

~Uh

]

< < <

The third derivative, denoted by « , is

]< < < /@1=36D^ªp~Unhp¥«nk «

. /2XM3D^] />XM3D[ªt~Unhy«nk

< < < < < <

. /2XN3

< <

and then <

D C

ªt~Unh

¬® . You can imagine a similar calculation starting with ]¬M/2143 , using the fourth derivative ,

which would give ¬®

. />XM3

D k

¬ |tªt~UMh

and so on. Using factorial notation,1 we write these expressions as

¬®

. /2XM3 . /2XN3

< < <

¥«D k°¬D C

ªV¯ |±¯

. . §

Writing for the §=²2³ derivative of , we have, for any positive integer

. /2XN3

C ¥¨´D

§ 1D[X

So we deﬁne the §=²>³ Taylor approximation at :

¡$# ¡ "!

Taylor Polynomial of Degree µ Approximating for near 0

.0/@1=365^] /@1=3

¬® ®

. />XM3 . /2XM3 . /2XN3 . /2XN3

< < < < <

« ¬

1 : 1 : 1 :^oty: 1 D[.0/>XM3:;. />XM3x1$:

~9¯ ªm¯ |m¯ ¯

] /2143 1D^X

We call the Taylor polynomial of degree § centered at , or the Taylor polynomial

about 1RD^X . 1

Example 3 Construct the Taylor polynomial of degree 7 approximating the function .0/@1=36D^f-¢s1 for near 0. 1D¶·ª Compare the value of the Taylor approximation with the true value of . at .

Solution We have

f-ws1 .0/2XM3D X

.0/@143D giving

. /@143D doeMf91 . /2XM3D h

< <

. /@143D¸A%f-¢¥1 . /2XM3D X

< < < <

. /@143D¹A£d?eMf}1 . /2XM3DAh

< < < < < <

¬® ¬®

. /@143D f-ws1 . /2XM3D X

® ®

.wº /@143D doeMf91 .wº /2XM3D h

® ®

.w» /@143D¸A%f-¢¥1 .w» /2XM3D X

® ®

.w¼ /@143DjA£d?eNf}1k .w¼ /2XM3DiAhnC

Âu¾t¿Â Èp¾t¿Â Recall that ½M¾o¿%½MÀ¢½ÁÃÂ-ÄMÅxÅxÅÇÆ0ÅuÂ . In addition, , and . 450 Chapter Ten APPROXIMATING FUNCTIONS

Using these values, we see that the Taylor polynomial approximation of degree 7 is

« ¬

1 1 1 14º 1±» 14¼

f-¢¥1%5^] /21436D[XÉ:hst1$:(X Ahs :(X :^hs :(X Ahs

~9¯ ªm¯ |m¯ lV¯ ¯ ¯

º ¼

1 1 1

D^1\A : A k 1 C

ªV¯ l9¯ ¯

Ë for near 0

/2XM3D^X fz¢s1 Notice that since .wÌ , the seventh and eighth Taylor approximations to are the same.

In Figure 10.4 we show the graphs of the sine function and the approximating polynomial of 1 degree 7 for 1 near 0. They are indistinguishable where is close to 0. However, as we look at values

of 1 farther away from 0 in either direction, the two graphs move apart. To check the accuracy of this

fz¢/@¶·ªM3D7Í ª}·~D[XVC r XM~Mla|vCtCoCzC

approximation numerically, we see how well it approximates ÊnÊ

¡=Ï

GVL

bÐÒÑ

¡Ï

GVL

bÐÒÑ

G GVL G

Figure 10.4: Graph of bÐÓÑ and its seventh degree Taylor polynomial, , for near

¶·aªDihnC X| htq CtCoC ] /2¶·aªN3D

Ë Ë Ê

When we substitute into the polynomial approximation, we obtain ¼

XVC r XN~9htªvCoCoCk

ÊMÊ which is extremely accurate—to about four parts in a million.

/21436D^d?eMfV1 r 1

Example 4 Graph the Taylor polynomial of degree approximating c for near 0.

Solution We ﬁnd the coefﬁcients of the Taylor polynomial by the method of the preceding example, giving

1 1 1±» 1±Ì

doeMf91R5[] /@1=36DihsA : A : C

~V¯ |±¯ ¯ rV¯

] /2143 1 Ì

Figure 10.5 shows that is close to the cosine function for a larger interval of -values than

/@143DhsAB1 ·~

the quadratic approximation ] in Example 2 on page 447.

K K

¡=Ô ¡4Ô

GVL GVL

G G

t t

K K

¡ ¡

GVL G9L

K K

¡ ¡

GVL GVL G G

Figure 10.5: approximates p better than for near .0/@143D[ÕyÖ Example 5 Construct the Taylor polynomial of degree 10 about 1RD^X for the function .

10.1 TAYLOR POLYNOMIALS 451

.0/2XM3Djh ÕyÖ ÕyÖ

Solution We have . Since the derivative of is equal® to , all the higher-order derivatives are equal

DØhnk~VkoCtCoCukohpX ÕyÖ .¢Ù /@1=3sDÕyÖ .¢Ù />XM3{DÕ Dh

to . Consequently, for any × , and . Therefore, the

Taylor polynomial approximation of degree 10 is given by

« ¬ _b

1 1 1 1

Õ 5]`_b}/@1=36Dih:©1$: : : :^ooy: k 1 C

ªm¯ |±¯ hpXV¯

~9¯ for near 0

ÕD[Õ D[~VC htrN~rVhprM~nrsCoCtC

To check the accuracy of this approximation, we use it to approximate Ë .

1;DZh ]0_xM/zhp3D~9C htrN~rmhtrnXmh ]`_b

Substituting gives Ë . Thus, yields the ﬁrst seven decimal places

1 ÕyÖ

for Õ . For large values of , however, the accuracy diminishes because grows faster than any .0/2143\DÝÕpÖ

polynomial as 1EÚÜÛ . Figure 10.6 shows graphs of and the Taylor polynomials of DXVkthnk~9k-ªmkz|

degree § . Notice that each successive approximation remains close to the exponential

curve for a larger interval of 1 -values.

¡=ß ¢¡=ßâ¡ ¡

¡=à ¡=à

QÞ Q Þ

á Figure 10.6: For G near , the value of is more closely approximated by higher-degree Taylor

polynomials

.0/2143D 1 hsAâ1

Example 6 Construct the Taylor polynomial of degree § approximating for near 0.

¬®

.0/2XN3sDØh . /2XM3sDãh . />XM3sDi~ . />XM3vDiªV¯ . /2XN3sD|m¯

< < < < <

Solution Differentiating gives , < , , , , and so on. This

means « ¬

5[] /21436Dih:©1´:©1 :©1 :1 :^otp:©1 k 1

hsAB1 for near 0,

Let us compare the Taylor polynomial with the formula obtained on page ?? for the sum of a ﬁnite ¨nä

geometric series: _

hsAB1

« ¬

Djh:1$:1 :1 :©1 :toa:©1 C

hsAB1

¨ä

_ 1 If 1 is close to 0 and is small enough to neglect, the formula for the sum of a ﬁnite geometric

series gives us the Taylor approximation of degree § :

« ¬

5h:1$:1 :1 :©1 :^ooy:©1 C hsAB1 452 Chapter Ten APPROXIMATING FUNCTIONS

Taylor Polynomials Around åæèç

Suppose we want to approximate .0/@1=3D^éws1 by a Taylor polynomial. This function has no Taylor 1;êiX polynomial about 1©D X because the function is not deﬁned for . However, it turns out that we can construct a polynomial centered about some other point, 1%D8 .

First, let’s look at the equation of the tangent line at 1D^8 :

D[.0/>8}3:;.=

This gives the ﬁrst Taylor approximation

1 84C

.0/@1=365^.0/>8}3:(.=

. />8}3o/@1%A893 . 1 The < term is a correction term which approximates the change in as moves away

from 8 .

17DÝ8 .0/>893

Similarly, the Taylor polynomial ]¨/2143 centered at is set up as plus correction /@1%A(893 terms which are zero for 1©D 8 . This is achieved by writing the polynomial in powers of

instead of powers of 1 :

¨ ¨

.0/@1=365] /21436D[6s:(s_/@1\A 893:( /@1\A 893 :^otp:; /@1\A 893 C

]'¨/@1=3 .0/2143

If we require § derivatives of the approximating polynomial and the original function

1D[8 1D[8 ²2³

to agree at , we get the following result for the § Taylor approximations at :

¡´# ¡

"! ë

Taylor Polynomial of Degree µ Approximating for near

.0/214365^] /@143

. />8}3 . /2893

< <

D[.0/2893:;.=<>/>893?/@1"AB893: /@1AB893 :[top: /@1\A 893

~V¯ ¯

]'¨/@1=3 1Di8

We call the Taylor polynomial of degree § centered at , or the Taylor polynomial about 1D[8 .

You can derive the formula for these coefﬁcients in the same way that we did for 8RDX . (See

Problem 28, page 454.) 1 Example 7 Construct the Taylor polynomial of degree 4 approximating the function .0/@143D£é¢s1 for near 1.

Solution We have

é¢s1 .0/xhp3D^é¢/xhy3D^X

.0/@1=3D so

. /@1=36D hy·a1 . /xhp3D h

< <

. /@1=36DìAhy·a1 . /xhp3D Ah

< < < <

. /@1=36D ~N·y1 . /xhp3D ~

< < < < < <

¬® ¬ ¬®

. /@1=36DjA ·y1 k . /xhp3D A C

Ê Ê

The Taylor polynomial is therefore

« ¬

/@1A;hy3 /@1"A;hy3 /@1"A;hy3

é¢s1R5]¬N/21436D[XÉ:[/@1\Ahp3'A :;~ A

~9¯ ªV¯ |m¯

« ¬

/@1"A;hy3 /@1\Ahp3 /21"Ahp3

: A k 1 C D/@1"A;hy3'A

ª | ~ for near 1

Graphs of é¢s1 and several of its Taylor polynomials are shown in Figure 10.7. Notice that

é¢s1 1 1

]¬N/2143 stays reasonably close to for near 1, but bends away as gets farther from 1. Also, éws1 note that the Taylor polynomials are deﬁned for 1ê;X , but is not.

