632 DR WALLACE ON A FUNCTIONAL EQUATION. 



by ^o » we may express the relation between -p- and x^ thus. 



^=/W- (1) 



Now, like as R is the resultant of two equal forces P, P, we may assume that 

 P and P are each the resultants of two equal forces p, p, which make equal angles 

 with them ; one pair acting in the du-ections AD, AD', and another pair in the 

 directions AD'', AD'". Let each of the four equal angles DAB, BAD', D"AB', 

 B'AD'", be denoted by x^, and we have, because P is the resultant oip, p, 



y =/(-.)• (2) 



therefore, taking the product of equations (1), (2), 



Now, the force R, which is equivalent to the equal forces P, P, must also be 

 equivalent to the four forces which compose P, P ; two of these are forces p, p, 

 which make with R angles each equal to x^ + x,, and the other two, p, p, make 

 also with R angles each equal to x^—x^. Let the resultant of the first pair be R', 

 and the resultant of the other pair R", we have then 



and — ■ — - = /(^o + ^/) +/ (^o - x) . 



Now, the forces R', R", which constitute the force R, lie in the same du'ec- 

 tion with it; therefore, R=R' + R", and so we have 



^=f{x^-^x:)+f{x-x). (4) 



We have now, from equations (3) and (4), 



/W/O^O =/(^o+^.)+/(^o-^.); 



and multiplying both sides by C^ a constant 



0/ W • Cf{x:) = C {C/ (.r, + X) + Cf{x-x) } ; 



and putting simply /(^o). and/(«.)> and f{x^ + x), and/(«o— ^/)' instead of the same 

 sjrmbols multiplied by the constant C (this, because of the indefinitude of the 

 symbol _/ is evidently allowable), we have this functional equation 



fix) .fix) = G{f{x^ + x)+f{x-x:)}, 



of which we have found two solutions (Art. 13) ; the first of these, however, only 

 will apply to the present case, because f{x) = — decreases while the angle x in- 

 creases ; thus we have 



y = /(*) = « cos (-^) ; 



