;~>60 PROFESSOR CHRYSTAL ON LAGRANGE'S RULE FOR THE 



Regard now x, y, z as functions of a fourth variable t ; then 



^x+yf y + ^ z = yjr(x,y,z) .... (B). 



Using (A), we get 



(xQ-yY)f y +(xB J -z?)xl,=-Fi(x,y,z) .... (C). 



We see from (C) that y^(x,y,z) will vanish if we establish between the three variables (x,y,z) 

 relations such that 



xQ-yP = 0, xR-z? = 0; 

 that is, 



dx/? = dy/Q = dz/K (D). 



These equations connecting x, y, z will determine x and y as functions of z ; it follows, 

 therefore, that by the substitution of these values of x and y, ^(x,y,z) will become a 

 function of z. Hence, since the differential coefficient of \J/-(x,y,z) must then vanish, it 

 follows that z must disappear of itself from \^(x,y,z) after the substitution. 



If, then, u{x,y,z) =a, v(x,y,z) = b, (E) be the integral system of (D), we must have as a 

 consequence of these equations 



f(x,y,z)=f(a,b) (F). 



But as a consequence of (F) and (E), we must have the identity 



^{x,y,z)=f{u{x,y,z\ v(x,y,z)} . 



Therefore, the supposed solution ^(x,y,z) = may be written in the form f(u,v) = 0. 



Overlooking the neglect of the possibility that P = may be a derivative of the 

 system (E), and thus (C) be satisfied, although ^(x,y,z)^z0, it is easy to show that the 

 above demonstration is quite fallacious. It is forgotten that the equation (C) is only 

 arrived at by means of (A) ; therefore, in order that ^(x,y,z) may vanish, it is not 

 sufficient merely to replace x and y in terms of z by means of (E), we must also use (A), 

 or its equivalent, y^(x,y,z) = 0. ^(x,y,z) does not vanish identically, and therefore 

 Lagrange's inference (in the words italicised above) that (F) results from (E) falls to the 

 ground. The simplest particular case will show the justice of this criticism. 



Consider the equation xp + yq = z and its solution zy — x 2 = 0. Here ^=zy — x 2 , and 

 we have 



\fs= —2xx+zy + yz. 



We may take u=xjz = a, v=ylz = b as the system (E). Therefore x = az, y = bz; so 

 that 



\Js= { — 2ax + bz + y}z . 



Replacing x and y by their values in terms of z, we find 



\Js = 2(b-a 2 )zz 



