Elements of Algebra

122 ELEMENTS SECT. II. fraction being $2401/400$, it exceeds the square of $\sqrt{6}$ only by $1/400$. Now, making $6 = \frac{2401}{400} - \frac{1}{400}$, so that $c = \frac{49}{20}$ and $b = -\frac{1}{400}$; and still taking only the two leading terms, we have $\sqrt{6} = \frac{49}{20} + \frac{1}{2} \cdot \frac{-1/400}{49/20} = \frac{49}{20} - \frac{1}{1960} = \frac{4801}{1960}$; the square of which is $\frac{23049601}{3841600}$; and 6, when reduced to the same denominator, is $=\frac{23049600}{3841600}$; the error therefore is only $\frac{1}{3841600}$. 367. In the same manner, we may express the cube root of $a+b$ by an infinite series; for since $\sqrt[3]{(a+b)} = (a+b)^{\frac{1}{3}}$, we shall have in the general formula, $n = \frac{1}{3}$, and for the coefficients, $\frac{n}{1} = \frac{1}{3}$; $\frac{n-1}{2} = -\frac{1}{3}$; $\frac{n-2}{3} = -\frac{5}{9}$; $\frac{n-3}{4} = -\frac{2}{3}$; $\frac{n-4}{5} = -\frac{11}{15}$, \&c. and, with regard to the powers of $a$, we shall have $a^n = \sqrt[3]{a}$; $a^{n-1} = \frac{\sqrt[3]{a}}{a}$; $a^{n-2} = \frac{\sqrt[3]{a}}{a^2}$; $a^{n-3} = \frac{\sqrt[3]{a}}{a^3}$, \&c. then $\sqrt[3]{(a+b)} = \sqrt[3]{a} + \frac{1}{3} \cdot b\frac{\sqrt[3]{a}}{a} - \frac{1}{9} \cdot b^2\frac{\sqrt[3]{a}}{a^2} + \frac{5}{81} \cdot b^3\frac{\sqrt[3]{a}}{a^3} - \frac{10}{243} \cdot b^4\frac{\sqrt[3]{a}}{a^4}$, \&c. 368. If $a$ therefore be a cube, or $a = c^3$, we have $\sqrt[3]{a}=c$, and the radical signs will vanish; for we shall have $\sqrt[3]{(c^3+b)} = c + \frac{1}{3} \cdot \frac{b}{c^2} - \frac{1}{9} \cdot \frac{b^2}{c^5} + \frac{5}{81} \cdot \frac{b^3}{c^8} - \frac{10}{243} \cdot \frac{b^4}{c^{11}} +$, \&c. 369. We have therefore arrived at a formula, which will enable us to find, by approximation, the cube root of any number; since every number may be resolved into two parts, as $c^3+b$, the first of which is a cube. If we wish, for example, to determine the cube root of 2, we represent 2 by $1+1$, so that $c=1$ and $b=1$; consequently, $\sqrt[3]{2} = 1 + \frac{1}{3} - \frac{1}{9} + \frac{5}{81}$, \&c. The two leading terms of this series make $1\frac{1}{3} = \frac{4}{3}$, the cube of which is $\frac{64}{27}$, too great by $\frac{10}{27}$: let us therefore make $2 = \frac{64}{27} - \frac{10}{27}$, we have $c = \frac{4}{3}$ and $b = -\frac{10}{27}$, and consequently $\sqrt[3]{2} = \frac{4}{3} + \frac{1}{3} \cdot \frac{-10/27}{16/9}$: these two terms give $\frac{4}{3} - \frac{5}{72} = \frac{91}{72}$, the cube of which is $\frac{753571}{373248}$: but, $2 = \frac{746496}{373248}$, so that the error is $\frac{7075}{373248}$; and in this way we might still approximate the faster in proportion as we take a greater number of terms *. * In the Philosophical Transactions for 1694, Dr. Halley has given a very elegant and general method for extracting roots of