Download Chapter Ten APPROXIMATING FUNCTIONS

Chapter Ten
APPROXIMATING
FUNCTIONS
In Section ??, for
, we found the sum
of a geometric series as a function of the
common ratio :
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
1DE8
The tangent line and the curve share the same slope at
. As Figure 10.1
suggests,
the tangent
1
8
line approximation to the function is generally more accurate for values of close to .
True value
J4K GVL
Approximate value of
J4K GVL
Tangent line
W
H
JNSTK FML K GUQFNL
H
P
GQRF
O IH
J4K FML
J4K FNL
I
I
F
G
G
Figure 10.1: Tangent line approximation of
J4K GVL
for G near F
8(DYX
We first focus on 1RD[X. The tangent line approximation at
Taylor approximation at
, or as follows:
Taylor Polynomial of Degree 1 Approximating
1DZX
\! $#
is referred to as the first
for
near 0
.0/@1=35^]`_a/21436D7.0/2XN3:(.=<T/2XN3b1
We/@1=now
look at approximations
by polynomials of higher degree. Let us compare approxima36D[d?eMfV1
1D[X
tions c
near
by linear and quadratic functions.
/21436D^d?eNf91
Example 1
Approximate c
Solution
The tangent line at
1RD^X
, with
1%DXVC XMl
in radians, by its tangent line at
is just the horizontal line g
c
If we take
1
, then
c
/21436D[d?eNf915jhMk
for
.
, as shown in Figure 10.2, so
1
near 0.
/2XmC XNln3`DdoeMfp/2XVC XMlM3`DXVC qnqMrsCoCtCuk
which is quite close to the approximation
c
Dih
1D[X
doeMf915ih
. Similarly, if
1DjAvXmCwh
/xAvXmCwhy30D[d?eNfo/zAvXVChp3`D^XmC qMqMl{CtCoC
, then
447
10.1 TAYLOR POLYNOMIALS
d?eNf91R5jh
is close to the approximation
. However, if
1RD^XVC |
, then
c
/2XmC |}3DdoeMfp/2XVC |N3`D[XVC qM~VhCoCtCuk
d?eMfV15h
so the approximation
is less accurate. The graph suggests that the farther a point
X
away from , the worse the approximation is likely to be.
‚
is
‚ƒ„
…‡†pˆ G
P
Q
1
G
€
€
Qs
Figure 10.2: Graph of …‡†pˆ
G
and its tangent line at G‰ƒŠ
Quadratic Approximations
To get a more accurate approximation, we use a quadratic function instead of a linear function.
/@143‹D^doeMf91
1
X
Example 2
Find the quadratic approximation to c
Solution
To ensure
that the quadratic,
, is a good approximation to c
at
, we require
doeMf91
that
and
the
quadratic
have
the
same
value,
the
same
slope,
and
the
same
second
derivative
at
1RD^X
]'Œn/>XM3D
/>XM3 ] Π< />XM3D
] Π< < /2XN3D
< /2XN3
< < />XM3
c
c
c
. That is, we require
,
, and
. We take the quadratic
Œ
polynomial
] Œ /@1=36D[‹:;s_‘1$:; Œ 1
k
for
near .
/@1=3DdoeMf91
]'ŒM/2143
and determine


,

_
, and
. Since
Œ /@1=36D76{:(s_?1$:(
]
]
Œ < /@1=36D7{_6:(~M
]
we have
Œ
Œ < < /@1=36D7~n
‹UD[]
Π/2XN3D
s_vD[]
Π< /2XN3D
~n
Œ
D[]
1ŽDX
Π< < />XM3D
Π1
Œ
and
Π1
< /@143DAf-’w“s1
c
Œ
< < /@143DAd?eNf91”k
c
/2XN3DdoeMfVX‰Djh
c
/@1=36DdoeMf91
c
‹UDjh
so
< /2XM3DjAfz’“vXD^X
c
c
s_vD^X
< < /2XN3DAd?eNfVXDjA•hMk
Œ

DjA
_
Œ
C
Consequently, the quadratic approximation is
doeMf91R5^]
Œ /@1=36Dih‹:X–?1"A
~
h
1
Œ
DihsA
Œ
1
~
k
for
1
near 0.
doeMf915]—ŒM/@1=3
Figure 10.3doeMsuggests
that the 1 quadratic approximation
is better than the linear
f91˜5E] _ /@1=3
approximation
for
near
0.
Let’s
compare
the
accuracy
of
the
two] Œapproximations.
]`_a/@1=3D™h
1
1BDšXVC |
doeMft/2XVC |N3D›XmC qN~9hCtCoC
/2XmC |}3vDšXVC qM~œX
Recall that
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
¡—¥ K G9L”ƒ„

Q‰ ž
…‘†tˆ G
Q
O
ž
QŸ
€
Ÿ
G
€
Qs
P
G
Figure 10.3: Graph of …‘†pˆ
and its linear,
¡—¥ K G9L
ž K GVLƒ„Q£¢už ¤
¡
, and quadratic,
ž K GVL
¡
, approximations for G near 0
Generalizing the computations in Example 2, we define the second Taylor approximation at
.
1RD^X
\! $#
Taylor Polynomial of Degree 2 Approximating
Π/@1=36D7.0/2XM3:;.
.0/214365[]
.
< /2XM3x1$:
< < /2XN3
~
for
near 0
Œ
1
Higher-Degree Polynomials
1šD¦X
In a small interval around
, the quadratic approximation to a function is usually a better
approximation than the linear (tangent line) approximation. However,
Figure 10.3 shows that the
1
quadratic can still bend away from the original function for large . We can attempt to fix this
by
.0/2143
using
an
approximating
polynomial
of
higher
degree.
Suppose
that
we
approximate
a
function
1
for near 0 by a polynomial of degree § :
.0/21435[]—¨/21436D7
:(

_ 1$:;‹Œo1
Œ
:^–o–t–y:;¨}©
_ 1
Œ ktCoCtC?k‹
‹nk{_œk6
¨}©
_
¨
:(¨91
¨
C
We need.0to
find the values of the constants:
. To do this, we] require
that
¨ /@143
/@1=3
the function
and
each
of
its
first
at
the
§ derivatives agree with those of the polynomial
1D[X
1RD[X
point
. In general, the more derivatives there are that agree at
, the larger the interval on
which the function and the polynomial remain close
to each other.
Dª
To see how to find the constants, let’s take §
as an example
Π1
.0/214365^]—«N/@1=36D[‹:;s_‘1$:;
Substituting
1%D^X
Differentiating
so substituting
gives
]'«N/@143
1RDX
yields
.0/>XM3D[]
« /2XN3D[
]•« < /@143‹D^s_:;~n
shows that
.
< /2XN3D]
Œ
:;‹«o1
 C
Œ 1$:(ªM«o1
Œ
k
« < />XM3D7{_aC
Differentiating and substituting again, we get
]
which gives
« < < /@1=36D7~U–œhp
.
< < /2XM3D[]
Œ
:(ª–t~U–nhy«?1”k
« < < />XM36D[~U–nhy‹Œnk
«
C
449
10.1 TAYLOR POLYNOMIALS
so that
Œ

]
The third derivative, denoted by
«< < <
.
D
< < /2XN3
C
~U–œh
, is
]•« < < < /@1=36D^ª–p~U–nhp«nk
so
.
< < < /2XM3D^]
« < < < />XM3D[ª–t~U–nhy‹«nk
and then

.
D
«
< < < /2XN3
C
ª–t~U–nh
]—¬M/2143
You can imagine a similar calculation starting with
¬‘®
which would give
.—­
/>XM3

D
¬
, using the fourth derivative
.—­
¬‡®
,
k
|•–tª–t~U–Mh
and so on. Using factorial notation,1 we write these expressions as
.
«D
Writing
.—­
¨ ®
< < < /2XM3
ªV¯
¬‘®
.—­
k°‹¬D
/2XN3
C
|±¯
.
for the §=²2³ derivative of , we have, for any positive integer §
.—­
¨´D
¨ ®
1D[X
C
¯
§
So we define the §=²>³ Taylor approximation at
/2XN3
:
Taylor Polynomial of Degree µ Approximating
.0/@1=365^]
¨
¨
for
near 0
/@1=3
D[.0/>XM3:;.
]
"! $#
< />XM3x1$:
.
< < /2XN3
~9¯
Œ
1
:
.
< < < />XM3
ªm¯
1
«
.—­
:
¬‘®
/2XM3
|m¯
/2143
We call1RD^X
the Taylor polynomial of degree § centered at
.
about
1
1D^X
¬
:^–o–t–y:
.—­
¨ ®
§
/2XN3
¯
.0/@1=36D^f-’“s1
Construct the Taylor polynomial of degree 7 approximating the function
.
1D¶—·œª
Compare the value of the Taylor approximation with the true value of at
.
Solution
We have
f-’w“s1
< /@143‹D
doeMf91
.
.
< < < /@143‹D¹Ad?eMf}1
¬‘®
/@143‹D
®
.—­wº
/@143‹D
®
.—­w»
.—­w¼
1 Recall
.
< < /@143‹D¸A%f-’“1
.
.—­
.0/2XM3D
giving
®
f-’w“s1
doeMf91
.—­wº
.—­w»
/@143‹DjAd?eNf}1”k
.—­w¼
Âu¾t¿ŽÂ
, and Èp¾t¿ŽÂ .
X
< < < /2XM3DA•h
¬‘®
/@143‹D¸A%f-’“1
that ½M¾o¿%½MÀ½‹ÁÃÂ-ÄMÅxÅxÅÇÆ0Åu . In addition,
h
< < /2XM3D
.
.—­
X
< /2XM3D
.
/2XM3D
®
X
/2XM3D
®
®
/2XM3D
¨
, or the Taylor polynomial
Example 3
.0/@143‹D
1
h
X
/2XM3DiA•hnC
for
1
near 0.
450
Chapter Ten APPROXIMATING FUNCTIONS
Using these values, we see that the Taylor polynomial approximation of degree 7 is
f-’“1%5^]
¼
1
/21436D[XÉ:hs–t1$:(X–
«
1
D^1\A
.—­wÌ
®
ªV¯
º
1
:
Œ
~9¯
A
l9¯
1
¼
Ë ¯
«
1
A„hs–
k
ªm¯
1
for
¬
1
:(X–
near 0
14º
:^hs–
|m¯
:(X–
lV¯
1±»
Ê ¯
A„hs–
14¼
Ë ¯
C
/2XM3D^X
fz’“s1
Notice that since
, 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
1
near
0.
They
are
indistinguishable
where
is
close
to
0.
However,
as
we
look
at
values
degree
7
for
1
of farther away from 0 in either direction, the two graphs move
apart. To check the accuracy of this
fz’“”/@¶—·œªM3D7Í ª}·œ~•D[XVC r ÊnÊ XM~Mla|vCtCoCzC
approximation numerically, we see how well it approximates
¡=Ï K GVL
G
ˆbÐÒÑ
Q

€
¡Ï K GVL
Qs
G
Figure 10.4: Graph of ˆbÐÓÑ
G
€
Î
ˆbÐÒÑ
G
and its seventh degree Taylor polynomial,
¶—·aªDihnC Xœ| Ë htq Ë Ê
Ï K GVL
¡
, for G near Š
CtCoC
When
we substitute
into the polynomial approximation, we obtain
XVC r ÊMÊ XN~9htªvCoCoC‡k
which is extremely accurate—to about four parts in a million.
/21436D^d?eMfV1
r
1
]
¼
/2¶—·aªN3D
Example 4
Graph the Taylor polynomial of degree approximating c
Solution
We find the coefficients of the Taylor polynomial by the method of the preceding example, giving
doeMf91R5[]
]
Ì
Œ
1
/@1=36DihsA
~V¯
:
¬
1
1±»
|±¯
A
Ê ¯
:
for
1±Ì
rV¯
near 0.
C
/2143
1
Figure 10.5 shows that Ì ] Œ is/@14close
to theŒ ·œcosine
function for a larger interval of -values than
3‹DhsAB1
~
the quadratic approximation
in Example 2 on page 447.
¡=Ô K GVL
¡4Ô K GVL

