Page:Elementary algebra (1896).djvu/208

190 Rh And therefore to five places of decimals 290 = 17.02938.

It will be noticed that in obtaining the second figure of the root, the division of 190 by 20 gives 9 for the next figure; this is too great, and the figure 7 has to be obtained tentatively.

208. If the cube root of a number consists of 2n + 2 figures, when the first n + 2 of these have been obtained by the ordinary method, the remaining n may be obtained by division.

Let Ndenote the given number; a the part of the cube root already found, that is, the first n+2 figures found by the common rule, with n ciphers annexed; x the remaining part of the root.

Then 3 N=a+x, N= a3 + 3a2x + 3ax2 + x^3, {N - a3}{3a2} = x + {x2}{a} + {x3}{3a2}(1)

Now N— a^3 is the remainder after n + 2 figures of the root, represented by a, have been found; and 3a^2 is the divisor at the same stage of the work. We see from (1) that N — a3 divided by 3 a^2 gives x, the rest of the quotient required, increased by {x2}{a} + {x3}{3a2}. We shall show that this expression is a proper fraction, so that by neglecting the remainder arising from the division, we obtain x, the rest of the root.

By supposition, x is less than 10^n, and a is greater than 10^{2n+1}; therefore {x2}{a} is less than {10^{2n}}{10^{2n+1}} : that is, less than {1}{10}; and {x2}{a} is less than {10^{3n}}{3 10^{4n+2}}; that is, less than {1}{3 10^{n +1}}; hence {x2}{a} + {x3}{3a2 is less than {1}{10} + {1}{3 10^{n +1}}, and is therefore a proper fraction.