3a2 is 0, and that the sum of the coefficients in the expansion is also 0. Ex. 2.-To expand (2x2 + 3x − 3) according to the descending powers of x. c for 2, 3, and — 3, we have (ax2 + Here, using a, b, and b4 x+12ab cx+ 6a2c2 46°c 12abc2 x2 + 4bc3x + c1, as required. Restoring the values of the letters, we get (22+3x-3)* = 16x96x + 120x - 216.5 351x+324x3 + 270x2 324x+81, which is clearly correct, since the sum of the coefficients (in the expansion) is 16, which is the fourth power of 2, the sum of the coefficients in 2x2 + 3x — 3. Ex. 3.-To expand (a + b + c + d)2. b9 2bc 2ac Here we have (a + b + c + d)2 = a2 + 2ab + + c2 2bd + 2cd + d2. 2ad + Ex. 4. To expand (ax + bxm-1 + cam-2 + dam-8 + etc.)", according to the descending powers of x. Because anamn is clearly the first term of the expansion, we easily get (axm + bxm-1 + cxm-2 + etc.)" = an xmn + -2 n(n−1)(n-2) an-sf3 xmn-2+ 1 2 3 X Remark.-If we put m = 8, a=813, b=216, c=0, d=336, e=56, f=224, g= 0, h = 64, i=16, k=0, 7=0, and so on; then, if we put n = our formula be 0.∞2 + 64x + 16)2 = ± (3x2 — 2x − 2) = the fourth root of 114 the first member of the equation, since the index clearly indicates that its fourth root is to be taken; and since the root is of an even degree, it may have either of the signs + or which is indicated by writing the ambiguous sign ± before the root. Thus, since am is represented by 3', ann, in the general expansion, is represented by 34× × × =3, the first term of the root; and the term na"-1bæmn−1, in the general expansion, becomes 1 (34)→1 × (— 216),3× ¦— 4 2x; and because c = 0, the third term of the general expansion becomes × (216) = 3 3 X x 2162, the remaining term of 32 812 the root; and it is easy to perceive that the root is exact. SECTION XI. EVOLUTION. (1.) EVOLUTION, or the extraction of roots, is the reverse of Involution, and consists in finding the root of any number or quantity regarded as a power of the root. Thus, since A× A× A × .... x A to n factors = A",. the n power of A, it follows that A is the n" root of A"; consequently, in order to get the nth root of a number or quantity, we must resolve (or separate) it into n equal factors, and take one of the factors for the root. (2.) According to 10 of Sec. I., VA" is used to signify the nth root of A", being called the surd sign, and ʼn written above it is called its index or exponent, since it indicates what root of the number or quantity under the surd is to be extracted. Because VA A, it follows that the extraction of the nth = n root of A" is equivalent to freeing A of its index n, or consists in reducing the index to unity; consequently, since A== = A1 = A, we may also express the n" root of A” by A”, or by (A")", supposing the meaning to be, that the index of 1 A is to be multiplied by; hence, VA and (A”)" are equiv n alent expressions, and so are (ab) and (ab), since ab is to be considered and treated as being an exact square; and in like manner ◊ (a + b — c)" and [(a + b — c)"]}, c)] are equivalent expressions, which signify the n" root of the m" power of a+bc or the mth power of the n' root of a + b clear by putting a + b − c = x2 or x = (a + b − c)" — the nth root of a + b — c. C, as is = (3.) If we represent the nth root of A" by x, we shall have "A" or "A"= 0, (1), and according to what has heretofore been shown (regarding a as the independent variable), we shall have non-1 = 0 or x2-1= 0, (2), for the first derived equation of (1); where it may be noticed that (1) is called a binomial equation, since its first member consists of only two terms. Since the first member of (2) is the (n - 1)" power of x, it clearly has n 1 roots, whose common value is 0; consequently, (1) can not have more than one positive nor more than one negative root; for if it can, then 0, a root of (2), must lie between every successive two of them, which is impossible. Hence, and from what has heretofore been shown, if n is an odd number, it results that (1) can have only one real root, whose sign is contrary to that of A", or the same as that of A"; consequently (1) must