OF ELLIPTIC FUNCTIONS. 
433 
This is in fact an identity ; to show it, writing for convenience 6 in place of uv, and 
observing that the terms 
-(l-0)(l-4 8 )+8(l-0) 2 , 
= (1— 0) 2 {8— (l + 0+0 2 + 0 3 + 5 4 +4 5 + 0 6 + 4 7 )} are 
= (l-^) 3 (7+6^+5^ + 4^ 3 +3^+2^+^), 
the whole equation divides by (1 — 5) 3 ; or throwing out this factor, it is 
_6(l_5) 5 -(l-0) 6 +7+65 + 50 2 +45 3 +35 4 +25 5 +5 6 
- 1 40(1 - 6)(1 + 5 3 ) - 2 85(1 - 5 + 0J = 0. 
The first line is =145(3— 55 + G5 2 — 35 3 +5 4 ) ; whence, throwing out the factor 145, the 
equation is 
3_ 55+65 2 — 35 3 -j-5 4 — (1— 5)(l+5 3 ) — 2(l-5+5 2 ) 2 , 
that is 
(1 - 6 + 0 2 )(3 _ 24+ P ) — (1 — tf 2 )(l - 6+ 0 2 ) - 2(1 - 0+ r-) 2 = 0 ; 
or throwing out the factor 1 — the equation is 
(3 — 25 + 5 2 )— (1— 5 2 ) — 2(1 — 5-f5 2 ) = 0, 
which is an identity. 
The other equation is 
'/+2j3y+(2 + 2(3) ^=«v( 2y+2/3y+2 £+/?) ; 
that is 
/+2/3 r - M Vj3 ! +2(l+(3)(^-y M v)-2«%=0, 
which might also be verified, but I have not done this. 
47. The conclusion is 
«=i, /3=i(n- 1 )’ Mr 
1 — 7w (1— uv){\— uv + vPv 2 ) 
where 
and of course 
1 —y 1— x /\—fix + yx- — l ,z 3 \ 2 
1 +y l+x\l+Px + 'yz‘ 2 + $z 3 ) ’ 
but the resulting form may admit of simplification. 
48. The endecadic transformation, n— 11. 
I have not completed the solution, but the results, so far as I have obtained them, 
are interesting. The coefficients are a, j3, y, S, s, £ ; and we have, as in general, 
5 = 1 
’ = H' 
2 e=u 7 v 
yM v 4 ) - 
1, 
