40-44] The Method of General Dynamics 41 



Comparison with equation (04) shews that K becomes equal to K a at the 

 point representing class A, equal to K b at the point representing class B, 

 and so on. We have now to examine the way in which values of K are 

 distributed in the new generalised space, and shall find that there is a single 

 point at which K has a minimum value K = 0, and that at all other points K 

 is positive. 



43. To find maximum or minimum values of K, if such exist, we proceed 

 to the variation of equation (68). The equation expressing this variation is 



' 



...^ 



Since, however, .X'i,X 2 ... cannot vary independently, but are subject to equa- 

 tion (b'7), their variations will be connected by the equation obtained by 

 the variation of (67), namely 



TX. = ................................. (70). 



s=l 



Following the usual method, we combine equations (69) and (70) in the 

 single equation 



(71), 



where X is an undetermined multiplier. The condition that BK shall vanish 

 is now given by the elimination of A, from the system of equations 



). 



^VVitt'S-S 

 The result of this elimination is 



( 



Xi X 2 ... X n , 



and if we combine this with equation (67) we find that 8K vanishes, subject 

 to equation (67), when 



Xl = x 2 =... = f ........................ . ..... (72). 



There is therefore one and only one point at which K has a stationary value. 



44. The next step is to expand K in the neighbourhood of the point 

 given by equations (72) at which 8K=0. A point adjacent to this point 

 may be taken to be 



(73), 



where e 1} e 2 ... are small, and where, since equation (67) must still be satisfied, 

 6j, e 2 ... are connected by the relation 



ei + e 2 +... + e n = ........................... (74). 



