finding "Fluents by Continuation . 4.35 
If J- = I , J', x*xs/ 2 + X=l X 2 + x\ " X 2 X i -— {Q -a. 
5 * ' 
* = 3 ij'x ! X \/ 2 + * = i x 2 + *1 3 X idl -4^ + 3« + 2Q > =jS. 
_____ » | r 
f =■ 5, / aAv + __ lH 4. 8**+ i /3 - ’ 
4/ h 2 A J 
IS "- 5 Q=y- 
AS 
j=7, /" a? 1 * v / 2 + * = tV X 2 + *1 1 x — - 4^! + — _ i6*ijL 
j 7 5 3 
2 iy-!f/ 3 + !f« + s6Q. 
25 s 5 5 
&C. &C. &C. 
II. Let F - 
a + bx m + x 2m 
Mx n ~ im x . w/"^ 1 • ^ 
* * = - ¥x n -i m x + Qtf—Vx - See. 
, then F == 
_ /- 2!H + I p iV B — 3 W— ‘ >r Qy»~4 w +i 
a +bx m +x* m a + bx m + x 2m ' ~n- 2 m+i rT^^T i + n - 4 w +1 
~&c.dbTdbU, where T and U are put for the fluents of the 
two laft terms, and P, Q, &c. for the co efficients arifing from 
the divifion. Now,^* 
n . 
X X 
a/ . « m . ztn 
v a + bx -\-x 
=/ 
x (g + ^ r/i + *t 
a + bx m + x zm ~ 
l — r . T .v-}~ i — 7* . 2 mx'" 
= (by 
Fx«+fe*+H' '- /'f+ 
fubftituting for F its value in the latter quantity, and putting 
A, B, C, &c for the co-efficients which arife in confequence 
thereof) Fx« + bx m + x 
»i ! _ A x 
/ 
- + B x 
a + bx m +x 2m \ 
“j- See < 
/ 
n—m . 
x x 
c + + 
Vol. LXXVI. 
- C x 
/ 
x n 2 m x 
. » m - 2ct 
+ A# ■+■ AT j 
L 1 1 