…‘†tˆ G
Q
¡
Figure 10.5:
Example 5
¡
Ô K GVL
€
…‘†tˆ G
G
€
Qs
ž K GVL
approximates …‘†pˆ
Construct the Taylor polynomial of degree 10 about
G
¡
ž K G9L
better than
1RD^X
¡
ž K GVL
for G near Š
for the function
.0/@143D[ÕyÖ
.
10.1 TAYLOR POLYNOMIALS
Solution
.0/2XM3Djh
ÕyÖ
451
ÕyÖ
®
®
 derivatives are equal
WeÕyhave
. Since the derivative
of is .—
equal
to , all the .—
higher-order
DØhnk‡~VkoCtCoCukohpX
Ö
­Ù /@1=3sDÕyÖ
­Ù />XM3{DÕ
Dšh
to . Consequently, for any ×
,
and
. Therefore, the
Taylor polynomial approximation of degree 10 is given by
Õ Ö
5]`_b}/@1=36Dih‹:1$:
1
Œ
~9¯
«
1
:
ªm¯
¬
1
:
:^–o–o–y:
|±¯
_b
1
k
hpXV¯
for
1
near 0
_
ÕD[Õ
C
D[~VC Ë htrN~œrVhprM~nrsCoCtC
To check1;
the
accuracy]0of_xMthis
approximation, we use it ]`
to_bapproximate
.
DZh
/zhp3D™~9C Ë htrN~œrmhtrnXmh

Substituting
gives
.
Thus,
yields
the
first
seven
decimal
places
Õ
1
ÕyÖ
for . For large1E
values
of , however, the accuracy diminishes
because
grows faster than any
ÚÜÛ
.0/2143\DÝÕpÖ
polynomial
as
.
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
1
curve for a larger interval of -values.
¡=ß
Î
á ¢‹¡=ßâ¡
ž
¡
oŠ
ž
¡
‘Š
¥
¡
¡=à
¡=à
G
á ¢
¡
QÞ
¥
¡
Q‹ Þ
Qs‘Š
Î
Figure 10.6: For G near Š , the value of á
¢
is more closely approximated by higher-degree Taylor
polynomials
Example 6
Construct the Taylor polynomial of degree § approximating
Solution
Differentiating gives
means
h
.0/2XN3sDØh
¨
5[]
hsAB1
,
.
< /2XM3sDãh
,
/21436Dih‹:1´:1
.
< < />XM3sDi~
Œ
«
:1
.
,
:^–o–t–p:1
for
hsAâ1
< < < />XM3vDiªV¯
¬
:1
h
.0/2143D
,
¨
¬‘®
.—­
k
1
near 0.
/2XN3sD|m¯
for
1
, and so on. This
near 0,
Let us compare the Taylor polynomial with the formula obtained on page ?? for the sum of a finite
¨nä _
geometric series:
hsAB1
hsAB1
1
1
¨œä
Djh‹:1$:1
Œ
:1
«
:1
¬
¨
:–t–o–a:1
C
_
If is close to 0 and
is small enough to neglect, the formula for the sum of a finite geometric
series gives us the Taylor approximation of degree § :
h
hsAB1
5h‹:1$:1
Œ
:1
«
:1
¬
:^–o–o–y:1
¨
C
452
Chapter Ten APPROXIMATING FUNCTIONS
Taylor Polynomials Around å„æèç
.0/@1=3‹D^éw“s1
Suppose we want to
approximate
by a Taylor polynomial.
This function has no Taylor
1DX
1;êiX
because the function is not defined for1%D8 . However, it turns out that
polynomial about
we can construct a polynomial centered about some other1point,
.
D^8
First, let’s look at the equation of the tangent line at
:
g
D[.0/>8}3:;.=<T/2893?/21"A8}3?C
This gives the first Taylor approximation
.0/@1=365^.0/>8}3—:(.=<T/2893?/21\A8}3
.
1
for
84C
near
< />8}3o/@1%A893
.
1
The 8
term is a correction term which approximates the change in as moves away
from .
]—¨/2143
17DÝ8
.0/>893
Similarly, the Taylor 1
polynomial
centered at
is set up as
plus correction
D8
/@1%A(893
. This is achieved by writing the polynomial in powers of
terms which are zero1 for
instead of powers of :
¨
.0/@1=365]
Œ
Π/@1\A893
/21436D[6s:(s_œ/@1\A893”:(
¨
:^–o–t–p:;
/@1\A893
¨
C
]'¨/@1=3
§
If we require
derivatives of the approximating polynomial
and the original
function
1D[8
1D[8
to agree at
, we get the following result for the § ²2³ Taylor approximations at
:
"! ´#
Taylor Polynomial of Degree µ Approximating
.0/214365^]
¨
for
near ë
/@143
.
D[.0/2893”:;.=<>/>893?/@1"AB893”:
< < />8}3
~V¯
/@1AB893
Œ
¨ ®
­
.
:[–t–o–p:
]'¨/@1=3
/2893
/@1\A893
¯
§
1ŽDi8
We call
the Taylor polynomial of degree § centered at
1D[8
nomial about
.
¨
, or the Taylor poly-
You can derive the formula for these coefficients in the same way that we did for
Problem 28, page 454.)
Example 7
Construct the Taylor polynomial of degree 4 approximating the function
Solution
We have
.0/@1=3D
é“s1
< /@1=36D
hy·a1
.
.
.
­
< < < /@1=36D
.
Œ
¬‘®
/@1=36DjA
.
«
~N·y1
Ê ·y1
.
¬
k
/@1A;hy3
é“s1R5]—¬N/21436D[XÉ:[/@1\A„hp3'A
D›/@1"A;hy3'A
/21"A„hp3
~
­
.
The Taylor polynomial is therefore
Œ
~9¯
:
.0/@143‹D£é“s1
8RD›X
for
1
. (See
near 1.
.0/xhp3D^铔/xhy3D^X
so
< < /@1=36DìA•hy·a1
.
é“s1
.0/2143
Œ
:;~
< /xhp3D
h
< < /xhp3D
A•h
< < < /xhp3D
~
¬‘®
/@1"A;hy3
/@1"A;hy3
ª
/xhp3D
«
ªV¯
A
A
C
Ê
«
A
Ê
/@1\A„hp3
|
/@1"A;hy3
¬
|m¯
¬
k
for
1
near 1
C
Graphs of
and several of
its Taylor
polynomials are shown 1 in Figure 10.7. Notice that
é“s1
1
stays reasonably close to
for near 1˜
1,ê;
but
bends
away as gets farther from 1. Also,
X
éw“s1
note that the Taylor polynomials are defined for
, but
is not.
]—¬N/2143
453
10.1 TAYLOR POLYNOMIALS
Î K G9L
¡
¡”¥ K GVL
î Ñ

G
ž K GVL
¡
G

¡—¥ K GVL
í
¡=ß K GVL
P
¡
ß K GVL
ž K GVL
¡
Î K GVL
¡
î
G
Ñ
Figure 10.7: Taylor polynomials approximate
î Ñ
G
closely for G 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
Taylor polynomials centered at
give good approximations to
for near . Farther
away, they may or may not be good.
The higher the degree of the Taylor polynomial, the larger the interval over which it fits the
function closely.
ï
ï
Exercises and Problems for Section 10.1
Exercises
For Problems 1–10, find the Taylor polynomials of degree ð
approximating the given functions for G near 0. (Assume ñ is
a constant.)

1.
3.
5.
7.
—òG
,
—òG
Ÿ
÷oø
Ÿ
ú
…-ù
÷
Ñ
`QRG
ƒÞ
ð
,
ƒ ð
G
,
,
ð
Ÿ
—òG
,
Ē
ƒ ð

9.

,ó ,ô
ð
ƒ 2.
,í ,Þ
,
QRG
…‘†pˆ G
4.
Þ
6.
,í ,Þ
,í ,Þ
8.
10.
Problems
¡
ž K
,
ù
÷
î
Ñ
Ñ
G
,
ð
,
K —ò˜G9L2ý
ƒ ð
K }òûGVL
Ē
,õ ,
,Þ ,ó
Ē
ð
,
ð
ö
,
For Problems 11–14, find the Taylor polynomial of degree
for G near the given point F .
Þ
ƒÞ
ð
ƒ
õ
,ö ,ü
13.
G
ˆbÐÒÑ
11.
á ¢
,
,
F
ƒ
€4þ ,
…‘†pˆ G
12.
Ē
ð
FUƒ„
,
ð
ƒÞ
Ÿ
14.
ð
,
ð
ƒ ,
'ò˜G
Ē
F
,
ƒ
F
ð
€=þ Þ
ƒ
,

,
,í ,Þ
ž
GVL‹ƒ[FÉòBÿ-G•ò
‡G
For Problems 15–18, suppose
is the
J
second degree Taylor polynomial for the function about
J
G‰ƒŠ
. What can you say about the signs of F , ÿ , if has the
graph given below?
15.
J4K GVL
J4K GVL
16.
17.
G
G
J4K GVL
18.
G
G
J4K G9L
19. Suppose the function
J4K G9L
is approximated near
GƒŠ
454
Chapter Ten APPROXIMATING FUNCTIONS
by a sixth degree Taylor polynomial
Î
K G9L”ƒíoGQ"ÞtG
¡
ò
G
õ
Give the value of
J4K ŠyL
(a)
JNSTK ŠpL
(b)
JNS S SÇK ŠpL
(c)
J
(d)
-K ŠyL
J
(e)
-K ŠpL
20. Suppose is a function which has continuous derivaK
S K LâƒQ‹ S S K LBƒ

tives, and that õ L⃠í
õ
,
õ
,
S S S K Lƒ£Q6í
õ
.
(a) What is the Taylor polynomial of degree for near
í
õ ? What is the Taylor polynomial of degree
for
near 5?
(b) Use the two polynomials that you found in part (a)
K
to approximate Þ ü L .
J4K GVLûƒ
J4K
23. (a) Based on your observations in Problems 21–22,
make a conjecture about Taylor approximations in
J
the case when is itself a polynomial.
(b) Show that your conjecture is true.
, for
G
í
G
near 0, to explain why
î Ð
¢
G
ˆbÐÒÑ
à
G
25. Use the ž fourth-degree
Taylor approximation
ß
ØQ
î Ð
¢
G
`Q
G
ò
à
…‘†pˆ G
ž
G
for
Þ
G
near
Š
G
ˆbÐÒÑ
ƒ„
.

