428 
PROFESSOR CAYLEY ON THE TRANSFORMATION 
38. Considering the modular equation as known, then by what precedes we have 
«+/3#+y# 2 +. . .=a ; 
that is, the given function A+B# + C# 2 + .. . has a factor l+^ + . . . + — of 
which one (the last) coefficient ~ is known, and we are hence able theoretically to deter- 
mine all the other coefficients rationally in terms of m, v ; that is, the modular equation 
being known, we can theoretically complete the solution of the transformation problem. 
I do not, however, see the way to obtaining a convenient solution in this manner. 
39. The formula in question for n= 3 is 
1 + sp39 _ l-sn9 / l+2sn9 — 2Fsn 3 9-Fsn 4 9 \ 2 
1 — sn 39 1 + sn 9 yl — 2sn 9 + 2A 2 sn 3 9— A 2 sn 4 9 J ’ 
which, putting therein x — sn 6, z= — sn 30, and replacing Tc by m 4 , may be written 
1 + z ( + ) = (1 +#)(1 - 2 #+ 2 mV - m 8 # 4 ) 2 ( ~ ), 
where the signs (- ) indicate denominators which are obtained from the numerators by 
changing the signs of z, x respectively. 
The theorem in regard to n= 3, thus is, is a factor of 1 — 2# + 2m 8 # 3 — mV; 
viz. writing in the last-mentioned function x= — A, we ought to have 
that is 
0-1 9 w4 
0-1 + 2 u 3 - 2 ~-- 4 . 
u 4 +2uv—2u 3 v 3 —v 4 =0, 
which is in fact the modular equation. 
40. And so for n = 5 if #=sn$, 2=sn50, and for n— 7 if #=sn0, z= — sn70, the 
formulae are : — 
n= 5, 
l+z=(l+x){ 1 
(-) + 2 
— 4 x 2 
-10 m 8 a - 3 
+ 5 u 3 x 4 
+ 4m 8 (3+2m 8 )# 5 
+4m 8 (1- m 8 )# 6 
— 4m 8 (2 + 3m 8 )# 7 
— 5m 16 x 8 
+ 10?* 16 x 9 
oi= 7 , 
l+*=(l+#){ 1 
(-) - 4 
X 
- 4 
X 2 
+ 4(2 + 7m 8 ) 
X 3 
-14m 8 
X 4 
— 28m 8 (3 + 2«£ 8 ) x 5 
+28m 8 (4+ u 3 ) x 6 
+ 4m 8 ( 16+ 51m 8 + 8 m 16 )# 7 
- m 8 (144 + 305m 8 + 16m 16 )# 8 
- 8 m 8 ( 4+ 25m 8 + 16m 16 )# 9 
