4.3. POWER SERIES 149
Example 4.58 Find the interval of convergence of the series:
1
X
nD1
x
n
n
2
Solution:
Another natural job for the ratio test.
lim
n!1
Λ
Λ
Λ
Λ
a
nC1
a
n
Λ
Λ
Λ
Λ
D lim
n!1
Λ
Λ
Λ
Λ
x
nC1
=.n C1/
2
x
n
=n
2
Λ
Λ
Λ
Λ
D lim
n!1
Λ
Λ
Λ
Λ
n
2
.n C1/
2
ξ x
Λ
Λ
Λ
Λ
D jxj ξ lim
n
!1
n
2
n
2
C 2n C 1
D jxj ξ1 D jxj
So ξ1 < x < 1, and the radius of convergence is r D 1. To ο¬nd the interval of convergence we
need to check the ends of the interval. ξ©ese are of the form
1
X
nD1
.Λ1/
n
n
2
ξ©e absolute value of each of these is a p-series with p D 2. Both endpoints correspond to
series that converge in absolute value β meaning they both converge. ξ©is makes the interval of
convergence Εξ1; 1ξ.
Λ
4.3.1 USING CALCULUS TO FIND SERIES
It is sometimes useful to represent functions as power series. ξ©e geometric series formula
(Knowledge Box 4.7) does this, for example, for the function f .x/ D
1
1 ξx
. We can use calculus
to derive power series for other functions.
150 4. SEQUENCES, SERIES, AND FUNCTION APPROXIMATION
Example 4.59 If jxj < 1, then the geometric series formula tells us that:
1
X
nD0
x
n
D
1
1 ξx
If we plug ξx into this identity we ο¬nd that:
1
X
nD0
.ξx/
n
D
1
1 ξ.ξx/
1
X
n
D
0
.ξ1/
n
x
n
D
1
1 Cx
or
1
x C 1
D 1 ξ x C x
2
ξ x
3
C x
4
ξ ξξξ
Integrate both sides and we get:
ln.x C 1/ C C D x ξ
1
2
x
2
C
1
3
x
3
ξ
1
4
x
4
C
1
5
x
5
ξ ξξξ
Plug in x D 0
ln.1/ CC D 0
C D 0
Which means, at least when jxj < 1,
ln.x C 1/ D
1
X
nD1
.ξ1/
nC1
x
n
n
Λ
ξ©is shows that we can ο¬nd power series that, when they converge, are equal to familiar tran-
scendental functions. ξ©e next section gives another technique for doing this, to be used when
the sequences you already know donβt give you enough power. Letβs encode this as a Knowledge
Box.
Knowledge Box 4.22
Using calculus to modify power series
Suppose that f .x/ D
1
X
nD0
a
n
x
n
. ξ²en:
Z
f .x/ ξ dx D
1
X
nD0
a
n
n C 1
x
nC1
and f
0
.x/ D
1
X
nD1
n ξ a
n
x
nξ1
:
4.3. POWER SERIES 151
Example 4.60 Find a power series for tan
ξ1
.x/.
Solution:
We already have seen that:
1
u C1
D 1 ξ u C u
2
ξ u
3
C u
4
ξ ξξξ
If we substitute u D x
2
we obtain:
1
x
2
C 1
D 1 ξ x
2
C x
4
ξ x
6
C x
8
ξ ξξξ
Integrate and we get:
tan
ξ1
.x/ C C D x ξ
1
3
x
3
C
1
5
x
5
ξ
1
7
x
7
C
1
9
x
9
ξ ξξξ
Substitute in x D 0 and we get C D 0. So we obtain the power series:
tan
ξ1
.x/ D
1
X
nD0
.ξ1/
n
x
2nC1
2n C1
Λ
Example 4.61 What is the interval of convergence for the series for
f .x/ D tan
ξ1
.x/
found in Example 4.60?
Solution:
Apply the ratio test:
lim
n!1
Λ
Λ
Λ
Λ
a
nC1
a
n
Λ
Λ
Λ
Λ
D lim
n!1
Λ
Λ
Λ
Λ
x
2nC3
=.2n C3/
x
2nC1
=.2n C1/
Λ
Λ
Λ
Λ
D lim
n!1
Λ
Λ
Λ
Λ
x
2
2n C1
2n C3
Λ
Λ
Λ
Λ
D jx
2
j lim
n!1
2n C1
2n C3
D jx
2
j ξ1 D x
2
152 4. SEQUENCES, SERIES, AND FUNCTION APPROXIMATION
So x
2
< 1 when ξ1 < x < 1, making the radius of convergence r D 1.
Now check the endpoints x D Λ1.
If x D ξ1 the resulting series is:
n
X
nD0
.ξ1/
n
.ξ1/
2nC1
2n C1
D
n
X
nD0
.ξ1/
nC1
2n C1
which converges by the alternating series test.
If x D 1 we get:
n
X
nD0
.ξ1/
n
2n C1
which also converges by the alternating series test.
ξ©e interval of convergence is thus Εξ1; 1ξ.
Λ
It is possible to ο¬nd a power series by just using algebra on a known series.
Example 4.62 Find a power series for:
f .x/ D
x
2
1 ξx
2
Solution:
We start with the known form:
1
1 ξu
D 1 C u C u
2
C u
3
C u
4
C ξξξ
Substitute in u D x
2
and we get:
1
1 ξx
2
D 1 C x
2
C x
4
C x
6
C x
8
C ξξξ
Now multiply both sides by x
2
and we get:
x
2
1 ξx
2
D x
2
C x
4
C x
6
C x
8
C x
10
C ξξξ
4.3. POWER SERIES 153
So
f .x/ D
1
X
nD0
x
2nC2
Λ
Example 4.63 Find a power series for:
g.x/ D ln.x
2
C 1/
Solution:
Start with the known result for
1
1 Cx
2
.
1
1 Cx
2
D 1 ξ x
2
C x
4
ξ x
6
C x
8
ξ ξξξ
2x
1 Cx
2
D 2x ξ 2x
3
C 2x
5
ξ 2x
7
C 2x
9
ξ ξξξ
Z
2x
1 Cx
2
ξ dx D
Z
ξ
2x ξ 2x
3
C 2x
5
ξ 2x
7
C 2x
9
ξ ξξξ
ξ
dx
ln.x
2
C 1/ C C D 2
ξ
1
2
x
2
ξ
1
4
x
4
C
1
6
x
6
ξ
1
8
x
8
C
1
10
x
10
ξ ξξξ
ξ
Set x D 0 and we get C D 0:
So: ln.x
2
C 1/ D
1
X
nD0
2 ξ.ξ1/
n
x
2nC2
2n C2
Λ
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