Unformatted text preview: 9.5 POWER SERIES AND INTERVAL OF CONVERGENCE !
if r > 1, then
! an diverges 8.5 Power Series
94h94 an could converge or diverge.
if r = 1, then
√
s test works since limn→∞ n an = r tells us that the ses comparable to a geometric series with ratio r.) Use this
o determine the behavior of the series. 94. ∞ # $
"
2 n 94h95 n n=1 95. 521 ∞ #
$
"
5n + 1 n
n=1 3 n2 Announcements:
"
• Webassign
due Friday
108.
(−1) cos(2 πn) is an alternating series.
!
The series
(−1) /n converges by the alternating
• Written "
HW due Sunday
series test.
!
109. The series
(−1) 2 converges.
•
Quiz
next
Monday
The series
1/(n + 1) converges by the ratio test. ngthen Your Understanding ∞ oblems ??, explain what is wrong with the statement.
94h108
2n 2 The series
!
2
1/n . ∞ 94h109 2 ! n n=0 n n n=0 1/n3/2 converges by comparison94h110
with 110. If ! a converges, then !(−1)n a converges.
n
n
94h111 111. If an alternating series converges by the alternating series
test, then the error in using the first n terms of the series
to approximate the entire series is less in magnitude than
!∞
that converges but
a

dithe
first term omitted.
n
n=1 oblems ??, give an example of: A series
verges. !∞ n=1 an 94h112 112. If an alternating series converges, then the error in using
the first n terms of the series to approximate the entire
series is less in magnitude than the first term omitted. 94h113 113. If 94h114 114. !
To find the sum of the alternating harmonic series
(−1)n−1 /n to within 0.01 of the true value, we can
sum the first 100 terms. An alternating series that does not converge. A series ! an such that
an+1 
lim
= 3.
n→∞ an  ! an  converges, then ! (−1)n an  converges. !
de if the statements in Problems ?? are true or false.94h115
Give
115. If
an is absolutely convergent, then it is convergent.
xplanation for your answer.
!
94h116 116. If
an is conditionally convergent, then it is absolutely
If the terms sn of a sequence alternate in sign, then the
convergent.
!
sequence converges.
94h117 117. If a > 0.5 b > 0 for all n and
bn diverges, then
n
!
!n
If
0
≤
a
≤
b
for
all
n
and
a
converges,
then
n
n
n
a
diverges.
n
!
bn converges.
94h118 118. Which test will help you determine if the series converges
!
!
If 0 ≤ anSection
≤ bn for
n and
bn
or diverges?
8.5 all
Power
Series an diverges, then
2010 Kiryl Tsishchanka
∞
The goal of this section is to show how series like below "
are
used
diverges.
1 to approximate Power series: ! x sinx, cosx,..
functions
such
If bn ≤ an ≤
0 for all n
and as ben ,converges,
thenPower Series
k3 + 1
!
k=1
an converges.
!
!
(a)
Integral
test
A power
series
is a series of the form and x is a variable, then a series of the form
DEFINITION:
If
an converges,
then
aIfn cconverges.
0 , c1 , c2 , . . . are constants(b)
Comparison test
! If 5 (c) Ratio test ∞
an converges, then lim an+1 /an!
 ̸= 1.
n→∞ c n xn = c 0 + c 1 x + c 2 x2 + . . . + c n x n + . . . n=0
POWER SERIES AND INTERVAL
OF CONVERGENCE is called a power series in x and a series of
! then form
In Section
saw that the
axcalled
converges
for −1 < xof<the
1and
diverges
where ??
x∞iswe
a variable
andgeometric
c ’s are series
constants
the coefficients
series.
! section studies then convergence of more general series constructed from powers.
otherwise. This
n
2
n
c
(x
−
x
)
=
c
+
c
(x
−
x
)
+
c
(x
−
x
)
+
.
.
.
+
c
(x
−
x
)
+
. .cos
. x,
x
n
0
0
1
0
2
0
n
0
Chapter
shows how such power series are used to approximate functions such as e , sin x,
More??generally:
n=0
and ln x. is called a power series in x − x0 . A power series about x = a is a sum of constants times powers of (x − a): EXAMPLES:
∞
∞Section 8.5 Power Series
"
!