ƒ
.
26. Use a fourth degree Taylor approximation for á , for
near 0, to evaluate the following limits. Would your
answer be different if you used a Taylor polynomial of
higher degree?
(b) 27. If
, and
,
, and
,
,
,
calculate the following limits. Explain your reasoning.
(a) (b) î
(a)
î
Ð
Ð
á
à
K pL{ƒ
S K pL•ƒš p î
¢
Ð
Î
Q
í
ƒŠ
¥
"! K
##$ J ¥
J ž
J Î
ž
¥
32. One of the two sets of functions, , , , or , ,
Î
, is graphed in Figure 10.8; the other set is graphed in
each have GƒØŠ . Taylor
Figure 10.9. Points and
polynomials of degree 2 approximating these functions
near G‰ƒŠ are as follows:
%
'
(
(
J Î K GVL
&
(
(
(
ž
( ò%Gòâ ?G
`ò˜GQG
ž
ž
`ò˜GÉòG
ž
¥ K GVL
„—ò˜GÉòŽ oG
ž K GVL
„—ò˜GÉòG
Î K GVL
„`QRGÉòG
ž
ž
(a) Which group of functions, the s or the s, is represented by each figure?
(b) What are the coordinates of the points and ?
(c) Match the functions with the graphs (I)-(III) in each
figure.
%
&
K pLsƒEŠ
J S K tLsƒ
J S S K tLĒ
K GVL
III
III
II
II
ž¤
J4K GVL
ž
G
and G•Q
á vQ˜Q
ž
à
QŽ`Q
J4K tLsƒ
J
J ž K GVL
to explain why
31. The integral à ˆbÐÒÑ þ L
is difficult to approximate
using, for example, left Riemann sums or the trapezoid
K
rule because the integrand ˆbÐÒÑ L þ is not defined at
ƒ
Š
. However, this integral converges; its value is
Š
Þ
Š
Estimate the integral using Taylor polynoü
ó
ô
mials for ˆbÐÓÑ about ƒŠ of
(a) Degree 3
(b) Degree 5
J ¥ K G9L
…‘†tˆ G
G‰ƒŠ
(a) How many zeros does each equation have?
(b) Which of the zeros of the two equations are approximately equal? Explain.
22. Find
thež third-degree Taylor polynomial for GVL„ƒ
Î
G
ò
G
Q
G`ò
about G‰ƒŠ . What do you notice?
ö
õ
GQ
30. Consider the equations ˆbÐÒÑ
21. Find
the second-degree Taylor polynomial for
ž
ÞpG
Q
GòŽ about G‰ƒŠ . What do you notice?
ö
Î
24. Show how
you can use the Taylor approximation
(b) Compare this result to the Taylor polynomial apJ4K
proximation of degree 2 for the function GVL—ƒ(á ¢
G‰ƒŠ
about
. What do you notice?
(c) Use your observation in part (b) to write out the Taylor polynomial approximation of degree 20 for the
function in part (a).
(d) What is the Taylor polynomial approximation
of dež
J4K
¢
?
gree 5 for the function GVLć
Î
¢
î
Ð
S S K pLƒ
õ
S K tLsƒEŠ
NS S K tLƒ
ö
J4K GVL
ž
K GVL
28. Derive the formulas given in the box on page 452 for
the coefficients of the Taylor polynomial approximating
J
a function for G near F .
29. (a) Find the Taylor polynomial approximation of degree
J4K
4 about G•ƒŠ for the function GVLƒá ¢ ¤ .
I
%
Figure 10.8
I
&
Figure 10.9
10.2 TAYLOR SERIES
10.2
455
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 )+*-,
1%DX
We have the following Taylor polynomials centered at
for
d?eNf91
:
doeMf91R5[]—M/@1=36Dih
doeMf91R5[]
~V¯
Œ
1
¬ /@1=36DihsA
doeMf91R5[]
doeMf91R5[]
Œ
1
Π/@1=36DihsA
doeMf91R5[]
/@1=36DihsA
»
:
Ê ¯
1±»
A
|±¯
Π/2143
]
1±»
A
|±¯
¬
1
:
~V¯
]—M/@1=3
|±¯
¬
1
~V¯
Œ
1
/@1=36DihsA
Ì
~V¯
Œ
1
¬
1
:
Ê ¯
]”¬N/@1=3
1±Ì
:
]
C
rV¯
/2143
]
/@143
Here we have a sequencedoeMoff91 polynomials,
,
,
, »
, Ì
, ..., each of which is a
1
better approximation
to
than
the
last,
for
near
0.
When
we
go
to
a
higher-degree
polynomial
]
]
1±Ìp·œrV¯
(say from » to Ì ), 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
doeMf91
the whole sequence of Taylor polynomials by writing the Taylor series for
:
h{A
1
Œ
~V¯
1
:
¬
1±»
A
|m¯
1±Ì
:
Ê ¯
A;–t–o–oC
rV¯
]'¨/@1=3
Notice that the partial sums of this series
are the
Taylor polynomials,
.
fz’“{1
ÕyÖ
We define the Taylor series for
and 1 similarly. It turns out that, for these functions, the
Taylor series converges to the function for all , so we can write the following:
1
f-’w“{1RD^1\A
d?eNf91RDihsA
Õ Ö
1
«
ªV¯
Œ
~V¯
Dih‹:1´:
1±º
:
1
:
1
Œ
~V¯
1±¼
l9¯
¬
A
Ë ¯
1±»
A
|m¯
:
1
«
ªm¯
/.
1
:
qm¯
1±Ì
:
Ê ¯
:
1
¬
|m¯
rV¯
A;–t–o–
A;–t–o–
:^–o–t–
f-’w“s1
d?eMfV1
ÕpÖ
1D[X
These series are also called Taylor expansions of the functions
,
, and about
.¨ The
1
·
¯
Πin the series. For example,
§
general term of a Taylor series is a formula which
gives
any
term
ÕyÖ
/zA•hp3-Ùp1
ÙM·9/T~
3?¯
×
is the general d?term
in the Taylor expansion for , and
is the general term in the
eNf91
expansion for
. We call § or × the index.
Taylor Series in General
.
Any. function , all of whose derivatives.0/@exist
at 0, has a Taylor
series. However, the
Taylor series
143
1
1
for does not necessarily
converge
to
for
all
values
of
.
For
the
values
of
for
which the
.0/2143
series does converge to
, we have the following formula:
456
Chapter Ten APPROXIMATING FUNCTIONS
Taylor Series for
"! $#
.0/2143D7.0/2XM3:;.
è
about
.
< /2XM3x1$:
< < /2XN3
0
Œ
~9¯
1
.
:
< < < /2XM3
ªV¯
1
«
:^–o–t–y:
.—­
¨ ®
/2XN3
§
¨
1
¯
:–t–o–
X
In addition, just as we have 1„
Taylor
polynomials centered at points . other than1„D™
, we
can also
D
8
8
(provided
all
the
derivatives
of
exist
at
).
For the
have a Taylor
series
centered
at
1
.0/@1=3
values of for which the series converges to
, we have the following formula:
Taylor Series for
J4K GVLƒ
J4K FML±ò
"! $#
è
about
J S S K FNL
J S K FML K GQFNLVò
ë
J S S S K FML
ž
K GQFML
ò
í
0#010
Î
K G•Q"FML
ò
32
J
uò
ð
K FML
K GQFML
2 40$0#0
ò
The Taylor series is a power series whose partial sums are the Taylor polynomials.
As we saw in
1[DZ8
Section ??, power series generally converge1 on an interval centered at
. The Taylor series
for such a function can be interpreted when 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
.0/@143
to
, though it does for most commonly1encountered
functions. Functions for which this is
DZ8
true for the Taylor series about every
point
in
their
domain are called analytic functions.
ÕyÖ
d?eMfV1
f-’“s1
1
, and
converge to the original function for all , so
Fortunately, the Taylor series for ,
they are analytic. See Section 10.4.
Intervals of Convergence of Taylor Series
é“s1
Let us look again at the Taylor polynomial for
about
page ??. A similar calculation gives the Taylor Series
éw“s1D›/@1\A„hp3'A
65
75 85
/@1"A;hy3
~
Œ
:
/@1\A„hp3
ª
«
A
/21"A„hp3
|
1D
h
that we derived in Example 7 on
¬
:–t–o–p:[/xA•hy3
¨}©
_
/@1"A;hy3
¨
:–t–o–tC
§
Example 2 onX page1š449
and Example 5 on page 451 show that this power series has interval of
êè~
convergence
. However,
although we know that the series converges in this interval,
éw“s1
we do notX yet
know
that
its
sum
is
. The fact that in Figure 10.10éw“s
the
polynomials
fit the curve
71
~
1
X
71(êj~
well
for
suggests
that
the
Taylor
series
does
converge
to
for
. For such
1
-values, a higher-degree
polynomial
gives,
in
general,
a
better
approximation.
1
~9k
However, when
the polynomials move away from the curve and the approximations get
worse as the degree oféw“sthe
polynomial
increases.
Thus, the Taylor polynomials are effective only
1
1
as approximations to
for values of between
0~ and 2; outside that interval, they should not
X
be used. Inside the interval, but near the ends, or , the polynomials converge very slowly.
This
é“s1
means we might have to take a polynomial of very high degree to get an accurate value for
.
9:
95
457
10.2 TAYLOR SERIES
P
O
Interval of convergence
/
¡
¡
¡
K V
G L
K GVL
Ï K V
G L
Ô K GVL
¡
K GVL
î
G
Ñ
G

¡
Ï K GVL
¡
í
Þ
O
O
O
î Ñ
O
¡4Ô K GVL
G
; ;
K GVL
¡
Figure 10.10: Taylor polynomials
;
K GVL
¡
¡=Ï K GVL
; ## converge to
Ô K G9L
¡
K GVL
¡
î Ñ
=< ?>
G
for Š
âG
B and
diverge outside that interval
To compute the interval of convergence exactly, we first
compute1the
radius of convergence
1D[X
D[~
using the method on page ??. Convergence at the endpoints,é“s1
and
, has to be determined
separately. However, proving that the series converges to
on its interval of convergence, as
Figure 10.10 suggests, requires the error term introduced in Section 10.4.
Example 1
Find the Taylor series for
Solution
Taking derivatives of
éw“”/zh:143
铔/zh‹:1=3
about
1RD^X
, and calculate its interval of convergence.
1D[X
and substituting
Œ
1
铔/xh{:1=3D1\A
:
~
«
1
leads to the Taylor series
ª
¬
1
A
Notice that this is the same series that we get by substituting
é“s1RDj/21"A„hp3'A
/@1\A„hp3
Œ
:
~
éw“s1
/@1\A„hp3
«
X
/xh‹:143
1
for
¬
in the series for
:^–o–t–
|
@5
A5
1âDh
/21\A;hy3
A
ª
:^–o–o–tC
|
?5
X
for
(1˜ê~
é“s1
:
.
[1ê7~
about 1RDconverges
for
, the interval of convergence for the
Since the series for
éw“”/zh‹:1=3
X
A•h ;1êEh
Taylor series for
about
is
. Thus we write
铔/zh:1=36D£1"A
Œ
1
~
:
«
1
ª
A
|
:^–o–t–
Notice that the series could not possibly converge to
defined there.
Interval of
convergence
P
A5
¬
1
for
铔/zhs:;1=3
B
¡
O
A•h
for
K GVL
(1˜êEh
1(êA•h
¡Ï K GVL
¡
.
since
铔/xhv:;143
K G9L
K —òGVL
î Ñ
G
Qs
¡
¡
K GVL
K GVL
¡Ï K V
G L
¡ Ô K GVL

O
O
O
¡
O
P
P
î Ñ
B
K GVL
K 'ò˜G9L
¡=Ô K GVL
Figure 10.11: Interval of convergence for the Taylor series for
¡
î Ñ
K GVL
K 0ò%GVL
is
C< ->
Qs
âG

is not
458
Chapter Ten APPROXIMATING FUNCTIONS
The Binomial Series Expansion
ED , with F
1RDX
.0/2143D›/xh‹:143
We now find the Taylor series about
for the function
not necessarily a positive integer. Taking derivatives:
.0/@143‹Di/zh‹:1=3
.
.
F
< /2143D
.
/
©
/
A„hp3?/
ED
F
.
©4Œ
ED
©
A~M3?/zh‹:1=3
«
k
D
]
« /@1=3uk-]
F
5[]'«M/21436Djh‹:
F F
/
A„hp3
Œ
1
~9¯
/
F F
< < < />XM3D
near 0 is
1$:
F F
< < />XM3D
.
1
F
< />XM3D
.
Thus, the third-degree Taylor polynomial for
/zh‹:1=3
.0/>XM3Dih
so
_
A„hp3?/zh‹:1=3
F F
< < < /2143D
ED
/xh‹:143
F F
< < /2143D
D
a constant, but
/
F F
/
:
A„hp3o/
A;hy3?/
ªm¯
F
?5 G5
¬ /@143?koCtCoC
A„hp3
F
F
A~n3?C
A~M3
«
1
C
Graphing .0/@1=3
forš
various
specific values of suggests that the Taylor polynomials
A•h E1
h
converge to
for
. (See Problems 24–23,
page 459.) This can
be confirmed using
.0/@1=3Di/zh‹:1=3
1RD[X
the radius of convergence test. The Taylor series for
about
is as follows:
ED
The Binomial Series
D
/xh‹:143
Djh‹:
F
F F
/
1û:
A„hp3
~9¯
1
Œ
F F
/
:
F
A„hp3o/
ªV¯
A~n3
«
1
:^–o–t–
for
=5 H5
A•h
„1
ID
7h
.
F
/xh:R143
In fact the binomial series gives the same result as multiplying
out when is a positive
integer. (Newton discovered that the binomial series can be used for noninteger exponents.)
Example 2
Use the binomial series with
Solution
The series is
«
/xh‹:143
1
F
D[ª
to expand
Dh‹:(ªœ1$:
ªU–p~
~9¯
Œ
1
/zh‹:1=3
«
.
ª–t~U–nh
:
1
ªV¯
«
ª–p~U–œhs–pX
:
|m¯
1
¬
:–t–o–oC
¬
The
term and all terms beyond it turn out to be zero, because each coefficient contains a factor
Œ
«
«
of 0. Simplifying gives
/xh‹:143
Djh‹:(ªœ1$:(ªœ1
:1
k
which is the usual expansion obtained by multiplying out
Example 3
Find the Taylor series about
Solution
Since
h
Di/zh‹:1=3
h‹:1
h
h‹:1
©
1D[X
for
_
Œ
F
/xA•hy3?/zAÉ~n3
Dih‹:^/zA•hp3b1û:
DihsAâ1$:1
A5 95
.
, use the binomial series with
©
.
h
h‹:1
_
D›/xh‹:143
«
/xh‹:143
~9¯
Aâ1
«
:–t–o–
DA•h
1
Œ
:
for
. Then
/xA•hp3o/xAÉ~M3?/xAvªN3
ªm¯
=5 H5
A•h
„1
7h
1
«
:–t–o–
.
This series is A•
both
a special
case of the binomial series and an example of a geometric series. It
h ;1
Eh
converges for
.
459
10.2 TAYLOR SERIES
Exercises and Problems for Section 10.2
Exercises
ž

For Problems 1–4, find the first four terms of the Taylor series
for the given function about Š .
QRG
13.
'ò˜G
1.
2.
`QG
Ÿ

