powers of a in the following manner; a" =√ā ; an¬1 = an a 323. This being laid down, the square root of a + b, may be expressed in the following manner : 324. If a, therefore, be a square number, we may assign the value of, and, consequently, the square root of a+b may be expressed by an infinite series, without any radical sign. Let, for example, acc, we shall have 1 b 1bb 1 b3 5 √ (cc + b) = c + X + X 8 c3 C a = c; then We see, therefore, that there is no number, whose square root we may not extract in the same way; since every number may be resolved into two parts, one of which is a square represented by cc. If we require, for example, the square root of 6, we make 64 +2, consequently cc 4, c=2, b = 2, whence results 62 +1-78 +84-18249 &C. 5 If we take only the two leading terms of this series, we shall have 24, the square of which, 25, is greater than 6; but if we consider three terms, we have 27, the square of which, 1521, is still too small. 256 325. Since, in this example, 1⁄2 approaches very nearly to the true value of 6, we shall take for 6 the equivalent quantity 2. Thus c c = 25; c=5; b=-4; and calculating only cc= the two leading terms, we find √6=+1×=-1× 5 3401 the square of this fraction, being 240, exceeds the square of 6 only by . Now, making 62401-, so that c= = = 13 and b = — and still taking only the two leading terms, we have 326. In the same manner, we may express the cube root cf 3 a+b by an infinite series. For since (a + b) = (a + b)3, we shall have in the general formula n, and for the coeffi &c., and with regard to the powers of a, we shall have an = √ā; an~!= 1 3 3 327. If a therefore be a cube, or a=c3, we have ✔ā = c, and the radical signs will vanish; for we shall have 328. We have, therefore, arrived at a formula, which will enable us to find by approximation, as it is called, the cube root of any number; since every number may be resolved into two parts, as c3+b, the first of which is a cube. If we wish, for example, to determine the cube root of 2, we represent 2 by 1+1, so that c = 1 and b = 1, consequently 3 1 1 √2 = 1 + } − +1, &c., the two leading terms of this series make 1 the cube of which, 4, is too great by 19. Let us then make 2 = 3 10 19, we have c = ÷ and b = — 17? 27 10 These two terms give 1,6 and consequently v+x 746496 3732489 753571 , the cube of which is 3431}. Now, 2 = }}}}}}, 373348 so that the error is 7. In this way we might still approx 7076 373248 imate, and the faster in proportion as we take a greater number of terms. CHAPTER XIII. Of the resolution of Negative Powers. 329. Wɛ have already shewn, that we may express by a→1; 1 α we may therefore also express by (a+b); so that the 1 a+b fraction may be considered as a power of a + b, namely, a+b that power whose exponent is -1; and from this it follows, that the series already found as the value of (a + b)" extends also to this case. a+b 330. Since, therefore, is the same as (a+b)-1, let us suppose, in the general formula, n=-1; and we shall first 1, &c. Then, for the powers of a; a" = a¬1 = reduce this quantity also to an infinite series. For this purpose, we must suppose n = — 2, and we shall first have for the coeffi 333. The different cases that have been considered enable us to conclude, with certainty, that we shall have, generally, for any negative power of a+b; And by means of this formula, we may transform all such fractions into infinite series, substituting fractions also, or fractional exponents, for m, in order to express irrational-quantities. 334. The following considerations will illustrate this subject further. If, therefore, we multiply this series by a + b, the product ought to be 1; and this is found to be true, as we shall see by performing the multiplication: 1 b b2 b3 b4 b5 a + 03 + a4 +, &c. ought also to be = 1. the operation: 2b 3bb 463 564 + a4 + ασ If, therefore, we multiply this series by (a+b), the product Now (a+b)=aa+2ab+bb. See |