Determinants and Their Applications in Mathematical Physics

120 4. Particular Determinants

method can be illustrated adequately by taking the particular case in which

(n, r)=(4,3) andφmis an Appell polynomial so thatF=1.

Let

T=

∣

∣C

3 C 4 C 5 C 6

∣

∣=T

3456.

Then

D(T)=3T 2456 ,

D

2

(T)/2!=3T 1456 +6T 2356 ,

D

3

(T)/3! =T 0456 +8T 1356 +10T 2346 ,

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

D

9

(T)/9! =T 0126 +8T 0135 +10T 0234 ,

D

10

(T)/10!=3T 0125 +6T 0134 ,

D

11

(T)/11!=3T 0124 ,

D

12

(T)/12! =T 0123 ,

=|φm| 4 , 0 ≤m≤ 6

= constant.

(4.9.8)

The array of coefficients is symmetric about the sixth derivative. This result

and several others of a similar nature suggest the following conjecture.

Conjecture.

D

nr

{T

(n,r)

}=(nr)!|φm|n, 0 ≤m≤ 2 n− 2

= constant.

Assuming this conjecture to be valid,T

(n,r)

is a polynomial of degree

nrand notn(n+r−1) as may be expected by examining the product of

the elements in the secondary diagonal. Hence, the loss of degree due to

cancellations isn(n−1).

Let

T=T

(n,r)

=

∣

∣C

rCr+1Cr+2···Cr+n− 1

∣

∣

n

,

where

Cj=

[

ψr+j− 1 ψr+jψr+j+1···ψr+j+n− 2

]T

n

ψm=

f

(m)

(x)

m!

,f(x) arbitrary

ψ

′

m=(m+1)ψm+1. (4.9.9)

Theorem 4.35.

T

′

=(2n−1+r)

∣

∣

CrCr+1···Cr+n− 2 Cr+n

∣

∣

n

=−(2n−1+r)T

(n+1,r)

n+1,n

.