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11.11 Taylor’sTheorem
291
EXAMPLE11.11.2
Findapolynomialapproximationforsinxaccurateto0:005.
FromTaylor’stheorem:
sinx=
XN
n=0
f
(n)
(a)
n!
(x a)
n
+
f
(N+1)
(z)
(N+1)!
(x a)
N+1
:
Whatcanwesayaboutthesizeoftheterm
f
(N+1)
(z)
(N+1)!
(x a)
N+1
?
Everyderivativeofsinxissinxorcosx,sojf
(N+1)
(z)j1.Thefactor(x a)
N+1
is
abitmoredicult,sincex acouldbequitelarge. . Let’spicka=0andjxj=2;ifwe
cancomputesinxforx2[ =2;=2],wecanofcoursecomputesinxforallx.
WeneedtopickN sothat
x
N+1
(N+1)!
<0:005:
Sincewehavelimitedxto[ =2;=2],
x
N+1
(N+1)!
<
2
N+1
(N+1)!
:
ThequantityontherightdecreaseswithincreasingN,soallweneedtodoisndanN
sothat
2
N+1
(N+1)!
<0:005:
AlittletrialanderrorshowsthatN=8works,andinfact2
9
=9!<0:0015,so
sinx=
X8
n=0
f(n)(0)
n!
x
n
0:0015
=x
x
3
6
+
x
5
120
x
7
5040
0:0015:
Figure 11.11.1showsthe graphsofsinx andand theapproximationon[0;3=2]. Asx
getslarger,theapproximationheadstonegativeinnityveryquickly,sinceitisessentially
actinglike x
7
.
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292
Chapter 11 1 SequencesandSeries
5
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2
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Figure 11.11.1
sinxandapolynomialapproximation. (AP)
Wecanextractabitmoreinformationfromthisexample.Ifwedonotlimitthevalue
ofx,westillhave
f
(N+1)
(z)
(N+1)!
x
N+1
x
N+1
(N+1)!
sothatsinxisrepresentedby
XN
n=0
f(n)(0)
n!
x
n
xN+1
(N+1)!
:
Ifwecanshowthat
lim
N!1
x
N+1
(N+1)!
=0
foreachxthen
sinx=
X1
n=0
f
(n)
(0)
n!
x
n
=
X1
n=0
( 1)
n
x
2n+1
(2n+1)!
;
thatis,thesinefunctionisactuallyequaltoitsMaclaurinseriesforallx. Howcanwe
provethatthelimitiszero? SupposethatN islargerthanjxj,andletM M bethelargest
integerlessthanjxj(ifM=0thefollowingiseveneasier). Then
jx
N+1
j
(N+1)!
=
jxj
N+1
jxj
N
jxj
N 1

jxj
M+1
jxj
M
jxj
M 1

jxj
2
jxj
1
jxj
N+1
111
jxj
M
jxj
M 1

jxj
2
jxj
1
=
jxj
N+1
jxj
M
M!
:
ThequantityjxjM=M!isaconstant,so
lim
N!1
jxj
N+1
jxj
M
M!
=0
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11.11 Taylor’sTheorem
293
andbytheSqueezeTheorem(11.1.3)
lim
N!1
xN+1
(N+1)!
=0
asdesired.Essentiallythesameargumentworksforcosxande
x
;unfortunately,itismore
diculttoshowthatmostfunctionsareequaltotheirMaclaurinseries.
EXAMPLE 11.11.3
Findapolynomialapproximationfore
x
nearx=2accurateto
0:005.
FromTaylor’stheorem:
e
x
=
XN
n=0
e
2
n!
(x 2)
n
+
e
z
(N+1)!
(x 2)
N+1
;
sincef
(n)
(x)=e
x
foralln.Weareinterestedinxnear2,andweneedtokeepj(x 2)
N+1
j
incheck,sowemayaswellspecifythatjx 2j1,sox2[1;3]. . Also
ez
(N+1)!
e3
(N+1)!