10.1 TAYLOR POLYNOMIALS 453

¡¥

G9L

GVL

¡¥

GVL

¡=ß

GVL

G G Figure 10.7: Taylor polynomials approximate Ñ closely for near , but not necessarily farther away

The examples in this section suggest that the following results are true for common functions:

1âDj8 .0/@1=3 1 8 ï Taylor polynomials centered at give good approximations to for near . Farther away, they may or may not be good.

ï The higher the degree of the Taylor polynomial, the larger the interval over which it ﬁts the function closely.

Exercises and Problems for Section 10.1 Exercises

For Problems 1–10, ﬁnd the Taylor polynomials of degree ð G

approximating the given functions for near 0. (Assume ñ is a constant.)

For Problems 11–14, ﬁnd the Taylor polynomial of degree ð

G F í

©Þ for near the given point .

ó ô ð õ ö

1. , ð , , 2. , , ,

ò£G QRG

G F G F Þ

4þ =þ

bÐÒÑ p

ò£G í Þ G Þ

ð ó

3. , ð , , 4. , , , 11. , , 12. , ,

Þ ©í

G ©í Þ G ©í Þ ð ð

-ù Ñ ù Ñ

÷oø ð ð ÷

5. ÷ , , 6. , ,

`QRG í Þ }òûGVL

á FU Þ 'òG F

ð ð õ ö ü 7. ú , , , 8. , , ,

13. , , ð 14. , ,

©í

í Þ òG9L2ý í Þ ð

9. , ð , , 10. , , ,

ò£G

Problems

GVL[FÉòBÿ-Gò¡ G

J4K

GVL GVL For Problems 15–18, suppose is the 15. GVL 16. 17. 18.

second degree Taylor polynomial for the function J about

G G G G

F ÿ

G© . What can you say about the signs of , , if has the J4K

graph given below? G9L

J4K G£ 19. Suppose the function G9L is approximated near 454 Chapter Ten APPROXIMATING FUNCTIONS

by a sixth degree Taylor polynomial (b) Compare this result to the Taylor polynomial ap-

J4K

GVL(á

Î ¢

K proximation of degree 2 for the function

¡£¢

G9L©íoGQ"ÞtG ò G

G ¤ õ about . What do you notice? (c) Use your observation in part (b) to write out the Tay- Give the value of

¢ lor polynomial approximation of degree 20 for the

J4K JNSTK JNS S SÇK J¦¥¨§ ©-K J¦¥ ©-K

pL pL yL pL (a) yL (b) (c) (d) (e) function in part (a).

(d) What is the Taylor polynomial approximation of de-

J4K

GVL©á

20. Suppose is a function which has continuous deriva- gree 5 for the function ?

K K K

S S S

LâQ LB Lâ í

õ õ õ

tives, and that , ,

G© GQ

S S S

bÐÒÑ

L£Q6í ¤

30. Consider the equations ¤ and

í õ

(a) What is the Taylor polynomial of degree for near (a) How many zeros does each equation have?

õ ? What is the Taylor polynomial of degree for (b) Which of the zeros of the two equations are approx-

near 5? imately equal? Explain. ¥

(b) Use the two polynomials that you found in part (a) K

L"!

bÐÒÑ

Þ L

31. The integral is difﬁcult to approximate

¤ ü to approximate .

using, for example, left Riemann sums or the trapezoid

J4K

GVLû

bÐÒÑ

21. Find the second-degree Taylor polynomial for rule because the integrand is not deﬁned at

ÞpG Q Gò G©

ö about . What do you notice? . However, this integral converges; its value is

J4K

GVL ¤ ¤#¤#¤$¤

ó ô

ü Estimate the integral using Taylor polyno-

22. FindÎ the third-degree Taylor polynomial for

bÐÓÑ

G ò G Q G`ò

G© mials for about of õ ö about . What do you notice? (a) Degree 3 (b) Degree 5

23. (a) Based on your observations in Problems 21–22,

J J J

¥ ¥

make a conjecture about Taylor approximations in 32. One of the two sets of functions, , , , or , ,

Î J

the case when is itself a polynomial. , is graphed in Figure 10.8; the other set is graphed in

G©Ø &

(b) Show that your conjecture is true. Figure 10.9. Points % and each have . Taylor

G bÐÒÑ polynomials of degree 2 approximating these functions

24. Show hoÎ w you can use the Taylor approximation

G©

G G

bÐÒÑ near are as follows:

GQ G

, for near 0, to explain why à . í G

K J K

¥ ¥

GVL(òGÉò oG G9L'( ò%Gòâ ?G

J K K

( `òGQG GVL(òGÉò£G

25. Use the fourth-degree Taylor approximation GVL

G G

ØQ ò G

J K K

Î Î

GVL( `òGÉò£G GVL(`QRGÉò£G

for near to explain why ¤

`Q G

Ð J

à .

(a) Which group of functions, the s or the s, is repre-

¢ á sented by each ﬁgure?

26. Use a fourth degree Taylor approximation for , for

& (b) What are the coordinates of the points % and ? near 0, to evaluate the following limits. Would your (c) Match the functions with the graphs (I)-(III) in each answer be different if you used a Taylor polynomial of ﬁgure.

higher degree?

ávQQ

Ð¨

(a) à

III II III II

á Q`Q Q ¤

(b) à

J4K K K J K K

S S

tLs pL{ pLsE tLs tLsE

27. If , and ,

K J K K K

S S S S S NS S

pL p tL í pL tL

õ , and , , ö ,

calculate the following limits. Explain your reasoning.

J4K J4K

GVL GVL î

î I

Ð Ð

K K

(a) (b)

GVL GVL

¢ ¢

& 28. Derive the formulas given in the box on page 452 for %

the coefﬁcients of the Taylor polynomial approximating I

J F a function for G near . Figure 10.8 Figure 10.9

29. (a) Find the Taylor polynomial approximation of degree

J4K

¤ GVL©á 4 about G for the function . 10.2 TAYLOR SERIES 455 10.2 TAYLOR SERIES

In the previous section we saw how to approximate a function near a point by Taylor polynomials. Now we deﬁne a Taylor series, which is a power series that can be thought of as a Taylor polynomial

that goes on forever.

¡ )+*-,

Taylor Series for cos )+* sin d?eNf91

We have the following Taylor polynomials centered at 1%DX for :

doeMf91R5[]M/@1=36Dih

doeMf91R5[] /@1=36DihsA

~V¯

1 1

doeMf91R5[] /@1=36DihsA :

~V¯ |±¯

1 1±» 1

: A doeMf91R5[] /@1=36DihsA

~V¯ |±¯ ¯

1 1±» 1 1±Ì

: A C doeMf91R5[] /@1=36DihsA :

~V¯ |±¯ rV¯ ¯

]M/@1=3 ] /2143 ]¬N/@1=3 ] /2143 ] /@143 Ì

Here we have a sequence of polynomials, , , , » , , ..., each of which is a 1

better approximation to doeMf91 than the last, for near 0. When we go to a higher-degree polynomial

] ] 1±Ìp·rV¯ Ì (say from » to ), we add more terms ( , for example), but the terms of lower degree don’t change. Thus, each polynomial includes the information from all the previous ones. We represent

the whole sequence of Taylor polynomials by writing the Taylor series for doeMf91 :

1 1±» 1 1±Ì

: A A;tooC h{A :

~V¯ |m¯ rV¯ ¯ Ê

Notice that the partial sums of this series are the Taylor polynomials, ]'¨/@1=3 . ÕyÖ We deﬁne the Taylor series for fz¢{1 and similarly. It turns out that, for these functions, the

Taylor series converges to the function for all 1 , so we can write the following:

1 1±º 1±¼ 1/.

f-w{1RD^1\A : A : A;to

ªV¯ l9¯ ¯ qm¯

1 1±» 1±Ì 1

: A : A;to d?eNf91RDihsA

~V¯ |m¯ ¯ rV¯

« ¬

1 1 1

Õ Dih:©1´: : : :^ot

~V¯ ªm¯ |m¯

f-ws1 d?eMfV1 ÕpÖ 1D[X

These series are also called Taylor expansions of the functions , , and about .¨ The

1 · ¯

general term of a Taylor series is a formula which gives any term in the series. For example, §

ÕyÖ /zAhp3-Ùp1 ÙM·9/T~ 3?¯

is the general term in the Taylor expansion for , and × is the general term in the

d?eNf91 × expansion for . We call § or the index.

Taylor Series in General

Any function . , all of whose derivatives exist at 0, has a Taylor series. However, the Taylor series

.0/@143 1 1 for . does not necessarily converge to for all values of . For the values of for which the series does converge to .0/2143 , we have the following formula:

456 Chapter Ten APPROXIMATING FUNCTIONS

¡$# ¡è

Taylor Series for "! about 0

. /2XN3 . /2XM3 . /2XN3

< < < < <

.0/2143D7.0/2XM3:;. /2XM3x1$: 1 : 1 :^oty: 1 :to

~9¯ ªV¯ ¯ §

In addition, just as we have Taylor polynomials centered at points other than X , we can also

8 . 1D8

have a Taylor series centered at 1D (provided all the derivatives of exist at ). For the .0/@1=3

values of 1 for which the series converges to , we have the following formula:

¡$# ¡è "!

Taylor Series for about ë

J K J K J K

S S S S S ¥32©

FNL FML FML

S 2

J4K J4K J K K K K K

GVL FML±ò FML GQFNLVò GQFML ò GQ"FML ò¡0#010uò GQFML ò40$0#0

í ð

The Taylor series is a power series whose partial sums are the Taylor polynomials. As we saw in Section ??, power series generally converge on an interval centered at 1[DZ8 . The Taylor series for such a function can be interpreted when 1 is replaced by a complex number. This extends the

domain of the function. See Problem 41. 1 For a given function . and a given , even if the Taylor series converges, it might not converge to .0/@143 , though it does for most commonly encountered functions. Functions for which this is

true for the Taylor series about every point 1DZ8 in their domain are called analytic functions.

d?eMfV1 f-¢s1 1 Fortunately, the Taylor series for ÕyÖ , , and converge to the original function for all , so they are analytic. See Section 10.4.

