4 series solutions
TRANSCRIPT
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Power Series Solutions of Linear DEs
Chapter 4
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Learning Objective
AtAt thethe endend ofof thethe section,section, youyou shouldshould bebeableable toto solve DE with Power Series assolutions..
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Power Series
A power series in is an infiniteseries of the form
ax
...)()()( 2210
0
axcaxccaxc n
nn
The above power series is centered at x = a.
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Power Series
n
n
x )1(0
center x = -1
n
n
n x
0
12 center x = 0
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Convergence
n
nn axc )(
0
A power series
is convergent at a specified value of x ifits sequence of partial sums converges.
NS
N
n
nn
NN
NaxcS
0
)(limlim exists.
If the above limit does not exists,the series is said to be divergent.
i.e.
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Interval of Convergence
n
nn axc )(
0
Every power series
has an interval of convergence. The interval of convergence is the set of allreal number x for which the seriesconverges.
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Radius of Convergence
n
nn axc )(
0
Every power series
has a radius of convergence R.
If R > 0, then the power series converges for and diverges
for
Rax
.Rax
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Ratio Test
Given a power series ,)(0
n
nn axc
Laxc
cax
axc
axc
n
n
nnn
nn
n
11
1 lim)(
)(lim
the series converges if
the series diverges
the test is inconclusive if
1 Lax
1 Lax
1 Lax
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A Power Series defines a function
A power series defines a function
n
nn axcxf )()(
0
whose domain is the interval of convergence of the series.
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Examples
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If the radius of convergence is R > 0, thenisf
• continuous • differentiable• integrable
over the interval (a-R, a+R).
Remark
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Example
Given
0n
nn xcy
Find yy and
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Example
...33
221
00
0
xcxcxcxcxcyn
nn
...320 2321 xcxccy
1
1
n
nn xncy
Since the first term is 0.
1
0
n
nn xncy
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...32 2321
1
1
xcxccxncyn
nn
...620 32 xccy
2
2
)1(
n
nn xcnny
Example
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Analytic at a Point
A function is analytic at a point aif it can be represented by a power seriesin x-a:
with a positive or infinite radiusof convergence.
n
nn axc )(
0
f
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Adding two Power Series
Example
0
12
n
nn xcS
Write as a single summation.
2
21 )1(
n
nn xcnnS
21 SS
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2 0
1221 )1(
n n
nn
nn xcxcnnSS
2 problems: exponents and starting indices
Example
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2 0
12)1(n n
nn
nn xcxcnn
Let 2 nk 1 nk
Example
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2
21 )1(
n
nn xcnnS
2 nk
021 )1)(2(S
k
kk xckk
Example
2 kn
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0
12
n
nn xcS
1 nk
112
k
kk xcS
Example
1 kn
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0 11221 )1)(2(
k k
kk
kk xcxckkSS
Now same exponentYet to solve: first term!
Example
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1 112
0221 )1)(2()1)(2(
k k
kk
kk xcxckkxcSS
1122 ])1)(2[(2
k
kkk xcckkc
1 1122 )1)(2(2
k k
kk
kk xcxckkc
Example
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12 2
2 3)1(2)1(n
nn
n n
nn
nn xncxcnnxcnn
Combine.
Exercise
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12 2
2 3)1(2)1(n
nn
n n
nn
nn xncxcnnxcnn
Let nk 2 nk nk
1 2 3
Solution
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2
)1(n
nnxcnn
nk
2
)1(k
kk xckk
1
Solution
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2
2)1(2n
nnxcnn
22 knnk
02)1)(2(2
k
kk xckk
2
Solution
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1
3n
nnxnc
nk
1
3k
kkxkc
3
Solution
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1
3k
kkxkc
02)1)(2(2
k
kk xckk
2
)1(k
kk xckk
Solution
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2
11 3)1(3
k
kk xkcxc
22
13
02 )1)(2(2)6(2)2(2
k
kk xckkxcxc
2
)1(k
kk xckk
Solution
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222
2321
3)1)(2(2
)1(1243
k
kk
k
kk
k
kk
xkcxckk
xckkxccxc
Solution
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kk
kkk xkcckkckk
xccc
]3)1)(2(2)1([
)123(6
22
312
Solution
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Ordinary and Singular Point
A point is said to be an ordinarypoint of the DE if both and are analytic at A point that is not anordinary point is said to be a singularpoint of the DE.
