over every second figure, both to the left hand in integers, and to the right in decimals. Always begin to point at the place of units; or, if the number to be extracted be entirely decimal, put a cipher in the unit's place, and over it put the first point. Find the greatest square in the first period on the left hand, and set its root on the right hand of the given number, after the manner of a quotient figure in Division. Subtract the square thus found from the same period, and to the remainder annex the two figures of the next following period, for a dividend. Double the root above mentioned for a divisor; and find how often it is contained in the said dividend, exclusive of its right-hand figure; and set that quotient figure both in the quotient and divisor. Multiply the whole augmented divisor by this last quotient figure, and subtract the product from the said dividend, bringing down to it the next period of the given number, for a new dividend. Repeat the same process over again, viz. find another new divisor, by doubling all the figures now found in the root; from which, and the last dividend, find the next figure of the root as before; and so on through all the periods, to the last. Note. The best way of doubling the root, to form the new divisors, is by adding the last figure always to the last divisor, as appears in the following examples. Also, after the figures belonging to the given number are all exhausted, the operation may be continued into decimals at pleasure, by adding any number of periods of ciphers, two in each period. is 2a+b, or double the first term increased by the second. And hence the manner of extraction is thus: Again, for a root of three parts, a, b, c, thus: (a+b+c)2 = a2 + 2ab + b2 + 2ặc + 2bc + c2 = a2 + (2a + b) b + (2a + 2b + c) c, the square of three terms, where a is the first term of the root, b the second, and c the third term; also a the first divisor, 2a+b the second, and 2a +26+c the third, each consisting of the double of the root increased by the next term of the same. In like manner (a+b+c+d)2=a2 +(2a+b) b+ (2a +2b+c) c + (2a + 2b + 2c + d) d; and so on, to whatever number of terms we proceed. And the mode of extraction agrees with the rule. See farther, Case 2, of Evolution in the Algebra. 4a2 + 3n For an approximation observe that √a2 + n = a. 4 u2 + n small in respect of a. nearly, in all cases where n is NOTE. When the root is to be extracted to many places of figures, the work may be considerably shortened, thus: Having proceeded in the extraction after the common method, till there be found half the required number of figures in the root, or one figure more; then, for the rest, divide the last remainder by its corresponding divisor, after the manner of the third contraction in division of decimals; thus: 2. To find the root of 2 to nine places of figures. The figures 1356, to the right of the vertical line, being obtained simply by FIRST prepare all vulgar fractions, by reducing them to their least terms, both for this and all other roots. Then 1. Take the root of the numerator and of the denominator for the respective terms of the root required; which is the best way if the denominator be a complete power: but if it be not, then 2. Multiply the numerator and denominator together; take the root of the product: this root being made the numerator to the denominator of the given fraction, or made the denominator to the numerator of it, will form the fractional root required. That is, √ √ This rule will serve, whether the root be finite or infinite: and sometimes one of these expressions will simplify the operation, sometimes another; as will be learnt from a little experience. 3. Or reduce the vulgar fraction to a decimal, and extract its root. This is generally the best method in practice when the terms of the fraction are not very low primes or very simple composite numbers. 4. Mixed numbers may be either reduced to improper fractions, and extracted by the first or second rule, or the vulgar fraction may be reduced to a decimal, then joined to the integer, and the root of the whole extracted. By means of the square root also may readily be found the 4th root, or the 8th root, or the 16th root, that is, the root of any power whose index is some power of the number 2; namely, by extracting so often the square root as is denoted by that power of 2; that is, two extractions for the 4th root, three for the 8th root, and so on. So, to find the 4th root of the number 21035'8, extract the square root two times in succession, as follows: 21035-8000 (145·037237 (12-0431407 the 4th root. 蒟物 4 285 5 29003 3 29006 108000 24083 75637 3 72249 24086 3388 687 980 17 20991 107 Ex. 2. What is the 8th root of 97·41 to four places of decimals? TO EXTRACT THE CUBE ROOT *. 