ALGEBRAIC Observe that, if there be more than two binary forms, the weight of the simplest perpetuant of degrees 0, 0, 0",...is 20+0+0+..-1-1, as can be seen by reasoning of a similar kind. To obtain information concerning the actual forms of the per- petuants, write where (1+0)(1+0)...(1+σx)=1+A₁x+Ax² + ... + Ax (1+1)(1+T)...(1+7)=1+ B+ B+ ... + B A₁+ B₁=0. For the case =1, 0=1, the condition is 071=AB=0, which, since A₁+ B₁ =0, is really a condition of weight unity. For w 1 the form is Aya + B₁b, which we may write ab₁-abo ao(1) (1)abo; the remaining perpetuants, enumerated by have been set forth above. 2 1-2' For the case 0-1, 0=2, the condition is ₁₁ = A₁B₁₂ = 0; and the simplest perpetuant, derived directly from the product AB is (1)(2)-(21); the remainder of those enumerated by 1-2.1-2 may be represented by the form (1+1)(2+1)-(11) (2+11)+... ±(2+11+1); λ and μ₂ each assuming all integer (including zero) values. For the case 0=0'-2, the condition is +++) (+) = -A3B,B,- A₁A,B=0. To represent the simplest perpetuant, of weight 7, we may take as base either AB,B2 or A₁A,B2, and since A₁+B₁ = 0 the former is equivalent to A,AB, and the latter to A,B,B; so that we have, apparently, a choice of four products. A B₁B, gives (2)(21) (21)(2), and AAB, (21)(2)(2)(21); these two merely differ in sign; and similarly A,B,B2 yields (2) (221), - (21)(22), and that due to A,A,B merely differs from it in sign. We will choose from the forms in such manner that the product of letters A is either a power of A₁, or does not contain A₁; this rule leaves us with AB,B, and A,B,B2; of these forms we will choose that one which in letters B is earliest in ascending die- tionary order; this is AB,B, and our earliest perpetuant is (2)(21), (221)a(2), - and thence the general form enumerated by the generating function 27 (1-2)(1-2) is (2+2)(2+114+1)-(2+21) (2μ₂+11 μ₁)+... ±(2+11+1)(2+1). For the case 0=1, 0'=3 the condition is (+1)₁+72)(01+73)=A,B+AB,B=0. By the rules adopted we take AB,B,, which gives (12)a(32) (1)a(321) + a(3212), FORMS 307 Selecting the product AB,B,B, we find the simplest per- petuant +a(432214), (14) a (4322)-(13)(43221)+(12) a (432212), -(1)a(432213) and thence the general form (1+4)+13+12+2) - ... ±α (4+1 3+1+2+4) due to the generating function 215 1-2.1-2.1-23.1-4. 101 The series may be continued, but the calculations soon become very laborious. Three Binary Forms.-Taking the partial degrees of a sem- invariant of three binary forms to be 0, 0, 0", an easy generaliza- tion of the foregoing leads to the generating function 1 (1-1-2)... (1-01-21-2)... (1-20X1-x1-2)... (1-20")' of the asyzygetic forms. If we place as numerator to this fraction 20+0+0"-1-1 we obtain the generator of the perpetu- ants. To obtain representative forms of perpetuants we require a general solution of the partial differential equation +α 1db₂ or say dodar d d +... bodby +b₁₂ +... .). dobą d codes + d +... ..) = 0, Pa+R+c=0. The general form of solution of +=0 we have seen to be (230) (12233...") -(12233...)(1-12233 ...00")+... the expression consisting of s+1 terms and the coefficients being Denote this by alternately +1 and -1. |(2238...)(123.... Now construct the expression (124298.00) (1223 Ms...") (1-10) -(1)(1+12... continued to s+1 terms and denote it by (1298) (1223...", and verify that, if it be operated upon by +e, the effect is to change t into t-1. The consequence of this is that if we form the simplest perpetuant of weight 7; and thence the general the expression form enumerated by the generating function 27 1-2.1-2.1-23' viz. :-(1+2)(3+1+1) - ... ±0 (33 +12+12+2) For the case 0=2, 0'=3, the condition is (+) (+) (+)(+73) (0+1)(02+72) (σ2+T3) x(1+T2) (1+T3)(T2+T3)=0. The calculation results in -A BBB+2A B,B2B-A B B B+ A B B₁-2A B&B.B₁ - A B B B+AB B B₁+ A,B B B+AB B-2A,BB,B + ABB₁ =0. By the rules we select the product A,B,B,B, giving the simplest perpetuant of weight 15, viz. :- (2)(3212) (21)a(321)+(2412) (32); and thence the general form (2^2+4) (38+12+114+2) - ... ± (2^2+11 μ1+2) (3+1+1) duc to the generating function 215 (1-7)(1— 23)9(1-29) * For the case 0-1, 0'4, the condition is (+) (+) (+) (+₁)II (T,+7)=0; the calculation gives A₁B (AB₂+ A,B,+B)( - B-A,B,B,- AB₁)=0. (2*3*...*) | (1'23...) (12333...") | - (12*29*s...0*) | (1ª-12^3^*...*^) (1°2233...0*0")] +(122*2*3...00) | (1-22^3^..^^0) (12233...") | continued to t+1 terms, we obtain a solution of a++8=0. To find the enumerating generating function of these forms sup- pose 0, 0, and " to be in ascending order of magnitude, and denote the form by (2*23*8.0*0) (11223.00) (112233.00"). In constructing this we have a choice of two places in which to place the part 1, three places in which to place the part 2, and so on; hence the generating function is 1 1-2.1-2.1-3...1-20.1-2.1-22.1- -20.1-2.1-23...1-20" precisely that of the asyzygetic forms. Hence the constructed expression may be taken to be the general expression of an asyzy- getic form. This idea is easily generalizable to the case of any number of binary forms. Stroh's form of seminvariant being w! (0a++...+σgα ++ B₁++₂ß₂+ ... + Teßo +1 +22+...+"Y"), the first case to consider is 0=1, 0'=1, 0"=1, leading to the con-