0x
)(xP )(xQ.0x
0)()( yxQyxPy
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A point that is not an ordinary pointis said to be a singular point of the DE.
Note: If at least one of the function and
fails to be analytic at then is a singular point.
0x)(xP
)(xQ0x
Ordinary and Singular Point
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1) Every finite value of is an ordinary point of the DE
x
.0)(sin)( yxyey x
Examples
0x
.0)(ln)( yxyey x
2) is a singular point of the DE
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Existence of Power Series Solutions
If is an ordinary point of the DE,we can always find two linearly independentsolutions in the form of a power seriescentered at ( ).
0xx
0xx
0
0 )(n
nn xxcy
Theorem
,0 Rxx
Each series solution converges at least on some interval defined by where R is the distance from to the closest singular point 0x
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Example
Find a power series solution centered at 0 for the following DE
.0 xyy
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.0 xyy
Ordinary points: All real numbers x.
Example
Since there are no finite singular points,The previous Theorem guarantees two power series solutions centered at 0, andconvergent for .x
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.0 xyy
Let the solution be n
nn
n
nn xcxcy
00
)0(
1
1
n
nnnxcy
2
2
)1(
n
nn xnncy
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0)1(
0
2 0
2
n n
nn
nn xcxxnnc
xyy
2 0
12 0)1(n n
nn
nn xcxnnc
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0)1(2 0
12
n n
nn
nn xcxnnc
2 nk1 nk
0)1)(2(0 1
12
k k
kk
kk xcxkkc
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0)1)(2(0 1
12
k k
kk
kk xcxkkc
0)1)(2(21 1
120
2
k k
kk
kk xcxkkcxc
0])1)(2([21
122
k
kkk xckkcc
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0)2)(1(.2
02.1
12
2
kk cckk
c
for ,...3,2,1k
Using the Identity Property:
The (recursive) relation generate consecutivecoefficients of the solution.
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0)2)(1( 12 kk cckk
)3)(2(0)3)(2( ,1 0
303
cccck
)4)(3(0)4)(3( ,2 1
414
cccck
0)5)(4(
0)5)(4( ,3 2525
cccck
)6)(5)(3)(2()6)(5(0)6)(5( ,4 03
636
ccccck
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...55
44
33
2210 xcxcxcxcxccy
...6.5.3.2
04.33.2
0 60413010 x
cx
cx
cxccy
n
nn xcy
0
..
4.3
1..
6.5.3.2
1
3.2
11 4
163
0 xxcxxcy
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...6.5.3.2
1
3.2
11)( 63
1 xxxy
...4.3
1)( 4
2 xxxy
)()()( 2110 xycxycxy
where
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Example
.0)1( 2 yyxyx
Find a power series solution centered at 0 for the following DE
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.0)1( 2 yyxyx
.0)1(
1
)1( 22
y
xy
x
xy
The standard form:
Ordinary Points: All real numbers x.Singular point: None.