Intervals of Convergence of Taylor Series

1©D h Let us look again at the Taylor polynomial for é¢s1 about that we derived in Example 7 on

page ??. A similar calculation gives the Taylor Series

« ¬

/@1"A;hy3 /@1\Ahp3 /21"Ahp3 /@1"A;hy3

¨}©

éws1D/@1\Ahp3'A : A :top:[/xAhy3 :totC

~ ª | §

Example 2 on page 449 and Example 5 on page 451 show that this power series has interval of

65 1êè~ convergence X . However, although we know that the series converges in this interval,

we do not yet know that its sum is éws1 . The fact that in Figure 10.10 the polynomials ﬁt the curve

757185~ éws1 X9571(êj~ well for X suggests that the Taylor series does converge to for . For such

1 -values, a higher-degree polynomial gives, in general, a better approximation.

9:~9k However, when 1 the polynomials move away from the curve and the approximations get

worse as the degree of the polynomial increases. Thus, the Taylor polynomials are effective only 1

as approximations to éws1 for values of between 0 and 2; outside that interval, they should not ~ be used. Inside the interval, but near the ends, X or , the polynomials converge very slowly. This means we might have to take a polynomial of very high degree to get an accurate value for é¢s1 .

10.2 TAYLOR SERIES 457

P O

K ¡

Interval of convergence GVL

GVL

í Þ

GVL

¡/¢

GVL

¡4Ô

GVL

¡£¢ î

GVL

K K K K

¢ Ô

¡ ¡ ¡=Ï ¡

G =<âG?>B GVL; GVL; GVL; G9L;

Ñ § Figure 10.10: Taylor polynomials ¤#¤#¤ converge to for and diverge outside that interval

To compute the interval of convergence exactly, we ﬁrst compute the radius of convergence 1D[~ using the method on page ??. Convergence at the endpoints, 1D[X and , has to be determined separately. However, proving that the series converges to é¢s1 on its interval of convergence, as

Figure 10.10 suggests, requires the error term introduced in Section 10.4. 1RD^X

Example 1 Find the Taylor series for éw/zh¥:143 about , and calculate its interval of convergence. 1D[X

Solution Taking derivatives of é¢/zh:©1=3 and substituting leads to the Taylor series

« ¬

1 1 1

é¢/xh{:©1=3D1\A : A :^ootC

~ ª |

1 é¢s1

Notice that this is the same series that we get by substituting /xh:143 for in the series for :

« ¬

/@1\Ahp3 /@1\Ahp3 /21\A;hy3

é¢s1RDj/21"Ahp3'A : A :^ot X?5(1ê~

ª |

~ for .

1âD h X@5[1©ê7~

Since the series for éws1 about converges for , the interval of convergence for the

1RDX AhA5;1£êEh

Taylor series for éw/zh:©1=3 about is . Thus we write

« ¬

1 1 1

é¢/zh¥:©1=36D£1"A : A :^ot AhA5(1êEh

ª |

~ for .

1(ê Ah é¢/xhv:;143 Notice that the series could not possibly converge to é¢/zhs:;1=3 for since is not

deﬁned there.

K K

¡Ï

¡ ¡£B

GVL

G9L GVL

§ P Interval of O

convergence

ò£GVL

GVL

K K

¡Ï ¡

GVL GVL

O P

GVL

¡=Ô

GVL

K K

'òG9L GVL

O P

0ò%GVL QsC<âG-> Figure 10.11: Interval of convergence for the Taylor series for Ñ is 458 Chapter Ten APPROXIMATING FUNCTIONS

The Binomial Series Expansion

1RDX .0/2143D/xh:143ED

We now ﬁnd the Taylor series about for the function , with F a constant, but

not necessarily a positive integer. Taking derivatives:

D .0/>XM3Dih

.0/@143Di/zh:©1=3 so

. /2143¥D /xh:143ED . />XM3D

< <

F F

. /2143¥D / Ahp3?/zh:©1=3ED . />XM3D / Ahp3

< < < <

F F F F

. /2143¥D / Ahp3?/ A ~M3?/zh:©1=3ED k . />XM3D / Ahp3o/ A©~n3?C

< < < < < <

F F F F F F

Thus, the third-degree Taylor polynomial for 1 near 0 is

/ Ahp3 / A;hy3?/ A ~M3

/zh:©1=3 5[]'«M/21436Djh: 1$: 1 : 1 C

F F F F F

~9¯ ªm¯

] /@1=3uk-] /@143?koCtCoC

« ¬

Graphing for various speciﬁc values of F suggests that the Taylor polynomials

Ah?5E1G5h

converge to .0/@1=3 for . (See Problems 24–23, page 459.) This can be conﬁrmed using

ED 1RD[X the radius of convergence test. The Taylor series for .0/@1=3Di/zh:©1=3 about is as follows:

The Binomial Series

/ Ahp3 / Ahp3o/ A©~n3

/xh:143 Djh: 1û: Ah=51H57h 1 : 1 :^ot

F F F F F

~9¯ ªV¯

F for .

/xh:R143ID

In fact the binomial series gives the same result as multiplying out when F is a positive

integer. (Newton discovered that the binomial series can be used for noninteger exponents.)

D[ª /zh:©1=3

Example 2 Use the binomial series with F to expand .

Solution The series is

ªUp~ ªt~Unh ªp~UhspX

« « ¬

/xh:143 Dh:(ª1$: 1 : 1 : 1 :tooC

~9¯ ªV¯ |m¯ ¬

The 1 term and all terms beyond it turn out to be zero, because each coefﬁcient contains a factor

of 0. Simplifying gives «

/xh:143 Djh:(ª1$:(ª1 :©1 k «

which is the usual expansion obtained by multiplying out /xh:143 .

h 1D[X

Example 3 Find the Taylor series about for h:©1 .

Di/zh:©1=3 DAh h:1

Solution Since , use the binomial series with F . Then

/xAhy3?/zAÉ~n3 /xAhp3o/xAÉ~M3?/xAvªN3 h

_ «

D/xh:143 Dih:^/zAhp3b1û: 1 : 1 :to

h:1 ~9¯ ªm¯

Aâ1 :to Ah=51H57h DihsAâ1$:1 for .

This series is both a special case of the binomial series and an example of a geometric series. It

A5;195Eh converges for Ah . 10.2 TAYLOR SERIES 459 Exercises and Problems for Section 10.2

Exercises

ò£G ò£G ò40#0$0

For Problems 1–4, ﬁnd the ﬁrst four terms of the Taylor series 12. ò£GÉòG QRG

for the given function about .

`QGÉòG QG ò£G QN0#0$0

13.

'òG

òG `Q

1. 2. 3. 4. ú

`QG

ò£G

G G G

QGVLQGQ Q Q QO0#010

14. Ñ

í Þ

For Problems 5–11, ﬁnd the ﬁrst four terms of the Taylor se-

G G G

ries for the function about the point . G

'ò£GVL GQ ò Q ò QH010#0

15. Ñ

í Þ

G Fû F

bÐÒÑ tKJ bÐÒÑLJ

5. , 6. , 7. , õ

Þ Þ FUQ Þ

4þ 4þ =þ

G G G

GGQ ò Q òP0#0#0

16. bÐÒÑ

G G Fè G F

þ þ

ù Ñ

õ ö

8. ÷ , 9. , 10. ,

FU Þ

4þ

G G G

G©GQ ò Q òP0#0#0

-ù Ñ

÷oø ÷

G FY

þ 17.

í ö

11. , õ

¢ Ô

G G G

á ò ò òP0#0#0 ò

18. ò£G

í Þ

Find an expression for the general term of the series in Prob-

¢ Ô

ð M

G G

lems 12–19 and give the starting value of the index ( or , G

ò Q òP0#010 G G G Q

for example). 19. p

Þ ó

Problems ó

20. (a) Find the terms up to degree of the Taylor series for By recognizing each series in Problems 27–35 as a Taylor se-

J4K K

GVL G L G© G

bÐÒÑ about by taking derivatives. ries evaluated at a particular value of , ﬁnd the sum of each G (b) Compare your result in part (a) to the series for bÐÒÑ . of the following convergent series.

How could you have obtained your answer to part (a) G

from the series for bÐÒÑ ?

2 K

27. 28. 2

Qs?L

J4K K

G9L' 'òB oG9L

ò ò òP0$010?ò ò¡010#0 'ò

ò¡0#010 ò Q ò40#0$0oò

21. (a) Find the Taylor series for about Q

" í

ò ?L; í"

ð ö by taking derivatives. õ

(b) Compare your result in part (a) to the series for

î òG9L

Ñ 29. 30.

2 2 2

Qs?L 0tu uptp

. How could you have obtained your an- p

'ò òRQ òRQ òP0#0#0?òRQ ò40#0$0

Q ò ò40$010?ò òP0

0ò%GVL

Þ Þ'S Þ'S Þ£S

" Þ L; swer to part (a) from the series for ?

î 'ò oGVL

the series for Ñ to be?

J4K

0ò%G GVL

ß ¥ ¥ ¥ ¥

2UT K K K K 2 K

nL aL L nL 0 22. By graphing the function and several of Qs?L

Q ò Q ò40$0#0oò òP0#0#0

its Taylor polynomials, estimate the interval of conver- 31. K

í Þ ò ?L

gence of the series you found in Problem 2. ð

J4K

Q% 0ò Q ò40$0#0

GVLU

¤ ¤

23. By graphing the function and several 32. ¤

'òG

ü ö ô

'òíò ò òP0#0#0

of its Taylor polynomials, estimate the interval of con- 33. ò

" í Þ

vergence of the series you found in Problem 3.

