standpoint. Preliminary to his solution of this problem, when speaking of the necessary near approach of comets to the sun in order that they may be visible, Laplace says: "One conceives that in order to approach so near the sun their velocity at the moment of their entrance into his sphere of activity must have a magnitude and direction comprised within narrow limits. It is required, therefore, to determine within these limits the ratio of the chances that give a sensible hyperbola to the chances that give an orbit which may be confounded with a parabola." But I cannot see that we gain much by the application of the calculus of probabilities to this question unless it be a somewhat clearer definition of the problem, since the conditions by which we restrict our solution are known beforehand, and as these conditions are varied our results will be changed. The condition of the question is this: We now have more than two hundred well known cometary orbits without a single example of a decided hyperbolic motion. No matter, therefore what may be the result of the application of the calculus of probabilities, it is quite certain that if comets enter our solar system they must do so under conditions that render very small the chance of hyperbolic orbits. On the other hand, we have a few, six or eight, orbits of meteors that show a hyperbolic motion. But in this case we have orbits that are determined from observations made with difficulty and which may be affected by large errors in the time of the flight of the meteor and in its position. The true course appears to be that our methods of observing meteors should be very much improved, so that we may obtain accurate observrtions of the duration of their flight and the position of their apparent paths. Then, having secure data, we can derive certain conclusions. . NOTE.-It may be interesting to see how Laplace gets the expression (a), given above. In addition to the notation already introduced, let a and e be the semi-axis major and the excentricity of the orbit; and let w be the direction which the comet's motion makes with PS. Then we shall have the well known equations Eliminating a and e from these equations, we have, Imagine a small sphere whose center is at the point P. The motion of the comet may be directed toward any point on the surface of the half of this sphere comprised within the sphere of the sun's activity. The probability of a direction forming the angle w with PS, the radius-vector, will be 27. sin w. Hence, dividing the integral 27./sin w. do by the surface of the semi-sphere we shall have the probability that the direction of the motion will be comprised within the limits zero and w. This probability is 1-cos w; and from the above value of sin'w we easily find the expression (a). REPETENDS. BY PROF. M. C. STEVENS, SALEM, OHIO. (Continued from page 15.) Let us illustrate by an example: 12. It is convenient Suppose is to be reduced; we know from the second principle that the last figure must be 7, and from (1) we get r2 to write r2 11, 72 l2 + m,, &c., in a column, thus: 11 = 7 2 5 and 23 7 and 4 9 and 15 2 and 6 1 and 97 4 and 18 11 and 19 3 and 210 6 and 11 4 and 212 2 and 13 = = = 4 6 7 1 = 1 = 4 9 2 = 5 == 3 2 =8 9 and 14 10 and 715 7 and 216 = 5 = 0 Here we stop, because the next figure will be 7, and it will commence to repeat. We have, therefore, = .058823529411764717. If the fraction had been, we should have had 21 = 3, r2 = 4, r2 I1 = 12, r2 12 + m1 = 9, r 2 1 3 + mz 2 2 2 The operation may be very much abbreviated by taking advantage of the well known property of repetends, that after one half of the figures are obtained the second half may be found by subtracting each figure of the first half successively from 9. We know when one-half of the digits are found because the remainder is then d1; and we know this in the above method from the fact that any number in the column divided by 1 gives the corresponding remainder. This property I demonstrate as follows: Let be a fraction which reduces to a repetend, [verse order, Let 21, 92, 93, 7 represent the successive digits of repetend in re Let 1, n 66 rn Let m, m2, M3 (103 4 + 102 3 + 10 2 + I1)^2 — 103 95 102 4-10 3 2 2 Substituting in t for m, its value from t2, in t for m, its value from (10 92+91)*2-92 10 -- (102 3 +10 % + 1)^2 — (10 75 + L2) 2 102 (103g1 +10% +102+91)~2 3 103 2 The law is now apparent. and we can write for the nth term .(10”—2qn-1+10”¬3gn-2+... 10g 2 +Z1)r2—(10”—3 In-1 +· ルー 10"-2 3 ..10g,+2). Substituting a for 10"-3n-1+10"-4In-2+... 10% 3 +12, we get (10′′~22n−1 +10n”¬3 9 m2 +........10%2+21)d+1 10-1 Dividing (A) by (B), member by member, we get 10-1 I also demonstrate the property above referred to, viz: That after onehalf of the digits are found the second half may be obtained by subtracting each figure already obtained successively from 9, as follows: Let be a fraction which when reduced to a decimal gives a repetend in which the remainder d-1 occurs We evidently have in which R contains the digits before reaching the remainder d—1, and R' the digits after reaching that remainder. But hence the sum of the second members of the equation must also equal 1. R' must contain the same number of places as R, since it contains the same sequence of figures differently arranged. Now since the fractional parts when added give 1, R + R' = 99. . . . . 9 repeated as many times as there are digits in R+R', whence R99...9 R'. Q. E. D. The second principle may be proved as follows: As has been shown, the remainder next to the last is In order that this may be an integer the units of d q1 must be 9; and it is evident that if the units of d be 1, then q, must be 9; if it be 3 then 91 must be 3; if it be 7 then q1 must be 7; but if it be 9 then g1 must be 1. In conclusion I will remark that it is improper to call repetends, as they are usually represented, decimals. For an example. .3 is not a decimal, but should be read three-ninths; so .16 should be read sixteen ninetyninths; .14 should be read 1 tenths, &c. SOLUTION OF A PROBLEM. BY G. W. HILL, ESQ., NYAOK TURNPIKE, N. Y. The following problem appeared in the Mathematical Monthly, (Vol. 1, p. 29), and no solution was published in that periodical: "Show that the product of six entire consecutive numbers cannot be the square of a commensurable number." |