Example
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Let the solution be n
nnxcy
0
1
1
n
nnnxcy
2
2
)1(
n
nn xnncy
.0)1( 2 yyxyx
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.0
)1()1(
0
1
1
2
22
n
nn
n
nn
n
nn
xc
xncxxcnnx
.0)1( 2 yyxyx
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0
)1()1(
01
2
2
2
n
nn
n
nn
n
nn
n
nn
xcxnc
xcnnxcnn
nk 2 nk
nk nk
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0
)1)(2()1(
01
02
2
k
kk
k
kk
k
kk
k
kk
xcxkc
xckkxckk
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0][1
)1)(2(62)1(
21
00
21
22
23
02
k
kk
k
kk
k
kk
k
kk
xcxcxcxkcxc
xckkxcxcxckk
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0
)1)(2()1(
62
22
22
2
32110
k
kk
k
kk
k
kk
k
kk
xcxkc
xckkxckk
xccxcxcc
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0
)1)(2()1(
62
22
22
2
320
k
kk
k
kk
k
kk
k
kk
xcxkc
xckkxckk
xccc
combine
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0]
)1)(2()1([
62
22
320
kkk
kkk
xckc
ckkckk
xccc
0])1)(2()1)(1[(
62
22
320
k
kkk xckkckk
xccc
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0])1)(2()1)(1[(
62
22
320
k
kkk xckkckk
xccc
0)1)(2()1)(1(
006
2
102
2
33
0220
kk ckkckk
cc
cccc
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kk
kk
ck
kc
ckk
kkc
c
cc
)2(
)1(
)1)(2(
)1)(1(
0
2
1
2
2
3
02
,...5,4,3,2k
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kk ck
kc
ccc
)2(
)1(
0;2
1
2
302
,...5,4,3,2k
0244.2
1
4
1,2 ccck
05
2,3 35
cck
0466.4.2
3
6
3,4 ccck
07
4,5 57
cck
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...55
44
33
2210 xcxcxcxcxccy
...6.4.2
3.1
4.2
1
2
1 60
40
2010 xcxcxcxccy
)()(
][...]6.4.2
3.1
4.2
1
2
11[
2110
1642
0
xycxycy
xcxxxcy
n
nn xcy
0
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xxy
xxxxy
)(
...6.4.2
3.1
4.2
1
2
11)(
2
6421
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Example
.0)1( yxy
Find a power series solution centered at 0 for the following DE
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Example
0)1( yxy
0)1()1(02
2
n
nn
n
nn xcxxcnn
0)1(0
1
02
2
n
nn
n
nn
n
nn xcxcxcnn
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Example
0)1)(2(1
100
2
k
kk
k
kk
k
kk xcxcxckk
0)1(0
1
02
2
n
nn
n
nn
n
nn xcxcxcnn
2 nknk
1nk
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Example
0)1)(2(1
100
2
k
kk
k
kk
k
kk xcxcxckk
0
)1)(2(2
11
1
00
12
02
k
kk
k
kk
k
kk
xc
xcxcxckkxc
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Example
0
)1)(2(2
11
11202
k
kk
k
kk
k
kk
xc
xcxckkcc
0])1)(2[(2 11
202
k
kkk
k xccckkcc
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Example
,...3,2,1,)2)(1(
0)2)(1(
2
102
12
12
0202
kkk
ccc
ccckk
cccc
kkk
kkk
Using Identity Property:
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Example ,...3,2,1,)2)(1(
2
1
12
02
k
kk
ccc
cc
kkk
3.2,1 01
3
ccck
4.3,2 12
4
ccck
5.4,3 23
5
ccck
6.5,4 34
6
ccck
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Example02
2
1cc
3.23.2,1 001
3
cccck
4.3.24.34.3,2 0212
4
ccccck
5.3.25.4.25.4.3.25.4,3 00023
5
cccccck
6.4.3.26.5.3.26.5.4.3.26.5,4 00034
6
cccccck
Case 1: 0,0 10 cc
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Example0
2
102 cc
3.23.2,1 101
3
cccck
4.34.3,2 112
4
cccck
5.4.3.25.4,3 123
5
cccck
6.5.46.5.3.26.5.4.36.5,4 01134
6
cccccck
Case 2: 0,0 10 cc
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Example
...55
44
33
2210 xcxcxcxcxccy
From case 1:
...5.3.24.3.23.22
504030200 x
cx
cx
cx
ccy
...]5.3.2
1
4.3.2
1
3.2
1
2
11[ 5432
0 xxxxcy
...5.3.2
1
4.3.2
1
3.2
1
2
11)( 5432
1 xxxxxy
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Example
...55
44
33
2210 xcxcxcxcxccy
From case 2:
...6.5.44.33.2
5141311 x
cx
cx
cxcy
...]6.5.4
1
4.3
1
3.2
1[ 543
1 xxxxcy
...6.5.4
1
4.3
1
3.2
1)( 543
2 xxxxxy
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End chapter 3
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