If all the terms containing a in the discriminant be collected, the portion involving a will be a; and we have to show that the last three terms are the discriminant of the quantic formed by putting a=0. by the quaternary one ·(7) Q = (bß' + v,) a' ”-1 + (pß'' + B'w,+w,) a'”~'+.....(8), where vw (7, 8)". The discriminant of U' may be obtained by eliminating (a', B') between U'=0, dU' / da' = 0 and dU'dẞ'=0, which is the same thing as eliminating (a', ') between Q=0, dQ/da'=0, dQ/ds=0. But since Qis a surface, the second form of the discriminant gives the tangent planes to Q which can be drawn through the edge AB of the tetrahedron of reference, and its degree is equal to the class m of the surface. In (8) put b= 0, then V2 is the tangent plane to the surface at A; and it is known from the theory of sufaces that when the vertex of the tangent cone lies on the surface, the complete tangent cone degrades into the tangent plane at A twice repeated and a cone of degree n(n-1)-2. Accordingly if the line AB lies in the tangent plane at A, the discriminant of Q gives the tangent plane at A twice repeated and m-2 other tangent planes through AB. Hence the discriminant of Q must contain v," as a factor. From this it follows that if in the original ternary quantic we put a=b= 0, its discriminant must be of the form c'; and in like manner if a=c=0 its discriminant must be b'Þ. The term bc is introduced for generality; since its omission would involve the assumption that the discriminant contains no terms of the form bc or be. 6. If A be the discriminant of a ternary quantic U, the conditions that the curve U=0 should be binodal are that the first differential coefficients of ▲ with respect to all the constants of U should vanish. Let U1, U,, U, be the first polars of U with respect to A, B, C; and let E,, be the binary quantic of (8, y), which is obtained by eliminating a between U, and U. Then the equations E1,=0, and E=0............. 12 ...(9) determine the lines drawn from A to the points of intersection of U1, U, and U1, U, respectively. But if U has a node, the three curves U, U, U, must pass through it, the condition for which is that the eliminant E of (9) must vanish. Hence where F is an extraneous factor with which we are not concerned. If U is a binodal curve, the first polars of every point must pass through the two nodes; hence (9) must have a pair of common factors, the conditions for which are that the differential coefficients of E with respect to each of the constants A, B, ..., and A', B', of E, and E, should vanish. Accordingly the conditions are expressed by the system of equations O = dE|dA = Fd▲]dA+▲dF|dA 13 and since A=0 and F will not vanish unless the vertex A is specially situated, it follows that d▲/da = 0. Let a, b, c,... be the constants of the original quantic U, then A, A', are functions of a, b, c, ... ..; ... and therefore +.... and since d▲ dA, &c., have been shown to be zero, it follows that dA/da, &c., are all zero. 7. The vanishing of any two of the differential coefficients. of A expresses the condition that the first polars of every point should pass through two fixed points, and they are therefore the necessary and sufficient conditions that a curve should be binodal; but although the number of equations is equal to the number of constants in U, only two of them are independent. The above conditions must also be satisfied * Salmon's Higher Algebra, §§ 96-103. His investigation concerning equalities between the roots of equations is by no means clear; but it seems scarcely worth while to rewrite it. when U has a cusp, because the first polars of every point must have bitactic contact with one another at a cusp; but they are not sufficient, because the condition that two curves should touch one another leads to a relation between their coefficients which is called their tact-invariant. The necessary and sufficient conditions that a curve should have a cusp are (i) that the tact-invariant of the first polars of two arbitrary points, such as A and B, should vanish; (ii) that any one of the quantities da/da should vanish. The conditions may also be expressed as follows. The quantics (9) must have a common square factor, which requires that their eliminant and that the discriminants of both should vanish, but these conditions are equivalent to only two. 8. The theory of tact-invariants, as expounded by Salmon,* is obscure and difficult to follow. I shall therefore give a proof of the principal theorem which I hope will place the subject in a clearer light. If U, U' be two curves of degrees n, n' and classes m, m' their tact-invariant is of degree 2n' (n-1)+m' in the coefficients of U and of degree 2n (n - 1) + m in those of U'. First. Let the curves be anautotomic; let (§, n, 4) be their points of contact and let U1=dU/d, &c. Then the conditions that U and U' should touch are that U1/ U,' = U2/U,' = U2/ U'.. ..........