1. DIVIDE the page into three columns, and call them L, M, N, in order from left to right, and so that м shall be double the breadth of L, and N double the breadth of M. At the head of N place the number whose root is to be extracted; This method was invented by the late Mr. W. G. Horner, and forms only one single and simple application of a universal method of extracting all roots whatever; and even his method, as applied to roots, is only a particular application of his general method of solving numerical equations of all orders. In the form here given it is rather more concise in the operation than as generally applied to equations of a higher order, and is put in this form for the use of those students who are not likely to proceed to the higher equations. It is recommended to those who intend to pursue the study of the subject to any extent, to adopt the form of work given in the equations in this volume, in preference to the present one, on account of the uniformity amongst the processes for all roots whatever. and mark off the place for the root as the quotient is marked in division, or the root in the extraction of the square root. Thus :— 2. Divide the number into periods of three figures each, by setting a point over the place of units, and also over every third figure from thence to the lefthand in whole numbers, and to the right in decimals. Find the nearest less cube number to the first (or left-hand) period, and having subtracted it therefrom, annex the next period to the remainder. Call this the resolvend. Also, set the root of the said cube in the place appropriated to the root. 3. In the column L put three times the root already found, and in м put three times its square. With this last number as a trial divisor of the resolvend, omitting the two figures to the right-hand, find the next figure of the root, and annex it to the former one, and also to the number in the column L. Multiply the number now in L by the new figure of the root, and place the product under that in м, but having its figures two places more to the right than the numbers already there. Add the two numbers together, and they will form the corrected divisor. 4. Multiply the corrected divisor by the root figure, and place it under the number in N, and subtract it therefrom t. Note. If the number last found should be greater than the number previously in N, the subtraction cannot be performed: the work dependent upon the root figure which produced it must, therefore, be cancelled, and a smaller root figure tried instead of it. If necessary, the same process must be again repeated, till a number is found which will admit of being subtracted according to the rule. 5. Write twice the last root figure under the number L, and the square of the last root figure under м. Add the two last written numbers in L together, and * The reason for pointing the given number into periods of three figures each, is, because the cube of one figure never amounts to more than three places; the cube of two figures to more than six, but always more than three; the cube of three figures never to more than nine, but always more than six; and so on to any extent. For a similar reason, a given number is pointed into periods of four figures, of five figures, of n figures, when the fourth, fifth, nth roots are to be extracted. A little attention to the composition of the algebraic expression for the cube of a binomial will render the truth of this rule very obvious. For (a + b)3=a3 + 3a2 + 3ab2 + b3=a3 + {3a2 +(3a + b) × b} b, the last form of which is exactly that whose composition is directed when B, C, &c. stand for other quantities to be afterwards brought down for continuing the pro cess. the three last written numbers in м*. These numbers will be respectively triple the new root and triple its square; and we may proceed with them to find a new root figure, and complete the operation as in rules 3 and 4. Note. After about one-third of the number of figures in the root have been obtained (or one-third of the number intended to be taken into account if the root be interminable) we may contract the work very considerably by the following method: Cut off one figure from м, and two from L, by vertical lines. Then with the new root figure work as in contracted multiplication, p. 57, keeping as the correction columns those immediately to the right of the vertical lines. We shall thus avoid all the work which does not contribute to the derivation of the figures of the root to the extent assigned. In the column м, the correction of the square of the last root figure falling to the right of the correction column is not put down, as in Ex. 2. Proceeding thus, we shall, in as many steps as we had taken before the contraction began, have cut off all the addends that would arise from L; and the process for obtaining the remaining third of the figures will be identical with that for Contracted Division, p. 59. Ex. 1. Extract the cube root of 4822854. Statement of the process. Here the number N being pointed as prescribed in the rule, the first period is 48, and the greatest integral cube contained in it is 27. Its root 3 is put in the root place, and the first resolvend is 21228. The trial divisor or number in м is 27, or three times the square of the root figure, and the number L is three times the root figure, or 9. The trial quotient, or integer part of 217, is 7. Annex this to L, which gives 97, and add 7 × 97 to м, the figures being carried two places to the right, which gives 3379, the product of which by 7 gives 23653. This cannot be subtracted from the resolvend: and hence the whole work dependent on 7 is to be obliterated, and a lower number, 6, tried. Performing the same operations with 6 as has been just described with 7, we get 96, 3276, and 19656 for L, M, and N, the last of which subtracted gives for the next resolvend 1572544. Adding 2 × 6 and 62 to the columns L and м respectively, taking in the 576 * That this method produces the triple of the new root in the column L, and the triple of its square in the column M, may be thus immediately shown : And we thus start from the last root a + b as we did at first from a. N (a + b. |