miller: modern higher algebra. 



135 



If we include in the theorem all gauche determinants of a degree 

 less than the even number ;/, it is also true of those of the degree //. 



d^D 



J ^j ^ , is a gauche determinant of the even degree n — 2, and there- 

 da"da^~} ° ° ' 



n n— 1 



fore a square. 



dD 



Now 



da^i-i 



dD 



-j-^ — ; that is to say 

 da^ ^ 



"n-l 



dD 



da^-i 



o — a 

 a 



5 

 1 ■ 



— an— 1 



.n-l 



-(-l)n-l 



o a 



o 

 a? 



We therefore find that equation (i) becomes 



d^D _ j dD ) -^ 

 ^ daiida^-i ^ ] da^-i f 



n n— 1 ' n • 



dD 



^an-i 



which clearly shows that D is the square of a rational function, (jP say, 

 of its elements; the function </» is an integral quantity, since the square 

 of a rational function cannot be an integral quantity except as </* itself 

 is an integral. 



III. 



Statement. — If u and v be eliminated from the equations 

 x=fj(uv), y=f2(uv), and z=f3(uv), 

 we shall obtain another function that may be expressed thns: F(xyz) 

 =0. 



The Problem then is to compute the numbers that are proportional to 

 the derivatives of the first order 0/ the function F(xyz). 



Suppose that in the list of values of // and v which are now under 

 consideration, f^(uv), f2(uv), and f3(uv), allow of partial derivatives 

 with regard to u and v; and in the list of values of .r, y, and z, which 

 correspond to the values of u and v, F(xyz) also allows of the partial 

 derivatives F',, F' , and F' . 



The supposition just made being allowed, we shall then have 



