in form with that for the potential of the propagation of sound in a uniform medium of indefinite extent, and leading of course to the same consequences so familiar in that case. For plane waves advancing in any common direction in the substance : since then x cos a + y cos B +z cos y = p, where a, B, y are the direction angles of propagation and p the distance of any wave plane from the origins, and since consequently d2_d2 d2 + + d, therefore the equation for dx2 dy2 dz2 dp2 = the determination of assumes the simplified form = d2 = a2 the dt2 dp2 complete integral of which in finite terms, viz., k.f(pat), where k is a small constant representing the absolute amplitude of the vibrations, and ƒ any arbitrary periodic function oscillating within finite limits as in vibratory motion generally, represents two systems of waves advancing in opposite directions with the common velocity a; the amplitudes of vibration undergoing no change for either system with its progress through the substance, and no wave of either system giving rise to a wave of the opposite system in its passage through any portion of the mass external to the region of the original disturbance giving rise to the motion. For spherical waves diverging from any common centre in the substance since then x2+ y2+z2r2, where r is the radius of any wave sphere of the system; and since consequently therefore the equation for the determination of assumes again the 2 db), , or, in consequence of the entire d2 (ro) d2 (rp). r independence of r and t, the equivalent form dt2 = a2 dr the com plete integral of which in finite terms, viz., ro = k . f (r ± at), where k and fare as before, represents again two systems of waves diverging in opposite directions from the common centre with the common velocity a, the amplitudes of vibration varying at considerable distances from the centre inversely as the distance for each, and no wave of either system giving rise to a wave of the opposite system in its passage through any portion of the mass external to the region of the original disturbance giving rise to the motion. 7170. (By the EDITOR.)-If 10 cards are taken at random from a pack, show that the respective probabilities (P1, P2) that they will contain (1) exactly 4 cards with hearts, (2) not more than 4 cards with hearts, are 27417 185932 1. The following considerations will enable us to build up the required fraction :-52 != possible permutations of the whole pack; 10! 42!-ways of arranging the pack when a particular set of 10 cards is partitioned off = ways in 13! and constantly taken first: e.g., 4 hearts and 6 other cards; 13! 9! different sets of 4 that can = which 4 cards can be taken out of 13; 9! 4! be made out of 13, disallowing mere differences of arrangements; corresponding numbers in regard to 6 cards out of 39. 39! 33! and 10! 42! 39! = = number of ways in which 4 cards of one kind and 94 336! 6 cards of other kinds can be arranged on one side of a partition, and the remainder of the pack on the other side, by allowing interchange of equivalent cards across the partition, and permutation on each side. Hence the probability that 10 cards drawn at random from a pack will contain 10! 42! 13! 39! exactly 4 hearts is This, reduced, is 14746 nearly, or 52! 9! 4! 33! 6! rather over. Consequently, a person accepting odds of 6 to 1, laid against the occurrence, should win in the long run. 2. The probability that the 10 cards contain not more than 4 hearts is the sum of the respective probabilities concerning 4, 3, 2, 1, 0 which (as found by the formula in Art. 1) are as follows:-4 hearts = 30334; 1 h. = 3 h. = •27807; 2 h. (4) 13 + 94307. This is slightly over 3. 39 [Putting C for the number of combinations of n things taken r together, the total number of ways and the number of favourable cases in (1) are respectively C), Cx C) (since each group of 4 hearts out of 13 may be combined with each group of 6 non-hearts out of 39); hence the probability in question is C x C(6) + C(1), which gives the first result; and the sum of 5 such probabilities gives the second result.] 1+ x2+ x1+ &c., and that of 2 sec2 x be 2+ 22x2 + 24, x2+ &c. ; prove 2! 