J4K

GVL

ò Q òP0$010 24. By graphing the function and several of Q

`QG 34.

Þ ó

its Taylor polynomials, estimate the interval of conver-

¤ ¤

Q% 0ò Q ò40#0$0

gence of the series you found in Problem 1. Compute the 35. ¤

í" radius of convergence analytically.

25. Find the radius of convergence of the Taylor series In Problems 36–37 solve exactly for the variable.

î `QGVL

around zero for Ñ .

'òGÉò£G ò£G ò40#0$0

26. (a) Write the general term of the binomial series for 36. õ

G òGVL2ý

about .

GQ G ò G òP010$0a©

37. ¤ í (b) Find the radius of convergence of this series.

460 Chapter Ten APPROXIMATING FUNCTIONS

J J J J4K

¥ ¥

G9L

38. One of the two sets of functions, , , , or , , 39. Suppose that you are told that the Taylor series of

G á G© is graphed in Figure 10.12; the other set is graphed in about is

Figure 10.13. Taylor series for the functions about a point

% &

¢ Ô ¥

corresponding to either or are as follows: à

G G G

J K K K K K K

¥ ¥

GVL©íò GQ?LQ GQ?L 0#010 GVL Q GQÞaL=Q GQÞaL 0#0#0

G ò£G ò ò ò òP0#0#0

í Þ

J K K K K K K

GVL©í¥Q GQ?L±ò GQ?L 0#010 GVL Q GQÞaLmò GQÞaL 0#0#0

J K K K K K K

Î Î

! !

GVL©ísQ% GQ?L±ò GQ?L 0#010 GVL ò GQÞaLmò GQÞaL 0#0#0

¢ ¢

¤ ¤

QtG á QoG á ¢

Find and ¤

SWV SYV

!aG !aG

à à

V V

X ¢X

(a) Which group of functions is represented in each ﬁg- ¢

V V V ure? V

40. Suppose you know that all the derivatives of some func-

J J

% & G (b) What are the coordinates of the points and ? tion exist at , and that Taylor series for about (c) Match the functions with the graphs (I)-(III) in each

ﬁgure. ß

G G G G

GÉò ò ò òP0#0#0uò ò40$0#0

I ¤

í Þ

J K J K J K J K

S S S S S S ¥ ©

pL yL yL yL

I Find , , , and ¤

Qs á[\

J Z

41. Let Z . We deﬁne by substituting in the

¢ á % Taylor series for . Use this deﬁnition2 to explain Eu- & ler’s formula

II II

á ò

pKJ bÐÒÑ]J

¤ Z III III Figure 10.12 Figure 10.13

10.3 FINDING AND USING TAYLOR SERIES

Finding a Taylor series for a function means finding the coefficients. Assuming the function has all its derivatives defined, finding the coefficients can always be done, in theory at least, by differen-

tiation. That is how we derived the four most important Taylor series, those for the functions ÕyÖ ,

doeMf91 /zh:£1=3ED f-ws1 , , and . For many functions, however, computing Taylor series coefﬁcients by differentiation can be a very laborious business. We now introduce easier ways of ﬁnding Taylor series, if the series we want is closely related to a series that we already know.

New Series by Substitution

Õ 17DèX

¤ ©

Suppose we want to find the Taylor© series for about . We could find the coefficients

Õ Ö AÉ~1±Õ

by differentiation.© Differentiating by the chain rule gives , and differentiating again

Ö Ö

AÉ~nÕ :£|M1 Õ

¤ ¤ ©

gives . Each time we differentiate we use© the product rule, and the number of

Õ Ö Õ Ö

¤ ¤

terms_x grows. Finding the tenth or twentieth derivative of , and thus the series for up to 1 the 1 or terms, by this method is tiresome (at least without a computer or calculator that can differentiate).

Fortunately, there’s a quicker way. Recall that

« ¬

Õ^Dh: : : : :^ot C

g g g

~V¯ ªV¯ |±¯ g

g for all

DjAs1

Substituting g tells us that

« ¬

/xAs1 3 /zAs1 3 /xAs1 3

Õ Dh:[/xAs1 3: : : :to 1C

ªV¯ |m¯ ~9¯ for all

2Complex numbers are discussed in Appendix B.

10.3 FINDING AND USING TAYLOR SERIES 461

Õ Ö

Simplifying shows that the Taylor series for ¤ is

» Ì

1 1 1

Õ DhsAB1 : A : :^ot 1

ªm¯ |m¯

~V¯ for all .

_b 1

Using this method, it is easy to ﬁnd the series up to the 1 or terms.

1D[X .0/@1=36D

Example 1 Find the Taylor series about for h:1 .

Solution The binomial series tells us that

_ « ¬

AhA5 5EhC Di/zh: 3 DjhsA : A : :to

g g g g g

g for

Substituting g gives

» Ì

Ah5(1H57h DihsAB1 :1 AB1 :©1 :to

h:1 for ,

which is the Taylor series for h:1 .

These examples demonstrate that we can get new series from old ones by substitution. More

advanced texts show that series obtained by this method are indeed correct. Dã1

In Example 1, we made the substitution g . We can also substitute an entire series into

another one, as in the next example.

D[X / 3D^ÕU`ba c'd

c _

Example 2 Find the Taylor series about _ for . _

Solution For all g and , we know that

« ¬

Õ Dih: : : : :^oo

g g g

~9¯ ªm¯ |m¯

and º

f-¢ D A : A;totC

_ _

ªV¯ l9¯

_ _

f-w g

Let’s substitute the series for _ for :

« « «

º h º h º

Õ Dhn:4e A : A£ot f: e A : Aot f : e A : Aot f :ÃotoC

`ba c£d

______

ªm¯ lV¯ ~V¯ ªm¯ lV¯ ªV¯ ªm¯ lV¯

_ _ _

To simplify, we multiply out and collect terms. The only constant term is the , and there’s only one

·~9¯

« «

_ _

_ term. The only term is the ﬁrst term we get by multiplying out the square, namely . There

A ·ªV¯

« _

are two contributors to the _ term: the from within the ﬁrst parentheses, and the ﬁrst term ·ªV¯

we get from multiplying out the cube, which is _ . Thus the series starts

« «

e f

`ba c/d

Õ Djh: : : A : :^oo

_ _ _

~9¯ ªV¯ ªV¯

:^oo Djh: : :X C

~9¯

_ _ for all _ 462 Chapter Ten APPROXIMATING FUNCTIONS New Series by Differentiation and Integration Just as we can get new series by substitution, we can also get new series by differentiation and integration. Here again, proof that this method gives the correct series and that the new series has

the same interval of convergence as the original series, can be found in more advanced texts.

h h

1D[X C

hsAB1

Example 3 Find the Taylor Series about for /xhsAB143 from the series for

h h

f[D e

h{AB1 1 g

Solution We know that /zhsAâ1=3 , so we start with the geometric series

« ¬

Ah=5195EhnC Djh:1$:1 :1 :©1 :to

hsAB1 for

Differentiation term by term gives the binomial series

h h

e AhA51957hMC D f7Dh:;~a1´:ªn1 :|n1 :to

1 hsAB1 g

/zhsAâ1=3 for

1D[X d1j {1 C

h:©1 h

Example 4 Find the Taylor series about for h"i from the series for

/ d1j {143 h

h"i h

1 h6:1

Solution We know that g , so we use the series from Example 1 on page 461:

/ dj ¥1=3 h

» Ì

D DhsAB1 :1 Aâ1 :©1 Aoo AhA5;1H57h

h i h

1 h:1

g for . g

Antidifferentiating term by term gives

1 14º 14¼ 1¦. h

1RD[:©1\A : A : Aot A;hA51957hMk dj 1RDlk

h:1 ª l q

Ë h

h i for

dj {XD[X ED[X h

where is the constant of integration. Since h i , we have , so

1±º 1±¼ 1 1¦.

: A A;to d1j {1D^1\A : Ah5;195EhnC

ª l q

h Ë

h"i for

dj ¥1 h The series for h i was discovered by James Gregory (1638–1675).

Applications of Taylor Series

d1j ¥1 ¶ h"i Example 5 Use the series for h to estimate the numerical value of .

10.3 FINDING AND USING TAYLOR SERIES 463

dj h;D ¶·y| d1j {1

h h"i h

Solution Since h i , we use the series for from Example 4. We assume—as is the 1£DÝh

case—that the series does converge to ¶·a| at , the endpoint of its interval of convergence.

dj ¥1 1%Djh h

Substituting into the series for h i gives

h h h h

¶D£| dj hÉD^|We=hsA : A : Aot f©C

ª l q

h i h Ë

dj 1 ¶ h Table 10.1 Approximating using the series for h i

ð 4 5 25 100 500 1000 10,000 m

2 2.895 3.340 3.182 3.132 3.140 3.141 3.141

²2³ n

Table 10.1 shows the value of the § partial sum, , obtained by summing the nonzero terms from

¶D^ªmCwht|mhCtCoCC n 1 through § . The values of do seem to converge to However, this series converges

very slowly, meaning that we have to take a large number of terms to get an accurate estimate for ¶ ¶ . So this way of calculating is not particularly practical. (A better one is given in Problem 1, page 475.) However, the expression for ¶ given by this series is surprising and elegant.

A basic question we can ask about two functions is which one gives larger values. Taylor series can often be used to answer this question over a small interval. If the constant terms of the series for two functions are the same, compare the linear terms; if the linear terms are the same, compare the quadratic terms, and so on.

