(ab + 1)c + a and [(ab + 1)c + a] × d + ab + 1 be + 1 (bc + 1) × d + b (1-x)2, which is the generating function of will be 1+2x+3x2+ etc. Hence, we shall clearly have 1 + 2x + 3x2 + 4x3 + etc. = (1-x)= the sought generating function. 8th. To find the generating function of the series 1 + 2x + 3x2 + 5x3 + 8x1 + 13x + etc., or of the fraction 1 1 + 2x + 3x2 + 5x3 + 8x2 + etc. Proceeding as in the last equation, we have a, b, c, d ex by taking the reciprocals, we shall have 1 + 2 + 3x2 + ; Remark. We have taken this question from p. 265 of the New York edition of Young's Algebra, where (1 - 1 is x) 2 given for the generating function of 1 + 2x + 3x2 + 5x3 + etc. 9th. To find the generating function of the series 3+5x+ 7x2 + 13x3 + 23x1 + 45x2 + 87x + etc. Proceeding as in the last two questions to find the generating function of the fraction 1 3+5x+7x2+ etc.' we get 13 2, and for the representatives of a, x the generating function of the fraction; and taking the reciprocals, we shall have 3+ 5x + 7x2 + 13x2 + 23x* + etc. 3- x 6x2 1- 2x — x2 + 23 as required. Remarks.-1. The answer which we have given to this question agrees with that given by Mr. Young, at the place cited above. 2. The method used in solving the three preceding questions is substantially the same as that given by Mr. Young at p. 174, etc., of his Theory of Algebraical Equations. Mr. Young ascribes the method to M. Le Barbier. 3. It is clear that the method is applicable to any recurring series, when the terms are rational, and the scale of relation consists of a finite number of terms. 10th. To find the approximate fractions of X3 3x 1 1 Here, dividing the numerator (arranged according to the descending powers of a) by the denominator, 1, and 1 by the remainder, and so on, we get a3, the representatives of a, b, c, and d. 1 1 9x2, and for 3x 3x 3 Hence, 3 and 3 9x2 3x 1 1 1 + 3x 3x 3x Again, putting the given expression in the form -1-3x+23, and proceeding as before, we get — 1, — for the representatives of a, b, c, and d; Remarks. If the fractions converge, with sufficient rapid satisfy the equation -3x-1=0 ought to reduce the converging fractions to 0 very nearly. Hence, if we put the numerators of the fractions which contain a, separately equal to 0, the roots of the resulting equations ought to be the approximate values of the roots of a 3x-1= 0. Thus, if we put the numerators of the second and third fractions (in the first set) equal to 0, we get the equations a2 = 3 and 3a-x=9. Solving these equations, the first gives x 1.732 and x = 1.732, and the second gives a = 1.904 and 1.57 for approximate values of the roots 1.879 and 1.532 of the equation a3-3x-1=0. - In like manner, if we take the second set of converging fractions, and put the numerators of the second and third equal to 0, we get the equations 3x + 1 = 0 and ∞2 + 9x + 30. Solving these equations, we get x = 1 3 0.333 and 0.347 for approximate values of the remaining root 0.3472 of the equation a 3x-10. Hence, we perceive in what manner we may (often) find the first figures of one or more of the roots of an equation. Thus, to find the approximate root of the equation a + 18x Then, by putting the numerators of the second and third fractions equal to 0, we get the equations 3-10 and a2 +5418. Solving these equations, we get a = =0.333 1 3 and a 0.33130; the second of these values being correct to four decimal places. (See Vol. I. of Bonnycastle's Algebra, pp. 147 and 148.) SECTION