qq

If al, a2, ...a, n be the roots of f=o, (1, R2, -Ai the roots of 0=o, the condition that some root of 0 =o may qq cause f to vanish is clearly R s, 5 =f (01)f (N2) ï¿½ ï¿½;f (Nn) = 0; so that Rf,q5 is the resultant of f and and expressed as a function of the roots, it is of degree m in each root 13, and of degree n in each root a, and also a symmetric function alike of the roots a and of the roots 1 3; hence, expressed in terms of the coefficients, it is homogeneous and of degree n in the coefficients of f, and homogeneous and of degree m in the coefficients of 4..

00Thus e 1 e 2 = - e2ei, and if q, q are any two quaternions, qq is generally different from q'q.

00The values of x and y are different, unless V (qq o) = o.

00The flame issuing from the furnace by (o) is always further utilized for boiling down the liquors obtained in a later stage, either in a pan (p) fired from the top and supported on pillars (qq) as shown in the drawing, or in pans heated from below.

00This has a reciprocal Q -1= p-r = qq-1 - wp1 rq1, and a conjugate KQ (such that K[QQ'] = KQ'KQ, K[KQ] = Q) given by KQ = Kq-}-rlKp+wKr; the product QQ' of Q and Q' is app'+nqq'+w(pr'+rq'); the quasi-vector RI - K) Q is Combebiac's linear element and may be regarded as a point on a line; the quasi-scalar (in a different sense from the rest of this article) 2(1+K)Q is Combebiac's scalar (Sp+Sq)+Combebiac's plane.

00If Q and Q' are commutative, that is, if QQ' = Q'Q, then Q and Q' have the same centre and the same radius.

00A(pp) =X, B(qq) =u, C(rr) =v, ~

00Draw Pp and Qq touching both catenaries, Pp and Qq will intersect at T, a point in the directrix; for since any catenary with its directrix is a similar figure to any other catenary with its directrix, if the directrix of the one coincides with that of the other the centre of similitude must lie on the common directrix.

00Similarly Qq must pass through the centre of similitude.

00If al, a2, ...a, n be the roots of f=o, (1, R2, -Ai the roots of 0=o, the condition that some root of 0 =o may qq cause f to vanish is clearly R s, 5 =f (01)f (N2) Ã¯¿½ Ã¯¿½;f (Nn) = 0; so that Rf,q5 is the resultant of f and and expressed as a function of the roots, it is of degree m in each root 13, and of degree n in each root a, and also a symmetric function alike of the roots a and of the roots 1 3; hence, expressed in terms of the coefficients, it is homogeneous and of degree n in the coefficients of f, and homogeneous and of degree m in the coefficients of 4..

00Hence T, the point of intersection of Pp and Qq, must be the centre of similitude and must be on the common directrix.

01

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