PHILLIPS AND MOORE. — LINEAR DISTANCE AND ANGLE. 73 



R hy a, product of points S we take from S all combinations D of 

 points such that D is complimentary to R, arrange the others in a 

 product C such that S = C D, and form the sum S {R D) C. To 

 obtain the product 



by resolving the second factor, we must take the sum of products, 

 of p~ by all but two letters of any term of the second, factor times the 

 product of those two. Those letters will occur in a combination p"^ 



and this combination may be selected in ( ) ways. Hence 



[p2.^»-i] = p"^^ A|^/.-i.;j=]_''2(''"-A_(,„_l'\p«.^, 



the second term being subtracted because in 



occur 2 I \_^ j terms of the form p'"- p, whereas there should be 



m — 1 in the expansion of [p^-jj"'' ^]. Simplifying the above expression 

 we get, since [p~-p'"'~^ = [p"''~^-p-], 



[p'"~'LP~] = 2~^p"'-p. 

 m 



Similarly 



[pm-l . ^/] ^ Q [pm-, . p2 . .pr-2^ _ [O Q _ r] p^ f'^ 



= r p"^-p^ ^ 



Since p^ is a scalar, we may solve this last equation for p''"^. Chang- 

 ing r into r + 1 in the result we have. 



f - "' ^P 



m-l „r+i1 



p"'{r+ 1) {771 -rY 



the equation holding for r = if we take p° = 1. Thus we have an 

 expression for p'' in terms of p'^^'^. Expressing p'"^'^ in terms of p""^-, 

 etc., we have finally 



p"' / [(r+ 1) (f-\-2) . . .m]{77i-r)l 



