and consequently the positions of the tangents are given by

the 2r+ 1 values of 4, viz.

V. "When two rhizic curves intersect at a point of odd multiplicity, the proposition III., stated in relation to even multiplicity, still holds good, 2r being replaced by 2r+1.

Prof. Cayley's addition (Quarterly Journal, Vol. X., p. 262) to my article on a Theorem in Maxima and Minima, makes

If the multiplicity of one curve, at an intersection of the two, be odd, that of the other is also odd and of the same degree.

it evident that the question was not left in its complete state by me. What he has shewn in relation to the particular case of double points may be generalized for all even multiplicities.

and therefore P'- P is alternately positive and negative as n

has successively the values 0, 1, 2, ... r — 1.

dr P

dur

and sin 2r0=0, whence 2r0=n, where n has the values.

and, if a be a value of 0 from the equation,

then 2r0=n+a, where n has the values 0, 1, 2, ......., 2r – 1,

and therefore

and therefore P'-P has alternately + and ceives its successive values.

This shews that a series of branches pass through the point in question on the surface w=P, the convexities of which are alternately up and down, the value of P being a maximum in regard to each of the former and a minimum in regard to each of the latter branches. The planes of these branches in the vicinity of the point, by what has been written above respecting rhizic curves, divide cylindrical space equiangularly.

I WRITE

ON A PROBLEM OF ELIMINATION.

By Prof. CAYLEY.

P= (α, ...Xx, y, z)*, Q = (a', ...Xx, y, z)*,

U= (a, ...Xx, y, z)TM, V=(b, ...Xx, y, z)",

and I seek for the form of the relation between the coefficients (a,...), (a', ...), (α, ...), (b,...), in order that there may exist in the pencil

λP+μQ=0

a curve passing through two of the intersections of the curves U=0, V=0.

The ratio u may be determined so as that the curve XP+Q=0 shall pass through one of the intersections of the curves U=0, V=0; or, what is the same thing, so as that the three curves shall have a common point; the condition for this is

Rest. (P+Q, U, V)=0,

a condition of the form

(λa + μa', ...)TM" (a, ...)11 (b, ...)TMm = 0 ;

or, what is the same thing,

(α, ......., a', ...)TM” (ɑ, ...)TM (b, .....)TMTM (λ, μ)TM” = 0,

which, for shortness, may be written

(A, .....Xλ, μ)”” = 0.

Suppose this equation has equal roots, then we have

Disct. Rest. (P+μQ, U, V)=0,

the discriminant being taken in regard to λ, μ. This is of the form

(α, ..., a', ...) 2.

.)2mn ̧mn−1) (α, ...) un ̧mn−1) (b, ...)umn(mn−1) = ().

=

It is moreover clear that the nilfactum is a combinant of the functions P, Q; and the form of the equation is therefore

Now the equation in question will be satisfied, 1°. if the curves U=0, V=0 touch each other; let the condition for this be = 0. 2°. If there exists a curve P+μQ=0 passing through two of the intersections of the curves U=0, V=0; let the condition be =0. There is reason to think that the equation contains the factor 2, and that the form thereof is 2 = 0).

Assuming that this is so, and observing that V, the osculant or discriminant of the functions U, V, is of the form ▼ = (a, ...)"(n+2m−3) (b, ...)m(m+2n−:

we have

)

which is the solution of the proposed question. Suppose for

instance n =

1, then

this is right, for writing P= ax + By + yz, Q = a'x + B'y + yz, V=bx+by+b'z, then if two of the intersections of the curve U=0, with the line V0 lie in a line with the point P=0, Q=0, then the point in question, that is the point (By' - B'y, ya' - y'a, aß'— a'ß), must lie in the line V=0; and the condition reduces itself to

"}} Imm

{(By' — B'y, ya' — y'a, aß' — a'ßXb, b', b")