—ò˜G
3.
4.
—òG
Ÿ
Ÿ
ú
`Q‚
K QGVLƒ„QGQ
î Ñ
K 'òGVLƒGQ
For Problems 5–11, find the first four terms of the Taylor series for the function about the point F .
8.
11.
G
ˆbÐÒÑ
5.
,
€4þ Þ
ù
÷
F ƒ
U
Ñ
 þ G
Qs
G
Fûƒ
,
9.
€4þ Þ
,
6.
KJ ,
…‘†tˆ
F•ƒ

Fèƒ
,
ˆbÐÒÑ
7.
€4þ Þ
 þ G
LJ ,
15.
FUƒ„Q
 þ G
10.
,
€=þ Þ
F
ƒ
16.
17.
FYƒ
M
Find an expression for the general term of the series in Problems 12–19 and give the starting value of the index (ð or ,
for example).
÷oø
á ¢ ¤
18.
19.
…-ù
ž
ƒ„—òG
ž
G
÷
G
ò
í
í
ò
Þ
G
ò
Ô
G
ò
ß
G
Q
ò
ö
G
ò
ƒG
Ï
G
Q
õ
ß
G
ò
ö
Î
G
ž
ž
…‘†pˆ G
Q
õ
Ï
G
Q
õ
ò
G
ò
Þ
P0#0#0
P0#0#0
P0#0#0
P0#010
G
ò
í
ß
G
Q
í
Q
Þ
Î
G
ò
O0#010
H010#0
ß
G
Q
í
ž
G‰ƒGQ
Ñ
N0#0$0
Q
Î
G
Q
ò
ß
òG
ž
G
Î
G
G•ƒGQ
ˆbÐÒÑ
QG
G
ß
òG
Î
ƒ„`QGÉò˜G
î Ñ
14.
òG
ž


40#0$0
Î
ƒ„—òGÉò˜G
12.
Þ
Ô
G
Q
ò
ó
Problems
20. (a) Find the termsž up to degree ó of the Taylor series for
J4K GVLƒ
K G
L
ˆbÐÒÑ
about G‰ƒŠ by taking derivatives.
(b) Compare your result in part (a) to the series for ˆbÐÒÑ G .
How could you have obtained your answer to part (a)
from the series for ˆbÐÒÑ G ?
J4K
K
î
21. (a) Find the Taylor series for G9L'ƒ Ñ 'òB oG9L about
G‰ƒŠ
by taking derivatives.
(b) Compare your result in part (a) to the series for
î Ñ K ‹òG9L
. How could you have obtained your anî K
swer to part (a) from the series for Ñ 0ò%GVL ?
(c) What do you expect the interval of convergence for
î K
the series for Ñ 'ò˜ oGVL to be?
By recognizing each series in Problems 27–35 as a Taylor series evaluated at a particular value of G , find the sum of each
of the following convergent series.
" P0$010
27.
'ò

29.
J4K GVLUƒ


31.
and several
32.
of its Taylor polynomials, estimate the interval of convergence of the series you found in Problem 3.
33.
23. By graphing the function
24. By graphing the function
J4K GVL‹ƒ
'ò˜G
Ÿ

and several of
34.
its Taylor polynomials, estimate the interval of convergence of the series you found in Problem 1. Compute the
radius of convergence analytically.
35.
`QG
Þ
ð
ò Q
ò
Î

ò
ò
Þ
¥ Î
K žaL
40$0#0
¥ ß
K žœL
Q
í
?ò Q
ò
Þ
40$0#0
" P0#0#0
P0$010
" 40#0$0
Q%Š
0ò˜Š
ü
'òŽíò
Q%Š
2 010#0
"
28.
?ò
ž
Þ
¥ ž
K žnL
Q
Q
ò
í

ò
J4K
22. By graphing the function GVLƒ Ÿ 0ò%G and several of
its Taylor polynomials, estimate the interval of convergence of the series you found in Problem 2.
ô
ò

í

ò
õ

QŠ

Þ
40#0$0
ò
oò
K
ð


ò
Þ
ò
ó
Š
Šœ
Š
Q
ŠtŠn
í
ò
26. (a) Write
the general term of the binomial series for
K —ò˜GVL2ý
about G‰ƒŠ .
(b) Find the radius of convergence of this series.
36.
ž
'ò˜GÉòG

G
ž
ò
í

40#0$0
P010$0 Î
òG
ò
G
Î
ò
œƒ
aƒŠ
õ
ò
?ò
2 0 U2 T P0#0#0
¥
K žnL
K Qs?L
ò
ô
ò
Q
Þ
0ò
ö
í

ò
Þ In Problems 36–37 solve exactly for the variable.
G•Q
ö
uŠpŠtŠpŠ
òò
Î

ò
ž
oò
 ‘ŠpŠ
25. Find the radius of convergence of the Taylor series
î K
around zero for Ñ `QGVL .
37.

Q
30.
RQ 'S RQ 'S P0#0#0 RQ "S 2 40#0$0 40$010

'ò
Þ
ò
ò?L
¥
ò
2 0#010
;
2 0 2 P0
;
K Qs?L
K ð
ò
ò?L
K Qs?L
K ð
tuŠ
L
ž
ò
460
Chapter Ten APPROXIMATING FUNCTIONS
J ¥
J ž
J Î
ž
¥
38. One of the two sets of functions, , , , or , ,
Î
is graphed in Figure 10.12; the other set is graphed in
Figure 10.13. Taylor series for the functions about a point
corresponding to either or arež as follows:
%
&
J ¥ K GVLƒíò
K GQŽ?LQ
K GQŽ?L
J ž K GVLƒíQ
K GQŽ?L±ò
K GQŽ?L
J Î K GVLƒísQ% K GQŽ?L±ò
K GQŽ?L
0#010
0#010
0#010
ž
ž
39. Suppose
that you are told that the Taylor series of
ž
G
á ¢ ¤
about G‰ƒŠ is
¥ K GVLƒ
õ
Q
õ
Î K GVLƒ
õ
&
ò
J
tion
is
J S K ŠpL
ž
G
GÉò
Find
I
,
¢
Î
ß
G
ò
ò
Þ
G
uò
J
à
about GƒŠ
ò
ð
J S S S K ŠyL
J
¥ à
K ŠyL
á
in the
. Use this definition2 to explain Eu-
á ¢
[\
á
ƒ
KJ Z ]J ò
…‡†pˆ
III
Figure 10.12
á ¢ ¤
oG
aG
à
í
II
III
ò
Þ
P0#0#0 2 40$0#0 ,
, and
. We define [\ by substituting Z J
ò
J S S K ŠyL
Z
&
í
¥ à
G
ò
ž
G
41. Let ƒ Ÿ Qs
Taylor series for
ler’s formula
II
10.3
Ô
G
ò
exist at Š , and that Taylor series for
I
%
G
¢
(a) Which group of functions is represented in each figure?
(b) What are the coordinates of the points and ?
(c) Match the functions with the graphs (I)-(III) in each
figure.
%
0#0#0 P0#0#0 0#0#0
! Q
! Q
#
0
#
0
0
Find !
SWVV X and !
SYVV X VV
VV
40. Suppose you know that all the derivatives of some funcž
ß ž
K G QÞaL
G òG
ò
ž
K G•QÞaLmò
K GQÞaL
ž
ž
K G•QÞaLmò
K G
QÞaL
tG
á ¢ ¤
aG
K G•QÞaL=Q
Q
ž K GVLƒ
J4K G9L”ƒ
ˆbÐÒÑ
Figure 10.13
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 differenÕyÖ
tiation.
That is how
we derived the four most important Taylor series, those for the functions ,
f-’w“s1
doeMf91
/zh:£1=3
,
, 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.
ED
New Series by Substitution
©
Õ
Ö
17DèX
¤
Suppose we want to find the Taylor© series for
about
. We
could find the coefficients
©
Ö ¤
Õ
Ö ¤
AÉ~œ1±Õ
by differentiation.
Differentiating
by
the
chain
rule
gives
, and differentiating again
©
Œ
©
Ö ¤ :£|M1
Ö ¤
AÉ~nÕ
Õ
gives
. Each time we differentiate we use
the product rule, and the number
of
©
©
Õ
Ö ¤
Õ
Ö ¤
Œ  Finding the tenth or twentieth derivative of
_x grows.
terms
,
and
thus
the
series
for
up
to
1
1
the
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
^
Œ
Õ •Dh‹:
DjAs1
Substituting g
Õ
2 Complex
©
Ö ¤
g
:
g
«
~V¯
:
g
¬
ªV¯
:
g
|±¯
:^–o–t–
for all g
C
Œ
tells us that
Dh‹:[/xAs1
Œ
3:
/xAs1
Œ
~9¯
numbers are discussed in Appendix B.
3
Œ
:
/zAs1
ªV¯
Œ
3
«
:
/xAs1
|m¯
Œ
3
¬
:–t–o–
for all
1—C
461
10.3 FINDING AND USING TAYLOR SERIES
Simplifying shows that the Taylor series for
©
Õ
Ö ¤
Œ
DhsAB1
Ö ¤
¬
1
:
©
Õ
A
~V¯
is
»
1
ªm¯
Ì
1
:
Example 1
Find the Taylor series about
Solution
The binomial series tells us that
h
Di/zh‹:
h‹:
g
D1
Substituting g
g
©
3
_
for
Œ
h‹:1
Œ
:
g
or
1
for all .
Œ 
1
terms.
h
.0/@1=36D
DjhsA
_b
1
Using this method, it is easy to find the series up to the
1D[X
:^–o–t–
|m¯
«
A
g
.
¬
:
g
:–t–o–
g
A5 5
A•h
for
EhœC
g
Œ
gives
h
Œ
DihsAB1
Œ
h‹:1
¬
:1
»
AB1
Ì
:1
:–t–o–
5 H5
A•h
for
(1
7h
,
h
which is the Taylor series for
Œ
h‹:1
.
These examples demonstrate that we can get new series from old ones by substitution. More
Œ
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.
_
Example 2
Find the Taylor series about
Solution
For all g and , we know that
D[X
_
^
Õ
Œ
Õ
`ba cd
4e _ _
Dhn:
A
«
ªm¯
:
_
_
A£–o–t– ‰:
lV¯
g
g
D
«
:
~9¯
A
g
¬
ªm¯
_
«
ªV¯
:
:
º
g
:^–o–o–
|m¯
A;–t–o–tC
l9¯
for g :
f
º
:
_ _ _
f-’“
f-’w“
3D^Õ
Dih‹:
and
Let’s substitute the series for
U`ba c'd .
_
/
for c
h
~V¯
e_ _
A
_
«
ªm¯
:
º
A„–o–t–
lV¯
Œ
f
:
h
ªV¯
e_ _
A
«
ªm¯
h
:
_
º
lV¯
A„–o–t–
f
«
:Öo–t–oC
Œ
Œ
To simplify, we multiply
out and collect terms. The only constant term is the , and there’s
only one
·œ~9¯
«
term. The only term is the
first termA we« ·œget
by
multiplying
out
the
square,
namely
. There
ªV¯
«
are two contributors to the term: the
from
within
the
first
parentheses,
and
the
first
term
·œªV¯
we get from multiplying out the cube, which is
. Thus the series starts
_
_
Õ
_
_
`ba c/d
Djh‹:
Djh‹:
_ _
:
_ _
:
_
Œ
~9¯
:
e _
Œ
~9¯
«
A
:X–
_
«
ªV¯
_ f
«
:
ªV¯
:^–o–o–
:^–o–o–
for all
_
C
_
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.
Example 3
Solution
Find the Taylor Series about
We know that
g e
for
/xhsAB143
Œ
/zhsAâ1=3
h
from the series for
C
hsAB1
, so we start with the geometric series
Œ
Djh‹:1$:1
hsAB1
h
Œ
h
[D
h{AB1
1
g
f
h
h
1D[X
«
:1
¬
:1
:–t–o–
=5 95
A„h
for
„1
EhnC
Differentiation term by term gives the binomial series
h
/zhsAâ1=3
g e
D
Œ
1
g
Example 4
Find the Taylor series about
Solution
We know that
1j
g "h i h
g
hg i j h
g
/
d
“{143
/
d
“1=3
1D[X
for
h"i 1j h
d
«
:|n1
:–t–o–
h
“{1
from the series for
hi j h
where

