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

36. These things being premised, it remains now to exhibit the Praxis of Resolution. Therefore let the Equation ${\displaystyle y^{3}+a^{2}y+axy-2a^{3}-x^{3}=0}$ be proposed to be resolved. And from its Terms ${\displaystyle y^{3}+a^{2}y-2a^{3}=0}$, being a fictitious Equation, by the third of the foregoing Premises, I obtain ${\displaystyle y-a=0}$, and therefore I write ${\displaystyle +a}$ in the Quote. Then because ${\displaystyle +a}$ is not the compleat Value of ${\displaystyle y}$, I put ${\displaystyle a+p=y}$, and instead of ${\displaystyle y}$, in the Terms of the Equation written in the Margin, I substitute ${\displaystyle a+p}$, and the Terms resulting (${\displaystyle p^{3}+3ap^{2}+axp}$, &c.) I again write in the Margin; from which again, according to the third of the Premises, I select the Terms ${\displaystyle +4a^{2}p+a^{2}x=0}$ for a fictitious Equation, which giving ${\displaystyle p={\tfrac {1}{4}}x}$, I write ${\displaystyle -{\tfrac {1}{4}}x}$ in the Quote. Then because ${\displaystyle -{\tfrac {1}{4}}x}$ is not the accurate Value of ${\displaystyle p}$, I put ${\displaystyle -{\tfrac {1}{4}}x+q=p}$, and in the marginal Terms for ${\displaystyle p}$ I substitute ${\displaystyle -{\tfrac {1}{4}}x+q}$, and the resulting Terms (${\displaystyle q^{3}-{\tfrac {3}{4}}xq^{2}+3aq^{2}}$, &c.) I again write in the Margin, out of which, according to the foregoing Rule, I again select the Terms ${\displaystyle 4a^{2}q-{\tfrac {1}{16}}ax^{2}=0}$ for a fictitious Equation, which giving ${\displaystyle q={\tfrac {xx}{64a}}}$, I write ${\displaystyle {\tfrac {xx}{64a}}}$ in the Quote. Again, since ${\displaystyle {\tfrac {xx}{64a}}}$ is not the accurate Value of ${\displaystyle q}$, I make ${\displaystyle {\tfrac {xx}{64a}}+r=q}$, and instead of ${\displaystyle q}$ I substitute ${\displaystyle {\tfrac {xx}{64a}}+r}$ in the marginal Terms. And thus I continue the Process at pleasure, as the following Diagram exhibits to view.

C 2

${\displaystyle y}$