KiRSTiNE Smith 
39 
V+l' 
1 
1 
2q-l ' 
1 
2^+3 
1 
2q+l 
1 
1 
2q + 5 
1 
+ a 
+ a 
1 
25- + 2^ — 3 
1 
2f+ 2p - 1 
1 
2q + 2p+l 
+ a 
+ a 
+ a 
1 
2(7 + 2/; - 1 29- + 2^ + 1 2(7 + 4^-3 
Leaving the first row unaltered and subtracting from each of the others the 
proceeding we get a determinant the first column of which is 
1, a;2- 1, x^{x^-l) ... .e2p-2(^2_ i)^ 
while the other columns are identical with those of the determinant 8 previously 
treated in the same way. When next the two first rows are left as they are and 
from each of the others is subtracted the proceeding one the result is 
n 
v+i(i= (-1) 
1 
1 
1-^2 
(1 - a;2 
2q-l 
2 
+ a 
{2q-l)(2q+l) 
2.4 
27TT + " 
2 
(2^+1) (2(7 + 3) 
2.4 
2q + 2^9-3 
2 
-t a 
{2q + 2p-d) {2q+2p~-l) 
2.4 
(2q - 1) i2q + I) {2q + S) {2q + 1) {2q + ^) {2q + 6) ••• (2*7+ 2];-3) ... (2(7 + 22>+ 1) 
.r2p-4(l_a;2)2 
2.4 
2.4 
2.4 
{2q+2p-5)...{2q + 2p-l) {2q + 2p-3) ...{2q + 2p+l)'" (2q + 4:p-l) ...{2q + ip-3) 
Leaving now three rows unaltered, next time four and so on, it is clear that we 
shall at last after p of these sets of operations get 
1 
i 
I 
I 
I (1 - 2;2)2 ^ - 
1 
2q- 1 
2 
+ a 
1 
1 
2q+ 1 
2 
2q + 2p-3 
2 
+ a 
(2q-l){2q + l) 
2.4 
(2(7+1) (2(^ + 3) 
2.4 
{2q + 2p-S) {2q + 2p-l) 
2.4 
{2q - 1) {2q + I) {2q + 3) {2q + I) {2q + S) {2q + 5) - {2q+2p-3) {2q + 2p + l) 
(l-a;2)f 
2.4...2y 
2.4... 2p 
2. 4. ..22) 
(2?-l) (2q+2p-l) (2(7+1) ...... {2q + 2p + }) ■■■ (2q + 2p-?,) (2(7 + 4^) -3) 
By treating the columns in the same way, leaving first two then three and so 
on unaltered, we find after the first set of operations 
