(532 DR WALLACE ON A FUNCTIONAL EQUATION. 



by ^o ' "^^ ^^J express the relation between -p- and x^ thus. 



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 fom- equal angles DAB, BAD', D"AB', 

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



|-=/W. (2) 



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



y =/ W -/W (3) 



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 pan- be R', 

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



and =/(«„ + ^/) +/ (^o - 2.-,) . 



P 



Now, the forces R', R", which constitute the force R, he in the same direc- 

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



y=/(-^„+^;;+/(^„-^.)- (4) 



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



and multiplying both sides by C a constant 



0/W ■ C/(:r,) = C {C/(.r„ + ^,) +C/(ar„-z,) } ; 

 and putting simply /(O. and/(«,), and /(«o + »^,), and/(a:„-«.), instead of the same 

 symbols multiphed by the constant C (this, because of the indefinitude of the 

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



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

 will apply to the present case, because /(x) = -p- decreases while the angle x in- 

 creases ; thus we have 



3. = f(x) = a cos (-^) ; 



