Page:The method of fluxions and infinite series.djvu/35

Rh of one Dimension only is affected, without the other indefinite Species, and by writing the Result in the Quote. So in the following Example, the Terms $$\tfrac x4$$, $$\tfrac{xx}{64a}$$, $$\tfrac{131x^3}{512a^2}$$, &c. are produced by dividing $$a^2x$$, $$\tfrac1{16} ax^2$$, $$\tfrac{131}{128} x^3$$, &c. by $$4aa$$.

36. These things being premised, it remains now to exhibit the Praxis of Resolution. Therefore let the Equation $$y^3 + a^2y + axy - 2a^3 - x^3 = 0$$ be proposed to be resolved. And from its Terms $$y^3 + a^2y - 2a^3 = 0$$, being a fictitious Equation, by the third of the foregoing Premises, I obtain $$y-a=0$$, and therefore I write $$+ a$$ in the Quote. Then because $$+ a$$ is not the compleat Value of $$y$$, I put $$a+p=y$$, and instead of $$y$$, in the Terms of the Equation written in the Margin, I substitute $$a+p$$, and the Terms resulting ($$p^3 + 3ap^2 + axp$$, &c.) I again write in the Margin; from which again, according to the third of the Premises, I select the Terms $$+4a^2p + a^2x = 0$$ for a fictitious Equation, which giving $$p = \tfrac14 x$$, I write $$- \tfrac14 x$$ in the Quote. Then because $$- \tfrac14 x$$ is not the accurate Value of $$p$$, I put $$- \tfrac14 x + q = p$$, and in the marginal Terms for $$p$$ I substitute $$- \tfrac14 x + q$$, and the resulting Terms ($$q^3 - \tfrac34 xq^2 + 3aq^2$$, &c.) I again write in the Margin, out of which, according to the foregoing Rule, I again select the Terms $$4a^2q - \tfrac1{16} ax^2 = 0$$ for a fictitious Equation, which giving $$q = \tfrac{xx}{64a}$$, I write $$\tfrac{xx}{64a}$$ in the Quote. Again, since $$\tfrac{xx}{64a}$$ is not the accurate Value of $$q$$, I make $$\tfrac{xx}{64a} + r = q$$, and instead of $$q$$ I substitute $$\tfrac{xx}{64a} + r$$ in the marginal Terms. And thus I continue the Process at pleasure, as the following Diagram exhibits to view. Rh