Binomial Biordinals. 251 



78. If we suppose this equation to be put under the form 

 we shall have /d»^^K=0, 



/(D) = D 3 + 3(a 2 -l)D 2 +{3(m 1 -a 2 ) + 2p + m s 



= (D + <v-l) 3 - {3(A-|) 2 + Jp}(D + « 2 -l) 



+ (2A-l){(A-i)^-Jp}. 



79. Put D + a 2 -l = A and A-i=B, then 



/(D)=A 3 -(3B^ + Jp)a + 2b(b^-Jp). 



80. In other words, 



/(D) = (A + 2B)(A-B + fI)(A-B-fI). 



81. The calculation of F(D) was still more troublesome 

 than that of /(D), and my original result was in defect by a 

 unit. But the masterly hand which corrected the defect 

 pointed out a shorter process of calculation, viz. that to be 

 given in arts. 88, 89. 



82. If we put 



D + <2 2 + 2ft> + 5 = V, A— w— i = G, 

 we shall at last find 



F(D)=(V + 2G)(V-G- + |J)(V-G— fJ). 



83. But a 2 is arbitrary; and if a 2 = l, then V becomes 

 D + 2ft> + 6, and V+2G- becomes V + 2A + 5, or V + 2B + 6, 

 while V-Gr becomes D — B + 3&) + 6. Hence, writing the 

 terordinal in the form 



we have /(»)?+ F(D-3)^=0, 



F(D-3) = (D + 2B + 3)(D-B + 3ft) + 3 + |J)(D-B + 3ft> + 3-|J), 



84. Moreover, turning to arts. 80, 82, 83, we see that if 



/(D) = <KD,B,I), 

 F(D-3) = </>(D + 2«; + 3, B-ft>, J), 

 and <£(D, B, I)£+<£(D + 2ft) + 3, B-ft>, J)£ 3 f=0. 



85. This result may be further simplified by taking a new 

 variable Z, given by Z = t~ B £ ; for we thus obtain 



<£(D + B, B, I)Z + <KD + B + 2ft> + 3, B-*>, J)* 3 Z = 0, 

 which (see arts. 80, 82-4) is equivalent to 



(D + 3B)(D + fI)(D-fI)Z 

 + (D + 3B + 3)(D + 3ft) + 3 + |J)(D + 3ft) + 3-|J)^ 3 Z=0. 