2
n
C01++Cx1 (x
Cn (x − a)n .
2 (x
1.
xn =
+−
x2a)
++x3C+
. .−
. a) + · · · + Cn (x − a) + · · · = 2010 Kiryl Tsishchanka n=0
Power Series n=0 ∞
!
xn x2 x 3
2. DEFINITION:
= 1 + x + If+c , c+, c. . ,. . . . are constants and x is a variable, then a series of the form
n!
2! 03! 1 2
n=0
3. ∞
! (−1)n n=0 ∞ x2n
x2 x4 !x6 n
2
n
x . .=
=1−
+
− cn+
. c0 + c1 x + c2 x + . . . + cn x + . . .
(2n)!
2!
4!
6!
n=0 ∞
!
(x − 1)n power series (x
− 1)2 (x
− 1)3 of the form
4. is called a =
1 + (x − 1) + in x and
+ a series +
...
n!
2!
3!
∞
n=0 5. ∞
! n (x (−1) ! n
2
n
2 − x0 ) + 3c2 (x − x0 ) + . . . + cn (x − x0 ) + . . .
0+
1 (x
+ 2)n cn (x − x0 ) = c(x
+c2)
(x + 2)
−
+ ...
n=0 = 1 − (x + 2) +
n!
2!
3! Some
examples:
is called a power series in x − x .
n=0 THEOREM: For any power series EXAMPLES:
∞ ∞
! 0 cn xn exactly one of the following is true: n=0 (a) The !
series nconverges only2for x3= 0. 1. x = 1 + x + x + x + ... (b) The n=0
series converges absolutely (and hence converges) for all real values of x.
∞ 3
(c) The !
seriesxnconverges absolutely
hence converges) for all x in some finite open interval
x2 x(and
= 1 +ifxx+< −R+or x +
. . At either of the values x = −R or x = R, the series
(−R,2.
R), andn!
diverges
> .R.
2!
3!
n=0
may converge absolutely, converge conditionally, or diverge, depending on the particular series.
∞ !
x∞
x
x
!
n x
3.
(−1)
1−
+
− )n , +
. . . one of the following is true:
THEOREM:
For a(2n)!
power=series
exactly
2! cn (x4!− x06!
n=0 2n 2 4 6 n=0 ∞
(a) The series
converges
!
(x
− 1)n only for x = x0 . (x − 1)2 (x − 1)3
= 1 + (x − 1) +
+
+ ...
n!
2!
(b) The series
converges absolutely (and hence converges) 3!
for all real values of x.
n=0
4. ∞ converges absolutely
n
(c) The series
(and hence converges)
in3 some finite open interval
!
(x + 2)2 for(xall+x2)
n (x + 2)
−x(x
+R2)or+x > x0 + R. −
+ . .values
.
(x0 −5.
R, x0 + (−1)
R), and diverges =
if x1 <
At either of the
x = x0 − R or
0−
n!
2!
3!
x = x0 + n=0
R, the series may converge absolutely, converge conditionally, or diverge, depending
on the particular series.
∞
! cn xn exactly one of the following is true: THEOREM: For any power series n=01 (a) The series converges only for x = 0. (b) The series converges absolutely (and hence converges) for all real values of x. the series alternates and is positive for n = 2, we multiply by (−1)n . For n = 2, we divide by 4, for
n = 3 we divide by 9, and in general, we divide by n2 . One way to write this series is
∞
"
(−1)n (x − 2)2n
.
n2
n=2 Practice: Chapter Nine SEQUENCES AND SERIES mple 2 ion !
∞
We think of a as a constant. For"
anyxnfixed x, the power series
Cn (x − a)n is a series of
Determine whether the power series
converges or diverges for
n
numbers like those considered in Section2??. To investigate the convergence of a power series, we
n=0
consider the partial sums, which in this case are the polynomials Sn (x) = C0 +C1 (x−a)+C2 (x−
(a)2 x = −1
(b) x = 3
a) + · · · + Cn (x − a)n . As before, we consider the sequence8 (a) Substituting x = −1, we S
have
0 (x), S1 (x), S2 (x), . . . , Sn (x), . . . .
$n
∞
∞
∞ #
"
"
"
xn
(−1)n
1
=
=
.
