Solution of a Functional Equation. 69 
forces are supposed to tend to diminish the distances p, q, 7, and 
by (4) we get Pip+Qsq+Ror=0, (5). 
Now since a, 6, a’, b’, are each invariable (because they be- 
lens to fixed points), we have 9p = — dz, dg = — or = 
= artiye Wy ’ PaEg oy, 
, and nod ee these values (5) be- 
= 
or+ i R — hence, since 6x, dy are arbitrary and 
y—v 
—R 
‘comes — Pdr — 
r “4 
independent of each other, we must have — R-—P=0, 
—-Q=0, or Pa . . R, a ha R, .:. P? +Q?=R/, (8), 
x-a’)? — 0b’)? Q y-b 
since ( ) mat ) =1, also pip (9) ; hence from (8) 
and (9) it is evident that if two forces are represented in quantity 
and directions by the two sides of a rectangle, their resultant is 
represented in quantity and direction by the diagonal which pass- 
es through the angle formed by the two sides that represent the 
forces. 
For other applications of (4) we shall refer to the Mécanique 
Analytique of Lagrange, and the Mécanique Céleste of La Place, 
especially to the first volume of the former work. 
Arr. VI.—Solution of a Functional Equation, which has been 
employed by Poisson in demonstrating the parallelogram of 
_ forces; by Greorce R. Perkins, A. M. 
~ Porsson, in his able Traité de Mécanique, (see second edition, 
Vol. I, p. 44 et seq.) has given a beautiful demonstration of the 
parallelogram of forces. He makes his demonstration rest upon 
-_ ppeenintton of yx, so as to satisfy the condition, 
groz=9(r+z)+9(t—Zz) 
Has says, pv =2cos. ax, will satisfy (1), and he further says, 
that no other value of gx will satisfy it; but he does not show 
how he determined this value of yz; but seems to have obtained 
it by induction. Neither does he show why there mers not be 
other values of yz, which will satisfy (1). 
