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