Page 135 - 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) and φ m is an Appell polynomial so that F =1.
Let

T = C 3 C 4 C 5 C 6 = T 3456 .

Then
D(T)=3T 2456 ,
2
D (T)/2!=3T 1456 +6T 2356 ,
3
D (T)/3! = T 0456 +8T 1356 +10T 2346 ,
.......................................
9
D (T)/9! = T 0126 +8T 0135 +10T 0234 ,
10
D (T)/10!=3T 0125 +6T 0134 , (4.9.8)
11
D (T)/11!=3T 0124 ,
12
D (T)/12! = T 0123 ,
= |φ m | 4 , 0 ≤ m ≤ 6
= constant.

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 {T (n,r) } =(nr)!|φ m | n , 0 ≤ m ≤ 2n − 2
nr
= constant.

Assuming this conjecture to be valid, T (n,r) is a polynomial of degree
nr and not n(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 is n(n − 1).
Let

T = T (n,r) = C r C r+1 C r+2 ··· C r+n−1 ,

n
where

T
C j = ψ r+j−1 ψ r+j ψ r+j+1 ··· ψ r+j+n−2
n
f (m) (x)
ψ m = , f(x) arbitrary
m!
ψ =(m +1)ψ m+1 . (4.9.9)
m
Theorem 4.35.

T =(2n − 1+ r) C r C r+1 ··· C r+n−2 C r+n
n
(n+1,r)
= −(2n − 1+ r)T .
n+1,n