Page:1902 Encyclopædia Britannica - Volume 25 - A-AUS.pdf/351

 307 ALGEBRAIC FORMS Selecting the product A^B4B3B|, we find the simplest perObserve that, if there be more than two binary forms, the petuant weight of the 1 simplest perpetuant of degrees d, 6', 6",...is 2O+0'+d"+.. - - i) as can be seen by reasoning of a similar kind. (l4)a(4322)& - (l3)a(4322l)6 + (l2)a(4322l2)i - (l)a(4322l2)J + ao(432-l )bi To obtain information concerning the actual forms of the perand thence the general form petuants, write (1 + Cp^Xl + o-^)... (1 + <tqx) = 1 + A-lX + Aga;2 + ... + Adx& (l*1+4)a(4^+Y3+V2+2)6 - ... ± a0(4^+13^+12fl2+2lAl+4)b, (1 Lr^X! +T2a3)...(l +T0-as) = l + B1a3 + B2;r2+ ... + Bg-cc#' due to the generating function where + Bj = 0. l-s.l-z2.l-z3.l-z4For the case 0 := 1, 0' = 1, the condition is (T jT ^ — A1B1 — 0, The series may be continued, hut the calculations soon become very laborious. which, since A1 + B1 = 0, is really a condition of weight unity. Three Binary Forms.—Taking the partial degrees ol a semFor w=l the form is AjOj + B^j, which we may write - a160 = invariant of three binary forms to be 0, 0(, 0', an easy generaliza(1 )afio 5 the remaining perpetuants, enumerated by, tion of the foregoing leads to the generating function l have been set forth above. 2 e 1 1 (i 2X1 z )... (i 3 X zX • •• (i - ^'Xi - *'2Xi - • (i - ’ For the case 0 = 1, 0' = 2, the condition is (r1r1r2 = A1B2 = 0 ; and the simplest perpetuant, derived directly from the product of the asyzygetic forms. If we place as numerator to this AnBa, 3 is (l)a(2)6-(21)6; the remainder of those enumerated by fraction 320+e +<,,,~1-l we obtain the generator of the perpetuants. To obtain representative forms of perpetuants we require -— z ^ _g2 may be represented by the form a general solution ol the partial differential equation (l^+lU2^+1)b - (l^)a(2^+1l)s +... ± (2^+1l^+1)i; + i ( 4+f’4++ X! and a<2 each assuming all integer (including zero) values. For d A ' C d' the case”0 — 0'= 2, the condition is °dc-Tc'Sa+-) • T r cricr2 iT2(c i + o'2)((ri + T1)(cri + r2)= — AIBJB2 — AXA2B| = 0. fia + + He — 0. To represent the simplest perpetuant, of weight 7, we may take or say we have seen to be as base either A|B1B2 or AjA^I, and since A1 + B1 = 0 the former The general form of solution of + is equivalent to AJAIB2 and the latter to A^jB? ; so that X2 A3 Ke S 3 (2 3 ... 6' 'X(l 2^ ... we2 have, apparently, a choice of four products. A|BaB2 gives (2 )„(21)& — (2'21)B^ yields (2)a(2-l)s(21)a(2 )6, and that due to AjA^I merely differs from it in sign. expression consisting of s +1 terms and the coefficients being We will choose from the forms in such manner that the product the alternately +1 and - 1. Denote this by of letters A is either a power of A1; or does not contain A this x rule leaves us with A^B^ and A^B2 ; of these forms we will choose that one which in letters B is earliest in ascending dictionary order ; this is A1B1B2, and our earliest perpetuant is Now construct the expression (22)a(21)6 - (22l)a(2)a, (i^*3.. 0'X0')b(is2fA23'A3...e,{le")c and thence the7 general form enumerated by the generating z ~l2^3-■ ■ e're"^ function ^ _^218 V s 2 /fl,, 1 2 1 )c + (i+2)(lt+22V*...0' ){)(l - 2^...0 (2A2+2)a(2^+iiM1+ )6 _ (2^+ l)a(2^+ lMi)i+... x 2 IA 1 1 + (2 2+ li i+ )(l(2^+ )6. continued to s +1 terms and denote it by For the case 0 = 1, 0' = 3 the condition is <r r r r < T <7 T <r T ♦ i 1 2' 3( r1 + 1)( i + 2)( i + 3) ^ AjBI + Af B2B3=0. By the rules adopted we take A|B2B3, which gives and verify that, if if he operated upon by H6 + Oc, the effect is to (l2)a(32)6 - (l)a(321)i + a0(3212)6, change t into t-l. The consequence of this is that it we form the simplest perpetuant of weight 7 ; and thence the general the expression form enumerated by the generating function (2K23K3...0'cfl)a | (l¥23X3../e(lS2^8.../0'')c | 1 - Z. 1 -32. 1 - S3’ -('l2'C23'C8...0Ke)a | (V“ V23X8...0'X0')b(lS2'i23'i8...e’^'X | viz. :-(lX‘+2)a(3«+12«+1)i - ... ±«0(3«+12ra+1lA'+2)1; + (l22K23K8...0Ke)a | (i*-22X23X3...0'Ae')b(lV^3.| For the case 0 = 2, 0' = 3, the condition is <r cr T r r 0 continued to 2 +1 terms, we obtain a solution of l121 2 l' 2 3( ’l + <r2)(<rl + Tl)(°’l 4" r2)(crl + T3)(<r2 + 7’l)(°’2 + T2)(<r2 + Ts) x(r1 + r2)(T1 + T3)(r2 + 73) = 0. Ha + fit + Hc = 0 . The calculation results in To find the enumerating generating function of these forms sup4 3 - A B3B2Bf3 +2A2 B3BlBf - A|B3B§Bf + AfB^Bj2 - 2A;]B1B B pose 0, 0', and 0" to be in ascending order of magnitude, and -AiBlB2B + A BIBIB1 + A2BlBIB| + AlBlB -2A2B3B22B21 denote the form by 4 + A2B B1 = 0. By the rules we select the product A4B3B2B2, giving the simplest perpetuant of weight 15, viz. :— In constructing this we have a choice of two places in which to (24)0(3212X- (24l)a(321)6 + (24l2)a(32)6; place the part 1, three places in which to place the part 2, and so on ; hence the generating function is and thence the general form l_2.l_22.l_z3. ..1-2:0.1-0.1-z2.1-Z0'.l-z2.1-z3. ..1_20" V2+V1+2)b - • • • ± (2X2+4l^+2)a(3'A3+12'i2+1)&, precisely that of the asyzygetic forms. Hence the constructed due to the generating function expression may he taken to be the general expression of an asyzygetic form. This idea is easily generalizable to the case of any (1 -z)(l -z2)2(l -z3) ■ number of binary forms. Stroh’s form of seminvariant being For the case 0 = 1, 0' = 4, the condition is —, (cr1a1 +  the calculation gives the first case to consider is 0 = 1, 0' = 1, 0" = 1, leading to the conA1B4(A2B2 + AjB, + B4)( - B! - AjBaBg - A2B4) = 0.
 * (l*2X23X3...0'X0')b(lS2^3'"3...^")c| ,
 * (2,C23K3...0'C0)a(lXl2X23X8...0'Xe')b(l'il2'i23,A3.../e")c |.