34
3 Study of Electrophysical Characteristics of Blood …
by [7, 8]
A
n,m,q
n
= C
(n,m,q)
n
−
k 1 d
n + 1
(n − m + 1)(n + m + 1)
(2n + 1)(2n + 3)
C
(n,m,q)
n +1 −
k 1 d
n
(n − m)(n + m)
(2n + 1)(2n − 1)
C
(n,m,q)
n −1 ,
(3.18)
B
n,m,q
n
=
−ik 1 md
n (n + 1)
C
(nm,q)
n
, C
(0,0,q)
n
=
(2n + 1) j
n (k 1 d),
C
(−1,0,q)
n
= −
(2n + 1) j
n (k 1 d),
(3.19)
C
(n+1,0,q)
n
=
1
n + 1
2n + 3
2n + 1
n
2n + 1
2n − 1
C
(n,0,q)
n −1 + n
2n + 1
2n − 1
C
(n−1,0,q)
n
−
1
n + 1
2n + 3
2n + 1
(n
+ 1)
2n + 1
2n + 3
C
(n,0,q)
n +1
,
(3.20)
C
(n,m,q)
n
=
√ (n − m + 1)(n + m)(2n + 1)
√ (n − m + 1)(n + m)(2n + 1)
C
(n,m−1,q)
n
−
−k 1 d
(n − m + 2)(n − m + 1)
(2n + 3)(n − m + 1)(n + m)(2n + 1)
C
(n,m−1,q)
n +1
−
−k 1 d
(n + m)(n + m − 1)
(2n − 1)(n − m + 1)(n + m)(2n + 1)
C
(n,m−1,q)
n −1
,
C
(n,m,q)
n
= C
(n,−m,q)
n
,
(3.21)
A
(n,m,3)
n
= A
(n,m,4)
n
= A
(n,−m,3)
n
= A
(n,m)
n
,
(3.22)
B
(n,m,3)
n
= B
(n,m,4)
n
= B
(n,−m,3)
n
= B
(n,m)
n
,
(3.23)
C
(n,m,3)
n
= C
(n,m,4)
n
= C
(n,−m,3)
n
= C
(n,m)
n
.
(3.24)
When d = 0, we have A
(n,m)
n
= δ n n , B
(n,m)
n
= 0. We substitute expressions (3.16)
and (3.17) in formulas (3.14) and (3.15) to obtain
3 Study of Electrophysical Characteristics of Blood …
by [7, 8]
A
n,m,q
n
= C
(n,m,q)
n
−
k 1 d
n + 1
(n − m + 1)(n + m + 1)
(2n + 1)(2n + 3)
C
(n,m,q)
n +1 −
k 1 d
n
(n − m)(n + m)
(2n + 1)(2n − 1)
C
(n,m,q)
n −1 ,
(3.18)
B
n,m,q
n
=
−ik 1 md
n (n + 1)
C
(nm,q)
n
, C
(0,0,q)
n
=
(2n + 1) j
n (k 1 d),
C
(−1,0,q)
n
= −
(2n + 1) j
n (k 1 d),
(3.19)
C
(n+1,0,q)
n
=
1
n + 1
2n + 3
2n + 1
n
2n + 1
2n − 1
C
(n,0,q)
n −1 + n
2n + 1
2n − 1
C
(n−1,0,q)
n
−
1
n + 1
2n + 3
2n + 1
(n
+ 1)
2n + 1
2n + 3
C
(n,0,q)
n +1
,
(3.20)
C
(n,m,q)
n
=
√ (n − m + 1)(n + m)(2n + 1)
√ (n − m + 1)(n + m)(2n + 1)
C
(n,m−1,q)
n
−
−k 1 d
(n − m + 2)(n − m + 1)
(2n + 3)(n − m + 1)(n + m)(2n + 1)
C
(n,m−1,q)
n +1
−
−k 1 d
(n + m)(n + m − 1)
(2n − 1)(n − m + 1)(n + m)(2n + 1)
C
(n,m−1,q)
n −1
,
C
(n,m,q)
n
= C
(n,−m,q)
n
,
(3.21)
A
(n,m,3)
n
= A
(n,m,4)
n
= A
(n,−m,3)
n
= A
(n,m)
n
,
(3.22)
B
(n,m,3)
n
= B
(n,m,4)
n
= B
(n,−m,3)
n
= B
(n,m)
n
,
(3.23)
C
(n,m,3)
n
= C
(n,m,4)
n
= C
(n,−m,3)
n
= C
(n,m)
n
.
(3.24)
When d = 0, we have A
(n,m)
n
= δ n n , B
(n,m)
n
= 0. We substitute expressions (3.16)
and (3.17) in formulas (3.14) and (3.15) to obtain
