Elements of Algebra

466 ADDITIONS. CHAP. I. in which the quantities $\alpha$, $\beta$, $\gamma$, $\delta$, $\ldots$ and $b$, $c$, $d$, $\ldots$ are integer numbers positive or negative; but at present we shall consider those Continued Fractions only, whose numerators $b$, $c$, $d$, $\ldots$ are unity; that is to say, fractions of this form, $$ \alpha + \frac{1}{\beta + \frac{1}{\gamma + \frac{1}{\delta} + \ldots}} $$ $\alpha$, $\beta$, $\gamma$, $\delta$, $\ldots$ being any integer numbers positive or negative; for these are, properly speaking, the only numbers, which are of great utility in analysis, the others being scarcely any thing more than objects of curiosity. 2. Lord Brouncker, I believe, was the first who thought of Continued Fractions; we know that the continued fraction, which he devised to express the ratio of the circumscribed square to the area of the circle was this: $$ 1 + \frac{1}{2} + \frac{9}{2} + \frac{25}{2} + \ldots $$ but we are ignorant of the means which led him to it. We only find in the *Arithmetica Infinitorum* some researches on this subject, in which Wallis demonstrates, in an indirect, though ingenious manner, the identity of Brouncker's expression to his, which is, $\frac{3 \times 3 \times 5 \times 5 \times 7, \ldots}{2 \times 4 \times 4 \times 6 \times 6, \ldots}$. He there also gives the general method of reducing all sorts of continued fractions to vulgar fractions; but it does not appear that either of those great mathematicians knew the principal properties and singular advantages of continued fractions; and we shall afterwards see, that the discovery of them is chiefly due to Huygens. 3. Continued fractions naturally present themselves, whenever it is required to express fractional, or imaginary quantities in numbers. In fact, suppose we have to assign the value of any given quantity $a$, which is not expressible by an integer number; the simplest way is, to begin by seeking the integer number, which will be nearest to the value of $a$, and which will differ from it only by a fraction less than unity. Let this number be $\alpha$, and we shall have $a-\alpha$ equal to a fraction less than unity; so that $\frac{1}{a-\alpha}$ will, on the contrary, be a number greater than unity: therefore let $\frac{1}{a-\alpha} = b$; and, as $b$ must be a number greater than unity,