that is, we may transfer to the numerator all the factors of the denominator, by giving to their exponents the sign Reciprocally, when a quantity contains factors, which have nega tive exponents, we may convert them into a denominator, observing merely to give to their exponents the sign+; thus the quantity Of the Formation of the Powers of Compound Quantities. 134. We shall begin this section by observing, that the powers of compound quantities are denoted by including these quantities in a parenthesis, to which is annexed the exponent of the power. The expression (4a3 — 2ab+ 5 b2)3, for example, denotes the third power of the quantity, 135. Binomials next to simple quantities are the least compli cated, yet if we undertake to form powers of these by successive multiplications, we in this way arrive only at particular results, as in art. 34, we obtained the second and third power; thus (x + α)2 = x2 + 2 a x 3 + a2, x2 + 3 a x2 + 3 a2 x + a3, + 4 ax3 + 6 a2x2 + 4 a3 x + a1, (x + a)3 = (x + α)1 = x2 &c. It is not easy from this table to fix upon the law, which determines the value of the numerical coefficients. But by considering how the terms are multiplied into each other, we perceive, that the coefficients have their origin in reductions depending on the equality of the factors, which form a power. This is rendered very evident by an arrangement, which prevents these reductions taking place. It is sufficient for this purpose to give to the several binomials to be multiplied different second terms. If we take, for example, x+a, x + b, x+c, x+d, &c. by performing the multiplications indicated below, and placing in the same column the terms, which involve the same power of x, we shall immediately find, that (x + α) (x + b (x + c) = x3 + ax2 + abx + abc + bx2 + acx + cx2 + bcx (x + a) (x + b) (x + c (x + d) = x2 + ax3 + abx2 + abcx+abcd + bx3 + acx2 + abdx +cx3 + adx2+acdx +bdx2 + cdx2 Without carrying these products any further, we may discover the law according to which they are formed. By supposing all the terms involving the same power of x, and placed in the same column, to form only one, as, for example, а x3 + bx3 + c x3 + dx3 = (a + b + c + d) x3, &c. 1. We find in each product one term more than there are units in the number of factors. 2. The exponent of x in the first term is the same as the number of factors, and goes on decreasing by unity in each of the following terms. 3. The greatest power of x has unity for its coefficient; the following, or that, whose exponent is one less, is multiplied by the sum of the second terms of the binomials; that, whose exponent is two less, is multiplied by the sum of the different products of the second terms of the binomials taken two and two; that whose exponent is three less, is multiplied by the sum of the different products of the second term of the binomials, taken three and three, and so on; in the last term, the exponent of x, being considered as zero (37), is equal to that of the first, diminished by as many units as there are factors employed, and this term contains the product of all the second terms of the binomials. It is manifest, that the form of these products must be subject to the same laws, whatever be the number of factors; as may be shown by other evidence beside that from analogy. 136. It will be seen immediately, that the products, of which we are speaking, must contain the successive powers of x, from that, whose exponent is equal to the number of factors employed, to that, whose exponent is zero. To present this proposition under a general form, we shall express the number of factors by the letter m; the successive powers of x will then be denoted by xm, xm-1, xm-2, &c. We shall employ the letters A, B, C, Y, to express the quantities, by which these powers, beginning with m-1, are to be multiplied; but as the number of terms, which depends on the particular value given to the exponent, will remain indeterminate, so long as this exponent has no particular value, we can write only the first and last terms of the expression, designating the intermediate terms by a series of points. The formula then .... x2 + A xm−1 + B xm−2 + Cxm-3 + Y, represents the product of any number m of factors, x + a, x + b, x + c, x + d, &c. If we multiply this by a new factor x+l, it becomes xm+1 + A xm + Bxm¬1+ Cxm-2 .... .... +1Y + 1xm + 1A xm−1 + lВ xm-2 It is evident, 1. that if A is the sum of the m second terms a, b, c, d, &c. A+ will be that of the m+1 second terms a, b, c, d, &c. l, and that consequently the expression employed to denote the coefficient will be true for the product of the degree m+1, if it is true for that of the degree m. 2. If B is the sum of the