-2 mth III. Having given an adequate number of terms of a series, to find the general term of the series. Suppose the general form of the term to be ax" + 0,2-1 + a2m2 + .... + Am_X + Am ; then there is required the index m and coefficients a, a,, ag, Am, Amti: 1. In the first place, we have seen (II. Schol.) that in an expression of the degree, the mth differences are all equal, and that all higher orders of differences become 0. Hence to find m, we have only to take the successive orders of differences, till we find one order all whose terms are equal. The number of these operations which are thus performed, gives the value of m. 2. Let Up, U2, U3, .... be the several given terms of the series : then, as these are the values of the general term when m is 1, 2, 3, ... we have 1"a + 1*-1 a, + 1*-? az + + lam-1 + 2m = U1 a, + + 2am-1 + 2m = U, m-1 (m + 1)"a + (m + 1)^, + ....... (m + 1) Qm-+ a, = Um+1; in which there are as many equations as there are unknown coefficients a, a,, Az, Qm, viz. m + 1. All these equations are, with respect to the unknowns, of the first degree ; and the method most readily applicable to the process of solution, is that pointed out at pp. 179, 180, of this work. Ex. 1. Let the series 7, 33, 79, 145, 231, be given to find its general term. 7|33|79|145 231... given series 20 20 201... second differences. Hence, as the second differences are all equal, we have m= 2, and the general term is of the form ax2 + 2x + az. Hence, substituting 1, 2, 3, in this for X, we have 2 a + a + Ag = 7 and, as at p. 180, we get 4a + 2a + ag = 33 a = 10, a, =- · 4, a, = 1. 9a + 3a, + a, = 79 The required expression is, therefore, u, = 10x2 - 4x + 1. Ex. 2. Find the general term of the series 2, 24, 108, 320. Also ascertain whether any of the terms 1512, 4668, 7290, and 11011, belong to the series ; and if so, assign their places. Ex. 3. Given 2, 14, 66, to find the general term, and hence the next term *. * This example was actually formed from the expression x4 — - 2-3 + x2 + x; that is, one of the fourth degree: but as the terms, so far as they are actually given, can be formed from one of the second degree, this latter ought to be considered the determinale solution of the question, in contradistinction to the indeterminate ones, which the solutions of the third, fourth, degrees do really become. These considerations suggest, that when a series of m terms is given, such that the mth difference is not constant, then we may fulfil the condition to which these given terms are subject, by taking Amtiu = 0, and therefore also Amuz=0: though at the same time the general term which results is only a particular case of a more general solution which would have been obtained by supposing the dimension of the general term to be higher. The assumption of the dimension of the general term is, therefore, in fact altogether arbitrary, and likewise 2, 14, 66, 212, 530 to Ex. 4. Given 2, 14, 66, 212, find the general term. IV. Having given a series of terms, to find the several orders of differences. 1. Subtract the first term from the second, the second from the third, and so on, to get the entire first order of differences ; employ the same process upon the first order to obtain the second; upon the second to obtain the third. The first terms of these several orders are those sought; as follows from definition 4. 2. Generally, however, it will be more convenient to employ the formula deduced in II. p. 274, making in all the expressions x = 1, and n= 1, 2, 3, ... in succession. The expression so modified becomes n m(m - 1) Unt th us Fui Ug 1 1.2 n Scholium. It is evident from both the methods, that there must be given one term more than the number indicated by the order of the difference sought, in order to render the problem determinate. EXAMPLES. =U U = 13 ... Ex. 1. To find the first term of the third order of differences of 1, 4, 8, 13, 19, .. Here U, = 1, Ug = 4, Ug = 8, U = 13, Ug = 19, 24 + 12 – 1= 0. Ex. 2. The first * term of the seventh order of differences of 1, 4,'8, 16, 32, 64, 128, Here Au = Ug — 729 + 2146 350g + 3544 21uz + 7u2 Uj; in which, inserting the given values of #j, U2, U3, we have ału, 256 – 896 + 1344 · 1120 + 560 168 + 28 1 = 3. Ex. 3. Given 1, 2, 4, 8, 16, ... to find the first term of the seventh order of differences. Ans. 1. Ex. 7. Find the first four orders of differences of the logarithms of 101, 102, 103, 104, 105, Ex. 8. Given 1, 6, 20, 50, 105, ... to find the first four orders of differences. so that the index be at least equal to the number of terms; these terms being always understood to be consecutive in the scale, beginning at unity. When, however, the calculation is of one single term, however distant, it will be effected more easily by the following process, without determining the general term of the series. If several be required, the general term, (that is, the most simple general term which can be found from the given numbers,) is indispensable. * When only the first term of a single order of differences is required, as in the first three examples, the second method is the preferable, as in the text; but when all are required in succession, it will be more convenient to employ the former method. Thus, if the several orders of this example were sought, the work would stand as below: Ex. 9. Given 1, 4, 8, 16, 32, 64, 128, 256, to find the several orders of differences. 