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Maclaurin and Taylor Series;Power Series
Objective: To take our knowledge of Maclaurin and Taylor polynomials and
extend it to series.
Maclaurin and Taylor Series
• We defined:
• the nth Maclaurin polynomial for a function as
• the nth Taylor polynomial for f about x = x0 as
nn
kn
k
k
xxn
xfxx
xfxxxfxfxx
k
xf)(
!
)(...)(
!2
)())(()()(
!
)(0
020
0//
00/
000
0
nn
kn
k
k
xn
fx
fxffx
k
f
!
)0(...
!2
)0()0()0(
!
)0( 2//
/
0
Maclaurin and Taylor Series
• It is not a big jump to extend the notions of Maclaurin and Taylor polynomials to series by not stopping the summation index at n. Thus, we have the following definition.
Example 1
• Find the Maclaurin series for
(a) (b) (c) (d)xsinxe xcosx11
Example 1
• Find the Maclaurin series for
(a) (b) (c) (d)
(a) We take the Maclaurin polynomial and extend it.
xsinxe xcosx11
...!
...!2
1!
)(2
0
k
xxx
k
xxp
k
k
k
n
Example 1
• Find the Maclaurin series for
(a) (b) (c) (d)
(b) We take the Maclaurin polynomial and extend it.
xsinxe xcosx11
...)!12(
)1(...!5!3)!12(
)1(1253
0
12
k
xxxx
k
x kk
k
kk
Example 1
• Find the Maclaurin series for
(a) (b) (c) (d)
(c) We take the Maclaurin polynomial and extend it.
xsinxe xcosx11
...)!2(
)1(...!5!3)!2(
)1(253
0
2
k
xxxx
k
x kk
k
kk
Example 1
• Find the Maclaurin series for
(a) (b) (c) (d)
(d) We take the Maclaurin polynomial and extend it.
xsinxe xcosx11
......1 2
0
k
k
k xxxx
Example 2
• Find the Taylor series for 1/x about x = 1.
Example 2
• Find the Taylor series for 1/x about x = 1.
• We found in the last section that the nth Taylor polynomial for 1/x about x = 1 is
nnn
k
kk xxxxx )1()1(...)1()1()1(1)1()1( 32
0
Example 2
• Find the Taylor series for 1/x about x = 1.
• We found in the last section that the nth Taylor polynomial for 1/x about x = 1 is
• Thus, the Taylor series for 1/x about x = 1 is
nnn
k
kk xxxxx )1()1(...)1()1()1(1)1()1( 32
0
...)1()1(...)1()1()1(1)1()1( 32
0
kk
k
kk xxxxx
Power Series in x
• Maclaurin and Taylor series differ from the series that we have considered in Sections 10.3-10.6 in that their terms are not merely constants, but instead involve a variable. These are examples of power series, which we now define.
Power Series in x
• If are constants and x is a variable, then a series of the form
is called a power series in x.
......2210
0
kk
k
kk xcxcxccxc
,...,, 210 ccc
Power Series in x
• If are constants and x is a variable, then a series of the form
is called a power series in x. Some examples are
......2210
0
kk
k
kk xcxcxccxc
,...,, 210 ccc
...11
1 32
0
xxxxx k
k
...!3!2
1!
32
0
xxx
k
xe
k
kx
...!6!4!2
1)!2(
)1(cos6422
0
xxx
k
xx
k
k
k
Radius and Interval of Convergence
• If a numerical value is substituted for x in a power series then the resulting series of numbers may either converge or diverge. This leads to the problem of determining the set of x-values for which a given power series converges; this is called its convergence set.
kk xc
Radius and Interval of Convergence
• If a numerical value is substituted for x in a power series then the resulting series of numbers may either converge or diverge. This leads to the problem of determining the set of x-values for which a given power series converges; this is called its convergence set.
• Observe that every power series in x converges at x = 0. In some cases, this may be the only number in the convergence set. In other cases the convergence set is some finite or infinite interval containing x = 0.
kk xc
Radius and Interval of Convergence
• This leads us to the following theorem.
Radius and Interval of Convergence
• This theorem states that the convergence set for a power series in x is always an interval centered at x = 0. For this reason, the convergence set of a power series in x is called the interval of convergence.
Radius and Interval of Convergence
• This theorem states that the convergence set for a power series in x is always an interval centered at x = 0. For this reason, the convergence set of a power series in x is called the interval of convergence.
• In the case where the convergence set is the single value x = 0 we say that the series has radius of convergence 0.
Radius and Interval of Convergence
• This theorem states that the convergence set for a power series in x is always an interval centered at x = 0. For this reason, the convergence set of a power series in x is called the interval of convergence.
• In the case where the convergence set is infinite, we say that it has a radius of convergence of infinity.
Radius and Interval of Convergence
• This theorem states that the convergence set for a power series in x is always an interval centered at x = 0. For this reason, the convergence set of a power series in x is called the interval of convergence.
• In the case where the convergence set extends between –R and R we say that the series has radius of convergence R.
Radius and Interval of Convergence
• Graphically, this is what that looks like.
Finding the Interval of Convergence
• The usual procedure for finding the interval of convergence of a power series is to apply the ratio test for absolute convergence. We will see that in the following examples.