;
so we need d to nd an n N N that t makes e
3
=(N+1)!   0:005. This s time N N = = 5 5 makes
e
3
=(N+1)!<0:0015,sotheapproximatingpolynomialis
e
x
=e
2
+e
2
(x 2)+
e
2
2
(x 2)
2
+
e
2
6
(x 2)
3
+
e
2
24
(x 2)
4
+
e
2
120
(x 2)
5
0:0015:
Thispresentsanadditionalproblemforapproximation,sincewealsoneedtoapproximate
e
2
, and any approximation n we e use e will increase e the e error, , but t we e will l not pursue this
complication.
Notewellthatintheseexampleswefoundpolynomialsofacertainaccuracyonlyon
asmallinterval,eventhoughtheseriesforsinxande
x
convergeforallx;thisistypical.
Togetthesameaccuracyonalargerintervalwouldrequiremoreterms.
Exercises11.11.
1. Findapolynomialapproximationforcosxon[0;],accurateto10
3
)
2. Howmanytermsoftheseriesforlnxcenteredat1arerequiredsothattheguaranteederror
on[1=2;3=2]isatmost10
3
? Whatiftheintervalisinstead[1;3=2]? )
3. FindtherstthreenonzerotermsintheTaylorseriesfortanxon[ =4;=4],andcompute
theguaranteederror termasgivenby Taylor’s theorem. (Youmaywanttouse e Sage or a
similaraid.) )
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294
Chapter11 Sequences s andSeries
4. Showthatcosxis s equaltoitsTaylorseriesforallxbyshowingthatthelimitoftheerror
termiszeroasN approachesinnity.
5. Showthate
x
isequaltoitsTaylorseriesforallxbyshowingthatthelimitoftheerrorterm
iszeroasN approachesinnity.
Theseproblemsrequirethetechniquesofthischapter,andareinnoparticularorder.Some
problemsmaybedoneinmorethanoneway.
Determinewhethertheseriesconverges.
1.
X1
n=0
n
n+4
)
2.
1
12
+
1
34
+
1
56
+
1
78
+)
3.
X1
n=0
n
(n+4)2
)
4.
X1
n=0
n!
8n
)
5. 1
3
4
+
5
8
7
12
+
9
16
+)
6.
X1
n=0
1
p
n+4
)
7.
X1
n=0
sin
3
(n)
n2
)
8.
X1
n=0
n
en
)
9.
X1
n=0
n!
135(2n 1)
)
10.
X1
n=1
1
n
p
n
)
11.
1
234
+
2
345
+
3
456
+
4
567
+)
12.
X1
n=1
135(2n 1)
(2n)!
)
13.
X1
n=0
6
n
n!
)
14.
X1
n=1
( 1)
n 1
p
n
)
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11.12 Additionalexercises
295
15.
X1
n=1
2
n
3
n 1
n!
)
16. 1+
5
2
22
+
5
4
(24)2
+
5
6
(246)2
+
5
8
(2468)2
+)
17.
X1
n=1
sin(1=n))
Findtheintervalandradiusofconvergence;youneednotchecktheendpointsoftheintervals.
18.
X1
n=0
2
n
n!
x
n
)
19.
X1
n=0
x
n
1+3n
)
20.
X1
n=1
x
n
n3n
)
21. x+
1
2
x
3
3
+
13
24
x
5
5
+
135
246
x
7
7
+)
22.
X1
n=1
n!
n2
x
n
)
23.
X1
n=1
( 1)
n
n23n
x
2n
)
24.
X1
n=0
(x 1)
n
n!
)
Findaseriesforeachfunction,usingtheformulaforMaclaurinseriesandalgebraicmanip-
ulationasappropriate.
25. 2
x
)
26. ln(1+x))
27. ln
1+x
1 x
)
28.
p
1+x)
29.
1
1+x2
)
30. arctan(x))
31. Usetheanswertothepreviousproblemtodiscoveraseriesforawell-knownmathematical
constant.)
