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88
Chapter 5
Partial Diﬀerential Equations in Spherical Coordinates
Solutions to Exercises 5.5
1.
Putting
n
= 0 in (9), we obtain
P
0
(
x
)=
1
2
0
0
±
m
=0
(

1)
m
(0

2
m
)!
m
!(0

m
)!(0

2
m
)!
x
0

2
m
.
The sum contains only one term corresponding to
m
= 0. Thus
P
0
(
x
)=(

1)
0
0!
0!0!0!
x
0
=1
,
because 0! = 1. For
n
= 1, formula (9) becomes
P
1
(
x
1
2
1
M
±
m
=0
(

1)
m
(2

2
m
)!
m
!(1

m
)!(1

2
m
)!
x
1

2
m
,
where
M
=
1

1
2
= 0. Thus the sum contains only one term corresponding to
m
=0
and so
P
1
(
x
1
2
1
(

1)
0
2!
0!1!1!
x
1
=
x.
For
n
= 2, we have
M
=
2
2
= 1 and (9) becomes
P
2
(
x
1
2
2
1
±
m
=0
(

1)
m
(4

2
m
)!
m
!(2

m
)!(2

2
m
)!
x
2

2
m
=
1
2
2
m
=0
²
³´
µ
(

1)
0
4!
0!2!2!
x
2
+
1
2
2
m
=1
²
³´
µ
(

1)
1
(4

2)!
1!1!0!
x
0
=
1
4
6
x
2
+
1
4
(

1)2 =
3
2
x
2

1
2
.
n
= 3, we have
M
=
3

1
2
= 1 and (9) becomes
P
3
(
x
1
2
3
1
±
m
=0
(

1)
m
(6

2
m
)!
m
!(3

m
)!(3

2
m
)!
x
3

2
m
=
1
2
3
(

1)
0
6!
0!3!3!
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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