62 Transactions of the Royal Canadian Institute 

 {v + b)-Xz-Xs= — — - — -— — o 



+ 3 



v.u + l.v + 2.u-\-S v+l.j'+2.f+3.j'+4 



j/ + 2.v + 3.j^+4.j'+5 



12vHv+S) ^ S{v-^2y{p-\-^) _|_ ^ 4(1^+4)3 (v+5) 



v.v-{-l.v + 2.v-\-3 v+l.u + 2.v+3.v-\-4: i^ + 2.i'+3.j'+4.j' + 5 



L.J/+1.Z/ + 2 j/ + l.j' + 2.j/+3 i/ + 2.v+3.i'+4J 



= 12X1 



Notice how the reduction from the fifth to the third powers of 

 V, v-\-2, v-\-4: is effected; also the manner in which the denominators 

 with four factors pass into denominators with three factors, and finally 

 the simultaneous conversion of the binomial coefficients 1, 3, 3, 1 into 

 the binomial coefficients 1, 2, 1. 



Suppose now that r > n in X^„ . Nielsen has not occasion to consider 

 this case, but his analysis applies here also. The X's will no longer vanish ; 

 moreover we know that when X^„ is expressed in partial fractions, these 

 partial fractions drop out leaving only an integral part. The order of 

 this integral part, putting m = 2(r — l), should apparently be (w+« — 1) — 

 (2w — 1), =m — n, but is actually m — 2n, owing to the vanishing, in the 

 numerator of the reduced fraction X^ , of the n terms x^, x^'^^, . . . . , 



The method for proving this need not be worked out in full; the 

 essential parts of the proof can be derived readily from the study of 

 particular examples. 



For instance Zt=(v+6)2 Xl- 12X1= -12X1, 



xl={v^-^yxl-%xl= -^xl, 



Xl=-4: 



SO that Xt= 12.8Z?= -12.8.4 



Xi = 8.4. 

 For Xl, m takes the value 6 and n = S; hence m — w = Oand the integral 

 part should reduce to a constant, as is in fact the case. 

 Again 



Xl= (v-]-Qyxt-12Xt= -12.SA{p-j-Qy-12Xt; 



Xt = {v-{-4:yxl-8Xl = 8A{y-{-4:y-8Xl 



X^=_8(.2+2. + 2), 

 so that X^ is a quadratic in v. By the theorem, since w = 8, n=3, the 

 order of the integral part should be w — 2w = 8 — 6 = 2, as is in fact the case. 



