Elements of Algebra

CHAP. XVI. OF ALGEBRA. 295 $r/p = 2$, or $r = 2p$, that is to say, $r = 0q + 2p$, whence would result the series $1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, \ldots$ from which we can draw no conclusion, because each term, divided by the preceding, gives always $x = 1$, or $x = 2$. But we may obviate this inconvenience, by making $x = y - 1$; for by these means we have $y^2 - 2y + 1 = 2$; and if we now make $y = q/p$, and $y^2 = r/p$, we shall obtain the same approximation that has been already given. 797. It would be the same with the equation $x^3 = 2$. This method would not furnish such a series of numbers as would express the value of $\sqrt[3]{2}$. But we have only to suppose $x = y - 1$, in order to have the equation $y^3 - 3y^2 + 3y - 1 = 2$, or $y^3 = 3y^2 - 3y + 3$; and then making $y = q/p$, $y^2 = r/p$, and $y^3 = s/p$, we have $s = 3r - 3q + 3p$, by means of which we see how three given terms determine the succeeding one. Assuming then any three terms for the first, for example $0, 0, 1$, we have the following series: $0, 0, 1, 3, 6, 12, 27, 63, 144, 324, \ldots$ The last two terms of this series give $y = 324/144$, and $x = 5/4$. This fraction approaches sufficiently near the cube root of $2$; for the cube of $5/4$ is $125/64$, and $2 = 128/64$; the difference, therefore, is only $3/64$. 798. We must farther observe, with regard to this method, that when the equation has a rational root, and the beginning of the period is chosen such, that this root may result from it, each term of the series, divided by the preceding term, will give the root with equal accuracy. To shew this, let there be given the equation $x^2 = x + 2$, one of the roots of which is $x = 2$; as we have here, for the series, the formula $r = q + 2p$, if we take $1, 2$, for the first two terms, we have the series $1, 2, 4, 8, 16, 32, 64, \ldots$ a geometrical progression, whose exponent $= 2$. The same property is proved by the equation of the third degree, $x^3 = x^2 + 3x + 9$, which has $x = 3$ for one of the roots. If we suppose the leading terms to be $1, 3, 9$, we shall find, by the formula, $s = r + 3q + 9p$, and the series $1, 3, 9, 27, 81, 243, \ldots$ which is likewise a geometrical progression.