Determinants and Their Applications in Mathematical Physics

128 4. Particular Determinants

=

n− 1

∑

i=0

(−1)

i

(

n− 1

i

)

(h+n+i)!

(h+i+ 1)!

x

h+i+1

=

n

∑

j=1

(−1)

j− 1

(

n− 1

j− 1

)

(h+n+j−1)!

(h+j)!

x

h+j

=

(n−1)!

2

h!x

h+1

(h+n)!

n

∑

j=1

K

1 j

x

j− 1

.

The theorem follows.

Let

Sn(x, h)=

n

∑

j=1

K

(n)

nj

(−x)

j− 1

=

∣

∣

∣

∣

∣

∣

∣

∣

∣

k 11 k 12 k 13 ··· k 1 n

k 21 k 22 k 23 ··· k 2 n

......................................

kn− 1 , 1 kn− 1 , 2 kn− 1 , 3 ··· kn− 1 ,n

1 −xx

2

··· (−x)

n− 1

∣

∣

∣

∣

∣ ∣ ∣ ∣ ∣ n

.

The column operations

C

′

j=Cj+xCj−^1 ,^2 ≤j≤n,

remove thex’s from the last row and yield the formula

Sn(x, h)=(−1)

n+1

∣

∣

∣

∣

x

h+i+j− 1

+

1

h+i+j

∣

∣

∣

∣

n− 1

.

Let

Tn(x, h)=(−1)

n+1

∣

∣

∣

∣

1+x

h+i+j− 1

−

1

h+i+j

∣

∣

∣

∣

n− 1

.

Theorem 4.38.

a.

(h+n−1)!

2

h!(n−1)!

Sn(x, h)

Sn(0,h)

=D

h+n− 1

[x

n− 1

(1 +x)

h+n− 1

].

b.

(h+n−1)!

h!(n−1)!

Tn(x, h)

Tn(0,h)

=D

h+n− 1

[x

h+n− 1

(1 +x)

n− 1

].

Proof.

Sn(0,h)=K

(n)

n 1

=

Kn(h)VnnVn 1

h+n

=(−1)

n+1

Kn(h)Vnn

(

h+n− 1

h

)

,