(1) gives 11 = (1 + 2) (1 EVOLUTION. 2 + 3) (1 + 2) ( 9 quired; consequently, the fifth root of 11 lies between the If we multiply the factors in the preceding value of 11 2 2 2 together, we shall have 11 = 1+(2+ + 3 5+7 + 279) +the 2222 sum of the products of every two of the terms 2,7 + the sum of the products of every three of them, etc. Now if we represent the 5th root of 11 by 1 + x, we shall have 11 = (1 + x)=1+5x+10x2+10x+5x+, which is clearly of a form analogous to the preceding value of 11; consequently, if we equate the corresponding terms, we shall 2 2 2 2 get 5x=2+ + + = 2 + 0.6666. + 0.4 + 0.2857 3 5 7 + + 0.2222. = 3.5745, or x = 9 3.5745 = 0.7149. Again, if we take the sum of the products of every two of 0.6052+0.20316+ 0.06344.0207; which clearly shows that the preceding value of x is a little too great, and that the true value of x lies between 0.6 and 0.7, so that 1.6 are (1.6) × clearly the first figures of the root. Since 11 = (1.6)5 × (1.049041 +), we have to extract the root of 1.049041 +, which, by our method, will be found to be 1.00962 +. = Hence, we shall have 11 = 1.6 x (1.00962 +) 1.61539+ for the fifth root of 11, correctly found to five decimal places. Ex. 5.-To extract the cube root of 48228544. Since 48228544=1000000 × 48228544 =(100)3 × 48.228544, 1000000 it is evident that the extraction of the root is reduced to that of 48.228544. Also, because 27 is the greatest integral cube in 48, and that 48.228544 = 27 x 48.228544 =33 x 1.786242 +, it is clear that the extraction of the root is reduced to that of 1.786 +. To extract the root of 1.786 +, we resolve it into the factors 1.262+, 1.207 +, and 1.171 +; consequently, we shall for the root. 0.640 have 1 + × (1+0.640) * = Hence, we get 48228544 = (100)3 × 33 × (1 + = (100) x (3.64) (364), which gives 148228544 = 364, the root being exact. Remark.-It is clear that the reduction of 48228544 to 48.228544 is equivalent to the separation of the number from the right into periods of three figures each, according to the common rule for the extraction of the cube root given in Arithmetic. Ex. 6.—To find the nTM root of 1 + 2. α By taking the n' root of. the members of (1), we shall n2a + (n-1)b, n2a + b) if b is so small in comparison to na that for n'a + b, n'a + 2b, etc., we may write n'a, n'a, etc.; then we shall have 1 + b na = If we take the product of the factors in the right member of the preceding equation, we shall clearly get (1 + = (2), very nearly; for it is clear that the products n2a b X etc., and so on, may be rejected on account of their minuteness in comparison to the terms which are series arranged according to the ascending powers of b, we Hence, by substituting the preceding values, we get b na If we expand (1+2) by the Binomial Theorem, we Since n is supposed to be a whole number not less than 2, and that in (2) b is regarded as being small in comparison to b b ince is generally much smaller than na that when is very small, it will be much more exact to find a (1 + 2) * its value. from (2) than to take the first four terms of (4) for Again, if we suppose n to be a very great number, it is sensible error, write-1, 1 n 3, etc., we may, without 2, 3, etc.; consequently, when n is very great, (3) or (4) will be reduced to (1 + + etc.) very nearly, (5), which shows that if n is infinitely great, (2) will be as exact as (4) continued to infinity. become n great. h =1+, (7), supposing n to be unlimitedly If we raise the members of (7) to the mnth power, it will ber of the equation as in (4), we shall get (1 + + m(mn—1)(mn—2) h3 m = 1+ +, etc.; or since 2 n 1 2 3 n2 1, mn for mn we must clearly write mn for mn 2, etc., so that the second member may (like the first member) be independent of n, we shall have (1+2) Exponential Theorem, in which any real positive or negative number may evidently be used for m, according to the nature of the case. In order to apply (8) to the extraction of a root of 1 + b it will be necessary to find h, which is called the Hyperbolic or Napierian Logarithm of (1+2) from (6), and to substitute it for h. |