therefore, for any function of Ao, A1, &c., the operations -A1, &c. &c.; give identical results. Now, if F (Ao, A,...) be a function of the differences of the roots a', B', &c. of the transformed equation Fo, the coefficient of the highest power of k will be a function of the differences of the roots a, ß... of the given equation. and therefore F is a covariant in k of the original equation. [See SALMON'S Higher Algebra, 2nd Edition, p. 119.] 7260. (By ELIZABETH BLACKWOOD.)—A pack of n different cards is laid face downwards on a table. A person names a certain card. That and all the cards above it are shown to him, and removed. He names another; and the process is repeated until there are no cards left. Find the chance that, in the course of the operation, a card was named which was (at the time) at the top of the pack. Solution by D. BIDDLE; BELLE EASTON; and others. The card named may with equal probability occupy any position in that portion of the pack not yet removed. There are three positions which instantly decide the issue, namely, the first (or top), the last, and the last but one; for the first and the last but one equally command success, since in the latter case only one card will remain, and this it is easy to name next time. The last or lowest position is the only one that insures failure. If the card named be in the lowest position but two, an equal chance of success or failure will be left for the next trial. If it be in the last but three, there will in the next trial be two chances of success to one of failure; and, if in the last but four, there will be five (or 2) chances of success to three (or 1) of failure. In other words, P1 = 1, P2 = 1, P3 = 3, P4 g. But, counting from the top of the pack, the probabilities attaching to the several positions are as follows:1, P-2, Pn-3, Pn-4 ... P1, 0; and but similarly = n n [(−1) P2-1+P-2]. But this resolves itself into [(n−1) Pn−2+ Pn−3]; .`. Pn-2-Pn-3 = (n − 1) (Pn-2-Pn-1), which shows that the probabilities are alternately greater and less, but that the differences between them rapidly become infinitesimal as n increases. Above Pio the probabilities are alike to six places of decimals, viz., 633388, or rather more than 8, each probability being resolvable, according to the foregoing statements, into terms of the series 6740. (By Professor ASAPH HALL, M.A.)—Given z = a sin (x + a) + b sin (y + B), reduce to the form z = = D sin (x+a+y+B+8). Solution by AsÛTOSH MUKHOPÂDHYÂY. We have za sin (x + a) + b sin (y + B); put a = p + q, b z = p[sin (x + a) + sin (y + B)] + q [sin (x + a) — sin (y + 6)] = +2q cos (x+a+ y + B) sin § (x + a−y−B). If we determine 0, D, so that = 2p cos § (x+a−y— ß) = D cos 10, 2q sin 1⁄2 (x + a− y — ß) = D sin 10, we have Z D sin(x+a+y+B+0). Finally, if we determine & from 8 = 10 − 1 (x + a +y+B), we get Z = D sin (x + a + y + B + d), which is the form required. 7643. (By Rev. H. G. DAY, M.A.)-A and B sit down to play for a shilling per game, the odds being k: 1 on B; they have m and n shillings respectively, and agree to play till one is ruined: find A's chance of success. Solution by the PROPOSER. Let ur be A's chance when he is r games ahead; then his probability k of scoring the next game is -; but in these cases 1+ 1 and of losing it his chances become ur+1 and u,-1 respectively; hence that is, the players' chances are directly as the sums they risk. 7677. (By W. E. JOHNSON, B.A.)—If p and n be any integers, and W1, W2 w-1 are all the nth roots of unity except unity itself, show that the remainder, when p is divided by n, is ... Solution by B. HANUMANTA RAU, M.A.; E. RUTTER; and others. wn-1 == − 1 ( n − 1 ) ; Corresponding to x = 1, we have y∞; the remaining values of y are given by ry"-1 + 1n (n − 1) yn −2+ =0; thus the sum of roots and hence, when p = mn, the given expression vanishes. (w p = mn, in which case it equals (n-1). Hence the proposition is completely proved. 7292. (By Dr. CURTIS.)-Two heavy particles P and Q are connected by a flexible and inextensible cord, which rests on a pulley of infinitesimal radius; P is restricted to the circumference of a smooth circle, whose centre is vertically under the pulley, or, more generally, of a smooth Cartesian oval, one of whose foci coincides with the pulley, and whose axis is vertical; it is required to prove that the curve to which Q should be restricted, in order that equilibrium should exist for all possible positions of P and Q, is a Cartesian oval. Solution by the PROPOSER. Let O, the centre of the pulley, be taken as origin, the axis of y being vertical; then, by the principle of virtual velocities, if y, y' be the coordinates of P and Q. Py+Qy' const., or y +λy' = A; also, r and r being the radii vectores of P and Q, r+r = const. = restricted as defined in the question, = B; but, as P is 22-2Lr + My = N, or (B-)- 2L (B—') + M (A—λy') = N, which establishes a linear relation between y' and r', and therefore represents a Cartesian oval as defined. 7615. (By W. NICOLLS, B.A.)—If u1 + U2 + Uz = c represent a surface of revolution, the origin being the centre of revolution, and u1, u, uz containing respectively all the terms of the first, second, and third degrees in x, y, z; prove that u, is perpendicular to the axis of revolution and a factor of 3. Solution by J. P. JOHNSTON, B.A.; ELIZABETH BLACKWOOD; and others. Considering the equation of the cubic surface in cylindrical coordinates (z, r, e) as an equation for r, and taking the axis of z as the axis of revolution, it is evident that it can only contain even powers of r since all sections perpendicular to the axis are circles. Therefore the equation is of the form z [2 +r2 ƒ1 (0)] + az2 + r2 ƒ1⁄2 (0) + bz + c = 0, where f1 (0) and ƒ (0) are quadratic functions of cose and sin 0. Hence it appears that if his be put in the form u3+2+U] c, u is perpendicular to the axis of revolution and a factor of 3. = [Mr. NICOLLS' theorem may be slightly generalised thus:-All cubics of revolution can be written in the form LC + C' + L = 0, where C and C' are cones and L a plane perpendicular to the axis of revolution.] 7445. (By C. LEUDESDORF, M.A.)-A particle, describing a circular orbit about a centre of attractive force μ (distance) -3 tending to a point on the circumference, is disturbed by a small force ƒ tending to the same point; prove that the variations of the diameter (2a) and of the inclination to a fixed straight line in the plane () of that diameter which passes through the centre of force are given by the equations Solution by D. EDWARDES; MARGARET T. MEYER; and others. If the attraction be μD-5 and the velocity of projection that from in finity, the orbit is u = (2a)-1 sec (0―w), where h2 Expressing that the part of = du which arises from the variation of the con de 7498. (By A. MARTIN, B.A.)-If a straight line be drawn from the focus of an ellipse to make a given angle a with the tangent, show that the locus of its intersection with the tangent will be a circle which touches or falls entirely without the ellipse according as cos a is less or greater than the excentricity of the ellipse. Solution by Rev. J. L. KITCHIN, M.A.; J. O'REGAN; and others. The equations of the tangent and of a line through the positive focus at angle a with tangent are or [kx (x-ae)+y (ky-ae)]2 = a2 (y—kx + ae k)2 + (a2 — a2e2) (ky + x— ae)3, which becomes k2 (x2 + y2)-2ae ky + a2e2 = a2 (1+k2) + + =0= b2 k k2 2 ..........(3). |