or 



( 42b ) 



'/. I c 1 — x da 



A — X — = X s 1 -f- 



a.r <i a dx 



The first member of equation (</'") becomes for x = : 



n — 1 



?< — 1 =p 



1/(1 +»i) 



and so, whether the sign - or the .sign -|- is chosen, always posi- 

 tive, if as in the case considered, the quantity e, is positive. For 

 x = l the first member of (<p'") becomes equal to: 



( . - 1) _ . j/i 



or 



n— 1 - 



This value would be negative when, as will be supposed in a 



following case, f 2 is negative bul il is also positive, if as is now 



the case, f 3 is positive. If the sign of the value of the first member 



of (<p'") is different for a? = and x = l, then there will be a 



value of x between and J satisfying {</'">. But in our case the 



first member of (#>'") has the same sign for x = and x = 1. From 



this it does not follow, of course, that there exists no root for {<p'"), 



but only that this equation either has no roots or an even number. 



This equation has no roots when the locus is imaginary — but if 



[/s l -\- w^/f, 



the latter exists, as is the case when 1 > , and when the 



n — 1 



locus is a closed figure, then there must be two. If the value of 



the first member of {<p'") is graphically represented between x = Ü 



and x = l, the curve representing this value, begins and ends positive. 



If this value passes to negative values, it must have an ordinate 



equal to at least twice, and so also assume a minimum value. 



Hence if there are two values of x satisfying {<p'") } the equation 



obtained by differentiating (</>'") with respect to x, must have a root. 



Now — — is equal to : 



dx 



d*A A d*A 



— x (1 — x) 



n dx* 1 dx" 2 



2 l dA\ 2 ( dA 



]/\A-x~[ v\a \ (1-*) — 



/ i lx I dx 



