TO EXTRACT THE CUBE ROOT. i. DIVIDE the page into three columns (1), (11), (111), in order, from left to right, so that the breadth of the columns may increase in the same order. In column (111) write the given number, and divide it into periods of three figures* each, by putting a point over the place of units, and also over every third figure, from thence to the left, in whole numbers, and to the right in decimals. 2. Find the nearest less cube number to the first or left-hand period; set its root in column (111), separating it from the right of the given number by a curve line, and also in column (1); then multiply the number in (1) by the root figure, thus giving the square of the first root figure, and write the result in (11); multiply the number in (11) by the root figure, thus giving the cube of the first root figure, and write the result below the first or left-hand period in (III); subtract it therefrom, and annex the next period to the remainder for a dividend. 3. In (1) write the root figure below the former, and multiply the sum of these by the root figure; place the product in (11), and add the two numbers together for a trial divisor. Again, write the root figure in (1), and add it to the former sum. 4. With the number in (11) as a trial divisor of the dividend, omitting the two figures to the right of it, find the next figure of the root, and annex it to the former, and also to the number in (1). Multiply the number now in (1) by the new figure of the root, and write the product as it arises in (11), but extended two places of figures more to the right, and the sum of these two numbers will be the corrected divisor; then multiply the corrected divisor by the last root figure, placing the product as it arises below the dividend; subtract it therefrom, annex another period, and proceed precisely as described in (3), for correcting the columns (1) and (11). Then with the new trial divisor in (11), and the new dividend in (111), proceed as before.† Note I. When the trial divisor is not contained in the dividend, after two figures are omitted on the right, the next root figure is 0, and therefore one cipher must be annexed to the number in (1); two ciphers to the number in (11); and another period to the dividend in (11). * The number is divided into periods of three figures each, because the cube of one figure never amounts to more than three figures; the cube of two figures to more than six, but always more than three; and so on. For a similar reason, a number is divided into periods of n figures, when the nth root is to be extracted. The truth of this rule will be obvious from the composition of the algebraic expression for the cube of a binomial. Thus (a+b)3±a3 +3 a2 b+3 ab2+b3; then by the rule Note II. When the root is interminable, we may contract the work very considerably, after obtaining a few figures in the decimal part of the root, if we omit to annex another period to the remainder in (111); cut off one figure from the right of (11), and two figures from (1), which will evidently have the effect of cutting off three figures from each column; and then work with the numbers on the left, as in contracted multiplication and division of decimals. EXAMPLE. Find the cube root of 21035-8 to ten places of decimals. TO EXTRACT ANY ROOT WHATEVER.* Let N be the given power or number, n the index of the power, A the assumed power, r its root, R the required root of N. Then, as the sum of n + 1 times A and n — 1 times N, is to the sum of n + 1 times N and n 1 times A, so is the assumed root r, to the required root R. Or, as half the said sum of n + 1 times A and n 1 times N, is to the difference between the given and assumed powers, so is the assumed root r, to the difference between the true and assumed roots; which difference, added or subtracted, as the case requires, gives the true root nearly. That is, n+I A+ n − 1 . N : n + 1 .N+n 1. A: 7: R. Or, n + 1. A + n 1.N: ANr: Rr. And the operation may be repeated as often as we please, by using always the last found root for the assumed root, and its nth power for the assumed power A. EXAMPLE. To extract the 5th root of 21035.8. Here it appears that the 5th root is between 7-3 and 74. Taking 7.3, its 5th power is 20730-71593. Hence then we have, N21035-8; r = 7·3; n = 5; n+1=3; 2. n 1=2 3A 62192.148 This is a very general approximating rule for the extraction of any root of a given number, and is the best adapted for practice, and for memory, of any that I have yet seen. It was first discovered by myself, and the investigation and use of it were given at large in my Tracts, vol. 1, p. 45, &c. Ans. 1.104089. Ans. 3470323. Ans. 1090508. Ans. 3022239. Ans. 1080059. OF RATIOS, PROPORTIONS, AND PROGRESSIONS. NUMBERS are compared to each other in two different ways: the one comparison considers the difference of the two numbers, and is named Arithmetical Relation; and the difference sometimes the Arithmetical Ratio: the other considers their quotient, and is called Geometrical Relation, and the quotient