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Review Problems
(b)
Substitution
t
=2
πn
into
x
p
(
t
)and
x
0
p
(
t
) yields
x
n
=
x
(2
πn
)=
x
h
(2
πn
)+
1
p
1+2(0
.
22)
2
sin
±
√
22
πn
+
ψ
²
,
v
n
=
v
(2
πn
)=
x
0
h
(2
πn
)+
1
0
.
22
+
√
2
p
1+2(0
.
22)
2
cos
±
√
22
πn
+
ψ
²
.
As
n
→∞
,
x
h
(2
πn
)
→
0and
x
0
h
(2
πn
)
→
0. Therefore, for
n
large,
x
n
≈
1
p
1+2(0
.
22)
2
sin
±
√
22
πn
+
ψ
²
=
a
sin
±
2
√
2
πn
+
ψ
²
,
v
n
≈
1
0
.
22
+
√
2
p
1+2(0
.
22)
2
cos
±
√
22
πn
+
ψ
²
=
c
+
√
2
a
cos
±
2
√
2
πn
+
ψ
²
.
(c)
From part (b) we conclude that, for
n
large
x
2
n
≈
a
2
sin
2
±
2
√
2
πn
+
ψ
²
and
(
v
n
−
c
)
2
≈
2
a
2
cos
2
±
2
√
2
πn
+
ψ
²
.
Dividing the latter by 2 and summing yields
x
2
n
+
(
v
n
−
c
)
2
2
≈
a
2
h
sin
2
±
2
√
2
πn
+
ψ
²
+cos
2
±
2
√
2
πn
+
ψ
²i
=
a
2
,
and the error (coming from the transient part) tends to zero as
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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