and the system is finally equivalent to only three independent conditions on account of the identity It may be noticed in passing that two of the conditions (9) may be thrown into the form as The conditions (9) may likewise be put in a form better known, follows. Developping the first pair of equations, and subtracting the results, we have (13) GH√F+HF√G+FG √H+√(FGH) adding the results, we have (14) 0; F√(GH)+G√(HF)+H√(FG)+FGH— A'B'C' transponing the last termes of (13) and quadring: FG2 H2 + GH2 F2+HF2 G2+2FGH(F√(GH)+G√(HF)+H√(FG)) =FGH; substituting from (14), and dividing throughout by A'B'C', there finally results: (15) A'B'C' — A'F2 — B'G2 — C'H2+2FGH = 0. By means of this reduction the system (9) admits of being expressed in a brief manner, as follows: Let and the two conditions resulting from the elimination of from any three of these equations will be those to which the quantities A, B, C, F, G, H, L, M, N, must be subject, in order that the transformation from (1) to (2) may be possible. But it does not seem worth, while to proceed to the development of the results, which would be a work of great labour. It remains to determine the geometrical results, ensuing upon the transformation in question; the principal peculiarities of which depend on the value of the quantity w. 1. Let w = 0; then the problem reduces itself to that considered in the latter part of your letter. 2. Let w be a real constant; then the surface will be cut circularly by either of the planes = @ lx +my+nz + kw (or, as they may be termed, the planes @, '). It may also be noticed that ,' and vanish together. 3. Let w be an imaginary constant; then the surface is cut circularly by either of the imaginary planes 0, or '=0, and also by the real @ = hyperboloïd (lx+my+nz) (l'x+m'y+n'z) = k2w2. 4. Let w be a homogeneous linear function of x, y, z; e. g. let i. e. by a plane making angles with the planes u and w, or v and w, the ratio of whose series is the surface k; it will be cut also in its intersection with will both intersect (1) in the same curve. This latter surface has a double contact with the sphere x2+ y2+x2 = const. and is cut circularly by the plane w1 = 0. a plane which cuts the former surface in the straight line It may further be remarked that the plane w, 0 cuts the original surface in the same curve as does the cone Or2 + uv = 0, whose cyclic planes are u=0, v = 0. London, March 26. 1853. W. Spottiswoode. 13. Lösung der Aufgaben C. und D. in Nr. 21. Band 45 dieses Journals S. 284. (Von dem Herrn Dr. phil. Lottner, Lehrer der Mathematik und Physik an der höhern Bürgerschule zu Lippstadt.) Aufgabe C. Es sei die aufzulösende Gleichung 4ten Grades, deren Coëfficienten, um die Brüche zu vermeiden, mit passenden Zahlen multiplicirt sind: x2-8ax3-12 bx2-48cx+12d = 0. (1.) Bezeichnet man: - (√ {6a2d + 18c2 — b3 +3bd+12abc +√[(6a2d+18c2 — b3 +3bd +12abc)2 — (b2 — Sac + d)3]} 3 √[A+√(A2— B3)] mit p und \V{6a2d + 18c2 — b3 +3bd+12abc - √[6a2d + 18c2 — b3 +3bd+12abc)2 — (b2 — 8ac + d)3]} = √[A— √(A2— B3)] mit 9, so sind, vorausgesetzt, dafs A2-B30 ist, die 4 Wurzeln der Gleichung: (2.) x1 = 2a+√(4a2 +2b+p+q) +√{8a2+4b—p−q+√[(8a2 + 4b − p − q)2 + 3 ( p − q)2]}, 2a+√(4a2+2b+p+y) — √ {8a2 + 4b — p − q + √[(8a2 + 4b − p − q)2 + 3 ( p − q)2]}, 2a−√(4a2+2b+p+q) +i√{—8a2—4b+p+q+√[(8a2+4b−p−q)2 +3 ( p − q)2]}, 2a-√(4a2+2b+p+q) — i√{—8a2-4b+p+q +√[(8a2+4b—p−q)2 +3 (p −q)2]}. Im andern Falle, wenn A2- B3 <0 ist, setze man: 2a+√(4a2+2b+2B1 cosα) + √(4a2 + 2b — 2B1 cos (л+α)) +√(4a2+26-2B cos (л-α)), 2a+√(4a2+26+2B cos fa) -√(4a2+2b-2B cos (л+α)) — √(4a2 + 2b — 2B1 cos ¦ (î — α)), 2a√(4a2+2b+2B cosa) -√(4a2+2b-2B cos (л+α)) +√(4a2+2b-2B1 cos (л—α)), X 4 = 2a−√(4a2+2b+2B1 cos}α) + √(4a2 + 2b — 2B1 cos 1⁄2 (л + α)) -√(4a2+26-2B cos (л-α)). Wie man sieht, müssen in diesem Falle die vier Wurzeln entweder alle reell, oder alle imaginär sein, je nachdem 4a2+2b-2B cos (л+α) und 4a2+26-2B*cos(л — α) beide positiv oder einer von beiden Ausdrücken negativ ist. Aufgabe D. Es ist nun zu zeigen, dafs wenn d0 ist, die Auflösung mit der der cubischen Gleichung (5.) √ {24a3+2ab+3c + √[( §‡ a3 +2ab+3c)2 — (4a2 + b)3]} √ {24a3+2ab+3c−√[( § 4a3+2ab+3c)2 — (a2 + b)3]} 3 = }{C—√(C2_D3}}} = v, so sind bekanntlich die drei Wurzeln derselben: 3 (7.) = [A + √(A2 — B3)] p und 9 auf: √ {18c2+12abc-b3 + √[(18c2 + 12abc — b3)2 — (b2 — 8ac)3]}, √/{18c2 + 12abc-b3 —√[(18c2 + 12abc — b3)2 — (b2 — 8ac)3]}. Es ist also zu zeigen, dafs nach der Substitution der Werthe (7.) in (2.), die Ausdrücke daselbst mit denen in (5 a.) übereinstimmen. |