lk
h
“‹1RD
h‹:1
Œ
Œ
¬
:1
g
d
The series for
“{1D^1\A
hi j h
d
“1
1
«
ª
:
1±º
l
«
1
1RD[:1\A
»
Aâ1
A
14º
:
ª
is the constant of integration. Since
h"i 1j h
Ì
:1
A„–o–o–
l
hi j h
d
1±¼
Ë
A
:
14¼
Ë
“{X‰D[X
.
1
q
.
1
:
q
Use the series for
h"i 1j h
d
“1
7hMC
C
Œ
A;–t–o–
for
A„–o–t–
, we have
for
A5 H5
A•h
¶
to estimate the numerical value of .
;1
ED[X
, so
;1
.
A5 95
5 95
A„h
7h
A;h
for
was discovered by James Gregory (1638–1675).
Applications of Taylor Series
Example 5
h‹:1
„1
, so we use the series from Example 1 on page 461:
DhsAB1
Œ
h‹:1
Œ
Antidifferentiating term by term gives
d
A5 95
A„h
for
h
h6:1
h
D
1
Œ
7Dh‹:;~a1´:ªn1
hsAB1
D
1
f
h
EhnC
„1
7hMk
Solution
j
Since hi h
10.3 FINDING AND USING TAYLOR SERIES
463
1j from Example 4. We assume—as is the
, we use the series for h"i h
case—that the series does converge to j at
, the endpoint of its interval of convergence.
Substituting
into the series for hi h
gives
We
f
hi j h
d
“h;D
¶—·y|
d
¶—·a|
1£DÝh
d
1%Djh
¶ŽD£|
d
“1
“hÉD^|
Table 10.1 Approximating
m2
“{1
h
=hsA
¶
h
:
ª
h
A
l
h
:
Ë
using the series for
A„–o–t– C
q
hi j h
d
“‹1
4
5
25
100
500
1000
10,000
2.895
3.340
3.182
3.132
3.140
3.141
3.141
ð
n
¨
Table 10.1 shows the value of¨ the § ²2³ partial sum, ,¶˜
obtained
by summing the nonzero terms from
D^ªmCwht|mhCtCoC‘C
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.
n
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
_
smallest, for near 0.
Solution
d
By looking at their Taylor series, decide which of the following functions
is largest, and which is
h
_
D[X
The Taylor expansion about
_
D[X
d
Õ
_
The Taylor expansion about
Í
h‹:
So, substituting
_
D›/xh‹:
_
3
D[X
po
©
_
Œ
_
d
Õ
_
_
º
_ _
hy· Í
_
h
_
~
º
_
:
«
Π3
ª
r
¼
Ë ¯
:^–o–t–oC
¬
:^–o–o–tC
|m¯
Œ
_
l
A
_
h Ê
Œ
_
~V¯
:
_
A
«
ªm¯
:–t–o–tC
Ë ¯
lV¯
_
3 /xA
Πo
:
~
¼
_
hsA~
Í
is
/zA
h
DihsA
_
:
_
h‹:
_
:
ªm¯
Œ
~V¯
_
A
«
A
:
(c)
l9¯
is
Dih‹:
for
:
ªV¯
_ _
Dih‹:
for
_
«
A
DihsA
AÉ~
is
_ _ _
h‹:f-’“
The Taylor expansion about
_
fz’“
D
So
Õ
(b)
for
f-’“
h
_
h‹:f-’w“
(a)
«
/zA
:
_
3 /xA
Πo
«
Π3?/xA
ªm¯
:–t–o–oC
Í
for :
h
hsA~
_
h
DihsA
Dih‹:
/xAÉ~
_
~
:
ª
~
_
ª
_
3”:
Œ
/xAÉ~
:
r
l
~
_
«
_
3
Œ
A
:^–o–t–?C
l
h Ê
/xAÉ~
_
3
«
:^–o–o–
ºŒ 3
_
«
:^–o–o–
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“
Õ
d
Since
_ 5
h‹:
_
Eh‹:
_
_
:
~
ª
Œ
~
Œ
_
~
_ 5
h
:
Œ
:
5ih‹:
_
hsA~
_ _
5ih‹:
h
Í
_
5ih‹:
C
_
[h‹:
ª
:
Œ
_
~
k
_
and since the approximations are valid for near 0, we conclude that, for near 0,
_ 5 d 5
h‹:(fz’“
Example 7
h
„Õ
C
_
hsA~
Í
Two electrical charges of equal magnitude
and opposite signs located near one another are called
A
and
are a distance apart. (See Figure 10.14.) The electric
an electrical dipole. The
charges
]
field, , at the point is given by
q
s
s
t
r
q
tq
D
A
Œ
t q r
/
:
C
Œ
3
t
«
Use series to investigate the behavior of the electric field far away from the dipole.
Show
that when
hy·
is large in comparison to , the electric field is approximately proportional to
.
r
u
¡
P
w
O
w
v
P
v
¡
w
w
Q
O
Figure 10.14: Approximating the electric field at due to a
a distance apart
dipole consisting of charges and Q
Solution
t
r t
In order to use a series approximation, we need a variable whose value is small. Although
we know
·
Πknow that
do
not
itself
is
small.
The
quantity
is,
however,
that is much smaller than ,hawe
·9/
:
3
·
very small. Hence we expand
in powers of
so that we can safely use only the first
few terms of the Taylor series. First we rewrite using algebra:
r
t
r
h
t
/
:
r
h
t
D
Œ
3
r
Œ
Now we use the binomial expansion for
t Q
h
Œ
h‹:
tr S
t e
©4Œ
D
h
t e
h
D
Œ
:(ª
·
3
Œ
Aâ|
Œ
tr
s
tq
Œ
A
/
t q r
:
3
Œ
D
Œ
D
s
Œ
~
A
h
hsA~
Œ
Aª
Œ
Œ
:|
DjAÉ~
:
ªm¯
Q tr S
C
, we have
q xt t e
y
t q e tr tr
h
C
/zAÉ~n3o/xAvªM3o/xAs|}3
:
:–t–o–
«
·
Œ
«
So, substituting the series into the expression for
D
1D
~9¯
©=Œ
h‹:
Œ
/xAÉ~M3?/zAvªM3
:
t Q tr S
r t and F
Q tr S
f
h
D
Œ
/zh‹:1=3
Q tr S
tr tr
hsA~
r t
ED with
/xh‹:
h‹:^/zAÉ~n3
Œ
r t
tr
tr
:ª
«
«
tr
Œ
Œ
AB|
A„–o–t–
f
C
tr
«
«
:^–o–t–
f{z
«
:^–o–t–
f
465
10.3 FINDING AND USING TAYLOR SERIES
t
Since r
r t converges. We are
is smaller than 1, the binomial expansion for
in
t
t interested
the electric field far away from the dipole. The quantity r
is small there, and r
and higher
·
/zh9:
·
©4Œ
3
·
/
·
Œ
3
powers are smaller still. Thus, we approximate by disregarding all terms except the first, giving
tq e t r f
s
~
Since
q
5
k
Œ
r
and are constants, this means that
s
r t
In the previous example,
we say that
·
in the expansion is
.
s
s
so
~
5
t qr
«
C
is approximately proportional to
is expanded in terms of
r t
·
hy·
t
«
C
, meaning that the variable
Exercises and Problems for Section 10.3
Exercises
Find the first four nonzero terms of the Taylor series about 0
for the functions in Problems 1–12.
1.
5.
9.
`Q% oG
Ÿ
î Ñ
|

á
K yQ ?‚NL
¤
2.
6.
10.
…‡†pˆ
÷?ø
€
K
J
ž
…‘ˆbÐÒÑ
K —ò
3. L
á
¢
4.

G
7.
11.
L ˆbÐÒÑ
Ÿ
}|
ž
`Q
8.
 ‚
á
…‘†pˆ
12.
14.
Î
…‘†pˆ
Ÿ
13.
—ò
~
lems 13–15. Give the general term.
—ò
K
~
15.
Î
K 'ò˜G9L
ž
K
L4Q
nˆbÐÒÑ
€
Î

ž
Q‚
ž
L
LJ
ˆbÐÒÑ
For Problems 16–17, expand the quantity about 0 in terms of
the variable given. Give four nonzero terms.

16.
G
in terms of
ò˜G
F
17.
Ÿ
F
ž
ò%G
ž
in terms of
where „ƒ
G
F
F
âŠ
Find the Taylor series around 0 for the functions in Prob-
Problems
18. For F , ÿ positive constants, the upper half of an ellipse
has equation
‚Uƒ
U…
J4K G9L”ƒÿ
Q
ž
G
F
ž
†J
J
‡J
small positive .
(a)
'ò
ˆbÐÒÑ
(b)

…‘†tˆ
(c)
`Q
J4K
(a) Find the second degree Taylor polynomial of GVL
about Š .
(b) Explain how you could have predicted the coefficient
of G in the Taylor polynomial from a graph of
J
.
(c) For G near Š , the ellipse is approximated by a
parabola. What is the equation of the parabola?
What are its G -intercepts?
(d) Taking FUƒí and ÿƒ , estimate the maximum difJ4K
ference between GVL and the second-degree Taylor
Q6Š  ŽG BŠ 
polynomial for
.
{> ?> 19. By looking at the Taylor series, decide which of the following functions is largest, and which is smallest, for
J
ž
20. For values of ‚ near 0, put the following functions in increasing order,
ž using their Taylor ž expansions.
K
î K
(a) Ñ tòv‚ L
(b) ˆbÐÒÑ ‚ L
(c) aQ …‘†pˆ ‚
21. Figure 10.15 shows the graphs of the four functions below for values of G near 0. Use Taylor series to match
graphs and formulas.

(a)
(d)
Q"G

Ÿ
ž
QRG
(b)
K uòGVL
‰ˆ
¥
ß
(c)
Š
'ò
G
466
Chapter Ten APPROXIMATING FUNCTIONS
M
(IV)
| €
vƒ
Expand
term.
Figure 10.15
22. Consider the functions ‚Uƒá
¢ ¤
and ‚Uƒ„
ž
þ K 0òG
L
(a) Write the Taylor expansions for the two functions
about Gƒ[Š . What is similar about the two series?
What is different?
(b) Looking at the series, which function do you predict
K
will be greater over the interval Qs ‘uL ? Graph both
and see.
(c) Are these functions even or odd? How might you see
this by looking at the series expansions?
(d) By looking at the coefficients, explain why it is rea¢ ¤
converges for
sonable that the series for ‚ƒá
ž
G
‚ƒ™ þ K {òG
L
all values of , but the series for
K
converges only on Qs ‘?L .
U
‹
‚‹
‹
23. The hyperbolic sine and cosine are differentiable and satisfy the conditions …‘†pˆ Švƒ„ and ˆbÐÒÑ ŠvƒŠ , and
!
K
aG
‹
GVLƒ
…‘†pˆ
ˆbÐÒÑ
!
! ‚‹
K
G
ˆbÐÒÑ
aG
GVLƒ
‹ …‘†pˆ
‹
K
K
u w
w
25. An electric dipole on the G -axis consists of a charge at
Gƒš
and a charge Q
at G˜ƒšQs . The electric field,
G‰ƒ
, at the point
on the G -axis is given (for  )
by
Œ
Œ
ƒ
K
u Mw
QâuL
ƒ; €
u
Show that, for large
ž
Q
K
u Mw
ò?L
ž
u ƒ
ž
F
ž
QG
as far as the second nonzero
u
Ž
K
Ž
€ u
ž
òF
ž
u
Q
L
,
 u Ž
ž
F
€
~
28. One of Einstein’s most amazing predictions was that light
traveling from distant stars would bend around the sun on
the way to earth. His calculations involved solving for
in the equation
~
~
ž
òÿ K —ò
…‘†pˆ
ò
…‘†pˆ
~
LƒŠ
where ÿ is a very small positive constant.
~
(a) Explain why the equation could have a solution for
which is near 0.
(b) Expand the left-hand side of the equation in Taylor
mž
series about ƒEŠ , disregarding terms of order
and higher. Solve for . (Your answer will involve
ÿ
.)
‹
J4K
G
Q


‹
`òâFœGVL þ
0òâÿ‡G9L
(a) Let G9L6ƒ
, where F and ÿ are
constants. Write J4down
the
first
three terms of the
K
Taylor series for G9L about G‰ƒŠ .
(b) By equating the first three terms of the Taylor seJ4K
ries about G˜ƒŠ for GVL and for á ¢ , find F and ÿ
J4K G9L
so that
approximates á ¢ as closely as possible
G‰ƒŠ
near
.
€
ž
òG
as a series in
ˆbÐÒÑ
24. Padé approximants are rational functions used to approximate more complicated functions. In this problem, you
will derive the Padé approximant to the exponential function.
ž
is given by
27. The electric potential, , at a distance along the axis
perpendicular to the center of a charged disc with radius
F
and constant charge density , is given by
G
(a) Using only this information, find the Taylor approxJ4K
imation of degree ð ƒ ô about GŽƒŠ for G9Lsƒ
G
…‘†pˆ
.
(b) Estimate the value of …‘†pˆ  .
(c) Use the result from part (a) to find a Taylor polynomial approximation of degree ð ƒ ö about G⃛Š
K
for G9L=ƒ ˆbÐÒÑ G .
|
F
|
26. Assume F is a positive constant. Suppose
the expression
(II)
!
u
Œ
where is a positive constant whose value depends on
the units. Expand as a series in  þ , giving the first
two nonzero terms.
(III)
(I)
~
~
~
29. The Michelson-Morley experiment, which contributed to
the formulation of the Theory of Relativity, involved the
ž
¥
difference between the two times and that light took
to travel between two points. If is the velocity of light;
ž
¥
¥
, , and are constants; and
, then the times
ž
and are given by
 
¥
ƒ
 <
 6
K Q


ž
ž
þ
ž
L
Q
€
6Q

¥

ž
þ
ž
‘ 
 ‘
¥
ž
ž
ƒ
€
`Q

ž

ƒ
Q
(a) Find an expression for
, and give its
ž
ž
Taylor expansion in terms of þ
up to the second
nonzero term.
(b) For small , to what power of is
proportional?
What is the constant of proportionality?

