Central Forces. 69 
Py e? e? 
ing quantities of the order e?, &c. dv’= (145 cos.2nt455 
e? 
(cos. nt — cos. 3nt)] xndt, or by integration v’ sant sin. 2nt-+- 
sin. 3nt sin. Snt- 4 
= 5 Sin. Pnt 5 a aer ombed 
e3 
ry (sin. me— ); but sin. nt — 
<a 2e? . 7 
ie sin. 2nt- “3 sin. Pat (20). Put ap’=c?*, then, since p’= 
Fa e? . ite 
a(1—e?), c?=a?(1—e?), or c=a (1 -5) neglecting quantities 
of the order e*, e%, &c. or 2e=a+a(1—e*?)=a+p’; hence by put- 
. e? 
tng ¢—p’=D, I have a—c=D, a? —c? =a2e?=(a+c)D, a= 
(a+e)D 2 = Aa+te)eD 4eD (a+c)D 
agge? a nes a nearly. Let —772— =sn. Y= 
pte > Ged ; 
¥ nearly,-5— = sin. Z=Z nearly, (Y and Z being small angles;) Y¥ 
sin. 2nt=V, Z sin.2nt=X3; then (20) becomes v’=nt+X+V, 
Which agrees with the second method of approximation given in the 
scholiam; Newton’s ap DO=c, AO=a,-PHB=v’, PSB=», 
dv 
and the angles X, V, are the same as his X, V. Again by (4), nat 
a 1+ecos.v 
=(1+e cos. v)? x(1—e2) ? also by (2) and (5), ian S. 
(1e?)? 
E=U ao and, since ndt=(1-ecos.¢) 
n ~ , 
1 
1—ecos. 93 hence 
5% (,90.\%' 1 Se SO fos 8 2)? 
7 (a) ~ (Ie cos. 9)?” ‘ndé ~ (i) Seer oh ia x 
I will now investigate a general formula, which will enable me to 
find v in terms of nt by (21). Let Q=Fy= any function of y, and 
y= any function of x; (x being a small variable quantity ;) then is Q 
@ function of x, which can generally be expressed by a series of the 
form Q=Q’+2Q ,4.27Q,420°Q,..42"Q,42"4' Qi, + etc. (a); 
QQ, Q., &c. being independent of 2. Put 2=0, them (a) be- 
comes Q=Q’, .*.Q/= the value of Q when c=0; and by taking the 
differential of (a) n times relativély to x, (supposing dx constant,) I 
have de" = 1.2.3...nQ, +2.3...(n-+1)7Q,., + ete. (6); by putting 