Example 6 By looking at their Taylor series, decide which of the following functions is largest, and which is

h:f-w Õ

_ _ hsA ~

smallest, for near 0. (a) (b) (c) Í

D[X fz¢ _

Solution The Taylor expansion about _ for is

º ¼

f-¢ D A : A :totC

_ _ _

ªV¯ l9¯ ¯

_ _

¼ So º

h:f-¢ Dih: A : A :^otoC

_ _ _

ªm¯ lV¯ ¯

_ _

D[X Õ d

The Taylor expansion about _ for is

« ¬

Õ Dih: : : : :^ootC

_ _ _

~V¯ ªm¯ |m¯

D[X hy· h:

Í _

The Taylor expansion about _ for is

_ _ « «

/zA /zA 3o/xA 3 3o/xA 3?/xA 3

h h

_po «

: D/xh: 3 DihsA : :^oo

~ ~V¯ ªm¯

_ _ _ _

h ª l

DihsA : A :tooC

~ r h

_ _ _

AÉ~ _

So, substituting for _ :

h ª l h

DihsA /xAÉ~ 3: /xAÉ~ 3 A /xAÉ~ 3 :^oo

~ r h

_ _ _

hsA ~

ª l

:^ot?C Dih: : :

~ ~

_ _ _ 464 Chapter Ten APPROXIMATING FUNCTIONS

For _ near 0, we can neglect terms beyond the second degree. We are left with the approximations:

h:(f-w 5ih:

_ _

Õ 5ih: :

h ª

C 5ih: :

_ _

hsA©~

h ª

Since

h: 5Eh: : 5[h: : k

~ ~

______

and since the approximations are valid for _ near 0, we conclude that, for near 0,

h:(fz¢ 5Õ 5 C

hsA ~

Í _

Example 7 Two electrical charges of equal magnitude and opposite signs located near one another are called

q r

an electrical dipole. The charges and q are a distance apart. (See Figure 10.14.) The electric ]

ﬁeld, s , at the point is given by

D A C

q q

/ : 3

t t

s r

Use series to investigate the behavior of the electric ﬁeld far away from the dipole. Sho« w that when

hy·

t t

is large in comparison to r , the electric ﬁeld is approximately proportional to .

¡ w w

P u O P v O ¡

Figure 10.14: Approximating the electric ﬁeld at duev to a

Q w dipole consisting of charges w and a distance apart

Solution In order to use a series approximation, we need a variable whose value is small. Although we know

t t

r r

that r is much smaller than , we do not know that itself is small. The quantity is, however,

· ha·9/ : 3

t t r very small. Hence we expand r in powers of so that we can safely use only the ﬁrst

few terms of the Taylor series. First we rewrite using algebra:

©=

h h h

D D h: C

/ : 3 /xh: · 3

t t t t t

r r

/zh:©1=3ED 1D · DjAÉ~

t r

Now we use the binomial expansion for with and F :

h /xAÉ~M3?/zAvªM3 h /zAÉ~n3o/xAvªM3o/xAs|}3

h: eh:^/zAÉ~n3 : :^ot f D :

r r r r

Q Q Q Q

~9¯ ªm¯

t t t t t t

S S S S

e f

D hsA ~ :(ª Aâ| :to C

r r r

t t t t

So, substituting the series into the expression for s , we have

h h

e f{z

D D A A hsA©~ :ª AB| :^ot

q q r r r

/ : 3

t t t t t t t

qyx s

e f

~ A ª :©| Aot C D

r r r q

t t t t

10.3 FINDING AND USING TAYLOR SERIES 465

· /zh9: · 3

t t

Since is smaller than 1, the binomial expansion for r converges. We are interested in

· / · 3

t t r the electric ﬁeld far away from the dipole. The quantity is small there, and r and higher

powers are smaller still. Thus, we approximate by disregarding all terms except the ﬁrst, giving

~ ~

5 e f©k 5 C

q r qr

t t s t

s so

hy· C

r s

Since q and are constants, this means that is approximately proportional to

t r

In the previous example, we say that s is expanded in terms of , meaning that the variable

· t

in the expansion is r .

Exercises and Problems for Section 10.3 Exercises

Find the ﬁrst four nonzero terms of the Taylor series about 0 lems 13–15. Give the general term. Î

for the functions in Problems 1–12. K

13. 'òG9L

L4Q

nbÐÒÑ

`Q% oG L á

1. 2. p 3. 4. 14.

15.

G yQ¥ ?NL

bÐÒÑ Ñ

÷?ø ~

5. 6. 7. 8. ~

`Q}|

K L p For Problems 16–17, expand the quantity about 0 in terms of

the variable given. Give four nonzero terms.

G F

9. 10. 11. 12.

á K

¤ 16. in terms of 17. in terms of

ò L ò òG

F ò%G

bÐÒÑ bÐÒÑLJ

Fâ

where F

Find the Taylor series around 0 for the functions in Prob-

Problems

F ÿ

18. For , positive constants, the upper half of an ellipse small positive J .

'ò

J tJ

has equation (a) bÐÒÑ (b) (c)

J4K

U G9L ÿU Q

J4K

GVL 20. For values of near 0, put the following functions in in-

(a) Find the second degree Taylor polynomial of

creasing order, using their Taylor expansions.

K K

about . î

tòv L L aQ

bÐÒÑ p (a) Ñ (b) (c) (b) Explain how you could have predicted the coefﬁ- cient of G in the Taylor polynomial from a graph of

J . (c) For G near , the ellipse is approximated by a

21. Figure 10.15 shows the graphs of the four functions be- parabola. What is the equation of the parabola? G

G low for values of near 0. Use Taylor series to match

What are its -intercepts? ÿ FU©í graphs and formulas.

(d) Taking and , estimate the maximum dif- J4K

ference between GVL and the second-degree Taylor

Q6 {>G?>B

¤ ¤

'ò uò¥GVL polynomial for .

(a) (b) (c)

Q"G

19. By looking at the Taylor series, decide which of the fol-

(d) QRG lowing functions is largest, and which is smallest, for

466 Chapter Ten APPROXIMATING FUNCTIONS u where M is a positive constant whose value depends on

(III) (I) þ the units. Expand as a series in , giving the ﬁrst two nonzero terms.

(IV) | 26. Assume F is a positive constant. Suppose is given by

the expression

|v F ò£G Q F QG

(II) ¤ G Expand | as a series in as far as the second nonzero

Figure 10.15 term. u

27. The electric potential, , at a distance along the axis

Uá U 0ò£G L þ perpendicular to the center of a charged disc with radius

22. Consider the functions and ¤ F

and constant charge density , is given by

(a) Write the Taylor expansions for the two functions

u u

G£[

about . What is similar about the two series? K

; ò£F Q L

¤ What is different?

(b) Looking at the series, which function do you predict u

U uL will be greater over the interval Qs ? Graph both Show that, for large ,

and see.

¤ this by looking at the series expansions?

(d) By looking at the coefﬁcients, explain why it is rea-

sonable that the series for converges for

G £ {ò G L

all values of , but the series for þ

?L converges only on Qs . 28. One of Einstein’s most amazing predictions was that light

traveling from distant stars would bend around the sun on~

23. The hyperbolic sine and cosine are differentiable and sat- the way to earth. His calculations involved solving for

v v©

bÐÒÑ

isfy the conditions p and , and in the equation

! !

~ ~ ~

K K

GVL G GVL G

p bÐÒÑ bÐÒÑ p

ò£ÿ ò ò L©

bÐÒÑ p p

!aG !aG

(a) Using only this information, ﬁnd the Taylor approx-

J4K ÿ

G G9Ls

where is a very small positive constant. ô

imation of degree ð about for

G p .

(a) Explain~ why the equation could have a solution for

p (b) Estimate the value of . which is near 0.

(b) Expand the left-hand~ side of the equation in Taylor

Gâ

ð ö

mial approximation of degree about E ~

K series about , disregarding terms of order

G9L= G

bÐÒÑ for . and higher. Solve for . (Your answer will involve 24. Pade´ approximants are rational functions used to approx- ÿ .) imate more complicated functions. In this problem, you will derive the Pade´ approximant to the exponential func- 29. The Michelson-Morley experiment, which contributed to

tion. the formulation of the Theory of Relativity, involved the

K K

J4K difference between the two times and that light took

F ÿ G9L6 `òâFGVL 0òâÿG9L (a) Let þ , where and are

to travel between two points. If is the velocity of light;

¥ ¥

constants. Write down the ﬁrst three terms of the

J4K , , and are constants; and , then the times

G9L G© Taylor series for about . and are given by

(b) By equating the ﬁrst three terms of the Taylor se-

J4K

G GVL á F ÿ

¥ ¥

ries about for and for , ﬁnd and

J4K

G9L á

Q Q

so that approximates as closely as possible K

Q L Q L

þ þ

6Q `Q

þ þ

G©

near .

25. An electric dipole on the -axis consists of a charge at

Q GQs

G£ (a) Find an expression for , and give its

u u

and a charge at . The electric ﬁeld, þ

G G Taylor expansion in terms of up to the second , at the point on the -axis is given (for ) nonzero term.

w (b) For small , to what power of is proportional?

u u

M M

K K ò ?L QâuL What is the constant of proportionality? 10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS 467

30. A hydrogen atom consists of an electron, of mass , or- 32. When a body is near the surface of the earth, we usually

biting a proton, of mass , where is much smaller assume that the force due to gravity on it is a constant

than . The reduced mass, , of the hydrogen atom is , where is the mass of the body and is the accel-

deﬁned by eration due to gravity at sea level. For a body at a distance

above the surface of the earth, a more accurate expres-

sion for the force is

(a) Show that . u

ò L

(b) To get a more accurate approximation for , express

as times a series in .

ing all but the constant term in the series. The ﬁrst- situation in which the body is close u to the surface of the

order correction is obtained by including the linear earth so that is much smaller than .

ô ó

term but no higher terms. If , by what

(a) Show that .

percentage does including the ﬁrst-order correction þ

(b) Express as multiplied by a series in .

change the estimate ?

31. A thin disk of radius and mass lies horizontally; a ries but no higher terms. How far above the surface

particle of mass is at a height directly above the cen- of the earth can you go before the ﬁrst-order correc-

ter of the disk. The gravitational force, , exerted by the

u tion changes the estimate by more than

disk on the mass is given by

u Þpp

? (Assume ó km.)