−
2n
2n
2
For a fixed value of x, if this
sequence
of partial
sums
converges
to a limit S, that is, if
n=0
n=0
n=0
522
Chapter Nine SEQUENCES
AND SERIES
lim Sn (x)
= S, then we say that the power series converges to S for this value of x.
n→∞
This is a geometric series with ratio −1/2, so the
to 1/(1 − (− 12 )) = 2/3.
! series converges
We think of a as a constant. For any fixed x, the power series
Cn (x − a)n is a series of
(b)
Substituting
x
=
3,
we
have
numbers like those considered in Section ??. To investigate the convergence of a power series, we
= C0 ∞
+C1#
(x−a)+C
consider the partial sums, which in this case are ∞
the polynomials
$n 2 (x−
∞Sn (x)
A power series may converge for
some
n values
n x and
"the
"
"not
8
3for others.
xsequence
3of
a)2 + · · · + Cn (x − a)n . As before, we consider
=
=
.
n
n
2
S0 (x), S1 (x), S2 (x), 2. . . , Sn (x), . .2. .
!
n=0
n=0
n=0
mple 1
Find an expression for the general term of the series and use it to write the series using
notation:
This is a geometric series withofratio
greater
than 1, so it diverges.
For a fixed value of x, if this sequence
partial sums
is, if
4
6 converges to8 a limit S, that 10
(x − 2)
(x − 2)
(x − 2)
lim Sn (x) = S, then we(x
say −
that2)
the power
− series converges
+ to S for this−value of x.
+ ···.
n→∞
4
9
16
25 A power series may converge for some values of x and not for others.
merical
and
View
Convergence
ion
The Graphical
series is about
x = 2ofand
the odd terms are zero. We use (x − 2)2n and begin with n = 2. Since Example 1 Solution !n . For n = 2, we divide by 4, for
the
alternates
is term
positive
for nand
=use
2, itwe
multiply
byusing
(−1)
Findseries
an expression
for the and
general
of the series
to write
the series
notation:
Consider
the
series
2
n = 3 we divide by
9,
and
in
general,
we
4
6
8divide by10n . One way to write this series is
(x − 2)
(x 2− 2)
(x − 2)3
(x − 2)
4
(x − 1)n
(x
−
1)
(x
−
1)
(x −− 1)
+
−
+ ···.
9+
16∞ − 25
+2n
· · · + (−1)n−1
+ ···.
(x − 1) − 4
n
2
3 " (−1) 4(x − 2)
n
.
2 and begin with n = 2. Since
The series is about x = 2 and the odd terms are zero. We use (x − 2)n2n
To
investigate the convergence of thisn=2
series,
we look at the sequence of partial sums graphed in
the series alternates and is positive for n = 2, we multiply
by (−1)n . For n = 2, we divide by 4, for
n=
8 3 we divide by 9, and in general, we divide by n2 . One way to write this series is Here we call the first term in the sequence S (x) rather than S (x) so that the last term of S (x) is C (x − a) .
Practice:
0 mple 2 Example 2 Determine whether the power series
∞ (a)
x =whether
−1 the power series " x
Determine
n=0 ion Solution 1 n n n ∞
"
(−1)n (x − 2)2n
.
n2
n=2 n 2n converges or diverges(b)
for (a) x = −1
(a)
Substituting x = −1, we have
(a) Substituting x = −1, we have ∞
"
xn
converges or diverges for
2n
n=0 ∞
" (b) x=3 x=3
∞
" n n x
(−1)
$n
∞
∞
∞ #
n= "
"
"
xn
(−1)
1
n =
=
− 2n .
2
n
n
2
2
2
n=0
n=0
n=0
n=0
n=0 = $n
∞ #
"
1
.
−
2
n=0 This
a geometric
series with
ratio with
−1/2, ratio
so the series
converges
1/(1 − converges
(− 12 )) = 2/3.to 1/(1 − (− 1 )) = 2/3.
