Elements of Algebra

CHAP. I. ADDITIONS. 475 $\alpha$, $\{\alpha + \frac{1}{\beta}\}$, $\{\alpha + \frac{1}{\beta + \frac{1}{\gamma}}, \ldots\}$ or, by reduction, $\alpha$, $\frac{\alpha\beta+1}{\beta}$, $\frac{\alpha\beta\gamma+\alpha+\gamma}{\beta\gamma+1}$, $\ldots$* which approach nearer and nearer to the value of $a$. In order to judge better of the law, and of the convergence of these quantities, it must be remarked, that, by the formulæ of Art. 3, we have $a = \alpha + \frac{1}{b}$, $b = \beta + \frac{1}{c}$, $c = \gamma + \frac{1}{d}$, $\ldots$ Whence we immediately perceive, that $\alpha$ is the first approximate value of $a$; that then, if we take the exact value of $a$, which is $\frac{ab+1}{b}$, and, in this, substitute for $b$ its approximate value $\beta$, we shall have this more approximate value $\frac{\alpha\beta+1}{\beta}$; that we shall, in the same manner, have a third more approximate value of $a$, by substituting for $b$ its exact value $\frac{\beta c+1}{c}$, which gives $a = \frac{(\alpha\beta+1)c+\alpha}{\beta c+1}$, and then taking for $c$ the approximate value $\gamma$; by these means the new approximate value of $a$ will be $\frac{(\alpha\beta+1)\gamma+\alpha}{\beta\gamma+1}$ Continuing the same reasoning, we may approximate nearer, by substituting, in the above expression of $a$, instead of $c$, its exact value, $\frac{\gamma d+1}{d}$, which will give $a = \frac{((\alpha\beta+1)\gamma+\alpha)d+\alpha\beta+1}{(\beta\gamma+1)d+\beta}$ and then taking for $d$ its approximate value $\delta$, we shall have, for the fourth approximation, the quantity $\frac{((\alpha\beta+1)\gamma+\alpha)\delta+\alpha\beta+1}{(\beta\gamma+1)\delta+\beta}$, and so on. Hence it is easy to perceive, that, if by means of the numbers $\alpha, \beta, \gamma, \delta, \ldots$ we form the following expressions, * See note, p. 471.