Page:The method of fluxions and infinite series.djvu/38

14 $$z^5$$; because we may observe, that the Quote ought always to ascend by the Interval of two Units, in this manner, $$z, z^3, z^5,$$ &c. Then at last we shall have $$y=z - \tfrac{1}{6} z^2 + \tfrac{1}{120} z^5 - \tfrac{1}{3040} z^7 + \tfrac{1}{362880} z^9,$$ &c.

41. And hence an Artifice is discover'd, by which Equations, tho' affected in infinitum, and consisting of an infinite number of Terms, may however be resolved. And that is, before the Work begins all the Terms are to be rejected, in which the Dimension of the indefinitely small Species, not affected by the radical Species, exceeds the greatest Dimension required in the Quote; or from which, by substituting instead of the radical Species, the first Term of the Quote found by the Parallelogram as before, none but such exceeding Terms can arise. Thus in the last Example I should have omitted all the Terms beyond $$y'$$, though they went on ad infinitum. And so in this Equation

$$ 0=\begin{cases} - 8 + z^2 - 4z^4 + 9z^6 - 16z^8, \&\text{c.}\\ + y \text{ in } z^2 - 2z^4 + 3z^6 - 4z^8, \&\text{c.}\\ - y^2 \text{ in } z^2 - z^4 + z^6 - z^8, \&\text{c.}\\ + y^3 \text{ in } z^2 - \tfrac12 z^4 + \tfrac13 z^6 - \tfrac14 z^8, \&\text{c.}\\ \end{cases} $$

that the Cubick Root may be extracted only to four Dimensions of $$z$$, I omit all the Terms in infinitum beyond $$+ y^3$$ in $$z^2 - \tfrac12 z^4 + \tfrac13 z^6$$, and all beyond $$- y ^2$$ in $$z^2 - z^4 + z^6$$, and all beyond $$+ y$$ in $$z^2 - 2z^4$$, and beyond $$- 8 + z^2 - 4z^4$$. And therefore I assume this Equation only to be resolved, $$\tfrac13 z^6y^3 - \tfrac12 z^4y^3 + z^2y^3 - z^6y^2 + z^4y^2 - z^2y^2 - 2z^4y + z^2y - 4z^4 + z^2 - 8 = 0$$. Because $$2z^{-2/3}$$, (the first Term of the Quote,) being substituted instead of $$y$$ in the rest of the Equation depress'd by $$z^{2/3}$$ gives every where more than four Dimensions.

42. What I have said of higher Equations may also be apply'd to Quadraticks. As if I desired the Root of this Equation

$$ 0 = \begin{cases} y^2\\ - y \text{ in } a + x + \tfrac{x^2}{a} + \tfrac{x^3}{a^2} + \tfrac{x^4}{a^3} \text{ }\&\text{c.}\\ +\tfrac{x^4}{4a^2} \end{cases} $$

as far as the Period $$x^6$$, I omit all the Terms in infinitum, beyond $$-y$$ in $$ a + x + \tfrac{x^2}a$$, and assume only this Equation, $$y^2-ay-xy-\tfrac{x^2}{a}y + \tfrac{x^4}{4a^2} = 0$$. This I resolve either in the usual manner, by making $y$