Thisis is
a geometric
series
−1/2,
so thetoseries
2
(b) Substituting x = 3, we have
(b)
Substituting x = 3, we have
∞
∞
∞ # $n
"
"
"
3
xn
3n
∞
∞ .n
∞ # $n
= "
=
n
"
3
2n
2nx n=0 "
2 3
n=0
n=0
=
=
.
n
n
2
2
2
This is a geometric series with ratio greater
than
1,
so
it
diverges.
n=0
n=0
n=0 This is a geometric series with ratio greater than 1, so it diverges. Numerical and Graphical View of Convergence
Consider the series (x − 1)
(x − 1)
(x − 1)
−
+ · · · + (−1)
(x − 1) − View+of Convergence
merical and Graphical
2
3
4 2 2 − 1)n
+ ···.
n
To investigate
convergence of this series, we look at the sequence of partial sums graphed in
Consider
thetheseries
8 Here 3 4 n−1 (x n
we call the first term in the sequence 2S0 (x) rather than S13(x) so that the last term
4 of Sn (x) is Cn (x − a) . (x − 1)n
(x − 1)
(x − 1)
(x − 1)
+
−
+ · · · + (−1)n−1
+ ···.
2
3
4
n
To investigate the convergence of this series, we look at the sequence of partial sums graphed in
(x − 1) − term in the sequence S (x) rather than S (x) so that the last term of S (x) is C (x − a) .
Convergence
of power series:
!
! 8 Here we call the first
Chapter Nine SEQUENCES AND SERIES 0 1 For
fixed
the
power
Cn (x
a)nfixed
is ax,series
of series Cn (x − a)n is a series of
22 any
Chapter
Ninex,
SEQUENCES
AND
SERIES
We
think
of
a series
as a constant.
For−any
the power
Section ??. To
investigate
theconsidered
convergence
of a power
we the convergence
!
numbers
like those
in Section
??. To series,
investigate
of a power
series, we
n
think
of a sums,
as a constant.
fixed
x, polynomials
the power series
CnC(x+C
− a)(x−a)+C
is a series(x−
of
the
partial
which
this
case
are the
Sn (x) =
= Cin0For
+Cany
this case areconsider
theWe
polynomials
Sn (x)
0
1
2
1 (x−a)+C
2 (x−
numbers
like
those
considered
in Section
??. To investigate
the8 convergence of a power series, we
2
n
8
+
·
·
·
+
C
(x
−
a)
.
As
before,
we
consider
the
sequence
a)
n
e, we consider
the sequence
consider
the partial sums, which in this case are the polynomials Sn (x) = C0 +C1 (x−a)+C2 (x−
8
S (x), S (x), S (x), . . . , S (x),
.... + · · · + Cn (x − a) . As before,
we1 consider
2 the sequence
n
), S1 (x), S2a)(x),
. . . , Sn (x), . . . . 0
2 n S0 (x), S1 (x), S2 (x), . . . , Sn (x), . . . .
For a fixed value of x, if this sequence of partial sums converges to a limit S, that is, if
sequence of partial
to athat
limit
S, that
is, if
lim Snsums
(x) =converges
S, then we say
the power
series
converges to S for this value of x.
n→∞
For a fixed value of x, if this sequence of partial sums converges to a limit S, that is, if
hat the power series
to S for this value of x.
lim Sconverges
n (x) = S, then we say that the power series converges to S for this value of x.
n→∞ A power series may converge for some values of x and not for others. mayothers.
converge for some values of x and not for others.
for some values Aofpower
x andseries
not for !
Find an expression for the general term of the series and use it to write the series using
notation:
!
4
6
8
10
!
Example 1
Find an expression for(x
the−general
term
it to−write
notation:
2)
(x −of
2)the series
(x −and
2) use (x
2) the series using
−
+ ···.
term of the series and use it to write the −series using+ notation:
4 2)4 (x −9 2)6 (x −
162)8 (x −25
(x −
2)10
−
+
−
+ ···.
(x − 2)6 (x − 2)8 (x − 2)104
9
16
25
+The series is −
+the
· · ·odd
. terms
olution
about x = 2 and
are zero. We use (x − 2)2n and begin with n = 2. Since
9
16
25
the series alternates and is positive for n = 2, we multiply by (−1)n2n
. For n = 2, we divide by 4, for
Solution
The series is about x = 2 and the odd terms are zero. We
use
(x
− 2) and begin with n = 2. Since
n = 3 we divide by 9, and in general, we divide by n2 . One way ton write this series is
the series alternates and is2npositive for n = 2, we multiply by (−1) . For n = 2, we divide by 4, for
dd terms arenzero.
