4n', etc., we shall clearly find two successive numbers which may be expressed by qn' and qn' + n', such that p shall be numerically greater than qn', and less than qn' + n', so that P will have qn' and qn'+n' for its limits. Hence, (b+ h) must clearly have (b + h)în' and (b + h)¶n'+n' for its limits; consequently, since n' may be taken so small that either limit shall differ insensibly from the other, it follows that (b+ h) may be considered as coinciding with either of them; hence, because (b + h) and (b + h)n'+n' can be expanded into series of the same form as (b + h)" or (b + h)−", it is clear that (b + h)o can also be expanded in a similar way, when p is put for n orn. 4. If n stands for any real positive or negative number, it follows from what has been done that we shall have (b + h)" n(n − 1)n-22 + etc, in which b and h are 1 2 =b" + nbn 1h + clearly arbitrary. If for h we put - h, we shall get (b − h)" = b” — nb”—1h+ 1)-- etc., since the odd powers of h are nega n(n 1 2 x bn-sh3 + which is such that n(n 1 n(n 1 2 -1) (n-2) (n-3)¿»-^* ± etc., (a), 2 3 4 if the number of any term is repre sented by m + 1, the term itself will be expressed by ± n(n − 1)(n − 2) × 2) x .... x (n − m + 1)¿n-mm, (b), in which m 1. 2 3 X +must be used for when m is an even number, and - must be used when m is an odd number; thus, if m =2 the term becomes (n-1)-2, and if m = 3 it be n(n − 1) (n − 2)fn−373; consequently, all the terms 1 2 3 of (a) can be deduced from (b), which is called the general term of (a). If in bh we call the first or leading term, and h the second or following term, then it is clear that if we multiply any term of (a) by the index of the leading term in it, diminish the index by unity, multiply the result by the following term (or ±h), and divide by the index of the following term in the product, the result will equal the next successive term of (a); consequently, all the terms of (a) can clearly be deduced from b", taken as the first term of the expansion in (a); and it may be added, that the method here given for finding the successive terms of (a) is substantially the same as what is called Newton's Binomial Theorem. EXAMPLES. 1. To find the third power or cube of a ±b. Putting a for b, and b for h, and 3 for n, in (a), we get a3¬3b3 = a3 ± 3a2b+3ab2±b3; noticing that a-3 = ao = 1, and that all the following terms of the series are reduced to 0, because they have 330 for a factor. 2. To expand (a + b) and (x − y)o. Here, we have (a + b) = a + 6a3b + 15a1b2 + 20a3b3 + 15a2b1 + 6ab3 + bo, and (x − y)3 = x3 — 5x1y + 10x3y2 — 10x2y3 + 5xy1 — y3. 3. To expand (3a + 4b)3, (5a — 2b)*. Here, putting 3a for b, and 4b for ±h, in (a), and using 3 for n, we get (3a + 4b)3 = (3a)3 + 3 × (3a)2 × 4b + 3 × (3a) x (4b) + (46)3 = 27a3 + 108ab + 144ab2 + 646; also × using 5a for b, 2b for h, and 4 for n, in (a), we have (5a — 2b)* = (5a)1 — 4 × (5a)3 × 2b + 6 × (5a)2 × (2b)2 — 4 × 5a × (2b)3 + (2b)* = 625a1 1000a3b + 600a2b2 +1661. 160ab3 Remarks.-If the coefficients in the first of these expansions are added, their sum will be found to equal 343 = 73; and the coefficients in the second expansion, when added according to their signs, will be found to equal 81 = 34; results which are clearly as they ought to be, as is evident by putting 1 for a and 1 for b. 4. To expand (a ‡ b)−1 and (2 + y)−2. Here, using a for b, ±b for ±h, and 1 for n, in (a), we get (a + b)-1 = a¬1‡ a−2b + a¬3b2 = a−1b3 + etc. = 1 b 7,2 a3 F + etc., which is clearly the same result that we shall a1 get from the division of 1 by a ±b, as it evidently ought to be; and in a similar way we get (x + y) is also the same result that the division of 1 by (x + y)2 + 2xy + y2 will give. = x2 (5.) We will now apply (a) to the expansion of any power of a polynomial expression. Let a + bx + cx2 + dx3 + etc., represent a polynomial, a being its first term, b the coefficient of x in its second term, c that of in its third term, and so on, according to the order of the letters in the alphabet; then the polynomial may clearly be supposed to be derived from a by changing a into a+bx, b into b + cx, c into c + dx, and so on. If we put a for b and bx + cx2 + etc., for ±h in (a), and represent the index of the power by n, we shall have (a + be + cx2 + etc.)" n(n − 1) a"+na"-1(bx + cx2 + etc.) + an-2 (bx + cx2 + etc.)3 + etc. = a" + nan-1x(b + cx + etc.) + etc., (c); and it is clear that the powers of b + cx + etc., in (e) admit of similar expansions, and so on. If the terms in (c) are arranged according to the ascending powers of a, and we express the coefficients of the powers of by using the vertical bar, we shall get (a + bx + cx2 If the terms of the polynomial are arranged according to the descending powers of x, or if the coefficients are detached, or 1 is put for a (provided any deficient terms are supplied by terms whose coefficients equal 0), it is clear that the expansion may be obtained as in (d); observing that the coefficients. of the powers of x, or the terms of the given polynomial, are supposed to be represented by the letters a, b, c, d, etc., taken alphabetically. If we have any coefficient in the expansion in (d), and suppose the letters of each of its terms to be written in alphabetical order, it is clear, from (c), that we can find the next successive coefficient by the following RULE. 1. When the last letter of any term of the given coefficient is not the last coefficient in the given polynomial. Multiply the term by the index of the last letter in the term, and subtract unity from the same exponent (of the letter) in the product, and multiply the result by the next (successive) letter of the alphabet; then the product will be a term of the required coefficient. 2. When the last two letters of any term of the given coefficient are successive letters of the alphabet. Find the term of the sought coefficient, which results from the last letter of the term, as in 1; then multiply the given term by the index of the last letter but one in the term, subtract one from the index of the same letter, and add one to the index of the following or last letter in the term; then the result, when divided by the increased index of the last letter (of the term), will be a (new) term of the sought coefficient. 3. All the terms which can be obtained by (1,) and (2,) from the given coefficient, when duly ordered and connected by their proper signs, will constitute the required coefficient, because the other letters which may enter the term of the given coefficient must (according to (c)) be considered as not giving any terms in the required coefficient. 4. In the case of (1), when the last letter is not raised to any power, it will be sufficient to change it into the next successive letter in the alphabet, in order to get the sought term. 5. If the last letter in any term of the given coefficient is the last coefficient in the given polynomial, it clearly can not give any term in the required coefficient, since the successive letters in the polynomial equal 0. Hence, we perceive how we may proceed in order to deduce the expansion in (d) from its first term, a". Remark. Our rule is substantially the same as that given by Arbogast, at p. 25 of his work entitled "Du Calcul. Des Derivations." EXAMPLES. Ex. 1.-To expand (1 + 2x — 3x2)3, according to the ascending powers of x. For 3ab, the coefficient of x, has been obtained from a3 regarded as the coefficient of ao 1, by (1) of the rule; the term 3ab2 of the coefficient of 2 has been obtained from 3ab by multiplying by 2 the index of a and subtracting 1 from the index, and multiplying the result by b, and dividing the product by 2, the increased index of b, according to (2) of the rule, and 3a2c, the remaining term of the coefficient of has been found from 3ab by changing b into c, according to (4); also, b3, the first term of the coefficient of a, has been found from 3ab2 of the coefficient of x, by changing a into b, and dividing by 3 the increased index of b agreeably to (2).; and the remaining term, 6abc, of the coefficient of a has been obtained from 3ab2 by (1); noticing that the term 3a2c of the coefficient of x2 can not give any term in the coefficient of a3, since the letters a and c are not successive letters of the alphabet, and that c is the last coefficient in a + bx + cx2, agreeably to (3) and (4), and so on. Restoring the values of the letters, we shall have (1 + 2x -3x2)=1+6x+3x2 - 28x3-9x+54x27x, which is clearly correct, since the sum of the coefficients in 1+2x |