1m
0
AND SERIES
CHAPTER
12 INFINITESEQUENCES
12.10 Taylor and Macla
Series
1. Using Theorem5 with ~
~ bn (x - 5)n . bn =
5.
~
n-O
J.- - -.j<8)(5)
VII
"so
n.
8!
2. (a) Using Fonnula6, a powerseriesexpansionof 1 at 1 musthavethefonn1(1} + f (1}(x - I) +.. '. Comparing
= -0.8. Butfromthegraph,f(l}
to thegivenseries,1.6- 0.8(x - I} + ..', wemusthavef(l}
1 centered at 1.
= 2, so 1"(2}
=-1.5; that is, 1"(2} is positive.But from the graph,1 is concavedownwardnearx
~f'(2}
+
2}2
we must have
-
~f'(2}(x
+
- 2}3 + .",
2}
f(2}(x
+
1(2}
fonn
-
(1 + X)-3 =
- 2} + 1.5(x - 2}2 - O.I(x
the
have
must
2
at
1
to the given series, 2.8 + 0.5(x
expansion"of
series
power
Comparing
A
(b)
Hence, the given series is not the Taylor series of
is positive.
must
be negative.Hence,the given seriesis not the Taylor seriesof 1 centeredat 2.
lim
I
~
n-+oo
soR= 1 (Rat
WeuseEquation7 with /(x) = CDS
x.
3.
n
f(n) (x)
0
COB
X
1
f(n) (o)
-sinx
~3!
cosx = 1(0)+ j'(O)x + L{22X2
+
2!
1
0
2
4
= 1- =2. + x4!
ao
-
-1
f4)
(0)
x 3 +~X
4
+...
6.
)nx2n
(2n)!
n
~
0
1r
(1
2
-cos:i
3
sin x
0
1
4
..
cosx
..
1
..
2
.
.
11m
. -an+1
I
I
-1
E(
n=O
,
.
=. lim
I
,
x 2n+2
-(2 n ) . = x 2. lim
n-oo an
n-oo (2n + 2)!
So R = 00 (Ratio Test).
x2n
n-oo (2n + 2) (2n + 1)
fn)(o)
4.
nj(n)(x)
= 0 if n is evenandf2n+l)
j(n) (0)
sm2x
.
0
sin2x
0
1
2cos2x
2
2
-22sin2x
0
3
-23cos2x
-23
4
..
24sin2x
..
0
..
.
.
.
= 0 < 1 for all x.
1
I
= OX) -j<n)(o
l::
n=O n!
lim
n-+oo
so R
l~
an
= 00
l
)
(0) = (-1)n22n+l, so
xn=
OX)
l::
f (2n+l)
(
0
) X2n+l
n=O (2n + 1)!
lim
.
22 Ixl2
= n-oo (2n + 3)(2n + 2) = 0 < 1 for all x,
(Ratio Test).
I :
an
7.
-(
3
2(:
4
-61
5
24(
12
ANDSERIES
-X
j<n)(O)
n=~~Xn=L."X
L.,
00
I
'.
n=l
=L.,(n-l)'"
n.
~00
Xn
12.
n=l
n=O
n
I"
'
Ix!
0
=
Ixln
n!
.-
]
-"
1m
I"
I =
lim I ~
n--+oo an
n--+oo
(n-1)!
0
[ I:i:ln+l
1
n.
n=O
~nn
00
n
lor
:5::00
-
<
-
=
xe
-
:t
8.
1m
984 0
n-+oo n
1
all x, so R = 00.
2
3
4
5
(n)
9.
f
n
j<n)
0
sinh x
.0
1
coshx
1
2
3
sinh x
. coshx
0
1
(x)
(0) =
{
J<~?(O)
0 if n is even
.
sinh x
0
.
.
.
.
.
.
.
.
.
x2n+l
~
n=o(2n+l)!
1 ifnisodd
.
x2n+l
13. Clearly,j<n)
Usethe RatioTestto find R. If an = (2n + 1)" then
1im I ~
Jim
I
~
n-+<x>an
4
IX)
so sinh x =
I
=
.