products of the m quantities a, b, c, d, &c. taken two and two, B+ A will express that of the products of the m+1 quantities a, b, c, d, &c. l, taken also two and two; for A being the sum of the first, IA will be that of their products by the new quantity introduced ; therefore the expression employed will be true for the degree m + 1, if it is for the degree m. If C is the sum of the products of the m quantities a, b, c, d, &c. taken three and three, C+ 1B will be that of the products of the m+1 quantities a, b, c, d, &c. l, taken also three and three, since B, from what has been said, will express the sum of the products of the first taken two and two, multiplied by the new quantity introduced 1; therefore, the expression employed will be true for the degree m + 1, if it is true for the degree m. It will be seen, that this mode of reasoning may be extended to all the terms, and that the last, Y, will be the product of m+1 second terms. The propositions laid down in art. 135, being true for expressions of the fourth degree, for example, will be so, according to what has just been proved, for those of the fifth, for those of the sixth, and, being extended thus from one degree to another, they may be shown to be true generally. It follows from this, that the product of any number whatever m, of binomial factors x+a, x + b, x + c, x + d, &c. being represented by xm + A xm-1 + В xm-2 + С xm-s+ &c. A will always be the sum of the m letters a, b, c, &c., B that of the products of these quantities, taken two and two, C that of the products of the quantities, taken three and three, and so on. To comprehend the law of this expression in a single term, 1 take one, whose place is indeterminate, and which may be represented by N xm―n. This term will be the second, if we make n = 1, the third, if we make n = 2, the eleventh, if we make n = 10, &c. In the first case, the letter N will be the sum of the m letters a, b, c, &c. in the second, that of their products, when taken two and two; in the third, that of their products, when taken ten and ten; and in general, that of their products, taken n and n. 137. To change the products (x + a) (x + b), (x + a) (x + b) (x + c), into powers of x+a, namely, into it is only necessary to make, in the development of these products, All the quantities, by which the same power of x is multiplied, become in this case equal; thus the coefficient of the second term, which in the product (x + a) (x + b) (x + c) (x + d) is a + b + c + d, Alg. 19 is changed into 4 a; that of the third term in the same product, which is, ab + acad+be+bd+cd, becomes 6 a2. Hence it is easy to see, that the coefficients of the different powers of x will be changed into a single power of a, repeated as many times as there are terms, and distinguished by the number of factors contained in each of these terms. Thus, the coefficient N, by which the power amn is multiplied, will, in the general development, be that power of a denoted by n, or a”, repeated as many times, as we can form different products by taking in every possible way a number n of letters from among a number m; to find the coefficient of the term containing "-" then is reduced to finding the number of these products. 138. In order to perform the problem just mentioned, it is necessary to distinguish arrangements or permutations from products or combinations. Two letters, a and b, give only one product, but admit of two arrangements, a b and ba; three letters, a, b, c, which give only one product, admit of six arrangements (88), and so on. To take a particular case, I will suppose the whole number of letters to be nine, namely, a, b, c, d, e, f, g, h, i, and that it is required to arrange them in sets of seven. It is evident, that if we take any arrangement we please, of six of these letters, a b c d e f, for example, we may join successively to it each of the three remaining letters, g, h, and i; we shall then have three arrangements of seven letters, namely, abcdefg, abcdefh, abcdefi. What has been said of a particular arrangement of six letters, is cqually true of all; we conclude, therefore, that each arrangement of six letters will give three of seven, that is, as many as there remain letters, which are not employed. If, therefore, the number of arrangements of six letters be represented by P, we shall obtain the number consisting of seven letters by multiplying P by 3 or 9-6. Representing the numbers 9 and 7 by m and n, and regarding P as expressing the number of arrangements, which can be furnished by m letters, taken n— 1 at a time, the same reasoning may be employed; we shall thus have for the number of arrangements of n letters, P(m(n-1)), or P(m-n+1). |