á !aG

e f

¥ 33. (a) Estimate the value of using Riemann

ò L

F sums for both left-hand and right-hand sums with

ð subdivisions.

J4K

FH< F

¢u¤

GVL©á Assume and think of as a function of , with (b) Approximate the function with a Tay-

the other quantities constant. lor polynomial of degree 6.

F þ (c) Estimate the integral in part (a) by integrating the

(a) Expand as a series in . Give the ﬁrst two Taylor polynomial from part (b). nonzero terms. (d) Indicate brieﬂy how you could improve the results

(b) Show that the approximation for obtained by us- in each case. ing only the ﬁrst nonzero term in the series is inde-

pendent of the radius, F . 34. Use Taylor series to explain how the following patterns

F$ y

y otÞy íy Þ Q ¤#¤1¤

imation in part (a) differ from the approximation in ¤

ó ó

(a) ô (b)

¤ ¤

üpô ütü

p tpíttÞp

part (b)?

¤ ¤#¤$¤

õ ó ö 10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS

In order to use an approximation intelligently, we need to be able to estimate the size of the error, which is the difference between the exact answer (which we do not know) and the approximate

value.

]¨/2143 .0/@143

When we use , the §=²2³ degree Taylor polynomial, to approximate , the error is the

difference

¨/@143D^.0/@1=30A ]¨/2143?C

¨ ¨ s

If s is positive, the approximation is smaller than the true value. If is negative, the approxima-

tion is too large. Often we are only interested in the magnitude of the error, .

] /@143

§ ¨

Recall that we constructed so that its ﬁrst derivatives equal ® the corresponding deriva-

¨ ¨

.0/@1=3 ot ] /@1=3 /2XN3D X /2XN3D X /2XN3DãX />XM3DãX

< < <

¨ ¨

s s s ¨ä s

_z® ¨ä

tives of . Therefore, , , , , _z® . Since is

/ :hy3 /@143ÃDZ. /2143

¨ä

_z®

§ ² s

an §=²2³ degree polynomial, its derivative is 0, so . In addition,

. /@1=3 1 X

suppose that is bounded by a positive constant , for all positive values of near ,

X´ê(1ê

V V V

say for V , so that

¨nä

_x®

A ê£. /2143{ê X´ê1£ê C

for

g ¨ä

This means that _z®

/@1=3{ê A ê X$ê;1£ê C

for g

468 Chapter Ten APPROXIMATING FUNCTIONS

X 1

Writing for the variable, we integrate this inequality from to , giving

Ö Ö Ö

¨ä

_x®

êk AOk êk / 3 X´ê1£ê k

for

g g g g ¨

so ®

A 1£ê /@143sê 1 X´ê(1ê C

for

g 1

We integrate this inequality again from X to , giving

Ö Ö Ö

êk AHk êk / 3 X´ê(1ê k

l l s

for

g g g g

h h

¨}©

so _z®

A 1 ê /@1=3sê 1 X´ê(1ê C

~ ~

for g

By repeated integration, we obtain the following estimate:

h h

¨ä ¨ä

_ _

A 1 ê ¨/@1=3{ê 1 X´ê1£ê k

/ :^hp3?¯ / :hy3u¯

s for

§ §

which means that

¨nä

¨ ¨

1 Xûê1£ê C /@143 D .0/21430A ] /@1=3 ê

/ :^hp3u¯

for

A êj1£êiX 8 DØX When 1 is to the left of 0, so , and when the Taylor series is centered at ,

similar calculations lead to the followingg result:

¡$#

Theorem 10.1: Bounding the Error in ¡

]¨/@143 .0/2143 8

If is the §=²2³ Taylor approximation to about , then

¨ä

¨/@1=3 D .0/@1=30AB]'¨/@1=3 ê 1\A 8 k

/ :^hp3?¯

¨nä

_x®

D . 8 1

where max on the interval between and .

Using the Error Bound for Taylor Polynomials

ÕyÖ ¬

Example 1 Give a bound on the error, s , when is approximated by its fourth-degree Taylor polynomial for

lûê;1£ê£XVC l

AvXVC .

.wº /@1=36DÕyÖ ÕyÖ

Solution Let .0/2143D[ÕyÖ . Then the ﬁfth derivative is . Since is increasing,

® ¢

¢º º

. /2143 êÕ D Õ5~ AvXVC lûê(1ê£XVC l}C

Í for

Thus ~

¬ D .0/@1=30AB]¬}/@143 ê 1 C

l9¯

lûê;1£ê£XVC l

This means, for example, that on AvXVC , the approximation

« ¬

1 1 1

Õ 5h:1$: : :

~V¯ ªV¯ |m¯

/2XmC lM3xºA5XVC XnXnX C

has an error of at most _ 10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS 469

The error formula for Taylor polynomials can be used to bound the error in a particular numer-

ical approximation, or to see how the accuracy of the approximation depends on the value of 1 or

/ :^hp3

§ §

the value of . Observe that the error for a Taylor polynomial of degree depends on the §

¨ä _

power of . That means, for example, with a Taylor polynomial of degree § centered at 0, if we ~ decrease 1 by a factor of 2, the error bound decreases by a factor of .

Example 2 Compare the errors in the approximations

h h

¢ _ ¢

Õ 5jh:XmCwh: /2XVC¢hp3 Õ 5jh:[/2XmC XNln3: />XVC XMlM3 C ~V¯

~9¯ and 1jD¦XmCwh

Solution We are approximating ÕyÖ by its second-degree Taylor polynomial, ﬁrst at , and then at

1RD^XVC XMl 1

« . Since we have decreased by a factor of 2, the error bound decreases by a factor of about D[r

~ . To see what actually happens to the errors, we compute them:

¢ _

Õ A£eh:XmCwh6: /2XmCwhy3 f[DihnC¢htXNl9h h6AhnC¢htXNlXnXMXUD^XVC XnXnXmh h

Ë Ë

~9¯

e f

Õ A h:(XVC XMls: />XVC XMlM3 DihnC XMlVhp~ h6AhnC XMlVhp~nlnXUD^XVC XnXnXMXM~Vh

~9¯

/2XmC XMXnXmh hy3-·V/2XVC XnXMXnXN~9hp3D[rVC¢h Since Ë , the error has also decreased by a factor of about 8.

Convergence of Taylor Series for cos ) doeMf91

We have already seen that the Taylor polynomials centered at 1%D[X for are good approxima- 1 tions for 1 near 0. (See Figure 10.16.) In fact, for any value of , if we take a Taylor polynomial centered at 1D[X of high enough degree, its graph is nearly indistinguishable from the graph of the

cosine near that point.

K K

Ô Ô

¡ ¡

GVL GVL

G G

p p

K K

¡ ¡

GVL GVL

G G Figure 10.16: Graph of p and two Taylor polynomials for near

Let’s see what happens numerically. Let 1D7¶·~ . The successive Taylor polynomial approxi- 1D[X

mations to d?eNfo/@¶·n~n3D[X about are

]N/@¶·n~n3D hsA/2¶·~M3 ·n~9¯ DjAvXmC ~nªnª XsCoCtC

]¬}/@¶·n~n3DihsA/2¶·~n3 ·n~9¯t:[/@¶·~M3 ·y|±¯MD XVC XVhpqnq CoCtC

] /@¶·n~n3D oo DjAvXmC XMXnXnrMqsCoCtC

] /@¶·n~n3D oo D XVC XnXMXnXN~{CoCtCC Ì 470 Chapter Ten APPROXIMATING FUNCTIONS

It appears that the approximations converge to the true value, d?eNfo/@¶·n~n3D[X , very rapidly. Now take

X 1RD^¶ d?eNf9¶DjAh

a value of 1 somewhat farther away from , say , then and

] /2¶36DihsA/2¶3 ·~9¯}D AvªmC qMª|NrnXsCtCoC

]¬M/2¶36D ot D XmCwhy~ªMqVhCtCoC

] /2¶36D ot D AhMC ~VhnhpªMl{CtCoC

] /2¶36D ot D AvXmC q XM~{CtCoC

]0_xM/2¶36D ot D AhMC XMXVhprnªsCtCoC

] n/2¶36D ot D AvXmC qMqnqMqnXsCtCoC

] /2¶36D ot DjAhMC XMXnXnXMX|vCtCoC

_x¬ ho|

We see that the rate of convergence is somewhat slower; it takes a ²2³ degree polynomial to approx-

doeMf9¶ r d?eNfo/@¶·n~n3 1

imate as accurately as an ²>³ degree polynomial approximates . If were taken still farther away from X , then we would need still more terms to obtain as accurate an approximation of

d?eNf91 . 1 Using the ratio test, we can show the Taylor series for d?eMfV1 converges for all values of . In addition, we will prove that it converges to doeMf91 using Theorem 10.1. Thus, we are justiﬁed in

writing the equality:

1 1 1±» 1±Ì

doeMf91DihsA 1 : A : A;to

~V¯ |m¯ ¯ rV¯

Ê for all . å Showing the Taylor Series for cos å Converges to cos

The error bound in Theorem 10.1 allows us to see if the Taylor series for a function converges to d?eMfV1

that function. In the series for , the odd powers are missing, so we assume § is even and write

1 1

¨ ¨ e f

:^otp:[/xAhp3 /@143D^doeMf91\A ] /@143D^doeMf91\A hsA k

~9¯ / 3u¯

§ ¨

giving

1 1

d?eMfV1Djh:1$: :^otp:^/zAhp3 : ¨/@1=3uC

~9¯ / 3?¯

doeMf91 Ú Û /21436Ú X §

Thus, for the Taylor series to converge to , we must have s as .

Showing 0 as _z®

doeMf}1 f-¢¥1 .0/@143vDid?eNf91 / :[hy3 . /@143

¨nä

_x®

§ ²

Proof Since , the § derivative is or , no matter what is.

. /2143 êEh X 1

So for all § , we have on the interval between and .

