124 A MEMOIR ON THE TRANSFORMATION OF ELLIPTIC FUNCTIONS. [578
18. One way of satisfying the equations is to write therein a=0, 8 = 0; the
equations thus become
k/3 2 = kly 2 ,
y 2 + 2/3y = klk (2/3y + /3 2 ) ;
or putting /3, y — oc, /3',
ka 2 = kl/3' 2 ,
/S' 2 + 2a / /3 / = klk (2a’¡3' + a' 2 ),
which (with a', /3' instead of a, /3) are the very equations which belong to the cubic
transformation ; hence a factor is {(il, l) 4 }.
Observe that for the values in question a = 0, 8 = 0, P = /3'ai 2 , Q = a',
(P ± Qx) 2 = x 2 (a' ± /3'x) 2 , = x 2 (P' + Q'x) 2 , if P' = cl', Q' = /3',
and therefore
1 — y _ 1 — x fP’ — Q'x\ 2
1 +y 1 4- x \P' + Q'x) ’
which is the formula for a cubic transformation.
19. The equations may also be satisfied by writing therein y = ka, 8 = k/3 ; in fact,
substituting these values, they become
k?a- = Qk-/3-,
2k 2 oi 2 + k (2a/3 + /3 3 ) = klk 2 (a 2 -f 2a/S) + 2H&/3 2 ,
k 2 a 2 + 2k (J3 2 + 2<x/3) = 2klk 2 (a 2 + 2a/3) + klk/3 2 ,
k 2 (/3 2 + 2a/3) = HP (a 2 + 2a/3) ;
the first and last of these are
ka 2 = kl/3 2 ,
¡3 2 + 2a/3 = klk (a 2 + 2a/3),
which being satisfied the second and third equations are satisfied identically; and these
are the formulae for a cubic transformation ; that is, we again have the factor {(il, l) 4 }.
Observe that for the values in question y = ka, 8 = k/3, we have P = a (1 + kx 2 ),
Q = /3(l+kx 2 ); so that, writing P' = a, Q' = /3, we have for y the value
1 — y (1 — x) (P' — Q'x) 2
1 + y (1 + x) (P' + Q'x) 2 ’
which is the formula for a cubic transformation.
20. It is important to notice that we cannot by writing a = 0 or 8 = 0 reduce the
transformation to a quintic one; in fact, the equation Ar 3 a 2 = i!8 2 shows that if either
of these equations is satisfied the other is also satisfied; and we have then the
foregoing case a = 0, 8 = 0, giving not a quintic but a cubic transformation.
And for the same reason we cannot by writing a = 0, /3=0, y = 0 or /3 = 0, y = 0,
8 = 0 reduce the transformation to the order 1. There is thus no factor kl— 1.
578]
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