We
use
(x
−
2)
and
begin
with
n
=
2.
Since
2
∞ divide n
2n way to write this series is
"
One
= 3 we divide by 9, and in general, we
(−1)by(xn−. 2)
for n = 2, we multiply by (−1)n . For n = 2, we
divide by
4, for.
∞
n n2
2n
"
2
(−1)
(x
−
2)
n=2
ral, we divide by n . One way to write this series is
.
n2
n=2 xample 1 ∞
"
(−1)n (x − 2)2n
2 . ∞ n n n n=0
n=0
n=0
Figure ??. The graph suggests that
the partial
sums converge
for x in
2). In Exam!the interval n(0,
1 is
9.5
POWER
SERIES
INTERVAL
OF
Weand
think
of we
a asshow
aseries
constant.
any
fixedsox,the
the
power
series
C1/(1
a) CONVERGENCE
series ofof 523
ples This
??
thatwith
theFor
series
converges
for
0 <
x
≤ AND
2. to
This
is −
called
thea interval
is a ??,
geometric
ratio
−1/2,
series
converges
(−
n (x
2 )) = 2/3.
numbers
likeofthose
convergence
this
series.
(b)
Substituting
x =considered
3, we havein Section ??. To investigate the convergence of a power series, we
Figure
??.
The
graph
suggests
that
the
partial
sums
converge
for
x
in
the
interval
(0,
In Exam(x) = C0 quite
+C1 (x−a)+C
consider
partial
sums,
which in
this
case
are
polynomials
$nnconverge
At x the
=1
.4, which
is inside
the"
interval,
the
series
appears
to
rapidly: 2).
2 (x−
∞
∞the
∞ # S
n
"
"
2 ?? and ??, we show
xconsider
3n sequence
ples
thewe
series
converges
for 0 83< x . ≤ 2. This is called the interval of
+ · · · + Cn (x − a)n . As that
before,
the
a)
=
=
. . 2nS7 (1.4) =20.33653 . . .
S5 (1.4) = 0.33698.
2n
convergence of this series.
n=0
S0 (x),
S1 (x), Sn=0
(x), . . . n=0
, Sn (x), . . . .
At x = 1.4, which Sis6 (1
inside
interval,
appears
to converge
quite rapidly:
.4) =the
0.33630
. .2.the series
S (1.4)
= 0.33645
...
This is a geometric series
with ratio greater than 1,8 so it diverges.
Table ?? shows the results
of using
x = 1.9
andSx (1
=.4)
2.3=in0.33653
the power
series. For x = 1.9,
(1
.4)
=
0.33698.
.
.
.
.
.
S
5 this sequence of partial sums
7
Forisa inside
fixed value
of x, ifof
convergesthe
to aseries
limitconverges,
S, that is, though
if
which
the interval
convergence but close to an endpoint,
S6which
(1
.4)that
=
0.33630
. series
. . interval
S8 (1of
.4)convergence,
=to0.33645
. the
. .value
lim
Sn (x)For
= S,
then
we
say
powerthe
converges
S for this
of x.
rather
slowly.
x =
2.3,
is the
outside
series
diverges: the
n→∞
Numerical and
ofmore
Convergence
largerGraphical
the value
ofView
n, the
wildly
the series
contribution
of theFor
twentyTable
?? shows
the
results
of using
x = oscillates.
1.9 and x In=fact,
2.3 the
in the
power series.
x = 1.9,
fifth
term
isinside
about
28;interval
the contribution
of the hundredth
about −2,500,000,000.
Figure though
??
which
is the
the
of convergence
but close term
to anisendpoint,
the series converges,
Consider
series
A
power
series
may
converge
for
some
values
of
x
and
not
for
others.
shows
interval
convergence
and is
theoutside
partial the
sums.
ratherthe
slowly.
Forofx(x
=
2.3, which
interval of convergence,
the series diverges: the
n
(x − 1)3
(x − 1)4
− 1)2
n−1 (x − 1)
+ wildly the
− series
+ · · Partial
· + (−1)
· . of the twenty− 1) of
− n, the
larger the(xvalue
more
oscillates.
