( 209 ) 



directly appear, that /• must be ^> n to satisfy the circumstance that 



dp 



— = shall pass through the top. Now we shall have to content 

 dx 



ourselves with calculating the value of k for different values of n with 



the aid of the table (p. 829). Thus = 0,3704 corresponds to 



n — l 



B 



C 



B \ 1 1 



— = , we find k = 1 + , n being = 1 A . Thus 



C h-1' ^0,0296 B ^0,3704 



we find : 



ttg n k 



1 1 



M 1 + ^ 1 + 



#,, = 0,4, and we find - then equal to 0,4 — 0,3704 = 0,0296. As 



0,37 ' 0,03 



2 1 



0,45 1 + - 1 + 



3 ' 0,075 



1 1 



0,46 1 + — - 1 + 



2,08 ' 0,105 



1 1 



0,47 1 + — - 1 + 



3,06 ' 0,15 



1 1 



0,48 1 -f 1 -f 



5,04 ' 0,24 



1 1 



0,49 1 + — — - 1 + 



10,91 ' 0,6 



k 

 So k always larger than n, but the ratio - decreasing with ?i. 



n 



B \ 11 



That — or - — must always be negative tor the 



C n — l k—l n -\ J 



dp d*\p 

 case that — = passes through the top of = 0, appears when 



dx dx* 



B 1 



we express the value of — in x 9 . This value is : 



C n — 1 



B 1 agfixg—l) 2 



C n—l l—2xg n—l 

 or 



B 1 _ Xg(l-x g ) i i^/ W^^Y 



C n-1 l—2*g J " + K Xg{l-Xy) 



