We can prove that if the three equations be satisfied by a system of values of the variable, the same system will also satisfy the Jacobian or functional determinant.
For if u, v, w be the polynomials of orders m, n, p respectively, the Jacobian is (u 1 v 2 w3), and by Euler's theorem of homogeneous functions xu i +yu 2 +zu 3 = mu xv1 +yv2 +zv3 = /IV xw 1+y w 2+ zw 3 = pw; denoting now the reciprocal determinant by (U 1 V2 W3) we obtain Jx =muUi+nvVi+pwWi; Jy=ï¿½.., Jz=..., and it appears that the vanishing of u, v, and w implies the vanishing of J.
CY The proof being of general application we may state that a system of values which causes the vanishing of k polynomials in k variables causes also the vanishing of the Jacobian, and in particular, when the forms are of the same degree, the vanishing also of the differential coefficients of the Jacobian in regard to each of the variables.
Since, If F = An, 4) = By, 1 = I (Df A4) Of A?) Ab A"'^1Bz 1=, (F, Mn Ax I Ax 2 Axe Ax1) J The First Transvectant Differs But By A Numerical Factor From The Jacobian Or Functional Determinant, Of The Two Forms. We Can Find An Expression For The First Transvectant Of (F, ï¿½) 1 Over Another Form Cp. For (M N)(F,4)), =Nf.4Y Mfy.4), And F,4, F 5.4)= (Axby A Y B X) A X B X 1= (Xy)(F,4))1; (F,Ct)1=F5.D' 7,(Xy)(F4)1.
Solving the equation by the Ordinary Theory Of Linear Partial Differential Equations, We Obtain P Q 1 Independent Solutions, Of Which P Appertain To S2Au = 0, Q To 12 B U =0; The Remaining One Is Ab =Aobl A 1 Bo, The Leading Coefficient Of The Jacobian Of The Two Forms. This Constitutes An Algebraically Complete System, And, In Terms Of Its Members, All Seminvariants Can Be Rationally Expressed.
1 Ab' Establishing The Ground Forms Of Degrees Order (I, O; I), (O, I; I), (I, I; O), Viz: The Linear Forms Themselves And Their Jacobian J Ab.
1 A2B' Where The Denominator Factors Indicate The Forms Themselves, Their Jacobian, The Invariant Of The Quadratic And Their Resultant; Connected, As Shown By The Numerator, By A Syzygy Of Degreesorder (2, 2; 2).
There is no linear covariant, since it is impossible to form a symbolic product which will contain x once and at the same time appertain to a quadratic. (v.) is the Jacobian; geometrically it denotes the bisectors of the angles between the lines ax, or, as we may say, the common harmonic conjugates of the lines and the lines x x .