ALGEBRAIC POINT FUNCTIONS IN N-SPACE. 419 



an arbitrary scalar. All terms vanish except those from 2 to q f -\- N— 2. 

 The equation takes the form 2c»bi»ai.Ei(x + g&i) = where i runs 

 from 2 to q' + N - 2. 



Consider the quantity Ei (x + jrai). It is a polynomial in g whose 

 absolute term is Ei(x). Since the above equation is to hold for all 

 values of a,- it must hold for all values of g. Hence the sum of the 

 absolute terms must vanish, that is 2cibi*8,iEi(x) = where i runs 

 through the q' + N — 3 values above noted. But this is impossible 

 unless the c's which have these subscripts are all zero : for the equation 

 now has the form 2L(x)£(x) = 0, the only a common to L and E is ai, 

 whence not fewer than q' + N — 2 E's satisfy such a relation, by the 

 corollary to theorem IV. 



Similarly we note that a m+2 occurs only in the second group of E's, 

 and prove in the same way that all c's corresponding to this group must 

 be zero, and so on, till finally only C\ remains. Hence C\ is also zero. 

 This is contrary to hypothesis that the c's be not zero. Hence the 

 determinant does not vanish identically. 



To show, finally, that the rule is always possible, we note, first, 

 that the rule observes the plan of grouping the E's so that no more 

 than q' + N — 2 belong to a standard set, E\ being common to all the 

 groups. The first group requires m + 1 of the a's. Each new group 

 requires one additional a. We have m + N — 2 a's available for 

 selection. It therefore suffices to show that the number of groups 

 is not greater than N — 2. We have, in other words to prove the 

 inequality 



q ^ (A - 2) (</ + N - 3) + 1 (58) 



Putting for q and q' their values in terms of N and K this becomes 



{K+\)(K+2)---(K+N-2) 



(N - 2) ! 



(N - 2)K(K + 1) (A' + 2) • • • (K + N - 3) 



(N - 2)! 



-A7--2 ^ 



- (N - 2) 



which simplifies to ^ (N - 2)K(K + N - 3). This holds when 

 N is 3 or more, and binary forms were above explicitly excluded from 

 consideration. The method is thus completely established. 



In agreement with these results we note that when N — 3 we have 

 but one group, as exemplified by (38). When A 7 = 4 we shall always 

 have two groups, but they are not both complete. If the first group 

 be always completed, a smaller and smaller proportion of the E's of 



