CONTRACTED METHOD. 658. 1. Find the square root of 8, correct to 6 decimal places. OPERATION. 12.828427+, Ans. 4 48. 400 384 562 1600 1124 5648 47600 45184 5656 2416* 2262 566 154 ANALYSIS. Extracting the square root in the usual way until we have obtained the 4 places, 2.828, the corresponding remainder is 2416, and the next trial divisor, with the cipher omitted, is 5656. We now omit to bring down a period of ciphers to the remainder, thus contracting the dividend 2 places; and we contract the divisor an equal number of places by omitting to annex the trial figure of the root, and regarding the right hand figure, 6, as a rejected or redundant figure. We now divide as in contracted division of decimals, (226), bringing down each divisor in its place, with one redundant figure increased by 1 when the rejected figure is 5 or more, and carrying the tens from the redundant figure in multiplication. We observe that the entire root, 2.828427+, contains as many places as there are places in the periods used. Hence the following RULE. I. If necessary, annex periods of ciphers to the given number, and assume as many figures as there are places required in the root; then proceed in the usual manner until all the assumed figures have been employed, omitting the remaining figures, if any. II. Form the next trial divisor as usual, but omit to annex to it the trial figure of the root, reject one figure from the right to form each subsequent divisor, and in multiplying regard the right hand figure of each contracted divisor as redundant. NOTES.-1. If the rejected figure is 5 or more, increase the next left hand figure by 1. 2. Always take full periods, both of decimals and integers. EXAMPLES FOR PRACTICE. 1. Find the square root of 32 correct to the seventh decimal Ans. 5.6568542+. place. 2. Find the square root of 12 correct to the seventh decimal place. Ans. 3.4641016+. 3. Find the square root of 3286.9835 correct to the fourth decimal place. Ans. 57.3322+. 4 Find the square root of .5 correct to the sixth decimal place. Ans. .745355+. 5. Find the square root of 64 correct to the sixth decimal place. Ans. 2.563479+. 6. Find the square root of 1.065 correct to the sixth decimal place. Ans. 1.156817+. 7. Find the value of 1.01253 correct to the fourth decimal place. Ans. 1.0188+. 8. Find the value of 1.023375 correct to the sixth decimal place. Ans. 1.011620±. CUBE ROOT. 659. The Cube Root of a number is one of the three equal factors that produce the number. Thus, the cube root of 343 is 7, since 7 x 7 x 7 = 343. To derive the method of extracting the cube root of a number, it is necessary to determine 1st. The relative number of places in a given number and its cube root. 2d. The relations of the figures of the root to the periods of the number. 3d. The law by which the parts of a number are combined in the formation of a cube; and 4th. The factors of these combinations. 660. The relative number of places in a given number and its cube, is shown in the following illustrations: 1st. That a root consisting of 1 place may have from 1 to 3 places in the cube. 2d. That in all cases the addition of 1 place to the root adds 3 places to the cube. Hence, I. If we point off a number into three-figure periods, commencing at the right hand, the number of full periods and the left hand full or partial period will indicate the number of places in the cube root. To ascertain the relations of the several figures of the root to the periods of the number, observe that if any number, as 5423, be decomposed, the cubes of the parts will be related in local value, as follows: 50003 125 000 000 000 157 464 000 000 = 159 220 088 000 II. The cube of the first figure of the root is contained wholly in the first period of the power; the cube of the first two figures of the root is contained wholly in the first two periods of the power; and so on To learn the combinations of tens and units in the formation of a cube, take any number consisting of two figures, as 54, and decompose it into two parts, 50+4; then having formed the square by 656, III, multiply each part of this square by the units and tens of 54 separately, thus, Of these combinations, the first is the cube of 50, the second is 3 times the square of 50 multiplied by 4, the third is 3 times 50 multiplied by the square of 4, and the fourth is the cube of 4. Hence, III. The cube of a number composed of tens and units is equal to the cube of the tens, plus three times the square of the tens multiplied by the units, plus three times the tens multiplied by the square of the units, plus the cube of the units. By observing the manner in which the cube is formed, we perceive that each of the last three parts contains the units as a factor; these parts, considered as one number, may therefore be separated into two factors, thus, (3 × 502 + 3 × 50 × 4 + 42) × 4. Hence, IV. If the cube of the tens be subtracted from the entire cube, the remainder will be composed of two factors, one of which will be three times the square of the tens plus three times the tens multiplied by the units plus the square of the units; and the other, the units. 1. What is the cube root of 145780726447 ? ANALYSIS. Pointing off the given number into periods of 3 figures each, the four periods show that there will be four figures in the root, (I). Since the cube of the first figure of the root is contained wholly in the first period of the power, (II), we seek the greatest cube in the first period, 145, which we find by trial to be 125, and we place its root, 5, for the first figure of the required root, and regard it as tens of the next inferior order, (654). We now subtract 125, the cube of this figure, from the first period, 145, and bringing down the next period, obtain 20780 for a dividend. And since the cube of the first two figures of the root is contained wholly in the first two periods of the power, (II), the dividend, 20780, must contain at least the product of the two factors, one of which is three times the square of the first figure (tens), plus three times the first figure multiplied by the second (units), plus the square of the second; and the other, the second figure (IV). Now if we could divide this dividend by the first of these factors, the quotient would be the other factor, or the second figure of the root. But as the first factor is composed in part of the second figure, which we have not yet found, we can not now obtain the complete divisor; and we therefore write three times the square of the first figure, regarded as tens, or 502 × 3 = 7500, at the left of the dividend, for a trial divisor. Dividing the dividend by the trial divisor, we obtain 2 for the second, or trial figure of the root. To complete the divisor, we must add to the trial divisor, as a correction, three times the tens of the root already found multiplied by the units, plus the square of the units, (IV). But as 50 × 3 × 2 + 22 = (50 × 3+2) × 2, we annex the second figure, 2, to three times the first figure, 5, and thus obtain 50 × 3 + 2 = 152, the first factor of the correction, which we write in the column marked I. Multiplying this result by the 2, we have 304, the correction, which we write in the column marked II. Adding the correction to the trial divisor, we obtain 7804, the complete divisor. Multiplying the complete divisor by the trial figure of the root, subtracting the product from the dividend, and bringing down the next period, we have 5172726 for a dividend. We have now taken the cube of the first two figures of the root considered as tens of the next inferior order, from the first three periods of the number; and since the cube of the first three figures of the root is contained wholly in the first three periods of the power, (II), the dividend, 5172726 must contain at least the product of the two factors, one of which is three times the square of the first two figures of the root (regarded as tens of the next order) plus three times the first two figures multiplied by the third, plus the square of the third; and the other, the third figure, (IV). Therefore, to obtain the third figure, we must use for a trial divisor three times the square of the first two figures, 52, considered as tens. And we observe that the significant part of this new trial divisor may be obtained by adding the last complete divisor, the last correction, and the square of the last figure of the root, thus: This number is obtained in the operation without re-writing the parts, by adding the square of the second root figure mentally, and combining units of like order, thus: 4, 4, and 4 are 12, and we write the unit figure, 2, in the new trial divisor; then 1 to carry and 0 is 1; then 3 and 8 are 11, etc. Annexing two ciphers to the 8112, because 52 is regarded as tens of the next order, and dividing by this new trial divisor, 811200, we obtain 6, the third figure in the root. To complete the second trial divisor, after the manner of completing the first, we should annex the third figure of the root, 6, to three times the former figures, 52, for the first factor of the correction. |