the Geometrical Ratio. So, of these two numbers 6 and 3, the difference, or arithmetical ratio, is 6 — 3 or 3; but the geometrical ratio is § or 2. There must be two numbers to form a comparison: the number which is compared, being placed first, is called the Antecedent; and that to which it is compared, the Consequent. So, in the two numbers above, 6 is the antecedent, and 3 the consequent. If two or more couplets of numbers have equal ratios, or equal differences, the equality is named Proportion, and the terms of the ratios Proportionals. So, the two couplets, 4,2 and 8,6, are arithmetical proportionals, because 4 — 2 = 8 — 6 = 2; and the two couplets 4,2 and 6,3, are geometrical proportionals, because == 2, the same ratio. To denote numbers as being geometrically proportional, a colon is set between the terms of each couplet, to denote their ratio; and a double colon, or else a mark of equality, between the couplets or ratios. So, the four proportionals, 4, 2, 6, 3, are set thus, 4:2:: 6 : 3, which means, that 4 is to 2 as 6 is to 3; or thus, 4: 26: 3; or thus, , both which mean, that the ratio of 4 to 2, is equal to the ratio of 6 to 3. Proportion is distinguished into Continued and Discontinued. When the difference or ratio of the consequent of one couplet and the antecedent of the next couplet, is not the same as the common difference or ratio of the couplets, the proportion is discontinued. So, 4, 2, 8, 6, are in discontinued arithmetical proportion, because 4 — 2 — 8 — 6 = 2, whereas, 2—8——6; and 4, 2, 6, 8, are in discontinued geometrical proportion, because == 2, but } = }, which is not the same. But when the difference or ratio of every two succeeding terms is the same quantity, the proportion is said to be continued, and the numbers themselves a series of continued proportionals, or a progression. So, 2, 4, 6, 8, form an arithmetical progression, because 4 2=6 48 62, all the same common difference; and 2, 4, 8, 16, a geometrical progression, because = { == 2, all the same ratio. When the following terms of a Progression exceed each other, it is called an Ascending Progression or Series; but if the terms decrease, it is a Descending one. So, 0, 1, 2, 3, 4, &c., is an ascending arithmetical progression, 1, &c., is a descending arithmetical progression : 16, &c., is an ascending geometrical progression, 1, &c., is a descending geometrical progression. but 9, 7, 5, 3, ARITHMETICAL PROPORTION AND PROGRESSION. THE first and last terms of a Progression, are called the Extremes; and the other terms, lying between them, the Means. The most useful part of arithmetical proportions, is contained in the following theorems: E THEOREM 1.—If four quantities be in arithmetical proportion, the sum of the two extremes will be equal to the sum of the two means. Thus, of the four 2, 4, 6, 8, here 2 +84 + 6 = 10. THEOREM 2.-In any continued arithmetical progression, the sum of the two extremes, is equal to the sum of any two means that are equally distant from them, or equal to double the middle term when there is an uneven number of terms. Thus, in the terms 1, 3, 5, it is 1 + 5 = 3 +3= : 6. And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 + 14 = 4 + 12 = 6 + 10 = 8+ 8 = 16. THEOREM 3.-The difference between the extreme terms of an arithmetical progression, is equal to the common difference of the series multiplied by one less than the number of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, the common difference is 2, and one less than the number of terms 9; then the difference of the extremes is 20 — 2 = 18, and 2 × 9 = 18 also. Consequently, the greatest term is equal to the least term added to the product of the common difference multiplied by 1 less than the number of terms. THEOREM 4.-The sum of all the terms, of any arithmetical progression, is equal to the sum of the two extremes multiplied by the number of terms, and divided by 2; or the sum of the two extremes multiplied by the number of the terms gives double the sum of all the terms in the series. This is made evident by setting the terms of the series in an inverted order under the same series in a direct order, and adding the corresponding terms together in that order. Thus, the sums are 16+16 +16 + 16 + 16 + 16 + 16 + 16, which must be double the sum of the single series, and is equal to the sum of the extremes repeated so often as are the number of the terms. From these theorems may readily be found any one of these five parts; the two extremes, the number of terms, the common difference, and the sum of all the terms, when any three of them are given; as in the following Problems: PROB. I. Given the extremes, and the number of terms; to find the sum of all the terms. RULE. ADD the extremes together, multiply the sum by the number of terms and divide by 2. EXAMPLES. 1. The extremes being 3 and 19, and the number of terms 9; required the sum of the terms? 2. It is required to find the number of all the strokes a clock strikes in one whole revolution of the index, or in 12 hours? Ans. 78. |