1/4/8/16/32 64 128 256 ...the given series ..fourth differences 3 seventh differences. 8 16 ... : n V. Having given the first terms of the first n orders of differences, to find the (n + 1)th term of the series : or in symbols, given Uy, Au,, A’u,, A*u, to find Un +1 m(m - 1) The solution is Untu = U, tau, + A’u, + ... carried to the 1 1.2 =u, + 204, + A'un = 1, + 3Δυ, + 3Δυ, + Δ3u ; and so on to any extent required, the coefficients being those arising from the expansion of the binomial (1 + 1)", and the necessary continuity of the law being capable of establishment nearly as in II. = = EXAMPLES. . Ex. 1. To find the twentieth term of the series, 2, 6, 12, 20, 30, ... w, uz u, u, Ug' .... the functions 2 2 2 .... second order of differences 0 0 third order of differences. Au, + Hence u, = 2, Au, = 4, 4ʻu, = 2, A3u, = 0, and so on, all the subsequent orders of differences being zero. Hence 19 19.18 U20 = U, + A’u, = 2 + 76 + 342 = 420, 1 1.2 which is the twentieth term of the series sought. Ex. 2. Find the tenth term of the series, 2, 5, 9, 14, 20, .. Ans. 65. Ex. 3. Required the fifth term of 1, 3, 6, 10, Ans. 15. Ex. 4. To find the tenth term of the series, 1, 4, 8, 13, 19, Ans. 64. Ex. 5. Required the twentieth term of 1, 8, 27, 64, 125, Ans. 8000. Ex. 6. Required the sixth term of 101, 1083, 118, 1293, . Ans. 1584 VI. To convert a given function in powers of x to one which shall have every term composed of factors x + a, x + b, and so on. Divide synthetically by x +a; the last coefficient is the coefficient of (x + a)0 of the transformed expression : divide again in the same manner by a + b, stopping one step sooner ; the last coefficient is that of (x + a) in the transM2 formed expression : proceed similarly with x + c, stopping one step sooner than in the preceding ; then the last coefficient is that of (x + a) (x + b) in the transformed expression. Proceed thus with all the factors, then the coefficients of the transformed expression will be all determined. For let Ax" + Ban–1 + Can+ + Lx+ Mx + N be the function given in powers of x: then dividing by the several given factors, we have in succession Ν, Ax-1 + B.2 + C "-" + + L,X + M + *+a N (x + a) (x+b) N + x+c+ (x+b)(x+c)(x+a) (x+b) (x+c)' () and so on, till we arrive at Bm-1 At + to. x + m (x + 1)(x + m) Multiplying now by (x + a) (x + b) (x + c) .... (x + 1)(x + m), we have ac x ) A(x + a) (x b) .... (x + m) + Br-1 (x + a) (x + b)... (x + 1) + +N for the transformed expression. If these factors be taken in arithmetical progression, the result is a transformation into an expression of factorials. See def. 7, p. 274. Cn_2 a 3 - 36 and the transformed function becomes, in the notation of Kramp, p. 274. 3(2 + 1)411 – 36 (2 + 1)311 + 113 (x + 1)2,1 - 98 (x + 1)111 + 7. Or, in the common notation, 3 (2 + 1) (x + 2) (x + 3) (2 + 4) - 36 (x + 1) (x + 2) (x + 3) + 113 (20 + 1) (v + 2) - 98 (x + 1) + 7. When the given expression itself is a combination of binomial factors, and it is required to transform it into some other combination, as a factorial one, the given expression may be first reduced to powers, and then transformed by the general rule; as in the next example. Ex. 2. Given (oC + 1) (2C 3) (x + 3) (0 + 5) to be converted into factors involving x + 1, x + 2, x + 3, x + 4. Since x + 1 is a factor of the given and the sought expressions, it need not be attended to, as it multiplies all the terms in both. However, for illustration a we shall work out as though no two factors in the two expressions agreed with each other. 1+1(-3 1 + 6 4 54 45 -3 · 3 -1 .1 5 + 9 + 45 Hence the given expression is x4 + 6x3 - 4x2 54x 45, and the factorial is (x + 1)411 - 4(x + 1)311 — 15(x + 1)2!1 15(x + 1). Ex. 3. Show that 23 = (x - 1) x (20 + 1) + x by this method. - 1. VII. To integrate the general term of a series, or to find the expression whose first difference constitutes that general term. 1. When the expression is composed of factors in arithmetical progression. Multiply the increment (or given general term) by the preceding value of the first factor, and divide the result by the number of terms thus obtained, and by the common difference of the factors. This result, when corrected, gives the sum required. 2. When the expression to be integrated is the reciprocal of such a series of factors in arithmetical progression. Expunge the last factor from the denominator; divide the resulting fraction by the number of factors remaining, and the common difference of the factors. Then this result, written minus, will be the integral sought. These being precisely the reverse processes by which the differences or increments were found, their truth is evident. The correction arises from this cause: that A (z + a) = Az, and hence it cannot à priori be ascertained whether the integral is z + a, or simply z. The correction is found from this consideration. If from any circumstance we can find what the aggregate value of a certain number of the terms is, and at the same time ascertain what value the integral gives of the same number; then the difference of these two results is the correction. Most frequently, putting x=0 is the best method : but examples will render this process much plainer than precept could do. The student will find little difficulty in reducing all expressions which involve only positive integer powers of x, that he commonly meets with, to one or more of these forms. : EXAMPLES. Ex. 1. Integrate 2411 or x (x+1) (x+2) (x+3). |