ž
þ
ž
Q
K Q


¥
ž
þ
ž
L
467
10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS
“
’
’
32. When a body is near the surface of the earth, we usually
assume that the force due to gravity on it is a constant
, where is the mass of the body and is the acceleration due to gravity at sea level. For a body at a distance
above the surface of the earth, a more accurate expression for the force is
30. A hydrogen atom consists of an electron, of mass , orbiting a proton, of mass , where
is much smaller
than . The reduced mass, , of the hydrogen atom is
defined by
“
”
ƒ
”
•’ “
’ “ ’˜
ò
(a) Show that ” ’ .
(b) To get a more accurate approximation for ” , express
” as ’ times a series in ’ “ .
(c) The approximation ” ’ is obtained by disregard-
’
–
’
U— 4
“ ’ e
–
and think of –
Assume H<
ƒ
F

ž
Q
’ “
’˜
F
the other quantities constant.
–
K F
ž
ò
L
u
– u
¥
à
á
¢ ¤ aG
sums for both left-hand and right-hand sums with
ƒ
ð
õ subdivisions.
J4K
(b) Approximate the function GVLƒá ¢u¤ with a Taylor polynomial of degree 6.
(c) Estimate the integral in part (a) by integrating the
Taylor polynomial from part (b).
(d) Indicate briefly how you could improve the results
in each case.
as a function of F , with
(a) Expand
as a series in F þ . Give the first two
nonzero terms.
(b) Show that the approximation for obtained by using only the first nonzero term in the series is independent of the radius, F .
(c) If F$ƒŠ Šy , by what percentage does the approximation in part (a) differ from the approximation in
part (b)?
–
10.4
ž
– ˜’ ! using Riemann
(a) Estimate the value of 33.
ž
¥
L
– ˜’ – ’˜
™
‰ˆ f 
ò
ž
(a) Show that
.
multiplied by a series in þ .
(b) Express as
(c) The first-order correction to the approximation
is obtained by taking the linear term in the series but no higher terms. How far above the surface
of the earth can you go before the first-order correction changes the estimate
by more than
uŠ
? (Assume ƒ ó ÞpŠpŠ km.)
–
ž
K
u
u ’„ u
31. A thin disk of radius F and mass
lies horizontally; a
particle of mass is at a height directly above the center of the disk. The gravitational force, , exerted by the
disk on the mass is given by
’
ƒ
where is the radius of the earth. We will consider the
situation in which the body is close to the surface of the
earth so that is much smaller than .
ing all but the constant term in the series. The firstorder correction is obtained by including the linear
í
þ 
ô
ó , by what
term but no higher terms. If
percentage does including the first-order correction
?
change the estimate
“
–
u
þ
” ’
34. Use Taylor series to explain how the following patterns
ž
arise: 

(a)

Š
üpô
Šp tŠpítŠtÞpŠ
ƒ„
õ
Š
Š
ó
Šy oŠtÞyŠ
ö
#$

ô
ó
íy ó
Þ
#1
(b)
Q S
Š
ütü
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
¨”/@143‹D^.0/@1=30A]—¨/2143?C
s
s
¨
s
¨
If
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¨­
¨ /2XN3D
.0/@1=3
X
X
–o–t–
/>XM3DãX
] ¨ /@1=3
<¨ /2XN3D
¨ < < /2XN3DãX
tives of
. Therefore,
,
,
, ä _z® ,
is
¨œ
¨œä _z® . Since
¨­
/
:›hy3
/@143ÃDZ.—­
/2143
¨œä _z®
its
.
In
addition,
an §=²2³ degree.—polynomial,
§
² derivative is 0, so
­
/@1=3
1
X
supposeX´that
is bounded by a positive constant , for all positive values of near ,
ê(1˜ê
say for
, so that
¨nä _x®
s
VV
This means that
g
VV
s`
›
A
A
›
ê
ê£.
s
s
›
­
/2143{ê
¨­
¨œä
_z®
/@1=3{ê
›
›
for
for
s
s
X´ê„1ê
X$ê;1ê
šs š
C
g
g
C
ƒ
468
Chapter Ten APPROXIMATING FUNCTIONS
œ
X
1
Writing for the variable, we integrate this inequality from to , giving
Ok › œ k s
g
›
s
Ö
Ö
A
so

A
¨œä
¨­
ê

/
3
ê
¨ ®
/@143sê
X
1
X´ê(1˜ê
1
Hk l
› œ g œ žk s
Ö
Ö
¨­
ê


so
h
A
›
~
Œ
1
s
ê
¨}©
¨­
œ g œ k l
› œgœ
h
/
›
:^hp3?¯
§
¨œä
1
_
s
ê
g
Ö
¨ ®
/
3
ê
for

_z®
h
/@1=3sê
›
~
Œ
1
X´ê(1˜ê
X´ê(1Žê
for
h
¨/@1=3{ê
/
›
:hy3u¯
§
¨œä
1
_
for
g
k
C
g
By repeated integration, we obtain the following estimate:
A
k
C
g
We integrate this inequality again from to , giving
A
X´ê„1ê
for

¨­
1ê
œ g œ žk › g œ
for
›
Ö
_x®
X´ê„1ê
g
k
which means that
šs
¨
š š
/@143
D
¨
.0/21430A]
1
A
h
š
/@1=3
ê
/
›
:^hp3u¯
§
¨nä
1
_
for
Xûê„1ê
êj1£êiX
g
g
C
When is to the left of 0, so
, and when the Taylor series is centered at
similar calculations lead to the following result:
Theorem 10.1: Bounding the Error in ¡
If
]—¨”/@143
is the §=²2³ Taylor approximation to
šs
where
›
D
max
¨/@1=3
¨nä
.—­
š š
D
š
.0/@1=30AB]'¨/@1=3
DØX
,
! $#
.0/2143
Ÿ
8
8
about , then
ê
›
/
_x®
8
on the interval between
§
š
:^hp3?¯
1\A8
š
¨œä
_
k
1
and .
Using the Error Bound for Taylor Polynomials
Example 1
Give
a bound on the error,
AvXVC lûê;1ê£XVC l
.
Solution
Let
.0/2143D[ÕyÖ
s
¬
ÕyÖ
, when
¢
is approximated by its fourth-degree Taylor polynomial for
.—­wº
. Then the fifth derivative is
š
.
­º
®
/2143
š
ê„Õ
Thus
º
D
šs š š
¬
This means, for example, that on
Œ
_ Œ 
A5
/2XmC lM3xº
D
Í
5
/@1=36DÕyÖ
Õ
~
.0/@1=30AB]—¬}/@143
5h‹:1$:
„XVC XnXnX Ê C
ÕyÖ
. Since
š
is increasing,
AvXVC lûê(1˜ê£XVC l}C
for
AvXVC lûê;1ê£XVC l
Õ Ö
has an error of at most

®
~
ê
š š
1
l9¯
º C
, the approximation
1
Œ
~V¯
:
1
«
ªV¯
:
1
¬
|m¯
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 numer1
ical approximation, or to see how the accuracy of the approximation depends on the value
of
or
/
:^hp3
²
§ . Observe that the error for a Taylor polynomial of degree § depends on the §
the value of
1
¨œä _
§
power of 1 . That means, for example, with a Taylor polynomial of~ degree
centered at 0, if we
decrease by a factor of 2, the error bound decreases by a factor of
.
`
Example 2
Compare the errors in the approximations
¢

Õ
Solution
_
h
5jh‹:XmCwh‹:
~9¯
Œ
/2XVChp3
Õ
and
¢


º
h
5jh‹:[/2XmC XNln3:
~V¯
ÕyÖ
/>XVC XMlM3
Œ
C
1jD¦XmCwh
We
are approximating
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:
¢ £e
¢ e

Õ
Õ
Since


º
_
A
h
A
h‹:XmCwh6:
h‹:(XVC XMls:
~9¯
h
/>XVC XMlM3
~9¯
/2XmC XMXnXmh Ë hy3-·V/2XVC XnXMXnXN~9hp3D[rVCh
/2XmCwhy3
f
Œ
[DihnChtXNl9h Ë h6A„hnChtXNlœXnXMXUD^XVC XnXnXmh Ë h
f
Œ
DihnC XMlVhp~ Ë h6A„hnC XMlVhp~nlnXUD^XVC XnXnXMXM~Vh
, the error has also decreased by a factor of about 8.
Convergence of Taylor Series for cos )
1%D[X
doeMf91
We have already
seen that the Taylor polynomials centered at 1
for
are good approxima1
near
0.
(See
Figure
10.16.)
In
fact,
for
any
value
of
,
if
we
take
a Taylor polynomial
tions for 1
D[X
centered at
of high enough degree, its graph is nearly indistinguishable from the graph of the
cosine near that point.
¡
Ô K GVL
¡

