SOLUTION OF PART FIRST. By GUSTAVUS FRANKENSTEIN, Springfield, Ohio. The plane triangle C 180 (A + B), = .. cotCcot (90 — 1 (A + B)) = = tan 1⁄2 (A + B) 1. On some properties of the powers of the same number. - The author announced the discovery of some general laws which regulate the series of the powers of any number. For instance, in the following series of the powers of 5, the number of digits in the several recurrent vertical series may be expressed by the powers of 2. termined. In the case of 5, S2 sums of the several series being tables of the powers of numbers whatever with very little labor. This discovery will enable certain calculations to be made with a degree of accuracy hitherto impossible.-J. POPE HENNESSY, of the Inner Temple, Report of the British Assoc. for the Adv. of Science. Leeds. 1858. = 2 (Sn−1 + 1), the consecutive (S-11), 7, 16, 34, 70, 142, &c. In this way may be constructed to any extent 2. The celestial part of the problem on page 204 of the March number of the Monthly is very simply solved in a general form. Employing the usual notation, S the integral with respect to the mass, the potential, ▲ the mutual distance of m and m', &c., the law of power of any system of fixed forces is Fixed forces might, it is true, be imagined which would not satisfy (2) and (3); but it can hardly be supposed that such would ever have been introduced into a system of nature. The problem here considered is to find the form of the function q, under the condition that two systems, whose linear dimensions are in the ratio n, shall present similar appearances at the same times. If (1) expresses the law of power of one system, that of the other is by this condition, Whence (4) } S [n3 m (n v)2] = μ S S [ no m m' q (n ▲)] + H2· ф SSmm' [no (n A) — & (^)] = 0 ... ... n q (n ▲) — & (A) = 0. The derivative of (4) with reference to n is (5) ❤ (n ▲) + n ▲ 4' (n ▲) = 0. The integral of (5) is q (n ▲) = a η Δ The only form of potential which can satisfy the present problem is therefore that which results in the law of gravity. — B. 3. Reply to query on page 186, March No. For the integral of see DE MORGAN'S Calculus, pp. 699, 700; BOOLE's Differential Equations, pp. 455, 459; or CARMICHAEL'S Calculus of Operations, p. 54. — B. NOTE ON EQUATIONS OF THE SECOND DEGREE. By Dr. R. C. MATTHEWSON, U. S. Deputy Surveyor, San Francisco, Cal. EVERY equation of the second degree, involving one unknown quantity, may be reduced to one of the following four forms, in which the coefficient of the second power of the unknown quantity will always be unity: 22 1st, x2 - 2rx = a2; 2d, x2+2rx = a2; 3d, x2-2rx — — a2; and 4th, 22+2 r x = — a2. The roots of the 2d form differ from those of the 1st, and the roots of the 4th from those of the 3d, only in the change of signs; because if the signs of the alternate terms of an equation be changed, the signs of all the roots will be changed, while their magnitudes remain unchanged. It is obvious, therefore, that the four forms may be reduced to two classes, the 1st class comprising the 1st and 2d forms, in which the absolute term is positive, and the 2d class comprising the 3d and 4th forms, in which the absolute term is negative. In both classes, considered as lineal equations, one half of the coefficient of the first power of the unknown quantity represents the radius of a circle, while the square root of the absolute term, in the 1st class, represents a tangent, and in the 2d class, a sine of that circle. In the first class one of the roots of the equation represents the external secant, and the other the diameter of the circle increased by the external secant, while in the 2d class one of the roots represents the versed sine and the other the diameter of the circle diminished by the versed sine. As there are no conditions in any of the four forms to indicate in what direction the lines are to be drawn, it follows that, in the 1st class, the external secant as well as the sum of the diameter and external secant, and in the 2d class, the versed sine as well as the difference of the diameter and versed sine, may be taken either positive or negative. Hence, it is plain, that in each class, the two roots of one of the forms are respectively equal to the two roots of the other in magnitude, but have contrary signs, as has been already shown, and may be seen at once by comparing the roots together, as follows: Construction of 1st Class. With the radius CB=r, half the coefficient of x, describe a circle; draw the tangent BA= a, the square root of the absolute term; and through A and C draw the secant A C produced, cutting the circumference of the circle in D and E; then AD and AE will be the roots of the equation, the former of which is the external secant and the latter the diameter of the circle increased by the external secant. Demonstration of 1st Class. Form 1st, AD= r−√r2+a2= rad-sec=-ext. secant, Form 2d, AD=−r+ √ r2+a2=-rad+sec=+ext. secant, 66 Form 3d, AE=-r-√+a2=-rad-sec-dia-ext. sec. شرح Roots of 2d Class. 2rx=- a2, or x = r = √r2 = a2, x2 + 2 rx = — a2, or xr ± √ r2 — a2. Construction of 2d Class. CD r, half the coefficient of x, describe a Demonstration of 2d Class. Form 3d, AD= r— 66 AE 2 rad COS +versed sine, r+√2-a2= rad+cos= +dia-ver. sine, Form 4th, AD=—r+√12 — a2=-rad+cos=-versed sine, 66 SOLUTION OF PROBLEMS IN PROBABILITIES. By Dr. JOEL E. HENDRICKS, Newville, Indiana. I HAVE recently had my attention directed to a question somewhat analogous to the Prize Question proposed in the Lady's and Gentle |