By the error bound formula in Theorem 10.1, we have

¨ä

¨ ¨

/2143 D d?eMfV1\AB] /2143 ê C

/ :^hp3?¯

s §

for every §

To show that the errors go to zero, we must show that for a ﬁxed 1 ,

¨ä

Ú X Ú ÛC

/ :hy3u¯

as §

§ 1

To see why this is true, think about what happens when § is much larger than . Suppose, for

1RDh C ª Ë

example, that . Let’s look at the value of the sequence for § more than twice as big as 17.3,

Dª Dª D^ªMr

§ §

say § , or , or :

Dª /xh C ªM3

ª ¯ Ë

For § :

h h C ª h

« «

Ì ¼

Dª /xh C ªM3 D /xh C ªN3 k

Ë Ë Ë

ªnrm¯ ªMr ª ¯ Ë

For § :

h h C ª h C ª h

Ë Ë

« «

. ¼

Dªnr /xh C ªM3 D /xh C ªM3 k CtCoC

Ë Ë

ªnqm¯ ªMq ªnr ª ¯ Ë For § :

10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS 471

h C ªN·aª

_ _ «

Since is less than , each time we increase § by 1, the term is multiplied by a number less

/zh C ªN3 ¼

¨ä

_ 1®

than . No matter what the value of ¼ is, if we keep on dividing it by two, the result gets

/zh C ªM3

¨nä

_x®

closer to zero. Thus goes to 0 as § goes to inﬁnity.

h C ª 1 : ~ 1

§ We can generalize this by replacing by an arbitrary . For , the following

sequence converges_ to 0 because each term is obtained from its predecessor by multiplying by a

¨nä ¨nä ¨ä «

number less than : _

1 1 1

k k koCtCoCoC

/ :hy3u¯ / :(~M3u¯ / :ªN3u¯

§ § §

Therefore, the Taylor series hsAB1 does converge to . ÕpÖ Problems 16 and 15 ask you to show that the Taylor series for fz¢s1 and converge to the

original function for all 1 . In each case, you again need the following limit:

é¢°¯ D^XmC

¨±³²

¯ §

Exercises and Problems for Section 10.4 Exercises

Use the methods of this section to show how you can estimate

the magnitude of the error in approximating the quantities in

L í

Ñ ù Ñ ¤

Problems 1–4 using a third-degree Taylor polynomial about ¤

Problems

J4K K

¡ à

L©á L

p õ

5. Suppose you approximate by a Taylor polyno- (a) The error in approximating õ by and by

õ1µ õ

mial of degree about on the interval ´ . . G

(b) The maximum error in approximating t by

(a) Is the approximation an overestimate or an underes- ¡

GVL G >( ¶

for ¶ .

timate? ¡¥

G G9L p (b) Estimate the magnitude of the largest possible error. (c) If we want to approximate by to an ac-

Check your answer graphically on a computer or cal- curacy of within 0.2, what is the largest interval of G culator. -values on which we can work? Give your answer to the nearest integer.

6. Repeat Problem 5 using the second-degree Taylor ap-

L á

K K

¡ ¡

à à

GVL

proximation, , to . GVL

LJ J

7. Consider the error in using the approximation bÐÒÑ

Qs u

í µ on the interval ´ . (a) Where is the approximation an overestimate, and

where is it an underestimate?

Q Q6í í ?

(b) Estimate the magnitude of the largest possible error. Qsu ó Check your answer graphically on a computer or cal- ó

culator.

Q6í

bÐÒÑLJ J

K K

¥ ¥

¡ à ¡ à G9L

8. Repeat Problem 7 for the approximation GVL

í þ

J .

â G

9. Use the graphs of t and its Taylor polynomi-

K K

¡ à ¡ à GVL als, GVL and , in Figure 10.17 to estimate the Figure 10.17 following quantities.

472 Chapter Ten APPROXIMATING FUNCTIONS

¥ ¸·º¹ 10. Give a bound for the maximum possible error for the ð What shape is the graph of ? Use the graph to

degree Taylor polynomial about GY approximating conﬁrm that

G u G

t bÐÓÑ

on the interval ´ . What is the bound for ?

>G QR {>G?>â

¤ ¤ ¤ ¶

¶ for

K K

¢ ¢

11. What degree Taylor polynomial about do you (b) Let þ . Choose

Q6 ¼>BGY> ¤

need to calculate to four decimal places? To six a suitable range and graph for ¤ . decimal places? Justify your answer using the results of What shape is the graph of ? Use the graph to

Problem 10. conﬁrm that

Î ¥

12. (a) Using a calculator, make a table of the values to four

>G QR {>G?>â

¤ ¤ ¤

bÐÒÑ ¶

decimal places of for ¶ for

GQ6 Q6 Þ Q6 Þ

¤ ¤$¤#¤ ¤ ¤ ¤#¤#¤ ¤ ¤

¤ (c) Explain why the graphs of and have the õ õ , , , , , , , , .

shapes they do.

(b) Add to your table the values of the error

G >â

GQG G

bÐÓÑ

¤ ¶

for these -values. 14. For ¶ , graph the error

¡=à

GQ G9L GQ

p p

GQG

quantity bÐÒÑ showing that

â pí Q% >G->B

< Explain the shape of the graph, using the Taylor expan-

¤ ¤ ¤ ¤

¶ ¶ õ õ

for

G G >â

¶ ¶ ¶

sion of . Find a bound for ¶ for .

¢ ¢

á á ·º¹

13. In this problem, you will investigate the error in the ð 15. Show that the Taylor series about 0 for converges to

G GVL'½Y

degree Taylor approximation to for various values of for every . Do this by showing that the error 2

½y¾ ð

. as ð .

K K

¥ ¥

¢ ¢

(a) Let . Using a cal- 16. Show that the Taylor series about 0 for bÐÒÑ converges

Q6 »>âG>â G G

bÐÓÑ ¤ culator or computer, graph for ¤ . to for every .

REVIEW PROBLEMS FOR CHAPTER TEN

Exercises

J pKJ bÐÒÑ¿

For Problems 1–4, ﬁnd the second-degree Taylor polynomial

6. 7. 8. 9.

QÞ| ÞsQRG

about the given point.

á G G/ G;

1. 2. Ñ 3.

G/ G

bÐÒÑ

Q Þ 4þ

For Problems 10–11, expand the quantity in a Taylor series

ù ÑLJ J

4. ÷ , around the origin in terms of the variable given. Give the ﬁrst Þ

4þ four nonzero terms.

J4K

GVL

5. Find the third-degree Taylor polynomial for v

F ÿ

G ò G Q Gò© G

ö õ

at . u Q

10. in terms of 11. in terms of

Fsò£ÿ F In Problems 6–9, ﬁnd the ﬁrst four nonzero terms of the Taylor series about the origin of the given functions.

Problems

12. Find an exact value for each of the following sums. 18.

K K K

p pL ò p pL ò p pL6ò ò ò

¤ ¤ ¤

ö ö ö ö

(a)

y tL

mQ ò Q ò Q òÞ=ò´ mò"}ò ò ò

13. 14. ô

í Þ

ö ö

ü ö ô

ò¡010#0?ò

010#0

àbà

K .

y tL p pL

0#010?ò

¤ ¤

K K

?L ?L

¤ ¤

ö ö

ò ?L ò ò ò40$0#0

¤ ¤ ö

(b) ö

í"

Þ íp ?

ô ó ô ô

Q ò òA0$0#0 mQ ò Q ò-0;010

15. yQ¥ yò 16.

" í Þ í ö Find the exact value of the sums of the series in Problems 13– õ

REVIEW PROBLEMS FOR CHAPTER TEN 473

í í í

ò ò ò ?L Q ò

íò"íò (b) The force on the particle at the point is given by ¤

17. 18. S2K

" í Þ í"

Q GVL G

¢ Ô

K K

í . For small , show that the force on the par-

uL uL

¤ ¤

òP0$010

ò0$0#0

Q ticle is proportional to its distance from the origin.

õ ö õ What is the sign of the proportionality constant? De-

scribe the direction in which the force points.

J J4K JNS>K JNS STK

ípLûY íyL ípL

19. A function has , õ and 28. The theory of relativity predicts that when an object

J4K

Qsu í uL

. Find the best estimate you can for ¤ . moves at speeds close to the speed of light, the object G 20. Suppose is positive but very small. Arrange the follow- appears heavier. The apparent, or relativistic, mass, ,

ing expressions in increasing order: of the object when it is moving at speed is given by the

formula

G/ G/ ?ò¥GVL; pQ G/ ¦á QsU G/ G `QG

bÐÓÑ Ñ t ù Ñ

÷?ø ÷

where is the speed of light and is the mass of the

21. By plotting several of its Taylor polynomials and the

J4K K

GVLR ò£GVL þ object when it is at rest.

function , estimate graphically the

interval of convergence of the series expansions for this (a) Use the formula for to decide what values of are

function about G . Compute the radius of conver- possible.

gence analytically. (b) Sketch a rough graph of against , labeling inter-

ò£Gò£G L=QG

Ñ cepts and asymptotes.

Ð¨ ¤

22. Use Taylor series to evaluate à G

bÐÓÑ (c) Write the ﬁrst three nonzero terms of the Taylor se-

L J

bÐÒÑ ries for in terms of .

23. (a) Find à . Explain your reasoning. (d) For what values of do you expect the series to con-

L J

bÐÓÑ verge?

J4K L$

(b) Use series to explain why J looks

J like a parabola near . What is the equation of 29. The potential v energy, , of two gas molecules separated

the parabola? by a distance is given by

J4K

v v

L á ©

¢ ¥ à

24. (a) Find the Taylor series for about . à

e f

v v

¼Q QQ Q

(b) Using your answer to part (a), ﬁnd a Taylor series

S S

expansion about G for

à à k

where and are positive constants.

á!

v v

(a) Show that if , then takes on its minimum

v value, . v

Q L

(b) Write as a series in up through the

ò ò ò ò ò40$010a v

quadraticv term.