…‘†pˆ G
Q
¡
¡
G
]
]
»
Ì
ž K GVL
and two Taylor polynomials for G near Š
1˜D7¶—·œ~
hsA„/2¶—·œ~M3
]”¬}/@¶—·n~n3DihsA„/2¶—·œ~n3
G
Qs
ž K GVL
Let’s see
what happens numerically.
Let
d?eNfo/@¶—·n~n3D[X
1D[X
mations to
about
are
]—ŒN/@¶—·n~n3D
…‘†pˆ G
€
€
Figure 10.16: Graph of …‘†pˆ
Ô K GVL
Œ
Œ
. The successive Taylor polynomial approxi-
·n~9¯
·n~9¯t:[/@¶—·œ~M3
DjAvXmC ~nªnª Ë XsCoCtC
¬
·y|±¯MD
XVC XVhpqnq Ë
CoCtC
/@¶—·n~n3D
–o–o–
DjAvXmC XMXnXnrMqsCoCtC
/@¶—·n~n3D
–o–o–
D
XVC XnXMXnXN~{CoCtC‡C
470
Chapter Ten APPROXIMATING FUNCTIONS
d?eNfo/@¶—·n~n3D[X
It appears that
the approximations converge
to the
true value,d?eNf9¶˜DjA•h , very rapidly. Now take
1
X
1RD^¶
a value of somewhat farther away from , say
, then
and
Œ /2¶”36DihsA„/2¶”3
]
Œ
·œ~9¯}D
AvªmC qMªœ|NrnXsCtCoC
]—¬M/2¶”36D
–o–t–
D
XmCwhy~ϻMqVhCtCoC
]
/2¶”36D
–o–t–
D
A•hMC ~VhnhpªMl{CtCoC
/2¶”36D
–o–t–
D
AvXmC q Ë Ê XM~{CtCoC
]0_xM/2¶”36D
–o–t–
D
A•hMC XMXVhprnªsCtCoC
]
_ Œn/2¶”36D
–o–t–
D
AvXmC qMqnqMqnXsCtCoC
]
_x¬ /2¶”36D
–o–t–
DjA•hMC XMXnXnXMXœ|vCtCoC
»
]
Ì
ho|
We seedoeMthat
the rate of convergence
is somewhat slower; it takes a ²2³ d?degree
polynomial
to approxf9¶
r
eNfo/@¶—·n~n3
1
²>³ degree polynomial approximates
imate
as accurately
as
an
.
If
were
taken still
X
farther
away
from
,
then
we
would
need
still
more
terms
to
obtain
as
accurate
an
approximation
of
d?eNf91
.
d?eMfV1
1
Using the ratio test, we can show the Taylor
series for
converges for all values of . In
doeMf91
addition, we will prove that it converges to
using Theorem 10.1. Thus, we are justified in
writing the equality:
Œ
¬
1
doeMf91DihsA
1
:
~V¯
1±»
A
|m¯
1±Ì
:
Ê ¯
1
A;–t–o–
rV¯
for all .
Showing the Taylor Series for cos å Converges to cos å
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
Œ
s
¨
¨
/@143‹D^doeMf91\A]
giving
1
hsA
~9¯
Thus, for the Taylor series to converge to
doeMf91
o
¨
:^–o–t–p:^/zA•hp3
¨
1
Œ
/
, we must have
©ª¨¬« `
š
š
§
:
3?¯
s
¨
s
/21436Ú
šs
/2143
­
š š
D
¨
d?eMfV1\AB]
/2143
X
š
­
š š
/
as §
f-’“1
for every §
:^hp3?¯
§
1
Œ
/
1
ê
o
§
3u¯
f
k
¨/@1=3uC
xå
Showing
0 as
¨œä _z®
doeMf}1
.0/@143vDid?eNf91
/
:[hy3
.—­
/@143
¨nä § _x®
Proof Since
, the
is
or1
² derivative
.—­
/2143
êEh
X
So for all § , we have
on the interval between and .
By the error bound formula in Theorem 10.1, we have
¨œä _
¨
¨
:^–o–t–p:[/xA•hp3
~9¯
Œ
1
d?eMfV1Djh‹:1$:
¤¥C¦ L§=¨
e
/@143‹D^doeMf91\A
Ú
Û
.
, no matter what § is.
C
1
To show that the errors go to zero, we must
show that for a fixed ,
¨œä _
š š
1
/
§
Ú
:hy3u¯
X
as
Ú
§
ÛC
1
To see why
this is true, think about what happens when § is much larger than . Suppose, for
1RDh Ë C ª
example,
that
. Let’sD^
look
at the value of the sequence for § more than twice as big as 17.3,
Dª Ê
Dª Ë
ªMr
say §
, or §
, or §
:
For §
For §
For §
Dª Ê
Dª Ë
Dªnr
h
:
ª Ë ¯
:
ªnrm¯
:
ªnqm¯
h
h
«
¼
/xh Ë C ªM3
«
/xh Ë C ªM3
h Ë Cª
Ì
/xh Ë C ªM3
«
.
D
h
–
D
ªMr
h Ë Cª
ªMq
–
ª Ë ¯
h Ë C ª
ªnr
/xh Ë C ªN3
–
h
ª Ë ¯
«
¼ k
/xh Ë C ªM3
«
¼ k
CtCoC
471
10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS
h Ë C ªN·aª Ê
Œ
_
_ we increase
«
§ by 1, the term is multiplied by a number less
Since_
is less than , each time
/zh Ë C ªN3 ¼
Œ
«
¨œ
ä _ ¼
than . No matter what_ the/zvalue
of
is,
if
we keep on dividing it by two, the result gets
h Ë C ªM3
¨nä _x®
closer to zero. Thus ­
goes to 0 as § goes to infinity.
h Ë 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
Œ
number less than :
¨nä _
¨näŒ
¨œä «
1®
®
1
/
Therefore, the Taylor series
: š š
š š
:hy3u¯
§
Œ
hsAB1
1
k
/
¬
·œ~V¯p:1
§
1
k
:(~M3u¯
/
·a|m¯œA„–o–t–
§
:ªN3u¯
koCtCoCoC
does converge to
d?eNf91
.
fz’“s1
Problems 16 and 151 ask you to show that the Taylor series for
and
original function for all . In each case, you again need the following limit:
±³°¯ ²
¨
converge to the
¨
1
é’
ÕpÖ
D^XmC
¯
§
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
Problems 1–4 using a third-degree Taylor polynomial about
G‰ƒŠ
.
Problems
J4K

´ 1µ
Lć
5. Suppose you approximate
by a Taylor polynoŠ
ƒŠ
on the interval Š zŠ õ .
mial of degree about
(a) Is the approximation an overestimate or an underestimate?
(b) Estimate the magnitude of the largest possible error.
Check your answer graphically on a computer or calculator.

1.
‰ˆ
¥
Š
´ µ
J K 
õ
L
3.
 þ Ÿ
í
4.
¡
ù
÷
Ñ

¥ à K
¶ ¶>
LJ J
LJ J
ˆbÐÒÑ
î Ñ
¡
ž à K GVL
¡
ž à K GVL
ˆbÐÒÑ
(a) Where is the approximation an overestimate, and
where is it an underestimate?
(b) Estimate the magnitude of the largest possible error.
Check your answer graphically on a computer or calculator.
Î
8. Repeat
Problem 7 for the approximation
þ í
.
2.
L
(a) The error in approximating …‘†pˆ õ by
and by
õ
¡ ž à K L
õ .
(b) The maximum error in approximating …‘†tˆ G by
¡ ž à K GVL
for G (‘Š .
¡—¥ à K G9L
(c) If we want to approximate …‘†pˆ G by
to an accuracy of within 0.2, what is the largest interval of
G
-values on which we can work? Give your answer
to the nearest integer.
6. Repeat ProblemK 5 using the second-degree Taylor ap¡ ž
L
, to á .
proximation,
7. Consider the error in using the approximation
on the interval Qs u .
Î
õ
Q
9. Use the graphs of ‚⃠…‘†tˆ G and its Taylor polynomi¡ ¥ à K GVL
¡ ž à K GVL
als,
and
, in Figure 10.17 to estimate the
following quantities.
í
G
Qsu Q
ó
Q6í
í
? ó
Q6í
¡
¥ à K G9L
¡
Figure 10.17
¥ à K GVL
472
Chapter Ten APPROXIMATING FUNCTIONS
¸·º¹
´ µ
¶Œ ¶ >
ž
, # # , , Œ
(b) Add to your table the values of the error
,
õ
, $ # , Q6Š
Þ
Q6Š

,Š ,Š

Š
Þ
Š
for these G -values.
(c) Using a calculator or computer, draw a graph of the
¥ ƒ
ˆbÐÒÑ G•QG
showing that
quantity
¶Œ ¶ < QRŠ
ž K GVLƒá ¢ Q
¡
ƒá ¢ Q
¶Œ ¶ >

ŽG
âŠ
ž
K tòvG±òsG
Q6Š
ž

þ pL

BG
Š

? Use the graph to
{> ?> Πand Πhave the
(c) Explain why the graphs of
ž
Î
ŽG
for
QRŠ

ŽG
âŠ

ž
¥
¶ ¶> Œ
ƒ
GQG
Œ
for
shapes they do.
¥
ˆbÐÓÑ
ž
ŽG
What shape is the graph of
confirm that
.
õ
? Use the graph to
‹ž
12. (a) Using a calculator, make a table of the values to four
decimal places of ˆbÐÒÑ G for
¥
{> ?> Œ
(b) Let
Œ for ¼> Y. Choose
> .
a suitable range and graph
Œ
¥
11. What degree Taylor polynomial about GƒŠ do you
need to calculate …‘†tˆ  to four decimal places? To six
decimal places? Justify your answer using the results of
Problem 10.
G‰ƒ„Q6Š
Œ
What shape is the graph of
confirm that
10. Give a bound for the maximum possible error for the ð
degree Taylor polynomial about G„ƒYŠ approximating
…‘†tˆ G
on the interval Š u . What is the bound for ˆbÐÓÑ G ?
G
14. For
> -> âŠ

, graph the error
à
ƒ
…‘†pˆ GQ
¡=à K G9Lƒ
…‡†pˆ GQŽ
¶Œ ¶ ¶ ¶ > 13. In this problem, you will investigate the error in the ·º¹
15. Show that the Taylor series about 0 for converges to
Œ '½
degree Taylor approximation to
for various values of
for every . Do this by showing that the error 2
y
½
¾
.
as
.
Œ
(a) Let
Œ for ». Using
> •> a cal- . 16. toShow thatfortheeveryTaylor. series about 0 for converges
culator or computer, graph
¥
âŠ
Špí
Q%Š
for
õ
ŽG
BŠ
Explain the shape of the graph, using the Taylor expanà
sion of …‘†tˆ G . Find a bound for
for G ⊠ .
õ
á ¢
ð
á ¢
á ¢
K GVL
G
ð
YŠ
ð
¥
ƒá ¢
¥ K GVLƒá ¢
¡
Q
K ò"GVL
Q
¥
Q6Š
G
ˆbÐÒÑ

âG
âŠ

G
ˆbÐÓÑ
G
REVIEW PROBLEMS FOR CHAPTER TEN
Exercises
For Problems 1–4, find the second-degree Taylor polynomial
about the given point.
á ¢
1.
4.
ù

G
ƒ
2.
î Ñ
/
G
G;ƒ
/
3.
G
ˆbÐÒÑ
Q
LJ , J
J
KJ
ž
¿
ž
…‘†pˆ

ž
ˆbÐÒÑ
7.
8.

9.
ÞsQRG
Ÿ
|
ž
QÞ
G‰ƒ
€4þ Þ
For Problems 10–11, expand the quantity in a Taylor series
around the origin in terms of the variable given. Give the first
four nonzero terms.
ƒ
Ñ
÷
€ þ Þ
4
6.
u v
J4K GVL„ƒ
5. Find
thež third-degree Taylor polynomial for
Î
G
ò
G
Q
Gò
at G•ƒ .
ö
õ
F
10.
ÿ
in terms of
Fsòÿ
F
Q
Ÿ
11.
u
in terms of
v
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.
ö
K 
(b)
K 
(a)
ö
ò
Î
Šp pL
ö
ž
Š tL
y
ö
K Š
ò
K 
ö
010#0
ò
?L
ž
ž
Šp pL
ò
ö
K 
Šp pL6ò
ö
?ò
¥ àbà
K  p
Š pL
ß
K Š ?L
K Š ?L
ö
ö
ò
ò
í
.
ö
ò
K 
" 40$0#0 ö
Šy tL
18.
ò
13.
010#0
mQ

í

ò

Q
ü

ò
ö
ô

Q
14.
ô

ò‰Þ=ò´ mò"}ò
0#010

?ò

ò
Þ
ò
¥ à
ò
15.
Find the exact value of the sums of the series in Problems 13–
yQ yò
" A0$0#0
Þ
Q
ô
í
ò

ó
Þ
ò
16.
mQ
ô
í
ò
íp õ
Q
? ö
-0;010
ô
ò
í”ò"í”ò
17.
í
õ
" í
P0$010
í
ò
í
ò
í
ò
Þ
K Š
18.
?L
K Š
ò
Q
uL
K Š
Q
J4K
(b) The force on the particle at the point G is given by
S2K GVL
Q
. For small G , show that the force on the particle is proportional to its distance from the origin.
What is the sign of the proportionality constant? Describe the direction in which the force points.

ò
Ô
ò
JNS STK ípLƒ
íyLƒ
19. A function has ípLûƒY ,
õ and
J4K
QsuŠ
. Find the best estimate you can for í uL .
20. Suppose G is positive but very small. Arrange the following expressions in increasing order:
/
G
/
;
K ?òGVL
î Ñ
G
ˆbÐÓÑ
/
pQ
U
¦á ¢ Qs
…‘†tˆ G
÷?ø
…‡ù
/
G
Ñ
÷
21. By plotting several of its Taylor polynomials and the
J4K
K
function GVLRƒ  þ ò£GVL , estimate graphically the
interval of convergence of the series expansions for this
function about Gƒ Š . Compute the radius of convergence analytically.
ž
22. Use Taylor series to evaluate
23. (a) Find
\ ˆbÐÒÑ
î Ð
à
J
J
K à
¢
L
K —òGòG
ž
G
ˆbÐÓÑ
î Ñ
î Ð
L=QG
J
. Explain your reasoning.
J4K
J
J
ˆbÐÓÑ
K L
L$ƒ
(b) Use series to explain why
looks
ƒ^Š
like a parabola near
. What is the equation of
the parabola?
J
J4K

28. The theory of relativity predicts that when an object
moves at speeds close to the speed of light, the object
appears heavier. The apparent, or relativistic, mass, ,
of the object when it is moving at speed is given by the
formula
à
G
Ÿ
`QG
’
k ! ¢
à
(c) Using your answer to part (b), show that

ò
í
"

ò
Þ K L
"

ò
K í
õ
L
40$010

ò
ó
K Þ
L
ò
aƒ„
25. (a) Find the first two nonzero
terms of the Taylor series
ž
J4K
for G9L”ƒ Ÿ Þ{QG about G‰ƒŠ
(b) Use your answer to part (a) and the Fundamental
Theorem of ¥ Calculus to calculate an approximate
value for
k €
Þ{QG
à
ž
!
aG
.
(c) Use the substitution G\ƒ^ ˆbÐÒÑ to calculate the exact value of the integral in part (b).
26. (a) Find the Taylor series expansion of ÷oø
(b) Use Taylor
series to find
G
÷oø
÷oø
…‡ù
÷
…‡ˆbÐÓÑ
Ñ
G
as
„½ G
…‡ˆbÐÓÑ
G
.
the
limit

Š
’
’

(a) Write the Taylor polynomial of degree 2 for about
G\ƒ£Š
. What can you say about the signs of the coefficients of each of the terms of the Taylor polynomial?
ž
ž
þ
’
à
is the mass of the

’




v
29. The potential energy, , of two gas molecules separated
by a distance is given by


à
v v f
e
v S Q v S ¼
Q

à
ƒ„Q
and
v
à
v
à
v
¥ ž
à
Q
are positive constants.

v v
à
(a) Show that if ƒ
, then takes on its minimum
à
.
value, Q
K
à L
Q
(b) Write
as a series in
up through the
quadratic term.
à
(c) For near , show that the difference between
and its minimum ž value is approximately proK
à L
Q
portional
to
. In other words, show that
à Lƒ
à
Q K Q
ò
is
approximately proportional
ž
K
à L
to Q
.
(d) The force, , between the molecules is given by
à
ƒ¦Q
þ
. What is
when ƒ
? For
à
near , show that is approximately proportional
K
à L
to Q
.
v



 v v
– v ¿! 
v v
v
v v
 
–!v
–
v
–
v
v
30. The gravitational field at a point in space is the gravitational force that would be exerted on a unit mass placed
there. We will assume that the gravitational field strength
at a distance away from a mass
is
!
G
27. A particle moving along the -axis has potential energy
K GVL
at the point G given by
. The potential energy has a
minimum at G‰ƒŠ .

