where n is the number of factors in the denominator of the traction. 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, 6, &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 onnitting the first of the last series of the additives of the quired 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 muinber, 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. (5.) Ex.1. Resolve (x+1)(x+3)(x+4) 1 into one or more terms of the (x+1)(x+3)(x+4) 1 B C form + + + &c. or find the co-efficients B, C, &c. 21 xall asle Here the additives a, b, c, &c. of the factors in the denominator of the given fraction, are 1, 3, 4, and the additives B, Y, &c.; of the factors in the denominators of the form required, are 0, 1, 2, &c. whence we have the operation, 0121416&c. 0 | 0(2) = 0 0(4) = 1 0.6 &c. 21416 1(3)=1.3 | 1.3(.5)=1.3.5 | 1.3.3(7)&c. (6.) The nth difference of the factorial.cmli is malīx (x+n)-n. For in the factorial mms, let x be increased by 1, then, (x+1)ms is the same function of x+1, that mi is of %. Now x+1)mi=(x+1)-(7 + m) and moli = x(x + 1)m-111 (see factorials); then the difference between these two similar functions is Azalt=[(+m) — x] (x+1)=-111=m(x+1)^III. Again in the function m(x+1)*-s!? let x be increased by 1 as before, then will m(x+2)- be the same function of x+2 that m(1+2) -liis of 1+1. Now m(x+2) --=m(+2) -(x+m) and m(x+1)--=m(1+1)(x+2) } see see factorials; then the difference between these two similar functions is Agml=mx A(x+1)-1=m[(x+m)-(x+1)](x+2)m=$ =mm-1)(x+2)-21 =mail x (x+2)m625, and so on ; therefore in taking the difference n times Arali=metix (+n)". Theorem 2. 07.) The nth difference of the fraction is (cm)*x For let the base = be increased by 1, then will = x (r + 1) maille ; ,-) - m in 1 1 therefore A mli (x+1)^-^(.+ m) 1(x+1)* 1 1 x Х (x + 1)^- 31.r+m)" (x+1) x** till By proceeding in the same manner we shall find 1 1 тха -MX-(m+1)x Therefore it is evident if this operation of differencing be continued n times that 1 (8.) the nth difference of any quantity of the form ap is alp-1)" x pe. For let z be increased by 1, then will ax pu to be the same function of x+1 that apt is of x. Whence ax Ape=ax pat-axp*=a'p-1)xpe: Now a(p-1) is a constant quantity, and pe is not changed by taking the difference. Whence ap-1)Ar=a (p-1) (p-1) p* = a (p-1)ox p", that is Aap"=(-1)ox p; 1 2 • The above is a factorial where all the factors are negative; thus, suppo : *=S, then (--)317 =(----1)(-m-2) amm(m+1)(+2). Therefore, every time the difference is taken, the exponent of the power of which the residual p-1, is the root, will be increased by anity; whence in laking the difference in times, the exponent of the power, of which the residual ral is the root, will be increased to n; whence Aap=a(p-1)" x pt as was to be demonstrated. (9.) Corollary. Hence if p be a fraction equal to then then 9 1 difference of a will be Х and this will be ob q* 1 ained by simply substituting for p, o if r= =p, then in all 9 ) (1) = P cases A ars=a[r— 1)" X 7*. Theorem 4. (10.) The difference of v w is equal to either of the factors multiplied into the difference of the other, plus the difference of the former multiplied into the succeeding value of the latter. Let v be increased by any quantity v, and w by any quantity w, and let v and w be either constant or variable. Then (u+v)(w+w) is the same function of v+v and wtw, which vw is of v and w; then the difference between these two similar functions is Av w=(v+v)(w+w)-vw, by definition, 1 that is A vw=vw trw+vwvw-VW =vw+vw+vw =vwtvíw+w) which was to be demonstrated, oi let w+w=W1, 1 then will Avw=vw+vwz. Problem S. Rule, Multiply the exponent m of the given factorial into another factorial, of which the first factor is one more, and the exponent one less, than that of the given factorial. K 1 By rule 4(x+3) =5(x+4)*. Problem 4. To find the nth difference of a factorial, of which the first factor is &, the number of factors m, and the common difference, unity. Rule. (12.) Form a factorial of which its first factor is equal to the exponent of the given factorial, and its exponent equal to number, ex. pressing tl.e number of differences; and the common difference of its factors is minus unity : multiply the factorial thus formed into another, which is the given factorial, having its first factor increased, and its exponent diminished by the number of differences to be taken. Examples. 1. Find the third difference of 2511. By rule Arsli=5317 x (x+3)211. 2. Find the nth difference of mis. By rule Axml=mnix(x+n mel. 3. Find the 4th difference of (x+3)612, By rule A(z + 3)=6471 x (x+7)21. Problem 5. 1 13 of the given factorial thus changed, into the same factorial, having its exponent diminished by unity. 1 3 Ar-3 3 3 Wy rule A r311 -3x3-111 I. |