(1 - 4) ( − 2) 2 3 a; consequently, subtracting this from 1 + x, and dividing the remainder (in a3) by n, we shall get 1 for the fourth term of the root. 1 Hence, we shall clearly have 1 + x + n + etc. for the nth root of 1 + x, which I is the same that the expansion of (1 + 2)", according to the ascending powers of a by the binomial theorem, will give; consequently, since the root of 1 + x can more readily be found by the binomial theorem than by our rule, it ought to be preferred in practice. 3d. To perceive the use of the binomial theorem in the extraction of the roots of expressions of the form 1 + x, take the following EXAMPLES. Ex. 1.-To extract the square root of 38. 1 Since +35=6(1+1)*, if in (1+x)2=1+ 1x+ 0.00038580+ 0.00001071 0.00000037 + etc.) 6.1644138 +, which is correct to seven decimal places. Ex. 2.-To extract the cube root of 124-125-1=125 × etc.) = 4.9866313, which is correct to seven decimal places. Thus, we have in this question found the value of 124 more readily than that of 38 in the preceding question, is much smaller than in comparison to 1 or because 1 125 1 18 unity, so that the terms of the series diminish much more rapidly in this than in the preceding question. Ex. 3. To extract the fourth root of 3. Since 3 =1+2 =(1+ =(1 1). (1 + 1). (1 + 1 ) ( 1 + 1). (1 + 1). (1+1), by extracting the fourth root we have +5 = (1 + 1)2. (1 + 2)2. (1 + 3)2 (1 + 2)2 · (1 + 2)2 (1 + 2)2. Hence, extracting the roots of the binomials in the right member of the equation by the binomial theorem to six decimal places, and then taking their products, we get 1/3 = 1.31607 correct to five decimal places. Remarks.-1st. We have given the preceding question for the purpose of showing how the binomial theorem may be applied to the extraction of any root of a number. For since any whole number can clearly be resolved into binomial factors whose first terms are 1, and second terms are very small in comparison to 1, it is evident that, by extracting the roots of the factors, and then taking their product, we shall get the root of the proposed number. Thus, since 20 = 16 + 4 =16(1 + 1) = (1 + 1)(1 + 1-1) ==> (1 + 1). (1 + 4). (1 + 3), we shall get 20 = (1 + 5 n I I n\5 (1 + 2)2. (1 + 2)2 = (1 + 3) *)': (1 + 5 1+ for an expression for the nth root of 20; consequently, if we put particular numbers for n, by finding the values of 14 (1 + 1)^", (1 + 2)2, and (1 + 1) *r (1 + 1)* by the binomial theorem, we shall readily get the roots of 20, which correspond to the particular values of n. 2d. It may not be improper to give in this place the following method of approximating to the root of a number of the form 1-9 in which is very small in comparison to 1. b α b a Thus, supposing n to denote the root to be extracted, we n b b shall, by the binomial theorem, have V1+=1+ + + etc., (2), which is of the form Ana + A1b + A2 na + etc., (3), such that the coefficient of is given by the equation A+ A-1(1) + 2 -m-: (1 + An m-3 n) (1 2n) (1 + 4 .... + A,((1 — n) (1 -1))) 3 2 (1 — n) (1 — 2n) (1 —- 3n) × ......... × (1 — (m − 1)n) 2 (1-n)(1 - 2n)(1 − 3n) × .... × (1 — mn) = 0, (4); then b (1) will be reduced to 1 + 1+ (5), since b divided b by the value of A will make the right member of (5) identi cal with that of (1). b a It is clear that we may put 1 for a in (2) and (5), provided is put for b. Because n is supposed to be a whole number not less than 2, if a and b have like signs, and the terms of (2) decrease in a high ratio from left to right, it is clear that if we take an odd number of the terms of (2) for A, the root as found from (5) will be too great, while an even number of the terms of (2) being taken for A will make the root too small. b It may be noticed if 1 + exceeds 1 by a very small num 1 α ber, that by putting for n, in (2) and (5), we shall clearly get n an approximate value of the nth power b of 1 + α To perceive the use of what has been done, take the fol Putting a = 1, b = 1, and n = 2, the first three terms of 2 = = (1.41) x 1.00598561 +, we of course 1.41 must be the first three figures of the root. Since 2 (1.41)2 x (1.41)2 get 1/2 1.41 1.00598561 + = 1.41 × 1.00298833 + = 1.4142135+, correct to seven decimals; noticing that V1.00598561 += 1.00298833 + has been found by putting a = 1, b = 0.00598561 +, and n = 2, in (2), etc. 2. To extract the thirteenth root of 1.02. Putting a = 1, b = 0.02, and n = 13, (2) gives A = 13 + 0.12-0.0004307 + etc.; consequently, the thirteenth root of 1.02 has 1 + 2 1312 =1.0015244 + for limits, and of course we have 1.0015243+ for the root, correct to seven decimals. 3. To find the first five figures of the seventh root of 1.5. which it is clear that we must have (1.1)2 = 1.21. 5. To find the twentieth power of 1.02 to four decimal 1 133 + 1 0.02, and n = (2) gives A = 20' + etc.; consequently, 19 133 20 2000 200000 20000000 using the sums of three and then of four terms of this series, of the power, 1.4858 must be the required figures. It is evident that by carrying the series (2) sufficiently far we may find the power to any required degree of exactness. 3d. It is clear that we can apply our rule, or the Binomial Theorem, to the extraction of the roots of series; but they can clearly be more readily found by the rule given for finding the powers of binomials. To understand the applications of the rule to the extraction of the roots of series, take the following EXAMPLES. 1. To extract the square root of the series 1 + 2x + 3x2 + 4+ 5+ etc. According to the rule, we shall have (a + bx + cx2 + |