Determinants and Their Applications in Mathematical Physics

(Chris Devlin) #1

300 6. Applications of Determinants in Mathematical Physics


Denote this particular solution byUr. Then,


tr=(−ω)

r
Ur,

where


Ur=

2 n

j=1

(−1)

j− 1
Mj(c)fr(xj)

ε


j

(6.10.54)

and the symbol∗denotes the complex conjugate. This function is of the


form (4.13.3), where


aj=

(−1)

j− 1
Mj(c)

ε

j

(6.10.55)

andN=2n. These choices ofajandNmodify the functionkrdefined in


(4.13.5). Denote the modifiedkrbywr, which is given explicitly in (6.10.3).


Since the results of Section 4.13.2 are unaltered by replacingωby (−ω),

it follows from (4.13.22) and (4.13.23) withn→mthat


Pm=(−1)

m(m−1)/ 2
2

m
2
− 1



wi+j+wi+j− 2



m

,

Qm=(−1)

m(m−1)/ 2
2

(m−1)

2 ∣

∣w
i+j− 2



m

. (6.10.56)

Applying the theorem in Section 6.10.4,


Pm=2

m
2
− 1
ρ

−m(m−1)

{

V 2 n(c)

}m− 1
H

(m)
2 n(ε),

Qm=2

(m−1)
2
ρ

−m(m−1)

{

V 2 n(c)

}m− 1
H

(m)
2 n

(

1

ε

)

. (6.10.57)

Hence,


Pn

Pn− 1

=2

2 n− 1
ρ

−2(n−1)
V 2 n(c)

H

(n)
2 n
(ε)

H

(n−1)
2 n (ε)

. (6.10.58)

Also, applying (6.10.32),


Qn+1

Qn

=2

2 n− 1
ρ

− 2 n
V 2 n(c)

H

(n+1)
2 n

(

1 /ε

)

H

(n)
2 n

(

1 /ε

)

=− 2

2 n− 1
ρ

− 2 n
V 2 n(c)

H

(n−1)
2 n


)

H

(n)
2 n
(ε∗)

. (6.10.59)

Sinceτj=ρεj, (the third line of (6.10.26)), the functionsFandGdefined


in Section 6.2.8 are given by


F=H

(n−1)
2 n (ρε)=ρ

n− 1
H

(n−1)
2 n (ε),

G=H

(n)
2 n
(ρε)=ρ

n
H

(n)
2 n
(ε). (6.10.60)
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