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A
Selected Answers
1.1.1. (2=3)x+(1=3)
1.1.2. y= 2x
1.1.3. ( 2=3)x+(1=3)
1.1.4. y=2x+2,2, 1
1.1.5. y= x+6,6,6
1.1.6. y=x=2+1=2,1=2, 1
1.1.7. y = 3=2, y-intercept: 3=2, , no
x-intercept
1.1.8. y=( 2=3)x 2, 2, 3
1.1.9. yes
1.1.10. y=0,y= 2x+2,y=2x+2
1.1.11. y=75t(tinhours);164minutes
1.1.12. y=(9=5)x+32,( 40; 40)
1.1.13. y=0:15x+10
1.1.14. 0:03x+1:2
1.1.15. (a) y
=
(
0
0x<100
(x=10) 10 100x1000
x 910
1000<x
1.1.16. y
=
(
0:15x
0x19450
0:28x 2528:50 19450<x47050
0:33x 4881
47050<x97620
1.1.17. (a)P P = 0:0001x+2
(b)x= 10000P+20000
1.1.18. (2=25)x (16=5)
1.2.1. (a)x
2
+y
2
=9
(b)(x 5)2+(y 6)=9
(c)(x+5)
2
+(y+6)
2
=9
1.2.2. (a) ) x = = 2,y y =3,m m = = 3=2,
y=(3=2)x 3,
p
13
(b) x= 1,y=3,m= 3,
y= 3x+2,
p
10
(c) x = 2,y y = 2,m=1,
y=x,
p
8
1.2.6. (x+2=7)
2
+(y 41=7)
2
=1300=49
1.3.1. fxjx3=2g
1.3.2. fxjx6= 1g
1.3.3. fxjx6=1andx6= 1g
1.3.4. fxjx<0g
1.3.5. fxjx2Rg,i.e.,allx
297
298
AppendixA SelectedAnswers
1.3.6. fxjx0g
1.3.7. fxjh rxh+rg
1.3.8. fxjx1orx<0g
1.3.9. fxj 1=3<x<1=3g
1.3.10. fxjx0andx6=1g
1.3.11. fxjx0andx6=1g
1.3.12. R
1.3.13. fxjx3g,fxjx0g
1.3.14. A=x(500 2x),fxj0x250g
1.3.15. V = = r(50 r
2
), fr r j j 0 0 <r r 
p
50=g
1.3.16. A=2r
2
+2000=r,frj0<r<1g
2.1.1.  5,  2:47106145,  2:4067927,
2:400676, 2:4
2.1.2.  4=3, 24=7,7=24,3=4
2.1.3.  0:107526881,  0:11074197,
0:1110741,
1
3(3+x)
!
1
9
2.1.4.
3+3x+x
2
1+x
!3
2.1.5. 3:31,3:003001,3:0000,
3+3x+x
2
!3
2.1.6. m
2.2.1. 10,25=2,20,15,25,35.
2.2.2. 5,4:1,4:01,4:001,4+t!4
2.2.3.  10:29, 9:849, 9:8049,
9:8 4:9t! 9:8
2.3.1. 7
2.3.2. 5
2.3.3. 0
2.3.4. undened
2.3.5. 1=6
2.3.6. 0
2.3.7. 3
2.3.8. 172
2.3.9. 0
2.3.10. 2
2.3.11. doesnotexist
2.3.12.
p
2
2.3.13. 3a
2
2.3.14. 512
2.3.15.  4
2.3.16. 0
2.3.18. (a)8,(b)6,(c)dne,(d) 2,(e) 1,
(f)8,(g)7,(h)6,(i) 3,(j) 3=2,
(k)6,(l)2
2.4.1.  x=
p
169 x2
2.4.2.  9:8t
2.4.3. 2x+1=x
2
2.4.4. 2ax+b
2.4.5. 3x
2
2.4.8.  2=(2x+1)
3=2
2.4.9. 5=(t+2)
2
2.4.10. y= 13x+17
2.4.11.  8
3.1.1. 100x
99
3.1.2.  100x
101
3.1.3.  5x
6
3.1.4. x
 1
3.1.5. (3=4)x
1=4
3.1.6.  (9=7)x
16=7
3.2.1. 15x
2
+24x
3.2.2.  20x
4
+6x+10=x
3
3.2.3.  30x+25
3.2.4. 6x
2
+2x 8