270 
Transactions of the Royal Society of South Africa. 
signs of the second n—1 variables in F(a, b, c, . . .) all the differences in the 
factorial development of F are changed into sums. 
9. From this as an immediate consequence we have 
¥(a, b, c, -d, -a, -b) = (d+a)(a+b)(b+c) 
. (d+b)(a+c) 
. (d+c) 
— product of the binomial sums of 
a, b, c, d : 
and similarly for every other such product. We can thus formulate the 
theorem that If in the new determinantal equivalent for the difference- product 
of n quantities we alter the signs of the second n — 1 variables in the functional 
symbol, the resulting determinant is equal to the product of the binomial sums 
of the said n quantities : for example, from the equality of § 4 
we deduce the companion equality 
F(a, b, c, d, -e, -a, -b, -c) = | a 0 b 2 c 4 d Q e 8 | ~ \ a 0 b 1 c 2 d z e 4 |. 
10. There is an entirely different way of viewing the product of binomial 
sums, namely, as the eliminant of a pair of equations in x. Taking, for 
example, the equations 
(a J r b-\-c J r d J r e)x^ + {abc-\- . . . -\-cde)x 2 -f abcde = 0\ 
x 4 + {ab+ . . . J r de)x 2 + {abcd+ . . . +bcde) = 0J 
where the coefficients are sums of combinations of a, b, c, d, e, it is not 
difficult to show that their eliminant is the 10-f actor product 
One sure basis for proof lies in the fact that, if we multiply both sides of 
the second equation by x and perform addition we obtain 
and that the substitution of any one of the five roots of this derived equation 
in the two original equations changes their left-hand members into the 
product of four of the factors of E. Another basis of proof is the fact 
that if in the original equations we put one of the teD factors equal to 0, 
say a+6 = 0, the equations are transformable into 
F(a, b, c, d, e, a, b, c) = \ a 0 b 1 c 2 d z e 4 | 
(a+b){a+c) . . . {d+e) 
(E) 
(x+a)(x+b)(x+c)(x+d)(x+e) = 0, 
and are thus seen to be consistent. 
