λ=1 If the method of differences is used for λ* = 1* + 2* + 3* + the +(−1)x−1 (x^ 1)2* + (−1)* 1* = «!. (x+2)* − (*+1) (x+1)* + (*+ 1). — ... + ( − 1 ) (*+1)2« +(−1)*+11* =0, K being a positive integer. 2 If the first given number is represented by a and the successive differences K Only in the last term of Sn, x n appears in x+1 factors, therefore the preceding terms disappear, and Also solved by Henry Heaton, and J. Scheffer. Several incorrect solutions were received. 209. Proposed by J. EDWARD SANDERS, Reinersville, Ohio. A thread makes n(=30) equidistant spiral turns around a rough cone whose altitude is h(=10 feet), and radius of base r(=11 inches). How far will a bird fly in unwinding the thread if the part unwound is at all times perpendicular to the axis of the cone? Solution by Professor B. F. FINKEL, A. M., 4038 Locust Street, Philadelphia, Pa. Let P and P' be two consecutive positions of the bird at any time; p and p' the two corresponding positions of the end of the string in contact with the cone, it being assum ed that the string adheres slightly to the cone in order that the conditions of the problem be fulfilled; s=pP-OEP, the length of the string unwound at any time; Op=w, pq'=dw; pp' ds; the angle between Ip and the line perpendicular to 00 at the beginning of the flight, the angle being measured in the direction in which the 0 = = bird flys around the cone; do the angle pIq=the angle ACD; s, the distance the bird has flown at any time; R the radius of the base of the cone; and l-its slant height. Then Since the string passes around the cone n times, it follows that .(1). where k=l/R. If 0=2an, this expression gives the complete length of the string. Now, ds,=PP' =[P' Q'2 +Q'P'"' 2+QQ'2 ]1 ...(5). -√(k2+02)do; and QQ'=II' —pq'cos / AOC_V (l2 — R2 ) dw Σπη Hence, by substituting these values of P'Q', Q'P", and QQ' and the value of s from (4) in (5), we have Then si == ι 4πη -loga f [3h* sinh 2(a log a)+keloga* +2k sinh (a log a)]* +4k2 sinh2 (x log a)+8k2 −4} cosh (z log a)dx. for each of the values of x=0, .1, .2, ..., .9, 1 and if these values be designated A,,, respectively, we have, by Cotes' Method of Approxi by A。, A1, A2, -------9 mate Quadrature,' 107 16067(A+A,,)+106300(A,+A,)−48525(A,+A ̧)+272400(A ̧+A,) 598752 + *See Roger Cotes' Opera Miscellanea, p. 83. −260550(▲ ̧+A ̧)+427368A, 598752 The ordinates A., A1, table of Hyperbolic Functions. ......... 9 A1, may be easily computed by means of a 10 In this solution it has been assumed that the unwinding of the string begins at the vertex of the cone. If the unwinding begins at the base of the cone, we replace, in (3), 0 by (2n-0) and take the negative sign of the radical since it is then a decreasing function of 0. This gives Substituting the values of s and ds in (7) and letting LR/4xn--the entire length of the string, and 0-2n-k sinh [(1-x) log a], where the value of which may be obtained by the foregoing method of approximation. DIOPHANTINE ANALYSIS. 126. Proposed by R. A. THOMPSON, M. A., C. E.. Engineer Railroad Commission of Texas. Eight persons wish to play a series of games of progressive duplicate whist. In one evening, 12 boards are played, 4 boards (and return) by one couple against each of the other three couples, the same partners being retained throughout one evening. How many evenings will be required to complete the series, and what is the order of play, it being required that each player shall play with every other player as partner, and that each couple shall play once and but once against every other couple. |