Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 9)

122 
A MEMOIR ON THE TRANSFORMATION OF ELLIPTIC FUNCTIONS. [578 
13. Reproducing for this case the general theory, it appears a priori that 11 is 
determined by a quartic equation; in fact, from the original equations eliminating 12, 
we have an equation 
U , 
U', 
where U, U' } V, V' are quartic functions of a, /3; that is, the ratio a : (3 has four 
values, and to each of these there corresponds a single value of 12; viz. 12 is deter 
mined by a quartic equation. 
14. Considering next the case n = 5, the quintic transformation; the elimination of 
12 gives the equations 
u_v_w 
V ~ V W' ’ 
where U, U', &c. are all quadric functions of a, /3, y. We have thence 4.4 — 2.2, =12 
sets of values of a : /3 : 7; viz. considering a, /3, 7 as coordinates in piano, the curves 
UV' — JJ'V = 0, UW' — U'W-0 are quartic curves intersecting in 16 points; but among 
these are included the four points U = 0, U' = 0 (in fact, the point a = 0, 7=0 four 
times), which are not points of the curve VW'—V'W = 0; there remain therefore 
16 — 4, =12 intersections, agreeing with the general value (n + 1). 2* ,n-8) . Hence 11 
is in the first instance determined by an equation of the order 12; but the proper 
order being = 6, there must be a factor of the order 6 to be rejected. To explain 
this and to determine the factor, observe that the equations in question are 
tea? (2ay + 2/3y + /3 2 ) — y 2 (2ay + 2 a/3 + /3 2 ) = 0, 
tea 3 (a + 2/3) — 7 3 (7 + 2/3) = 0 ; 
at the point a = 0, 7 = 0, the first of these has a double point, the second a triple 
point; or there are at the point in question 6 intersections; but 4 of these are the 
points which give the foregoing reduction 16 — 4 = 12; we have thus the point a = 0, 
7 = 0, counting twice among the twelve points. Writing in the two equations /3 = 0, 
the equations become tea 3 y — ay 3 =0, tea 4 — y 4 = 0, viz. these will be satisfied if tea?—y 2 =0, 
that is, the curves pass through each of the two points (/3 = 0, 7 = ± ka), and these 
values satisfy (as in fact they should) the third equation 
k 2 (2ay + 2a/3 + /3-) a (a + 2/3) — 7 (7 + 2/3) (2aey + 2/3 + /3 2 ) = 0. 
It is moreover easily shown that the three curves have at each of the points in question 
a common tangent; viz. taking A, B, C as current coordinates, the tangent at the 
point (a, /3, 7) of the second curve has for its equation 
A (2a 3 -1- 3a 2 /3) te + B (tea 3 — 7 3 ) — C (27 3 + 3y 2 /3) = 0 ; 
and for /3 = 0, y = ±ka, this becomes 2kA + B (k + 1) + 2G = 0, viz. this is the line from 
the point (¡3 = 0, 7 = ± ka) to the point (1, — 2, 1). And similarly for the other two 
curves we find the same equation for the tangent.
	        
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