numbers comprised between 1 and 10; for the na power of 9 is the largest perfect power which can be expressed by n figures. When 2V contains more than n figures, there will be more than one figure in the root, which may then be considered as composed of tens and units. Designating the tens by a, and the units by b, we have (Art. 204), n—1 iV=(a+6)"=a"+na"-,6+n——-a"-262+, &c.; that is, the proposed number contains the n''1 power of the tens, plus n limes the product of the n—1 power of the tens by the units, plus a series of other parts which it is not necessary to consider. Now, as the n'h power of the tens cannot give units of an order inferior to unity followed by n ciphers, the last n figures on the right, cannot make a part of it. They must then be pointed off, and the root of the greatest n'h power contained in the figures on the left should be extracted; this root will be the tens of the required root. If this part on the left should contain more than n figures, the n figures on the right of it, must be separated from the rest, and the root of the greatest n'h power contained in the part on the left extracted, and so on. Hence the following RULE. I. Divide the number N into periods of n figures each, beginning at the right hand ; extract the root of the greatest n"' power contained in the left hand period, and subtract the nth power of this figure from the left hand period. II. Bring down to the right of the remainder corresponding to the first period, the first figure of the second period, and call this number the dividend. HI. Form the n—1 power of thefirst figure of the root, multiply it by n, and see how often the product is contained in the dividend: the quotient will be the second figure of the root, or something greater. IV. Raise the number thus formed to the n'h power, then subtract this result from the two first periods, and to the new remainder bring down the first figure of the third period: then divide the number thus formed, by n times the n—1 power of the two figures of the root already found, and continue this operation until all the periods are brought down. Extract the 4th root of 531441. 53.1441 | 27 24= 16 4x23=32 | 371 (27)4= 531441. We first divide off, from the right hand, the period of four figures, and then find the greatest fourth root contained in 53, the first period to the left, which is 2. We next subtract the 4th power of 2, which is 16, from 53, and to the remainder 37 we bring down the first figure of the next period. We then divide 371 by 4 times the cube of 2, which gives 11 for a quotient: but this we know is too large. By trying the numbers 9 and 8, we find them also too large: then trying 7, we find the exact root to be 27. 214. Remark. When the degree of the root to be extracted is a multiple of two or more numbers, as 4, 6, . . . ., the root can be ob. tained by extracting the roots of more simple degrees, successively. To explain this, we will remark that, (tfy^a3 X a3 X o3 X a3=a3+3+3+3=a3 * 4=ai2. and that in general (a*)*=ii"Xfl"Xo"'Xa" • • - =am*n (Art. 13). Hence, the nth power of the m,h power of a number, is equal to the mntb power of this number. Reciprocally, the mnth root qf a number is equal to the nth root of the mth root of this number, or algebraically there will result... ^/~a~a'n; for from the definition of a root, we EXAMPLES. V 2985984 = S/Y 2985984 = Vl728=12; V 1771561 =\/ V 1771561 =11; V 1679616 = Vl296 = \// Vl296=6. Remark. Although the successive roots may be extracted in any order whatever, it is better to extract the roots of the lowest degree first, for then the extraction of the roots of the higher degrees, which is a more complicated operation, is effected upon numbers containing fewer figures than the proposed number. Extraction of Roots by approximation. 215. When it is required to extract the n'h root of a number which is not a perfect power, the method of (Art. 213), will give only the entire part of the root, or the root to within unity. As to the fraction which is to be added, in order to complete the root, it cannot be obtained exactly, but we can approximate as near as we please to the required root. Let it be required to extract the n'h root of the whole number a, to within a fraction —; that is, so near it, that the error shall be p less than —. p We will observe that a can be put under the form ———. If denote the root of ap" to within unity, by r, the number r* (r+l)n or a, will be comprehended between — and —; there p* p" foro the Va~ will be comprised between the two numbers, r r+1 r — and . Hence — is the required root, to within the P P p H fraction —. p Hence, to extract the na root of a whole number to within a fraction —, multiply the number by p"; extract the n'h root of the product to within unity, and divide the result by p. 216. Again, suppose it is required to extract the n'h root of the fraction -7-. o Multiply each term of the fraction by b"~'; it becomes ———r~ Let r denote the na root of ab"~x, to within unity; a6"-' a if , (r+lf —— or —, will be comprised between — and ;— 6" b r b" b" Therefore, after having made the denominator of the fraction a perfect power of the n"' degree, extract the 11th root of the numerator, io within unity, and divide the result by the root of the new denominator. When a greater degree of exactness is required than that indi 1 1 cated by ~r\ extract the root of ab"^, to within any fraction —; r' r and designate this root by —. Now, since — is the root of the JP p 1 r' numerator to within —, it follows, that -7- is the time root of p bp the fraction to within bp 217. Suppose it is required to extract the cube root of 15, to within ^r. We have 15xl23=:15xl728=25920. Now the ItO cube root of 25920, to within unity, is 29; hence the required root 29 5 * 12 °r ?I2 Again, extract the cube root of 47, to within —. We have 47x20'=47x8000=376000. Now the cube root 72_ 12 within Find the value of V25 to within 0,Q01, To do this, multiply 25 by the cube of 1000, or by 1000000000, which gives 25000000000. Now, the cube root of this number, is 2920; hence V25=:2,920 to within 0,001. In general, in order to extract the cube root of a whole number to within a given decimal fraction, annex three times as many ciphers to the number, as there are decimal places in the required root; extract the cube root of the number thus formed to within unity, and point off from the right of this root the required number of decimals. 218. We will now explain the method of extracting the cube root of a decimal fraction. Suppose it is required to extract the cube root of 3,1415. As the denominator 10000, of this fraction, is not a perfect cube, it is necessary to make it one, by multiplying it by 100, which amounts to annexing two ciphers to the proposed decimal, and we have 3,141500. Extract the cube root of 3141500, that is, of the num. ber considered independent of the comma, to within unity; this gives 146. Then divide by 100, or VlO00000, and we find V3,1415=1,46 to within 0,01. |