XVI. SOLUTION OF RECIPROCAL OR RECURRING EQUATIONS. -2 (1.) LET A2 + A ̧1⁄2-1 +A ̧1⁄2-2 + ... + Am-222 + Am-13 + Am = 0, (1), stand for any equation of the mth degree; then, if A, Am, A1 = Am-1, A2 = Am-2, and so on, the equation is called a reciprocal or recurring equation. = (2.) Supposing (1) to be a reciprocal or recurring equation, by putting A, A1, A2, etc., for Am, Am-1, Am-2, etc., it is clear that the equation may be written in the form A(+1) + A1(2m2 + 1) + A22(2-4 + 1) + etc. = 0, (2). -4 If m is an odd number, since m2, m 4, and so on, are odd numbers, it is clear that 2+1, 2-2 + 1, 2m−4 + 1, and so on, will each be exactly divisible by z + 1; conse quently, 21 must be a root of (2) or (1). To find the remaining roots of (1), we divide it by z + 1, and put the quotient equal to 0, and we shall get the equation A ̧m-1 + (A1 — A1)2−2 + (A2 — A1 + Ao) × 2-3 + .... + (Am-3 Am-1 + Am)2 + (Am-1 — Am)2 + Am = 0, (3); and since A1 = Am, A1 = Am-19 A2 = Am-2, and so on, it is clear that (3) is a reciprocal or recurring equation, of the degree represented by the even number m 1. До m - -- 1,2m-4 (3.) If m is an odd number, and A。 = — Am, A1 = Am-19 A2 = — Am-2, and so on, then it is clear that (1) may A2 be written in the form A(21) + A12(2-1) + A2(2-1) + etc. = 0; consequently, since m, m — 2, 4, and so on, are odd numbers, 2” — 1, 2m-2 1, and so on, must each be exactly divisible by z- 1, and of course z = 1 must be a root of (1). Dividing (1) by z 1, and putting the quotient equal to 0, we shall get the reciprocal or recurring equation A ̧m-1 + (A1 + Aŋ)2m−2 + (A2+ A1 + Ap)zm−3 + . . . . + (A2 + A1 + A1)22 + (A1 + A1)2 +A, 0, of the even degree m - 1; and solving this equation, we shall get the remaining roots of (1). = = (4.) If m is an even number, and A, Am, A1 = Am-1, and so on, and if the middle term of (1) is wanting, then by writing (1) in the form A。(1⁄2” − 1) + Å ̧2(1⁄2”—2 — 1) + A2(2-4 -1)+ etc. = 0, since 2-1, 2-2-1, and so on, are exactly divisible by 22-1, it follows that 2 = 1 and 2= 1 must be roots of (1). Dividing (1) by 22 — 1, and putting the quotient equal to 0, we get the reciprocal or recurring equation A-2 + A12m3 + (A。 + A2)1⁄2ï¬−4 + (A1 + A ̧)2m-5 + (A。 + A2 + A1)2−6 + .... + (A。 + A + A ̧)≈1 + (A1 + Ag)≈3 + (A。 + A1)1⁄2o + A12 + A, = 0; consequently, from the solution of this equation, we shall get the remaining roots of (1). (5.) Hence, supposing (1) to be a reciprocal or recurring equation of an even degrée, and that m 2n, then (1) may be written in the form A(22n + 1) + A12(22n−2 + 1) + A222 (22-4+1)+A ̧3(22n-6+1) + . . . . + An-12"-1(22 + 1) +A1e" = 0; or dividing the terms of this by 2", we shall have the equa tion A,(2+2)+A, (21+)+(+-) +An-1(2 + 1) + A‚ = 0, (a). Putting z+ An 1 1 1 2 = x, (b), we 1 easily get 22 + ==∞2 — 2, 23 + 3 = x23 — 3x, 2a + == 3x) × x − (x2 — 2) = x* — 4x2 + 2, 23 − — 1 تاج — (23 — 3x) = x1 — 5a3 + 5x, and so on. it will be reduced to an equation of the form Ax2 + Bæ2-1 +Can-2+ etc. = 0, (c); which being of the n" degree, its degree is half that of (1). Solving (c) we shall get the values of x, which will of course be then roots of the equation; then putting these roots separately, for a in (b), the values of z and correspond 2 ing to each root, will be two of the roots of (1); consequently, the 2n roots of (1) will become known. Because the values of z and are the roots of (1) which 2 correspond to each root of (c), it is plain that the roots of (1) |