Jim
n-+(X)
X2n+3
I
(2n
n-+(X)
.
(2n+
an
1)!
1
+ 3)!
x2n+l
14.
Ii
1
=x 2 . n-oo
m (2n+ 3)(2~+ 2) = 0 < 1
for all x, soR = 00.
10.
rn)(O) = { ~
if n is even
~
so cosh x =
if n is odd
Usethe RatioTestto find R. If an =
~'X2n
x2n
l:: -.
n=O (2n)!
then
f(n){2)= i
lim I ~
"-00 an
I=
X2n+2
.
lim.n-+(X) (2n + 2)!
I
=x 2 . lim
n-+oo
(2n)!
X2n
lim I~
I
n-+oo an I
1=0<1
(2n + 2)(2n + 1)
15.
for all x, so R = 00.
11.
f(x)
= 7+ 5(x -
IX)
,(x-
0
2 - 2)2+ L 2) + 2i(x
2)n
n=3n.
= 7 + 5(x - 2) + (x - 2)2
Sincean = 0 for largen, R = 00.
~
cosx = E :
k=O
Jim I~
n-.~
an I
~
-
0
CHAPTER
12
INFINITE SEQUENCES AND SERIES
. j<k>(7r/2)(X- ~)k
SlllX - k~O k!
2
=
00
16.
=1-
(x
-
19. If f(x)
1r/2)2 + (X - 1r/2)4 - (X - 1r/2)6 + . . .
2!
.
4!
andM=:
6!
by Theorer
20.
If
=f
f(x)
andM=1
lim [ Ix -
l =
lim l ~
n-+~
"--+(X) a,.
1I"/212n+2
lim R..(x
.
(2n + 2)!
(2n)!
] =
Ix - 11"
/2fn
lim
Ix - 1I"/2f
= 0 <'1 for all x,
n-"~ (2n + 2)(2n + 1)
n-+(X)
21. If f(x)
soR = 00.
=s
haveIf(ft+1
so by Forml
asn-+oof
represents si
22.If f(x) = c(
haveI/(n+l:
so by Formu
fi1 -.!
- 3 - ~1
(x
f: (-l)nl.3.5.
= n=O
lim l ~
l
n-+CX) an
3 (x - 9)2
- 9) + 22-:-,3"5
2! - ~3. 5
... .(2n-l)
2n . 32n+l . n!
.
[ 1.3.5
= lim.
"-00
.
(x - 9)
n
(x - 9)3
3!
as n -+ 00 f(
+ . ..
represents co
(XJ
23. coax = L (
.,=0
.
(2n - 1)[2(n + 1) - 1] Ix - 9,"+1
2"+1 .3[2(,,+1)+1]. (n + I)!
+ 1) Ix - 91
]
= n~~ (2n
2. 32(n+ 1)
[
2" .32"+1 . n!
"]
1 .3.5. ... . (2n - 1) Ix - 91
24. ex =
(XJ x"
L -;.,=0 n.
25.tan-l x =
= ! Ix- 91 < 1
9
0
}:
n=
(X)
26.sinx=E(n=O
for convergence,
so Ix - 91< 9 and R = 9.
18.
X-2
=1-
2(x - 1) + 6 .
~.!f
2!
-
24 .
~~
3!
+ 120.!=-=.!t - . . .
4!
27. eZ =
= 1 - 2(x - 1) + 3(x - 1)2- 4(x - 1)3+ 5(x - 1)4- ...
(X) xn
E -;
n=O n.
(X)
28. cosx = E (
n=O
~
=L
(-l)"(n
+ l)(x
- 1)".
n=O
f(x)
= xcos~
29. sin2 x = .!(l
2
lim
n-+oo
I~
I
an
I"
(n + 2) Ix - Iln+l
- n--oo
1m (n + 1) Ix - Iln
-
=
.Ix
Jim [ ~
n-cx> n + 1
-
11 ]
=
.
(X) (
Ix -11
< 1 for convergence, so R = 1.
Ln=l