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Power Series Solution for the Harmonic Oscillator Equation
:
ODE:
0
y
dx
y
d
2
2
2
=
λ
+
Transform independent variable
x
λ
=
ς
Transform derivatives
ς
λ
=
ς
ς
=
d
dy
dx
d
d
dy
dx
dy
2
2
2
2
2
d
y
d
dx
d
d
dy
d
d
d
dy
dx
d
dx
dy
dx
d
dx
y
d
ς
λ
=
ς
ς
λ
ς
=
ς
λ
=
=
Substitute in ODE:
0
y
d
y
d
y
dx
y
d
2
2
2
2
2
2
2
=
λ
+
ς
λ
=
λ
+
Simplify
0
y
d
y
d
2
2
=
+
ς
Express dependent variable as a power series
Function
(
29
⋅
⋅
⋅
+
+
+
+
+
+
+
=
=
∑
∞
=
7
7
6
6
5
5
4
4
3
3
2
2
1
0
0
n
n
n
ς
c
ς
c
ς
c
ς
c
ς
c
ς
c
ς
c
c
ς
c
ς
y
Derivatives:
(
29
⋅
⋅
⋅
+
+
+
+
+
+
=
=
∑
∞
=

5
6
4
5
3
4
2
3
2
1
1
n
1
n
n
ς
c
6
ς
c
5
ς
c
4
ς
c
3
ς
c
2
ς
c
ς
nc
ς
d
ς
dy
(
29
(
29
∑
∞
=


=
⋅ ⋅
⋅
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
2
n
2
n
n
5
7
4
6
3
5
2
4
3
2
2
2
ς
1
n
n
c
ς
c
6
7
ς
c
5
6
ς
c
4
5
ς
c
3
4
ς
c
2
3
c
1
2
ς
d
ς
y
d
(
29
(
29
0
ς
c
ς
c
ς
c
ς
c
ς
c
ς
c
ς
c
c
ς
c
6
7
ς
c
5
6
ς
c
4
5
ς
c
3
4
ς
c
2
3
c
1
2
y
ς
d
ς
y
d
7
7
6
6
5
5
4
4
3
3
2
2
1
0
5
7
4
6
3
5
2
4
3
2
2
2
=
⋅
⋅
⋅
+
+
+
+
+
+
+
+
⋅
⋅
⋅
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
+
Substitute in ODE and collect terms on powers of
ς
(
29
(
29
(
29
(
29
(
29
0
ς
c
c
6
7
ς
c
c
5
6
ς
c
c
4
5
ς
c
c
3
4
ς
c
c
2
3
c
c
1
2
5
5
7
4
4
6
3
3
5
2
2
4
1
3
0
2
=
⋅
⋅
⋅
+
+
⋅
+
+
⋅
+
+
⋅
+
+
⋅
+
+
⋅
+
+
⋅
The coefficients of each power of
ς
must equal 0 to satisfy the equation for all
ς
Equating these terms
0
ς
:
1
2
c
c
0
c
c
1
2
0
2
0
2
⋅

=
⇒
=
+
⋅
1
ς
:
2
3
c
c
0
c
c
2
3
1
3
1
3
⋅

=
⇒
=
+
⋅
2
ς
:
1
2
3
4
c
3
4
c
c
0
c
c
3
4
0
2
4
2
4
⋅
⋅
⋅
=
⋅

=
⇒
=
+
⋅
3
ς
:
1
2
3
4
5
c
4
5
c
c
0
c
c
4
5
1
3
5
3
5
⋅
⋅
⋅
⋅
=
⋅

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This note was uploaded on 04/11/2008 for the course CHNE 525 taught by Professor Prinja during the Fall '08 term at New Mexico.
 Fall '08
 Prinja

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