2! 4! that the coefficients u2, 4, 6, &c. may be found from the relations taking care in the last series to stop before the first suffix exceeds the second, and to halve the coefficient when they become equal. Apply this method to verify the value of 10, found (by another process) in DE MORGAN'S Differential Calculus, to be 50521. Solution by the Rev. T. C. SIMMONS, M. A. Differentiating twice with respect to x the identity coefficient of x2n-2 in the expansion of sec≈ (2 sec2 x-1) = same coefficient in + U2n-2 U2n-4 V2 u2n-6 + + whence follows the first required relation. = + U2n U2n-2 + ·) +... Иг U2n-4 U4 2n! 2n-2! 2! 2n-4! 4! with the reservation stated in the question. ... 2! 4! Now put n successively equal to 2, 3, 4, 5; then, observing that u。 and no each 1, we find 5, v4 = 32, U6 = 61, 16 = 544, = 1385, V8 4 (1385 +1708+875) = 15872, 1385 +6832+11200 + 15232 + 15872 50521. = 7733. (By H. J. READ, M.A.) — Transform means of the geometry of the ellipse. Solutions by (1) D. EDWARDES; (2) H. L. ORCHARD, B.Sc., M.A. 1. Let P be any point on an ellipse, Q an adjacent point, P', Q' corres ponding points on the auxiliary circle; then, drawing QL perpendicular r = SP. = a ex = a(1- — e cos p). to SP, But ra (1-2) 1He cos 0; .'. a (1 − e2) = a (1-e cos p) (1-e cos 0)... (a). PQ TQ b' = P'Q' T'Q' where b' is the semi-diameter parallel to TQ. But [The formula (6) seems to be useful; e.g., area of ellipse = 2. S 2. Otherwise r2 de by the geometry, eMS+eSR, i.e., r = er cos 0 + LS =-er cos 0+1, 1 + e cos 0 = = SP ePN = = r 1) + 2lr — 12] $ r since 1-e2 5635. (By ELIZABETH BLACKWOOD.)-Two excursion trains, each m yards in length, may start with equal probability from their respective stations at any time between 2 o'clock and 10 minutes past 2, in directions at right angles to each other, each at a uniform rate ; find the chances of a collision, each being n yards distant from the point at which their lines cross, and both being ignorant of the risk they are running. Solution by D. BIDDLE. = distance in of train in yards, If v = velocity of each train in miles per hour, then v miles covered in 10 minutes; and, if m = length Tom length referred to a mile as unit. Also = 6m 1= fraction of 1760v before or after train A. 1760v 10 2.6m = time 1760v starts as much as time during which the chance is 10 minutes occupied in clearing the level crossing. during which a collision can occur when train B 6m 2.6m 1760v curtailed owing to limitation in the time of starting of the two trains, and when the average factor is reduced from 2 to (2+1). Consequently, 2.6m 2.6m 2.6m 3.6m 1760v 1760v 1760v 2.1760v 10 {(10. = + } = chance required 1 (20-1760) 1760v 10 Thus, let m = 200, and v = 35, then the chance of a collision The distance n of each train from the level crossing does not appear to affect the probability of a collision, under the other specified conditions. And the moral to be drawn is, that in such cases, the higher the speed and the shorter the train, the less the chance of disaster. 7730. (By W. J. GREENSTREET, B.A.)—Prove that (1) the polar of a fixed point with regard to a series of circles having the same radical axis passes through another fixed point; and (2) these two points subtend a right angle at either limiting point. Solution by A. H. CURTIS, LL.D., D.Sc.; T. BRILL, B.A.; and others. Lemma. The line joining a point A to any point B, situated on the polar of A with respect to any circle, (1) is equal to the sum of the tangents to the circles from A and B, and (2) is equal to the double of the tangent drawn from its middle point H. Let C be the centre of the circle and a its radius. = AC2+ BC2-2a2 = 2CH2+2AH2 – 2a2, therefore AH2 = CH2-a2 = square of tangent from H, therefore AB = the double of such tangent. Let now the polars of A taken with respect to any two circles meet in B, bisect AB in H, then the tangents from H to the two circles are each = AH, therefore equal to each other; therefore H is on the radical axis of the circles, and, as AH=BH, A and B are equidistant from it; if we find then on DE a point B, whose distance from the radical axis = the distance |