connected with the Theory of Probabilities. 169 



V 



equations of the system (1) are satisfied, the function =^- in the 



final equation varies continuously from to 1. In such varia- 

 tions it must at least once coincide with the value r. Hence 

 there must be at least one solution of the entire system (1) in 

 positive values of x, y, ... z. It remains then to show that there 

 cannot be more than one such solution. If there'be more than 



V 

 one such solution, the value of ==- must, in the variation above 



described, more than once coincide with r ; and therefore, vary- 

 ing continuously, it must between such points of coincidence 

 admit of a maximum or minimum value. But it may be shown 

 that there exists no such value. The nature of the proof of this 

 proposition will be best illustrated by a particular example. 



Let us take the particular case in which n=2. Our equa- 

 tions then are 



axy + bx ._. 



t i =p, (5) 



axy + bx + cy + a 



y + °y =q; (6) 



axy + bx + cy +d 



and we are to inquire if it is possible to satisfy, by positive values 

 of x and y, the conditions 



axy + bx 



axy + bx + cy + d ~"' 



axy + cu . . . * 



—-{- ,= maximum or minimum. 



axy + bx + cy + d 



Differentiating in the usual way, we find, after slight reductions, 

 [ay + b) {cy + d)dx + (ad—bc)xdy — 0, 

 (ad—bc)ydx + (ax + c)(bx + d)dy = 0, 

 from which, eliminating dx and dy, 



(ay + b) (cy + d) (ax + c)(bx + d) — (ad—bc)*xy=0 ; 

 or developing, 



(Pbcxhf + ab(ad+ bc)x i y -f ac(ad + bc)xy' 2 + ab^dx* 2 

 labcdxy + ac l dif + bd(ad + be) x + cd(ad +bc)y + bed 2 = 0. (7) 

 Now this equation, since its first member consists wholly of 

 terms of positive sign, cannot be satisfied by any set of positive 

 values of x and y. Hence, by the previous reasoning, the system 

 (5), (6) admits of one, and of oidy one, solution in positive values 

 of x and y. 



The class of functional determinants, to use Jacobi's expression, 

 exemplified in the first member of (7), and obtained in like 