`Q
(a) Use the formula for to decide what values of are
possible.
(b) Sketch a rough graph of against , labeling intercepts and asymptotes.
(c) Write the first three nonzero terms of the Taylor series for in terms of .
(d) For what values of do you expect the series to converge?
where
á
€ ’
ƒ
’

where is the speed of light and
object when it is at rest.
L”ƒ
á
24. (a) Find the Taylor series for
about ƒŠ .
(b) Using your answer to part (a), find a Taylor series
expansion about G‰ƒŠ for

473
REVIEW PROBLEMS FOR CHAPTER TEN
ß
ž0$0#0
uL
ö
JNS>K
?L
í
õ
J
" K Š
ž
—
—“
!
“
ž
–
where is constant. In this problem you will investigate
the gravitational field strength, , exerted by a system
consisting of a large mass
and a small mass , with a
distance between them. (See Figure 10.18.)
v
“
’
474
Chapter Ten APPROXIMATING FUNCTIONS
u
¡
P
“
O
v
P
O
’
Figure 10.18
–
(a) Write an expression for the gravitational field
¡
strength, , at the point .
v u
v
u
v u
–
“
’
J4K Gvò
L
K Gsò
’
!
aG
K2J4K GVL
32. Use Taylor expansions for
firm the chain rule:
!
!
K2J4K
aG
J S K GVL
K GVLbL4ƒ
K GVL9ò
M
J4K ‚”ò
K GVLbLbLƒ
J S K
33. Suppose all the derivatives of
has a critical point at G‰ƒŠ .
L
J4K GVL
and
/0 K GVLbL
S K GVL
K Gò
L
ÁÀ
Q
< ?>
=< ?>
ŽG
€
½y¾

Š
âŠ
ŽG
€
to explain why the following sum must approach €=þ
ð
:

`Q

ò
í

Q
õ
ö
0#010
ò
K Qs?L
?ò
2UT
ž
¥
ð
¼> ?<
ò
36. Find
a Fourier polynomial of degree three for
ž
á  ¢
, for Š ŽG ( .
as

Þ
J4K GVL$ƒ
J4K
to con-
(a) Write the ð
Taylor polynomial for at G‰ƒŠ .
(b) What does the Second Derivative test for local maxima and minima say?
(c) Use the Taylor polynomial to explain why the Second Derivative test works.
Â
J
1Â
(a) If the Fourier coefficients of are F and ÿ , show
J S
are
that the Fourier coefficients of its derivative
ÿ
Q
F
and
.
J
(b) How are the amplitudes of the harmonics of and
JNS
related?
J
J S
(c) How are the energy spectra of and related?
M #Â
exist at GûƒŠ and that
·º¹
Qs
37. Suppose that G9L is a differentiable periodic function
J
of period € . Assume the Fourier series of is differentiable term by term.
S K GVL
J4K GVLƒ
L
and
in Taylor series and take
31. Expand
a limit to confirm the product rule:
!
35. Use the Fourier polynomials for the square wave
(b) Assuming is small in comparison to , expand
in a series in þ .
ž
K
(c) By discarding terms in þ L and higher powers,
explain why you can view the field as resulting from
ò
, plus a correction
a single particle of mass
term. What is the position of the particle of mass
ò
? Explain the sign of the correction term.
“
34. (Continuation of Problem 33) You may remember that
the Second Derivative test tells us nothing when the second derivative is zero at the critical point. In this problem
you will investigate that special case.
Assume has the same properties as in Problem 33,
S S K ŠyLɃjŠ
. What does the Taylor
and that, in addition,
polynomial tell you about whether has a local maximum or minimum at G‰ƒŠ ?
M Â
 1Â
1Â !Â
% &
%
»& % Â & #Â % 1Â & !Â
39. Suppose that is a periodic function of period
Pand.
that is a horizontal shift of , say Show that and have the same energy.
J
38. If the Fourier coefficients of are F and ÿ , and the
and , and if and
Fourier coefficients of are
J ò
are real, show that the Fourier coefficients of
F `ò
ÿ `ò
are
and
.
J
€
J
K GVLvƒ
J4K Gûò
uL
J
PROJECTS
Exercises
1. Machin’s Formula and the Value of
(a) Use the tangent addition formula
j h ÄÃ GÅ
“”/ „:
with
Ã
D
$3D
h"i 1j h
d
¶
j h »Ã Pj h »Å
Nj h CÃ9j h »Å
“
hsA
„:
“
“/xhy~œX}·}hnhpqM3
to show that
hi j h
h"i 1j h 9e f hi j h 9e f h"i 1j h
à Å h"i 1j h
(b) Use
to show that
hi j h e f h"i 1j h e f
A
d
“
“/zhy·n~œªnqN3
hp~nX
“
hnhpq
ED
“
and
d
Å
D
~
A
›D
d
d
“
d
h
“
~nªnq
[D
d
“/xha·œln3
h
l
D
d
“
l
hp~
C
“hnC
475
PROJECTS
Use a similar method to show that
h"i 1j h 7e f
|
d
[D
h"i 1j h }e
h"i 1j h
formula:
“/xha·œ~œªMqM3
h
“
l
h"i 1j h
d
(c) Obtain
Machin’s d
|
d
“/xhy·nln3'A
“
C
hnhtq
¶—·y|
D
.
(d) Use the Taylor polynomial approximation of
degree 5 to the arctangent
function to ap¶
proximate the value of . (Note: In 1873
William
Shanks used this approach to calculate
¶
to 707 decimal places. Unfortunately, inlM1946
~œr
it was found that he made an error in the
²>³
place.)
(e) Why do the two series for arctangent converge
so rapidly here while the series used in Example 5 on page 462 converges so slowly?
3
2. Approximating the Derivative
In applications,
.0/@1=3
frequently
known
the values of a function 1  1 are
1 
~
CoCtC?C

only at discrete values ,
,
,
Suppose we
are interested in approximating the
. < /@1  3
derivative
. The definition
­ÇÆ
É È± ¯
.=<>/21ma36D™éw’

suggests that for small
as follows:
.
Æ
­ Æ
Æ
Æ
.
3'A.0/21ma3
Æ
< /@1=3
Æ
36D^.0/21ma3”:;.
< /@1±y3
into the approximation for
.0/21m:
Æ
3`A.0/21mp3
Æ
D7.
.
Æ
< /21
:
 3
< /@1ma3”:
.
< < /21ma3
Œ
Æ
~
:–t–o–
, we find
.
< < /21ma3
Æ
~
:^–o–t–
This suggests (and it can be proved) that the error
in the approximation is bounded as follows:
.0/@1
VV

Æ
:
VV
30A.0/@1
 3
Æ
A.=<>/21
where
š
.
< < /2143
š ›
ê
3 From
Ë i
e
Mark Kunka
Table 10.2
"
K2J4K G à
uŠ
uŠ
š
 3
VV ›ÊÆ
VV
1\AB1m
ê
~

ß

š š Æ̚
ê
C

õ
ŠtŠ
ö
ŠtŠpŠ

õ
Šy õ
ŠtŠpŠtŠ

ö
õ
õ
õ
ž
"uŠ
Šy õ
Š
õ
ÎÍ
=Í
AÍ
AÍ
Error

‘Š
Š
‘Š
Š
‘Š
Î
ß
(a) Using Taylor series, suggest an error bound for
each of the following finite-difference approximations.
.
< /21
.0/21mp3`A.0/21mvA
 35
.0/21 
.=<>/21ma35
(ii)
.
:
Æ
Æ
3'A.0/@1
< /21ma35
AÉ.0/21m‹:;~
~
Æ
Æ
Æ

3
A
Æ
3
Æ
3:rN.0/@1m{:
3'ABrN.0/@1±vA
hp~
Æ
(iv) Use each of the formulas in part (a)
to
ÕyÖ
© _ derivative
©=Œ
© of
© at
«
¬
approximate
the
first
1èD¹X
DihtX
kohpX
kohpX
kohtX
for
.
As is decreased by a factor of 10, how
does the error decrease? Does this agree
with the error bounds found in part (a)?
Which is the most accurate formula?
Æ
Æ
.0/@1=3^D
ha·y1
©
(v) Repeat
part
(b) using
and
1mBD
htX
º
. Why are these formulas not
good approximations anymore? Continue
to decrease by factors of 10. How small
does have to be before formula (iii) is
the best approximation? At these smaller
values of , what changed to make the formulas accurate again?
Æ
Æ
Æ
]`_xN/2143
3. (a) Use a computer
algebra system to find
_bM/@143
Πpolynomials
Œ
and
,
the
Taylor
of degree
1RD[X
fz’“
1
doeMf
1
10 about
for
and
.
(b) What similarities do you observe between the
two poynomials? Explain your observation in
terms of properties of sine and cosine.
q
k

Î
Š
J4K G à LbL þ
L=Q

ž
uŠ
uŠ
ò
¥
(iii)
C
Æ
›ÊÆ
(i)
Such finite-difference approximations are used frequently in programming a computer to solve differential equations.
Taylor series can be used to analyze the error
in this approximation. Substituting
.0/@1m{:
Æ
30A.0/@1±p3
we can approximate
.0/21m:
< /21ma35
Æ
.0/@1±:
Æ
X
Notice that as
, the error also goes to zero,
provided
is bounded.
.0/@143DšÕ Ö
1  D
As
an example, we take
and
X
. < /21  3´D¦h
, so
. The error for various values of
are given in Table 10.2. We see that decreasing
by a factor of 10 decreases the error by a factor
of
·œ~
about 10, as predicted by the error bound
.
›
f
hp~œX
Ú
Æ
3:„.0/@1±
476
Chapter Ten APPROXIMATING FUNCTIONS
4. (a) Use
your computer
algebra system to find
/@1=3
]
/2143
¼
and ¼ 1›, D¦
theX Taylor
polynomials
of
.0/@1=3%D
fz’“s1
degree
7
about
for
and
/@1=3D^f-’“s1Éd?eNf91
«
c
.
1
(b) Find the ratio between the coefficient of
in
the two polynomials.
Do
the
same
for
the
coef14º
1±¼
ficients of
and .
(c) Describe the pattern in the ratios that you computed
in part (b). Explain it using the identity
f-’“”/>~œ143D7~`fz’“{1Éd?eNf91
.
Ö© _ :
֌
.0/2143D
.
5. Let
. Although the formula
for
1D7X
.
is not defined
at
, we can make . continuous
.0/2XN3D
h
by setting
.
If we do this, has Taylor
1D[X
series around
.
] _b /@1=3
(a) Use a computer algebra system to find
,
q
ÏpÐ
1âD
the
Taylor
polynomial of degree 10 about
X
.
for .
(b) What do you notice about the degrees of the
.
terms in the polynomial? What property of
does this suggest?
.
(c) Prove that has the property suggested by part
(b).
n
/@143‹D
Ö
f-’w“—/
œ gœ
Œ
3

6. Let
.
]`_-_n/@1=3
(a) Use a computer algebra system to find
,
1âD
the
Taylor
polynomial
of
degree
11
about
X
/@1=3
, for
.
(b) What is the
percentage
error in the approxima/xhy3
]0_‡_a/zhp3
tion of
by />~M3
? What about the ap] _-_ />~M3
proximation of
by
?
n
n
n