Integrating (5) and (6), (supposing 8 = 0 and =0 when t= 0,) and from (8) 0; hence a sphere placed on a smooth inclined plane will not roll but slide, and the foot note, page 35, from Peck's Mechanics, is not correct. See Walton's Mechanical Problems, prob. 7, p. 446, 3d Edition. NOTE ON RIGHT-ANGLED TRIANGLES.-BY JOSEPH B. MOTT, NEOSHO, Mo.-In any right-angled triangle ABC, A C being the base, CB the perpendicular and A B the hypothenuse, if we put a2 + 1 = the hypothenuse and a2-1 = the base, we shall have (Eud 47, I,) 2 a= the perpendicular. Hence if we maintain the above relation between the sides, we may assign any value whatever to a, either whole or fractional, and all the sides of the triangle will be rational. For instance, if a = 1, (h, b and p representing respectively the hypothenuse, base and perpendicular) we have And if a = any other value, as 4, 5, 6, 7, 8, &c., the corresponding sides are found to be 8, 5, 17; 10, 24, 26; 12, 35, 37; 14, 48, 50; 16, 63, 65, &c. If a be assumed fractional, say = , and in succession, we shall have for the sides (41, 9, 43); † (34, 16, 30), and † (29, 21, 20). Multiplying these sides respectively by 16, 9 and 4, we obtain the three triangles whose sides are respectively 41, 9, 40; 34, 16, 36, and 29, 21, 20. If we desire to determine a particular kind of triangle, having for instance, the difference between their base and perpendiculars = 1; put a = 12, and multiply the resulting sides by 25, and we get 120, 119, 169. Put a, and multiply by 289, and we get 697, 696, 985. Put a = 1,52, and multiply by 4900, and we get 23660, 23661, 33461, &c. This last class of triangles I consider among the most interesting. THE CORRECT METHOD OF LEAST SQUARES.-BY R. J. ADCOCK, MONMOUTH, ILL.—1. The point determined by two given points is their center of gravity. The point determined by any number of given points is their center of gravity. For combining them in sets of two, so as to take each point twice, gives the same number of other points, which being combined continually, as before, finally terminates at the center of gravity of the given points. 2. Hence a straight line which is to pass through a given point, having its direction to be determined by other given points, must pass through their center of gravity; because that is the point determined by the other given points. 3. Hence also, a straight line determined by any given points, passes through their center of gravity, and has such a direction that it passes through the two most remote centers of gravity of the two sets of points on opposite sides of the common center of gravity; and is that principal axis of least moment of inertia of the given points. 4. Hence in general, any point, line or surface, determined by any given points, is that point, line or surface which has the sum of the squares of the normals to it from the given points, a minimum. And is the point, line or surface which has the mean position of all that can be taken. The common application of the method is incorrect, not agreeing with 2, 3, and 4. ANSWER TO PROF. BROOKS' QUERY.-No rigorous demonstration can be given of the formula mentioned, it must be regarded as empirical. The function (x), which denotes the number of primes less than x, is discontinuous, and the problem of finding functions representing it approximately is an indeterminate one. Legendre was the first to give a formula of this kind, (Essai sur la Theorie des Nombres, 2d Edition, Paris, 1808, p. 394, or 3d Edition, Vol. II, p. 65); it is log x being Napierian. The constant 1.08366 evidently has been assumed so as to represent as well as possible the actual count of primes below 400,000, the limit of tables of primes at that time. But Tchebichef has shown since, (Liouville's Journal, 1st Series, Vol. XVII, pp. 341–390), that if we wish to represent (x) as nearly as possible by a function of where we do not set any limits to the largeness of x, we must put A and B both equal to unity. Between finite limits for x however it is possible to represent (x) much nearer by assigning to A and B other values. Tchebichef has also proved that, regard being had to the whole infinite series of integers, (x) is better represented by the definite integral than by the formula in question. Gauss also, (Werke, Vol. II, p. 434), prefers it. Tchebichef has succeeded in finding two functions of x, one of which is always less than (x) and the other greater, and he proposes to take the mean of the two as the approximate value of (x), but this does not give as good results as Legendre's formula or the definite integral just mentioned. If two functions of a limiting (x) and differing less than unity could be found, this would settle the question of how many primes there are less than x, and consequently whether a given number is a prime or not--the great desideratum of the Theory of Numbers. But such functions will probably never be obtained; how complex they ought to be will be evident from some facts relative to the irregular distribution of primes, which we shall shortly mention. Endeavoring to represent the distribution of primes below 3,000,000 by the formula The following is a comparison of all the formulas here mentioned, with the actual number of primes below each half million up to three millions: dre's Error last Error Formula given Primes 500,000 41,556 41,606.4+ 50.4 41,532.7-23.3 41,545.9-10.1 1,000,000 78,501 78,627.5+126.5 78,543.2+ 42.2 78,526.7+25.7 1,500,000 114,112 114,263.1+151.1 114,180.1 + 68.1 114,122.4 +10.4 2,000,000 148,883 149,054.8 +171.8 148,975.8+ 92.8 148,874.18.9 2,500,000 183,016 183,245.0+229.0 183,175.1+159.1 183,024.1+ 8.1 3.000.000 216 745 216.970.64-225.6 216,912.6+167.6 216,709.3-35.7 That primes are quite irregularly distributed, will be plain from this: The 26,379th hundred contains no prime. The 27,050th hundred contains 17 primes. Even when they are taken in large masses, the irregularity is still apparent, thus: The 27th hundred thousand contains 6,762 primes. For the numerical data here given I am indebted to Gauss, (Vol. above quoted.) G. W. HILL. referred to by Prof. Brooks in No. 3, requires that the logarithm should be hyperbolic, which he neglects to state. This formula is fully and ably discussed in Vol. II, § VIII, p. 65 of Legendre's Theorie des Nombres, 3d Edition, 1830. Many years since I became interested in this matter, and I then used the formula Solutions have been received as follows: R. J. Adcock solved 7; Irving P. Church solved 6; G. W. Hill solved 9; Prof. A. Hall solved 10; H. Heaton solved 6, 7, 8, 9 and 10; Prof. D. Kirkwood solved 7; Artemas Martin solved 5, 6 and 7; Prof. J. Scheffer solved 5 and 7; James Stott solved 6; Walter Siverly solved 8; and Prof. W. Wylie solved 7. David Wickersham and William Hoover sent solutions to Problems in No. 1 too late for notice in No. 3. 5. "In a plane triangle there are given the three lines bisecting the angles, a, b and c, to find the sides." SOLUTION BY PROF. J. SCHEFFER, COLLEGE OF ST. JAMES, MD. Let CF-a, B E = b, A D = c be the three bisecting lines, and let , y and represent the three corresponding sides. According to a well known theorem we have y: 2::AF: B F:: A F: x-AF, whence Substituting the values for A F and B F, we obtain These three elegant expressions can be very usefully employed for the computation of the three bisecting lines, if the sides are given; the reverse operation, however, leads to an equation in x, or y, or 2, the degree of which renders the solution, in a general form, a matter of impossibility. Even if a, b and c have numerical values, the solution would prove very fatiguing for the most patient of all reckoners. We may, therefore, consider the solution for three different bisecting lines impracticable, and we propose to give a full solution for the case of an isosceles triangle. Let ABC represent an isosceles triangle (A CBC) and BEAD=b, CF-a, the bisecting lines. Denoting the angle E BA =DAR by 4, we have in the triangle A BE, = b: A B:: sin 2: sin 34, but A B=2A F-2 a cotang 2. hence we have the equation 2 a cos 24 = b sin 3 4. After an easy transformation we get |