Determinants and Their Applications in Mathematical Physics

156 4. Particular Determinants

=

n+r

∑

σ=r

(

n

n−σ

)(

n

n−σ+r

)

u

σ

v

n−σ+r

=(uv)

r

n+r

∑

σ=r

(

n

σ

)(

n

σ−r

)

u

σ−r

v

n−σ

. (4.12.8)

Part (a) follows after interchanginguandvand replacingnbyi. Part (b)

then follows easily.

It follows from Lemma 4.56 thatPi+j(x) is equal to the coefficient of

t

i+j

in the expansion of the polynomial

[(u+t)(v+t)]

i+j

=[(u+t)(v+t)]

i

[(u+t)(v+t)]

j

=

2 i

∑

r=0

λirt

r

2 j

∑

s=0

λjst

s

. (4.12.9)

Each sum consists of an odd number of terms, the center terms beingλiit

i

andλjjt

j

respectively. Hence, referring to Lemma 4.57,

Pi+j(x)=

min(i,j)

∑

r=1

λi,i−rλj,j+r+λiiλjj+

min(i,j)

∑

r=1

†

λi,i+rλj,j−r

=2

min(i,j)

∑

r=0

λi,i+rλj,j−r, (4.12.10)

where the symbol†denotes that the factor 2 is omitted from ther=0

term. Replacingibyi−1 andjbyj−1,

Pi+j− 2 (x)=2

min(i,j)

∑

r=0

†

λi− 1 ,i−1+rλj− 1 ,j− 1 −r. (4.12.11)

Preparations for the second proof are now complete. Adjusting the dummy

variable and applying, in reverse, the formula for the product of two

determinants (Section 1.4),

|Pi+j− 2 |n=

∣ ∣ ∣ ∣ ∣ ∣

2

min(i,j)

∑

s=1

†

λi− 1 ,i+s− 2 λj− 1 ,j−s

∣ ∣ ∣ ∣ ∣ ∣ n =

∣

∣

2 λ

∗

i− 1 ,i+j− 2

∣

∣

n

∣

∣

λj− 1 ,j−i

∣

∣

n

, (4.12.12)

where the symbol∗denotes that the factor 2 is omitted whenj= 1. Note

thatλnp=0ifp<0orp> 2 n. The first determinant is lower triangular

and the second is upper triangular so that the value of each determinant

is given by the product of the elements in its principal diagonal:

|Pi+j− 2 |n=2

n− 1

n

∏

i=1

λi− 1 , 2 i− 2 λj− 1 , 0