Elements of Algebra

290 ELEMENTS SECT. IV. $8p = 20$, or $8p = 4$. This gives $p = \frac{1}{2}$, and $x = 4\frac{1}{2}$, which already approaches nearer the true root. If, therefore, we now suppose $x = 4\frac{1}{2} + p'$; we are sure that $p'$ expresses a fraction much smaller than before, and that we may neglect $p'^2$ with greater propriety. We have, therefore, $x^2 = 20\frac{1}{4} + 9p' = 20$, or $9p' = -\frac{1}{4}$; and consequently, $p' = -\frac{1}{36}$; therefore $x = 4\frac{1}{2} - \frac{1}{36} = 4\frac{17}{36}$. And if we wished to approximate still nearer to the true value, we must make $x = 4\frac{17}{36} + p''$, and should thus have $x^2 = 20\frac{1}{1296} + 8\frac{34}{36}p'' = 20$; so that $8\frac{34}{36}p'' = -\frac{1}{1296}$, or $\frac{322}{36}p'' = -\frac{1}{1296}$, and $p'' = -\frac{36}{322 \times 1296} = -\frac{1}{36 \times 322} = -\frac{1}{11592}$; therefore $x = 4\frac{17}{36} - \frac{1}{11592} = 4\frac{5473}{11592}$, a value which is so near the truth, that we may consider the error as of no importance. 787. Now, in order to generalise what we have here laid down, let us suppose the given equation to be $x^2 = a$, and that we previously know $x$ to be greater than $n$, but less than $n+1$. If we now make $x = n+p$, $p$ must be a fraction, and $p^2$ may be neglected as a very small quantity, so that we shall have $x^2 = n^2 + 2np = a$; or $2np = a-n^2$, and $p = \frac{a-n^2}{2n}$; consequently, $x = n + \frac{a-n^2}{2n} = \frac{n^2+a}{2n}$. Now, if $n$ approximated towards the true value, this new value $\frac{n^2+a}{2n}$ will approximate much nearer; and, by substituting it for $n$, we shall find the result much nearer the truth; that is, we shall obtain a new value, which may again be substituted, in order to approach still nearer; and the same operation may be continued as long as we please. For example, let $x^2 = 2$; that is to say, let the square root of 2 be required; and as we already know a value sufficiently near, which is expressed by $n$, we shall have a still nearer value of the root expressed by $\frac{n^2+2}{2n}$. Let, therefore, 1. $n=1$, and we shall have $x = \frac{3}{2}$. 2. $n=\frac{3}{2}$, and we shall have $x = \frac{17}{12}$. 3. $n=\frac{17}{12}$, and we shall have $x = \frac{577}{408}$. This last value approaches so near $\sqrt{2}$, that its square $\frac{332929}{166464}$ differs from the number 2 only by the small quantity $\frac{1}{166464}$, by which it exceeds it. 788. We may proceed in the same manner, when it is