Специальные функции. Мицик М.Ф. - 12 стр.

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    ,   
ν
,
.
     , 
 Maple 9.5.
 2.

( )
xJ
2

( )
xN
2
.
.
.

( )
( )
2
1
2
21;
+= zxzzxω
(37)

x

z
,     ,
. 
1<z
,

1x
   
021
2
+ zxz
. ,

( )
zx;ω

1<z

( )
( )
1,1,21
0
2
1
2
<=+
=
xzzxPzxz
n
n
n
, (38)

( )
xP
n

( )
( )
dzzzzx
i
xP
z
n
n
=
+=
δ
π
1
2
1
2
21
2
1
, (39)
(
)
xN
2
(
)
xJ
2
                                                            ,                                 ν,
                                                                                   .
                                                                               ,
                  Maple 9.5.


                      J 2 (x)




                    N 2 (x)




                                                           2.
                                                     J 2 (x )                N 2 (x ) .

                                                                         .
                                                                                       .



                                    (                )
                                                       1
                  ω ( x; z ) = 1 − 2 xz + z         2 −2
                                                                                           (37)
                       x      z,                                                              ,
 –                       .                                                                 z <1,
x ≤1                                                        1 − 2 xz + z 2 ≠ 0 .                  ,
     ω ( x; z )                                    z <1

        (1 − 2 xz + z )
                                1        ∞
                             2 −2
                                      = ∑ Pn ( x ) z n ,         z < 1, x ≤ 1 ,            (38)
                                        n =0

              Pn ( x )

                                ∫ (1 − 2 x z + z )
                                                       1
         Pn ( x ) =
                     1                              2 −2
                                                            z −n−1dz ,                     (39)
                    2π i       z =δ