(11). The elimination of (§, n, ) between U, U' and any one of (11) gives the condition that the three curves should intersect at a point; but the eliminant is not the tact-invariant, and we shall proceed to show how the latter may be obtained in a form free from extraneous factors. Let A be the point of contact of U and U'; let u,=pß+qr; and consider the three curves U=an-1u, + an−2u, +.............. (12), Multiply (13) by p (n−n') a"-2 and subtract from (14) and we obtain an equation of the form W = an+n'-1w,+..., * Higher Plane Curves, §§ 96-8. which shows that W has a node at A. Its first polar with respect to B, which may be any arbitrary point, is W,=0 which passes through A; hence the tact-invariant may be obtained by eliminating (E, n, ) between U, U' and W. Now the coefficients of both curves appear in W, in the form of the product aa'; and since W, is of degree n+n' — 3, the eliminant is of degree n' (n+n' - 3) in the coefficients of U, of degree n (n+n'-3) in those of U', and of degree nn' in those of W, Accordingly the tact-invariant is of degree n' (2n + n' — 3) = 2n' (n − 1) + m' in the coefficients of U, and of degree in those of U. = Secondly. Let U be autotomic. If U'λa+μß + vy, β νγ, the tact-invariant R of U and U' is the tangential equation of U; and by the theory of tangential equations the degree of Rin (,, v) is equal to m the class of U. Since the tangential equation may be obtained by eliminating a between U and U' and equating to zero the discriminant of the resulting binary quantic of (B, y), it follows that the degree of R in the coefficients of U is 2 (n-1). Putting n'= 1, m'=0, it will be found that the previous results agree when one of the curves is autotomic and the other is a straight line. We therefore conclude that when one of the curves, say U, has & nodes and κ cusps, the degree of its coefficients in the tact-invariant is unaltered, but the degree of the coefficient of the other curve U' is reduced by 28 + 3. 9. When both curves are of the same degree the following proof may be given by means of the properties of the Steinerian; and since the effect produced by a node or a cusp cannot depend on the degree of the curve on which it lies, it follows that the theorem is true for autotomic curves of different degrees. Let O, O' be two points whose coordinates are (f, 9, h) and (f', g', h'); P, P' the first polars of the primitive curve U with respect to O, O'. Then since .... P=ƒU1+gU2+h U2 = 0...................................... .(15), where UdUda, &c., it follows that every coefficient of P is multiplied by f, g or h; hence the degree of every invariant in the coefficients of P is equal to its degree in (f, g, h). Let f', g', h', f, g, h) be the tact-invariant of P and P'; and let O' be a fixed point; then P and P' will touch one another whenever O lies on a tangent drawn to the Steinerian from 0. Accordingly the equation 0 gives a relation between (f, g, h) which is the equation of the tangent drawn from O' to the Steinerian; and since m tangents can be drawn from O', where m 3 (n-1) (n-2), m is the degree of 4 in (f, g, h) and therefore in the coefficients of P. In like manner by taking as a fixed point and O' as a variable one, it follows that is of degree m in the coefficients of P'. = Let O' be a point on the Steinerian, then only m-2 tangents can be drawn from O' to this curve; also P' has a node at the point on the Hessian which corresponds to O'. It therefore follows that the line joining O' to the node on P' is a square factor of which must be rejected; and that the tact-invariant, which is the equation of the tangents drawn from O' to the Steinerian, is of degree m-2 in the coefficients of P. Accordingly a node on P' reduces the degree of the tact-invariant by 2 in the coefficients of P, but leaves the degree in those of P' unchanged. If O' be a node on the Steinerian, only m-4 tangents can be drawn from O' to this curve; and since P' is a binodal curve two nodes on P' reduce the degree of the coefficients of Pin the tact-invariant by 4, in accordance with what has just been proved. If O' is a cusp on the Steinerian, m-3 tangents can be drawn from O' to this curve; also P' has a cusp at the point on the Hessian which corresponds to O'. It therefore follows that the line joining O' to the cusp on P' is a cubic factor of o which must be rejected; and that a cusp on P' reduces the degree of the tact-invariant by 3 in the coefficients of P, leaving its degree in those of P' unchanged. From these results we conclude that if U, U' are two curves of degrees n, n'; and if U is anautotomic and U' has S'nodes and cusps, the degree of the tact-invariant in the coefficients of U' is n (2n'+n-3); but its degree in those of U is n' (2n+n' - 3) — 28′ — 3к' = 2n' (n − 1) + m'. Accordingly if U has & nodes and cusps the degree of the tact-invariant in the coefficients of both curves is as stated above. 10. Salmon has enunciated the following theorem,* but I fail to see that he has given anything which can be called a valid proof. * Higher Plane Curves, §§ 398-9. Higher Algebra, §§ 186–7. Ꭱ |