In fact,
S
Table
9.1
sumsthe
for contribution
series+
in· ·!
2 14 (x) S5 (x)3
4
n
Interval for the
xample 1
Find
an
expression
term of of
thethe
series
and use
to write
using
✲ thegeneral
fifth✛
term
isofabout 28;
contribution
hundredth
term
isx about
−2,500,000,000.
Figure ??
Example
??it
with
=the
1.9series
inside
intervalnotation:
convergence
To investigate
the convergence of this series, we look
at
the
sequence
of
partial
sums
in
9.5sums.
POWER SERIES
AND INTERVAL graphed
OF CONVERGENCE
523
shows the interval of(x
convergence
6 partial
− 2)4
(xand
− 2)the
(xof−convergence
2)8
(x − and
2)10 x = 2.3 outside
8 Here we call the first term in the sequence
n
−
+ than S1 (x)−so that the last term
+ · ·of· .Sn (x) is Cn (x − a) .
S0 (x) rather
(x)
S
Table
9.1
Partial
sums
for
series
in
4 14
9
16
25
S5 (x)
Interval
n
S
(1
.9)
n
S
(2
.3)
n
n
✛ graph
✲
Figure ??. The
that the partial sums
converge
the interval
of suggests
Example
?? with for
x =x
1.9ininside
interval (0, 2). In Examconvergence
2.3
2We use
0.495
0.455
2n
x terms are
olution
The series
about
x = 2that
and the
odd
zero.of
(x0− 2)
with
nis
= called
2. Since the interval of
convergence
and
=2.
2.3This
outside
ples ?? and
??, iswe
show
the
series
converges
for
<
xand2x≤begin
1
1.9
3
n
5
0.69207
1.21589
the series alternates and is positive for n sec2tab1
= 2, we multiply
by
(−1) . For 5n = 2, we
divide by 4, for
convergence
of
this
series.
2n
Sn (1
way
to.9)
write 8thisn series
isSn (2.3)
n = 3 we divide by 9, and in general, we divide by n8 . One0.61802
0.28817 Numerical and Graphical view of
convergence: At x = 1.4, which is inside
the
series appears
to
converge
quite rapidly:
2.3the interval,
0.495
0.455
11 2 1.71710
n 11 2 2n0.65473
x∞
1 fig9h11 1.9 " (−1) (x − 2)
S8 (x)
3
. 0.69207 14 5 −0.70701
1.21589
14 5 0.63440
S7 (1.4) = 0.33653 . . .
8
0.61802
8
0.28817 sec2tab1 n2
.4) = 0.33698.
..
n=2
x =S
25S(1
11 (x) S6for(1
.4) =
Figure 9.11: Partial sums
series
in 0.33630 . . .
S8 (x)
Example ?? converge for 0 < x < 2 .4) =
11S8 (1
0.65473
14 0.33645
. . 1.71710
.
11 14
−0.70701
∞
Table ?? shows thex =
results
of using
x =POWER
1.9SERIES
and AND
x INTERVAL
= 2.3OFinCONVERGENCE
the power
series. For x = 1.9,
"
xn 9.5
2 S11of(x)
Notice whether
that the the
interval
convergence,
0<x≤
is centered
on x = 1. Since the523
Determine
power
series
converges
or2,diverges
for
n
isextends
inside
the
interval
of
convergence
but
close
to
an
endpoint,
the
seriesinterval
converges, though
2
one
unit
on sums
eitherfor
side,
weinsay
the radius of convergence of this series is 1.
n=0
Figure
9.11
:
Partial
series
Figure ??.
Thexgraph
suggests
that theispartial
sums the
converge
for x inoftheconvergence,
interval (0, 2). In the
Examrather slowly.
For
= 2.3,
interval
series diverges: the
(a)
x=
Example
????,
converge
for that
0which
< the
x <series
2 outside
ples
??
and−1
we show
converges(b)
for 0x<= x3 ≤ 2. This is called the interval of
tervals
Convergence
largerofthe
value ofofn,
more wildly the series oscillates. In fact, the contribution of the twentyconvergence
thisthe
series.
Notice that
interval
of convergence, 0 < x ≤ 2, is centered on x = 1. Since ...
View
Full Document