K K K

í Þ " L í" L Þ L

à ó

õ (c) For near , show that the difference between

and its minimumv value is approximately pro-

Q L

25. (a) Find the ﬁrst two nonzero terms of the Taylor series portional to . In other words, show that

J4K

G9L Þ{QG G©

à à

Q L ò

for about Q

v is approximately proportional

à L (b) Use your answer to part (a) and the Fundamental to Q .

Theorem of ¥ Calculus to calculate an approximate

v v v

(d) The force, , between the moleculesv is given by

¦Q¿! !

Þ{QG !aG

value for . v . What is when ? For

nearv , show that is approximately proportional

G\^

bÐÒÑ

à L (c) Use the substitution to calculate the ex- to Q .

act value of the integral in part (b). G bÐÓÑ 30. The gravitational ﬁeld at a point in space is the gravita-

26. (a) Find the Taylor series expansion of ÷oø .

(b) Use Taylor series to ﬁnd the limit tional force that would be exerted on a unit mass placed

ù Ñ ÷

÷oø there. We will assume that the gravitational ﬁeld strength

G½ ¤

as !

G bÐÓÑ

at a distance away from a mass is ÷oø

27. A particle moving along the G -axis has potential energy

G GVL

at the point given by . The potential energy has a !

minimum at G .

(a) Write the Taylor polynomial of degree 2 for about where is constant. In this problem you will investigate G\£

. What can you say about the signs of the co- the gravitational ﬁeld strength, , exerted by a system

efﬁcients of each of the terms of the Taylor polyno- consistingv of a large mass and a small mass , with a mial? distance between them. (See Figure 10.18.)

474 Chapter Ten APPROXIMATING FUNCTIONS

34. (Continuation of Problem 33) You may remember that

P O P O

u v the Second Derivative test tells us nothing when the second derivative is zero at the critical point. In this problem Figure 10.18 you will investigate that special case.

Assume has the same properties as in Problem 33,

S S yLÉj

and that, in addition, . What does the Taylor

¡ mum or minimum at ?

strength, , at the point . u

v 35. Use the Fourier polynomials for the square wave

(b) Assuming isv small in comparison to , expand

Qs Q <G?>â

J4K

GVLÁÀ

in a series in . v

=<G?>

K L

explain why you can view the ﬁeld as resulting from to explain why the following sum must approach =þ as

½y¾ a single particle of mass , plus a correction ð :

term. What is the position of the particle of mass

2UT

ò `Q Q ò¡0#010?ò Qs?L

? Explain the sign of the correction term.

í ò

ð õ ö

J4K K

Gvò L Gsò L

31. Expand and in Taylor series and take

a limit to conﬁrm the product rule: J4K GVL$

36. Find a Fourier polynomial of degree three for

á ¼>G?<(

S S

K K J K K J4K

K2J4K , for .

GVL GVL GVLbL4 GVL GVL9ò GVL

J4K

!aG

37. Suppose that G9L is a differentiable periodic function

of period . Assume the Fourier series of is differen-

tiable term by term.

J4K K

ò L Gò L

32. Use Taylor expansions for M and to con-

Â ÿ1Â

(a) If the Fourier coefﬁcients of are F and , show J ﬁrm the chain rule: S

that the Fourier coefﬁcients of its derivative are

ÿ#Â Q FÂ

M M

S S

K K J K K

K2J4K and .

GVL GVLbLbL GVLbL/0

aG ! (b) How are the amplitudes of the harmonics of and

JNS related?

J J

Gû

Â ÿ1Â 33. Suppose all the derivatives of exist at and that F

38. If the Fourier coefﬁcients of are and , and the

G© ! Â

has a critical point at . 1Â

% &

Fourier coefﬁcients of are and , and if and

G©

% &»

·º¹

(a) Write the ð Taylor polynomial for at . are real, show that the Fourier coefﬁcients of

FÂ`ò #Â ÿ1Â`ò ! Â

& % &

(b) What does the Second Derivative test for local max- are % and .

ima and minima say? 39. Suppose that is a periodic function of period and

J K J4K

GûòP uL

that is a horizontal shift of , say . ond Derivative test works. J Show that and have the same energy.

PROJECTS

Exercises

¶ A dj /zhy·n~ªnqN3 h

1. Machin’s Formula and the Value of h i to show that h

(a) Use the tangent addition formula hp~nX

d1j 9e fA dj 9e f[D d1j hnC

hnhpq ~nªnq

h"i h h i h h"i h

j »Ã:Pj »Å

j /ÄÃ:GÅ$3D

h h

ÃEDÅD d1j /xha·ln3

hsANj CÃ9j »Å

h h

(b) Use h"i to show that

h h

l h

f e f e

D d1j C ~ dj

l hp~

Ã D d1j /xhy~X}·}hnhpqM3 Å D

h"i h h i h h with h"i and

PROJECTS 475

Ú X

Use a similar method to show that Notice that as Æ , the error also goes to zero,

h hp~X

provided is bounded.

.0/@143DÕ 1 D

l hnhtq

h h"i h

h"i As an example, we take and

X . /21 3´D¦h

, so < . The error for various values of Æ

Æ are given in Table 10.2. We see that decreasing

¶·y| D

| d1j /xhy·nln3'A d1j /xha·~ªMqM3

·~

h"i h h"i h

. about 10, as predicted by the error bound ÊÆ . (d) Use the Taylor polynomial approximation of

degree 5 to the arctangent function to ap- ¶ proximate the value of . (Note: In 1873 Table 10.2

William Shanks used this approach to calculate

K2J4K J4K

à à

G ò L=Q G LbL

to 707 decimal places. Unfortunately, in 1946 þ Error

lM~r

u" ÎÍ"u

²>³

¤ ¤

ö õ ö

it was found that he made an error in the õ

u t y y =Í

place.)

¤ ¤

õ õ Î

(e) Why do the two series for arctangent converge ß

u tp AÍ

¤ ¤

õ õ

so rapidly here while the series used in Exam- ß

u tpt AÍ

¤ ¤ õ ple 5 on page 462 converges so slowly? õ 2. Approximating the Derivative3 In applications,

the values of a function .0/@1=3 are frequently known

1 1 1 ~ CoCtC?C

Æ only at discrete values , ÇÆ , , (a) Using Taylor series, suggest an error bound for

Suppose we are interested in approximating the each of the following ﬁnite-difference approxi-

. /@1 3

derivative . The deﬁnition mations.

.0/21mp3`A .0/21mvA 3

. /21 35

.=<>/21ma36DéwÈ¯

Æ ±

(i)

.=<>/21ma35

Æ Æ

. /@1=3

< ~

suggests that for small Æ we can approximate (ii)

3:rN.0/@1m{: 3'ABrN.0/@1±vA 3:.0/@1±

as follows: AÉ.0/21m:;~

. /21ma35

Æ Æ Æ

.0/21m¥: 3'A .0/21ma3

(iii) hp~

. /21ma35 C

Æ Æ

Æ (iv) Use each of the formulas in part (a) to

ÕyÖ

« ¬

Such ﬁnite-difference approximations are used fre- approximate the ﬁrst _ derivative of at 1èD¹X DihtX kohpX kohpX kohtX

quently in programming a computer to solve differ- for Æ .

ential equations. As Æ is decreased by a factor of 10, how Taylor series can be used to analyze the error does the error decrease? Does this agree in this approximation. Substituting with the error bounds found in part (a)?

Which is the most accurate formula?

. /21ma3

< <

.0/@1m{: 36D^.0/21ma3:;. /@1±y3 : :to .0/@1=3^D ha·y1

Æ Æ

Æ (v) Repeat part (b) using and

htX º

1mBD . Why are these formulas not

. /21 3 into the approximation for < , we ﬁnd good approximations anymore? Continue

to decrease Æ by factors of 10. How small

.0/21m¥: 3`A .0/21mp3 . /21ma3

< <

D7. /@1ma3: :^ot

< Æ

does Æ have to be before formula (iii) is

~ Æ

Æ the best approximation? At these smaller

This suggests (and it can be proved) that the error values of Æ , what changed to make the for-

in the approximation is bounded as follows: mulas accurate again?

]`_xN/2143

3. (a) Use a computer algebra system to ﬁnd

Æ ÊÆ

_bM/@143

q V

V and , the Taylor polynomials of degree

1RD[X fz¢ 1 doeMf 1

V V V

V 10 about for and . V whereV (b) What similarities do you observe between the

two poynomials? Explain your observation in

. /2143 ê e 1\AB1m ê C

< <

Ë i ÆÌ terms of properties of sine and cosine. 3From Mark Kunka 476 Chapter Ten APPROXIMATING FUNCTIONS

4. (a) Use your computer algebra system to ﬁnd the Taylor polynomial of degree 10 about 1âD

/@1=3 ] /2143 X .

¼ ¼

and q , the Taylor polynomials of for .

.0/@1=3%D fz¢s1

degree 7 about 1D¦X for and (b) What do you notice about the degrees of the

/@1=3D^f-¢s1Éd?eNf91 . « c . terms in the polynomial? What property of (b) Find the ratio between the coefﬁcient of 1 in does this suggest?

the two polynomials. Do the same for the coef- (c) Prove that . has the property suggested by part 1±¼

ﬁcients of 14º and . (b).

/@143D f-w/ 3

6. Let n . g

puted in part (b). Explain it using the identity ]`_-_n/@1=3 f-¢/>~143D7~`fz¢{1Éd?eNf91 (a) Use a computer algebra system to ﬁnd ,

. 1âD

Ö Ö

.0/2143¥D . :

_ the Taylor polynomial of degree 11 about

X /@1=3

5. Let ÏpÐ . Although the formula for

1£D7X . , for n .

is not deﬁned at , we can make continuous

h .

.0/2XN3D (b) What is the percentage error in the approxima- ]0__a/zhp3

by setting . If we do this, has Taylor /xhy3 1D[X

tion of n by ? What about the ap-

/>~M3 ] />~M3

series around . _-_

] /@1=3

_b proximation of by ? (a) Use a computer algebra system to ﬁnd ,