nid of the form (x+a)(x+2)(x+c) &c. into a series of factorials which shall have r alone, or x with any given additive number for the first factor of each, and of which the factors shall have any given common difference. Let a, b, y, &c. be the given quantities in arithmetical progression, which are respectively to be added to x, in order to form the series of factorials, and let d be their common difference ; find n, the number of given factors, then the general form of the series will be (x+)*4 + B(r+a)-1d+C(r+ ad+ &c. where we have only to find the co-efficients of the 2d, 3d, &c. terms, which will be found by the following Rule. (2.) 1. From the given quantities a, b, c, a, &c. in the whole number of binomial factors, subtract a, b, y, &c. respectively, to the same number, and call the remainders the first order of differences. 2. From 1, c, d, &c. the whole number of factors, except the first, subtract a, b, &c. respectively, to the same number, and call these the second order of differences. 3. Proceed in the same manner, omitting one every time, until one only remain. 4. Then form a series of successive sums, by taking the first difference of the first order, for the first term, the sum of the first and second for the second term, and so on, and the last term of this series, is the co-efficient B, of the second factorial. 5. Multiply the terms of the series of successive sums, respectively, by the second order of differences, and with the products thus found, form a second series of successive sums, in the same manner as before, and the last term or sum of this series will be the co-efficient of the third factorial. 6. Proceed in the same manner, until there be only a single product which will be the last co-efficient or term. In these operations when negative quantities occur the sign is placed above, in order to obtain a more convenient arrangement, as is done in the indices of Logarithms. The following tables exhibit the form of the operation, as now expressed by the rule. The first table simply contains the additive part of the required binomial factors, and the other columns consist of the additive parts, which are to be found in the binomial factors of the given Algebraic product. In these last, we must observe that the first term of every new column, is the same as the second term of the preceding column, and the other terms follow in consecutive order, and end with the additive number to be found in the last factor of the given Algebraic product. The operation of differences is carried from the first table to the second, and the first order of differences e, f', g', h', &c. are placed in the first column of the second table, and the successive sums are placed in a column on the right with a line between them. The factors e, f", g", &c. are the second differences; A, B, C, are the real products of these factors in numbers : ill, £"', '", &c. are the suecessive sums of the products, A, B, C, &c. and so on. Erample 1. Resolve (x+1)(x+3)(x+5) into factorials, of which the first factor shall be x, and the factors shall increase by unity. Here a=i, b=3 and c=5, also a=0, B=1 and y=2, for as the factors proposed to be resolved are limited in their number, so is each of the series a, b, c, &c. as also Q, B, %. &c. Whence the operation, where for convenience the negations are marked above. gives (x-1)(+3)(x+3)=2311+ 4.x211 +11 - 15. Example 2. Resolve (2-1)(x+3)(x+4)(x+5) into factorials, so that the first factor of each may be x, and the factors to increase by 2. Here a=i, b=3, c=4, and d=5. Also a=0, B=2, y=4, and d=6. whence 1 3 4 5 i(3)=3 | 3 (4) 3 (4) = 12 | 12(5)=60 3 1 0 (2)=0 3 (3) = 921 0 0(1)=0 6 5 2 4 4 3 а gives (2-1)(x+3)(2+4)(x+5)=242 — 231232212--212—60. If we wish to resolve an algebraic product of binomial factors into powers, we have only to make the first column a series of cyphers. Problem 2. (3.) To resolve the reciprocal of an algebraic product, consisting of С binomial factors, or a quantity of the (r+a)(r+ b)(1+c) &c. A B С into a series of the form :+ + &c. where n is the number of factors in the denominator of the traction. n+ ild * + 214 Here, as in the former case, we have only to find the co-efficients, or numerators A, B, C, &c. which will be obtained by the following Rule. (4.) 1. From the additives a, b, y, d, &c. in the required forms subtract a, b, c, d, &c. the additives of the binomials of the given fraction to the same number, and call the remainders the first order of differences. 2. From B, 9, 8, 5, &c. omitting the first of the last series, and taking in the next term, following the last, subtract a, b, c, d, respectively, the same additives as before, and all the remainders, the second order of differences. 3. Proceed, in the same manner, always omitting the first of the Jast series of the additives of the required form, and adding the succeeding term to the last term of the sums, and subtracting the terms of the constant series a, b, c, &c. in order to obtain a new order of differences, which will be the same in number, as at first order of differences, until the differences become each equal to Zero, or till a sufficient number of the terms are found. 4. Proceed as in the former rule to find the successive sums and products, which will always be the same in number as the additive a, b, c, d, &c. in the factors of the given algebraic fraction ; then the last successive sum of every column will be the co-efficient of each respective term of the required form, omitting the first, which has unity for its co-efficient. The following tables exhibit the form of the operation which is similar to the former, except that a, b, y, d, change places with a,b,c,d, and that all the columns of this operation must contain the same number of terms. Examples. 1 (5.) Er.1. Resolve (x+1)(x+3)(x+4) into one or more terms of tho 2 1 B с form: + &c. or find the co-efficients B, C, &c. + + > Here the additives a, b, c, &c. of the factors in the denominator of the given fraction, are 1, 3, 4, and the additives a, b, y, &c.; of the factors in the denominators of the form required, are 0, 1, 2, &c. whence we have the operation, Proceeding in the operation 0101246&c. 0 | 0(2) = 0 0(4) = &c. 2 | | (7) *c. 11(3)=1.3 1.3(.5)=1.3 0.6 (6.) The nth difference of the factorial zmit is malix (x+n)nr. For in the factorial zmie, let x be increased by 1, then, (x+1)ms is the same function of 2+1, that rmi is of x. Now x+1)mi=(x+1)=(x + m) and cli=x(x+1)--11 (see factorials); then the difference between these two similar functions is Ázail=[(x+m) – ] *(x+1)=-11=m(x+1HI, (m-. Again in the function m(x+1)*-s13 let x be increased by 1 as before, then will m(x+2)m- be the same function of x+2 that m(+2) -1 is of x+1. Now m(x+2) --**=m( +2)(m) } see factorials ; and m(x+1)--=m(x+1)(x+2)=-1 m المعه then the difference between these two similar functions is =mm-1)(r+2)*- =malt x (x+2)+75, and so on ; therefore in taking the difference n times Azali mulix (I + n)maalle. For let the base : be increased by 1, then will 1 be 1 1 G#m-) Now (x+1) =(x+1)-1(x + m) and = x (x + 1) mill; Ime 1 1 therefore A moli (x+1)+-11(x+m) 1(x+1)" + 1 Х Х (x+1)-11 x(x + m)“ (x+1) 2974111 By proceeding in the same manner we shall find 1 1 — тXA :-MX-(m+1)x -m in 1 antils Therefore it is evident if this operation of differencing be continued n times that 1 1 or A = F mall x according as n is an odd or even number. mli ***11 (8.) the nth difference of any quantity of the form ap* is a(p-1)" x pa. For let z be increased by 1, then will ax puts be the same function of x+1 that aps is of x. Whence ax Ape=axpo+imQxp=ap-1)Xpe: Now a(p-1) is a constant quantity, and pe is not changed by taking the difference. Whence alp-I) Arsa (p-1) x (p-1)xp* = a (p-1)exp", that is Aap"=(-1)'p'; 1 2 • The above is a factorial where all the factors are negative; thus, suppo:: *=S, then (mm)31ī=(-Xm-1)(---)a-m(m + 1)^2+2). |