have n-1 impossible or imaginary roots. Similarly, if n is an even number and A" positive, (1) has only two real roots, one of which is positive and the other negative; consequently, the remaining n 2 roots must be imaginary; also, n still being an even number, if A" is negative; then, since (1) can not have an even number of positive or negative roots, it follows that its roots must all be imaginary. - = = (4.) If n = pq, (1) becomes - Ap=0 or P = API, (3); which, by extracting the 9th root, gives a Ao, and thence by extracting the p" root we get a A; and if we first extract the pth root, and then the qth root of the result, we shall still have x = A; the same result that we shall get from the extraction of the pq" root of the members of (3). Hence, when the index of the root is composite or consists of factors, we may clearly extract the root denoted by any one of the factors, and then taking any one of the remaining factors as an index, we may extract the root of the root found, and so on until all the factors have been used; then the final root will be the same that we shall get from the extraction of the root denoted by the given index. Thus, for 16 we may write √√√16 = 2 = the fourth root of 16; and for 64 we may write 64 182 or √ √64 = √4: =2 the sixth root of 64. (5.) It is clear that A, which expresses the nth root of A, p ра may be written in the form A = An+p × A(n+p)2 × A(n+p)3 × etc., (1); for by adding the indices of A, in the right mem ber of the equation, their sum becomes 1 + + n + p (n + p)2 so that the members of (1) 1 3 Α' If n = 3 and p = 1, (1) becomes A = × Α, x etc., consequently the extraction of the cube root can be reduced to that of fourth roots, and of course to that of square roots, and their product continued indefinitely; thus, if A = 2, we get (by using eight factors in the preceding product) 23 = 1.2599 = the cube root of 2, correct to four decimals. It is evident that p, in (1), may be positive or negative, provided that the indices of A are sensibly less than unity. It is clear, from what has been done, that the extraction of the nth root may be reduced to the extraction of square roots and their products. 1 (6.) Supposing to be an exact power of r (r being rational, at least in one of its letters), and that is not the product of rational factors, then if for r we put r+y, we shall get (r+y)" = p2 + npr−ly + etc.; consequently, if we find the greatest common divisor of " and n-1 (the coefficient of y, which is the first derived function of "), we shall get -1, and then if we divide " by p-1, we shall get r, the root of pr a3 EXAMPLES. Ex. 1.-To find the square roots of a2+2ab + b2 and 2ab+ b2. Putting ay for a in these expressions, we get 2(a + b) and 2(ab) for the coefficients of y; consequently, we shall get a + b and ab for the square roots, which may clearly be taken with the sign+ or -. Ex. 2.-To find the square root of a1-2a3x+3a2x2 - 2ax3 + x1. Putting ay for a, we get 4a3 - 6a2x + 6ax2 - 2x3 for the coefficient of y; also, putting x + y for æ, we get 46ax2+6a2x - 2a. Because the sought root is clearly the greatest common divisor of these coefficients, we add half the second taken in a reverse order to the first, and thence 6a2x + 6ax2- 2x3-a3 +3a2x · get 4a3 3αx2 + 2x3 = 3a3 — 3a2x+3ax2=3a(a2-ax+x2); consequently, ±(a2—ax+x2) equals the required root. Ex. 3.-To extract the cube root of a +15a2 6a+ 1. - 6a+ 15a1 — 20a3 - Putting ay for a, we get 6(a5a+10a3- 10a2 + 5a -1) for the coefficient of y; consequently, dividing the given expression by a 5a+10a3 10a2 + 5a − 1, we get a 1 for the quotient, which is clearly the sixth root of the given expression, and of course (a — 1)2 = a2 — 2a + 1 must be the cube root, as required. Ex. 4.-To find the square root of a 2x z + x2 - 2x22 + 4xz 2x + 22 2z+ 1. xy Putting y for x, we get 2xz2-4x2 + 2x-222 + 422=2(xz2 2xz+x 2222-1) for the coefficient of y; consequently, dividing the given expression by xz-2xz+x z2 + 2z −1, we get x 1 for one of the factors of the root; and putting z+y for 2, by proceeding as before, we |