366 MR. IVORY ON THE THEORY OF THE ELLIPTIC TRANSCENDENTS. 
And if R be transformed into a function of x 2 , we shall obtain 
M 2 + ,3 2 N 2 = (1 + x 2 ) (1 + c 2 2 x 2 ) 2 (1 + c 2 x 2 ) 2 .... (1 + cV/) 2 , 
the new symbol c 2n being determined by this formula, 
2 = 1 — & 2 sin 2 X 2n . 
The last equation may be resolved into these two, 
M + = (1 +xj~\) (1 + c 2 xj^l) 2 
M - pxN^J— 1 = (1 — x,J— 1) (1 — c 2 x,J— l) 2 
• (1 + c p _ 2 x ,J — l) 2 , 
• (1 ~ Cp-zxJ- l) 2 : 
the second of which being divided by the first, there will result. 
1 — tan rp V' — 1 1 — x */ — I ( 
'1 — c, z x */ — 1\ 2 / 
1 — Cp_ 2 x V' — 1 \ 
1 + tan \/ — 1 1 + x */ — 1 * ' 
K 1 + c 2 x \/ — 1 / V 
.1 + Cp _2 X v — 
Now u being an arc of a circle, we have this well known formula. 
u = 
2 */- l 
X log. 
( 1 — tan u V — 1 \ 
1 + tan u V — 1 / 
+ tan u V 
wherefore, if we take the logarithms of the factors of the foregoing expression, 
and substitute the equivalent circular arcs, we shall obtain, 
p being an odd number, 
^ = <p + 2 <p 2 + 2 <p 4 + 2 cp 6 . . . 2 (10) 
the arc <p 2n being determined by the equation, 
tan <p 2n z=c 2n X tanp. 
When p is an even number, we have this equation in § 5, 
(1 -k 2 z 2 ) R 2 = ft 2 z 2 (1 - z 2 ) P 2 + Q 2 . 
And, using N and M to denote the products of the binomials in the numerator 
and denominator of the equation (8), we have 
. , B tan <p N 
tan \L = - — - — . 
r M 
By the substitution of 
QzJ\-z 2 . P = 
; for z as before, it will be found that, 
V 1 + x- 
(3iN _ M 
(1 + x a ) 
V 
Q = 
(i + * a )4 