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(a)
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1|;|||limlim1
1
convergesxx
x
x
u
ukk
k
k
k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(a)
• This series converges when |x| < 1. At a value of 1, the test is inconclusive. We now need to check the endpoints individually to see if they are included.
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1|;|||limlim1
1
convergesxx
x
x
u
ukk
k
k
k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(a)
• These both diverge, so the interval of convergence is (-1, 1), and the radius of convergence is R = 1.
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
0
.....11111k
k
0
.....1111)1(k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(b)
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1;01
lim!
)!1(lim
11
converges
k
x
x
k
k
x
u
ukk
k
k
k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(b)
• This ratio is always less than 1, so the series converges absolutely for all values of x. Thus the interval of convergence is and R =
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1;01
lim!
)!1(lim
11
converges
k
x
x
k
k
x
u
ukk
k
k
k
k
),(
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(c)
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1;|)1(|lim!
)!1(lim
11
convergesxk
xk
xk
u
ukk
k
k
k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(c)
• This ratio is never less than 1, so the series diverges for all nonzero values of x. Thus the interval of convergence is x = 0 and R = 0.
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1;|)1(|lim!
)!1(lim
11
convergesxk
xk
xk
u
ukk
k
k
k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(d)
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1;3
||
2
1
3
||lim
)1(3
)2(3lim
1
11
converges
x
k
kx
x
k
k
x
u
ukk
k
k
k
k
k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(d)
• This ratio is less than 1 when |x| < 3. Again, the test provides no information when x = +3, so we need to check them separately.
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1;3
||
2
1
3
||lim
)1(3
)2(3lim
1
11
converges
x
k
kx
x
k
k
x
u
ukk
k
k
k
k
k
k
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(d)
• This is the conditionally convergent harmonic series, so x = 3 is good.
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1
)1(
)1(3
3)1(
0
kk
k
kk
kk
Example 3
• Find the interval of convergence and radius of convergence of the following power series.
(a) (b) (c) (d)
(d)
• This is the divergent harmonic series, so x = -3 is bad.• Interval (-3, 3] and R = 3.
0 !k
k
k
x
0k
kx
0
!k
kxk
0 )1(3
)1(
kk
kk
k
x
1
1
13
3)1()1(
)1(3
)3()1(
0
kkk k
kkk
kk
kk
Power Series in x – x0
• If xo is a constant and if x is replaced by x – xo in the power series expansion, then the resulting series has the form
• This is called a power series in x – xo.
0
02
020100 ...)(...)()()(k
kk
kk xxcxxcxxccxxc
Power Series in x – x0
• If xo is a constant and if x is replaced by x – xo in the power series expansion, then the resulting series has the form
• This is called a power series in x – xo. Some examples are
0
02
020100 ...)(...)()()(k
kk
kk xxcxxcxxccxxc
1...;4
)1(
3
)1(
2
11
1
)1(0
32
0
xxxx
k
x
k
k
3...;!3
)3(
!2
)3()3(1
!
)3()1(0
32
0
xxx
xk
x
k
kk
Power Series in x – x0
• The first of the previous series is a power series in x – 1 and the second is a power series in x + 3. Note that a power series in x is a power series in x – xo in which xo = 0. More generally, the Taylor Series
• is a power series in x – xo.
k
k
k
xxk
xf)(
!
)(0
0
0
Power Series in x – x0
• The main result on convergence of a power series in x – xo can be obtained by substituting x – xo for x. This leads to the following theorem.
Power Series in x – x0
• It follows from the theorem that the set of values for which a power series in x – xo converges is always an interval centered at x = xo; we call this the interval of convergence.
• We can also have convergence only at the point xo and we would say that the series has a radius of convergence of R = 0.
• The series could also converge everywhere. We say that this has a radius of convergence of infinity.
Power Series in x – x0
• Graphically, this is what that looks like.
Example 4
• Find the interval of convergence and radius of convergence of the series
12
)5(
k
k
k
x
Example 4
• Find the interval of convergence and radius of convergence of the series
|5|1
|5|)5()1(
)5(limlim
22
2
11
x
k
kx
x
k
k
x
u
uk
k
kk
k
k
12
)5(
k
k
k
x
Example 4
• Find the interval of convergence and radius of convergence of the series
|5|1
|5|)5()1(
)5(limlim
22
2
11
x
k
kx
x
k
k
x
u
uk
k
kk
k
k
12
)5(
k
k
k
x
64
1|5|
x
x
Example 4
• Find the interval of convergence and radius of convergence of the series
• Checking the endpoint at 4 by replacing x with 4.
• This converges absolutely. It also converges by the alternating series test.
12
)5(
k
k
k
x64
1|5|
x
x
21
2
)1()5(
kk
x k
k
k
Example 4
• Find the interval of convergence and radius of convergence of the series
• Checking the endpoint by replacing x with 6.
• This is a convergent p-series. The interval of convergence is [4, 6] and the radius of convergence is R = 1.
12
)5(
k
k
k
x64
1|5|
x
x
21
2
1)5(
kk
x
k
k
Homework
• Section 9.8• Page 667• 1, 3, 5, 11, 13, 15, 19, 21,• 30 – 48 mult. of 3
