Page:The method of fluxions and infinite series.djvu/32

to be added to the Quote in this manner; that is, if you extract the lesser Root out of the three last Terms of its secondary Equation: For thus you will obtain, at every time, as many Figures again in the Quote.

23. And now from the Resolution of numeral Equations, I shall proceed to explain the like Operations in Species; concerning which, it is necessary to observe what follows.

24. First, that some one of the specious or literal Coefficients, if there are more than one, should be distinguish'd from the rest, which either is, or may be suppos'd to be, much the least or greatest of all, or nearest to a given Quantity. The reason of which is, that because of its Dimensions continually increasing in the Numerators, or the Denominators of the Terms of the Quote, those Terms may grow less and less, and therefore the Quote may constantly approach to the Root required; as may appear from what is said before of the Species $x$, in the Examples of Reduction by Division and Extraction of Roots. And for this Species, in what follows, I shall generally make use of $x$ or $z$; as also I shall use $y$, $p$, $q$, $r$, $s$, &c. for the Radical Species to be extracted.

25. Secondly, when any complex Fractions, or surd Quantities, happen to occur in the proposed Equation, or to arise afterwards in the Process, they ought to be removed by such Methods as are sufficiently known to Analysts. As if we should have $y^{3}+{\tfrac {bb}{b-x}}y^{2}-x^{5}=0$, multiply by $b-x$, and from the Product $by^{3}-xy^{3}+b^{2}y^{2}-bx^{3}+x^{4}=0$ extract the Root $y$. Or we might suppose $y\times b-x=v$, and then writing ${\tfrac {v}{b-x}}$ for $y$, we should have $v^{3}+b^{2}v^{2}-b^{3}v^{3}+3b^{2}x^{4}-3bx^{5}+x^{6}=0$, whence extracting the Root $v$, we might divide the Quote by $b-x$, in order to obtain $y$. Also if the Equation $y^{3}-xy^{1/2}+x^{4/3}=0$ were proposed, we might put $y^{1/2}=v$, and $x^{1/3}=z$, and so writing $vv$ for $y$, and $z^{3}$ for $x$, there will arise $v^{6}-z^{3}v+z^{4}=0$; which Equation being resolved, $y$ and $x$ may be restored. For the Root will be found $v=z^{3}+6z^{5}$, &c. and restoring $y$ and $x$, we have $y^{1/2}=x^{1/3}+x+6x^{5/3}$, &c. then squaring, $y=x^{2/3}+2x^{4/3}+13x^{2}$, &c.

26. After the same manner if there should be found negative Dimensions of $x$ and $y$, they may be removed by multiplying by the fame $x$ and $y$. As if we had the Equation $x^{3}+3x^{2}y^{-1}-2x^{-1}-16y^{-3}=0$,

multiply by $x$ and $y^{3}$, and there would arise $x^{4}y^{3}+3x^{3}y^{2}-2y^{3}-16x=0$. And if the Equation were $x={\tfrac {aa}{y}}-{\tfrac {2a^{3}}{y^{2}}}+{\tfrac {3a^{4}}{y^{3}}}$,