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

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 ${\tfrac {x}{4}}$, ${\tfrac {xx}{64a}}$, ${\tfrac {131x^{3}}{512a^{2}}}$, &c. are produced by dividing $a^{2}x$, ${\tfrac {1}{16}}ax^{2}$, ${\tfrac {131}{128}}x^{3}$, &c. by $4aa$.

36. These things being premised, it remains now to exhibit the Praxis of Resolution. Therefore let the Equation $y^{3}+a^{2}y+axy-2a^{3}-x^{3}=0$ be proposed to be resolved. And from its Terms $y^{3}+a^{2}y-2a^{3}=0$, 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 $+4a^{2}p+a^{2}x=0$ for a fictitious Equation, which giving $p={\tfrac {1}{4}}x$, I write $-{\tfrac {1}{4}}x$ in the Quote. Then because $-{\tfrac {1}{4}}x$ is not the accurate Value of $p$, I put $-{\tfrac {1}{4}}x+q=p$, and in the marginal Terms for $p$ I substitute $-{\tfrac {1}{4}}x+q$, and the resulting Terms ($q^{3}-{\tfrac {3}{4}}xq^{2}+3aq^{2}$, &c.) I again write in the Margin, out of which, according to the foregoing Rule, I again select the Terms $4a^{2}q-{\tfrac {1}{16}}ax^{2}=0$ 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.