= Adding these equations, we get I ¿[(7 + 4√/3)" — (7 — 41/3)"]√ abc(a + b + c) + 1/2 [(7 14√3° + 4√/3)" + (7 − 4√/3)"](ab + ac + bc) } ÷ abc. Substituting this value in (5), and taking reciprocals, we find x = abc ÷ I { 27/3 [(7+4V/3)" — (7—4√/3)"]Vabc(a+b+c) + {[(7 + 4√/3)"+(7—4√ 3)"— 2](ab+ac) + {[(7 + 4√/3)" + (7 −4v/3)" + 4]bc }. Similarly we find Yn I =abc÷ { 2√/3 [(7+4V/3)" — (7—4V/3)*]Vabc(a+b+c) + b[(7 + 4√/31" +(7—4V/3)" — 2](ab+bc) + f[(7+4√/3)" + (7 −4√/3)* +4]ac }· Zn = abc ÷ 213 [(7 + 4√/3)" — (7 −4√ 3}"]√ abc(a+b+c) + {[(7 + 4v′3)" + (7—41/3)" — 2](ac+bc) + }['7+4√/3)* + (7 − 4√/3)" + 4)ab }· NOTE ON THE BINOMIAL THEOREM. BY GAETANO LANZA, ASSITANT PROF. OF MATH. AND MECH., MASS. INSTITUTE OF TECNOLOGY. MR. EDITOR: -Allow me to call your attention to a demonstration of the Binomial Formula for positive integral exponents, given by Bobillier, which, though very simple and lucid, is not very generally known. It does not involve at all the subjects of Permutations and Combinations, and hence in order to understand the Binomial Theorem the student is not obliged previously to master a number of propositions on the above subjects. It would therefore seem very desirable that this demonstration should find a place in our text books and be taught in our schools. For the benefit of any of your readers who may not be familiar with the notation employed, I will state that n (=factorial n) is an abbreviated way of writing 1.2.3.4. n, that is the product of the natural numbers from 1 to n inclusive; thus | 3= 1.2.3 6; 4 1.2.3.4 = 24, &c. ........ The demonstration is as follows: = = The following three equations will be evident on inspection, viz.: It is now proposed first to find an expression equal to (x + a)" n where n is any positive integer. In the above three results the following laws held: I. The number of terms on the right hand side is greater by one than the exponent of (x + a) on the left hand. II. The exponent of x in the first term is the same as the exponent of the power to which (x+a) is raised, and in each succeeding term the exponent of x is less by unity than that in the preceding term; while that of a follows the same law beginning at the other end of the series. III. Each power of x or of a has for denominator its exponent taken factorially. We shall now prove that these laws hold whatever positive integral value be assigned to n. Suppose the laws to hold when the exponent is n -- 1, that is I, that is suppose Now multiply this identical equation by the following also identical, member by member, viz.: and as the result of the multiplication we shall obtain the following: Hence, if the above laws hold for an exponent - 1, they hold for the exponent n, which is greater than the former by unity; but they have been proved to hold for the exponent 3, hence they hold for 4, hence for 5, and so on, ad infiinitum; therefore they are true in the case of any positive integral exponent. Now multiply both members of equation (1) by n and we obtain SOLUTIONS OF PROBLEMS IN NO. 8. Solutions of problems in No. 8 have been received as follows: From Prof. A. B. Evans, 34, 35 & 36; R. M. DeFrance, Esq., 34: Henry Gunder, 34 & 35; L. Regan, 34 & 35; James Stott, 35; and Walter Siverly, 35 & 36. SOLUTION BY R. M. DEFRANCE, ESQ., MERCER, PA. Put x + y + z =s and xy + xz + yz=m, and substitute in (3) and it becomes (s 3ms +54) whence we get s* 3ms = ±81.... 108(s 3ms +54) 3645; = By substitution (2) becomes $2 2m = 33. .(4). Eliminating m from (4) and (5) we get s'-998 = 162. Multiply ing this eq. by s we have s 1625, or, by transposition, 18s2 + 81 81s 1625 + 81, whence 99s 9 and s = 9 or 1(9±√153) s Substituting for s in (5) we get m = 24 or 1(51±9√153). - Put x+y t and xy = u, and, neglecting the irrational and m, we have ₺ + z = 9, u + tz = 24 and uz = 18. and u from these three equations we get Put zv+3 and substitute in (8), we get v3 and v = 13: ...23±√3. But xyz="9 (by (6)), ... = 18 From (1) xy = 18 ÷ 2 (3±√3 1=6=√3. --- x − y = ±√3. and consequently 35. "A man let a stone weighing 40 lbs. to a neighbor-the latter broke it accidentally into four parts-and upon returning the fragments consoled the owner by remarking that now he could weigh all numbers = 40, between one and forty. In other words, given a+b+c+d: to determine such values for a, b, c and d as will, by association, produce all numbers from one to forty.' SOLUTION BY PROF. A. B. EVANS, LOCKPORT, N. Y. 36. and a+b+c+d=7a I, b "A perfectly flexible cord of given length is suspended from two points whose coordinates are x', y' and x", y". How must the weight of the cord vary from point to point in order that it may hang in the arc of a circle." SOLUTION BY WALTER SIVERLY, OIL CITY, PA. Let H K be be the two given points of suspension, √(x' — x'')2 + (y' — y'"')2 = 26 A = the distance between them, 27 = the length of the cord, r the radius of the cir from which may be determined and the circle drawn passing through H and K. Let AB be the horizontal diameter of the circle, O its lowest point to which transfer the origin making the axis of x horizontal and of y vertical, and let m be the unit of mass at any point P; and c the tension at O. It is shown in works on statics that ries inversely as the square of its distance below the horizontal diame ter of the circle. |