<Sol. 1>
1. x  0
(a).
2. x  0
when
x0
3.  x   x ,   R
4. x  y  x  y
1. x, x  0
2. x, x  0
(b).
x0
when
3. x, y  y, x
4.  x, y   x, y
5. x  y, z  x, z  y, z
6. x, y
2
 x, x y , y
(c). Yes， x  x, x
1/ 2
The inner product space contains angle information but the normed space does Not
(d). A Hibert space is a complete inner product space.
x 
x, x
d ( x, y )  x  y 
x  y, x  y
(e). A Banach space is a complete normed space
(f).
when n    x n+1  xn 
xn 1

2 xn
此 cauchy sequence 不具完備性
2 xn2  xn2  2  xn  2
(g).
(h).
c1 x
(1)
 x
(2)
 c2 x
(1)
c1 & c2  0, x  R n
25 49

16 25
<Sol. 2>
Set of functions un ( x)  x n , n=01,2,…
The interval is  1  x  1 and w( x)  1
In accordance with the Gram-Schmidt orthogonalization process descried,
u0  1 , hence
0 
1
2
Then
 1  x  a1,0
1
2
and
1
a1,0  
1
x
dx  0
2
by symmetry. We normalize  1 to obtain
3
x
2
Then we continue the Gram-Schmidt procedure with
1 
 2  x 2  a2,0
1
3
 a2,1 x
2
2
where
a2 , 0   
1
1
a2,1   
x2
2
dx  
3
2
3 3
x dx  0
2
1
1
again by symmetry. Therefore
 2  x2 
1
3
and on normalizing to unity, we have
2 
5 1 2
 (3x  1)
2 2
The next function, 3 ( x) , becomes
3 
7 1
 (5 x3  3x)
2 2
Show that
n 
2n  1
Pn ( x)
2
So
P0 ( x)  1
P1 ( x)  x
1
(3 x 2  1)
2
1
P3 ( x)  x(5 x 2  3)
2
P2 ( x) 
<Sol. 3>
Because
 2 (r , , z ) 
 r2
 2 1  1  2  2



0
r 2 r r r 2  2 z 2
2
 2
  2
2  

r


r
0
r 2
r  2
z 2
So
r
   2
 2
[r
]  2  r2 2  0
r r

z
∴ That Laplacian operator  2 is self-adjoint
Set
  1  2
 1 (r, , z)  R(r )Q( )Z ( z)
So
1
1
R QZ  R QZ  2 RQZ  RQ Z  0
r
r
R  1 R  1 Q Z




0
R r R r2 Q Z
1
R   R 
r  1 Q  Z  0

R
r2 Q Z
Set
 Z
2
 Z    Z ( z )  A cos z  B sin z
 
 Q    2  Q( )  C cos   D sin 
 Q
1
R   R 
r  1 (  2 )  2  0

R
r2
 r 2 R  rR  (2 r 2   2 ) R  0
 R(r )  C I  (r )  D K  (r )
Then lim  bounded
r 0
B.C.
 R ( r )  C  I  ( r )


1 (r  a, , z )  V1 ( , z ) 


 21 (r , , z )  0   1 (r , , z  0)  0  
A0
  ( r , , z  L)  0
0  B sin L  L  n    n , n  1,2,3...
1

L

∵ 1 (r , , z )  (C cos   D sin  ) I  (a) sin z


∴ V1 ( , z )  [Cn cos  Dn sin  ]I  (
 0 n1
n
n
a) sin(
z)
L
L
Orthogonal：
L

0
2
0

cos   sin(

  [Cn I  (
  0 n 1
n
z )V1 ( , z )ddz
L
L
2
n
n
n
a )  (  cos  cos  d ) sin(
z ) sin(
z )dz ]
0
0
L
L
L
n
z )V1 ( , z )ddz
0 0
L

2
L
n
n
I (
a)  cos 2 d  sin 2 (
z )dz
0
0
L
L
L
 Cn
 
2
cos  sin(
2
for
  0 ，  cos2 d  2
for
  0 ，  cos2 d  
0
2
0
L
n  0 ，  sin 2 (
for
0
 C0 n 
 Cn 
2
L
0
L
2
0
0


2
0
n
L
z )dz 
L
2
n
z )V1 ( , z )ddz
L
n
LI 0 ( a)
L
sin(
n
z )V1 ( , z )ddz
L
n
LI  ( a)
L
cos  sin(
n
z )V1 ( , z )ddz
L
 Dn 
2
L
n
n
I  ( a)  sin 2 d  sin 2 (
z )dz
0
0
L
L
L
2
0
0

sin  sin(
So that


1 (r , , z )  [Cn cos  Dn sin  ]I  (
 0 n1
n
n
r ) sin(
z)
L
L
And so
 2 (r, , z)  R(r )Q( )Z ( z)
So
1
1
R QZ  R QZ  2 RQZ  RQ Z  0
r
r
R  1 R  1 Q Z




0
R r R r2 Q Z
1
R   R 
r  1 Q  Z  0

R
r2 Q Z
Set
 Z
2
 Z    Z ( z )  A cosh z  B sinh z
 
 Q    2  Q( )  C cos   D sin 
 Q
1
R   R 
r  1 (  2 )  2  0

R
r2
 r 2 R  rR  (2 r 2   2 ) R  0
 R(r )  C J  (r )  D N  (r )
Then lim  bounded
r 0
 R ( r )  C  J  (r )
B.C.
  2 ( r  a,  , z )  0
 J  (a)  0   a   n , n  1,2,3...


  2 (r ,  , z )  0   2 (r ,  , z  0)  V2 (r ,  )  
  (r ,  , z  L)  0

2


2
∵  2 (r , , z )  ( A cos   B sin  ) J  ( r ) sinh  z
∴ V2 (r , ) 
 An 


[ A
  0 n 1
n
cos  Bn sin ]J  (n r ) sinh( n L)
2 cos ech(n L)

2
a
d  drrV2 (r ,  ) J  (n r ) sin 
0
a I  1 (n a) 0
a
2 cos ech(n L) 2
 Bn 
d  drrV2 (r , ) J  (n r ) cos 
2 2

0
a I  1 (n a) 0
2 2
So that


 2 (r , , z )  [ An cos  Bn sin  ]J  (n r ) sinh( n z)
 0 n1