ANALYSIS OF EQUATIONS, &c.

THEOREM.

Let any quantity = g be taken as the first member; I say that this quantity, whatever it may be, converges (by our method) at length to the true root.

The demonstration is evident from the following construction.

Let the equation ba—aaa=c be resumed.

A mathematical derivation uses large curly braces to group expressions. On the left, two cases for the variable 'a' are defined: "g + x = a" and "If, however, g were greater than a { g - x = a". These lead to a second group containing the expanded algebraic expression "bg—ggg+b—3gg × x — 3gxx — xxx = ba — aaa = c" and the resulting value for x expressed as a fraction "x = (c + ggg - bg) / (b - 3gg) + (3gxx + xxx) / (b - 3gg)".

From there also x = (c+ggg-bg) / (b-3gg) + (3gxx+xxx) / (b-3gg), all of which are evident by analytical reduction.

A similar construction is to be used in all equations of all kinds.
From the set (t) = (c+ggg-bg) / (b-3gg) and (m) = (3gxx+xxx) / (b-3gg), it will be x=t+m.

Whence it is clear that as many cases emerge as there are variations that can be made of (...x=...t...m), which are eight.


+	x	=	+	t + m
—	x	=	—	t — m
+	x	=	+	t — m
—	x	=	—	t + m
+	x	=	—	t + m
—	x	=	+	t — m
+	x	=	—	t — m
—	x	=	+	t + m

The theorem (x = .. t .. m) I call absolute (and its x absolute) so that they may be distinguished from (x) = .. t), to which I previously gave the names of Convergents.

These things being so positioned, it is clearly evident.

1. In x=+t+m (which is the case of ba—aaa=c), the convergent x is less than the absolute by the quantity (m), that is, a part of its whole.

Therefore, with the convergent x (by hypothesis, or by the precepts) added to its (g), it is clear that this new (g), increased by the quantity (t), is greater than the preceding, and in this way (g) is increased and consequently (m) is diminished to infinity. After infinite convergence, however, the difference or (m) is rendered smaller than any assignable quantity. By the same process, the second case is proven.

Let +x=+t—m be proposed, or aa=c, x = (c—gg)/2g. It is clear that the convergent (x) is greater than the absolute, by the quantity (m) to be subtracted that is, not subtracted; therefore, when added to (g), it makes (g) greater than (a), hence (-gg greater than c). Therefore, the theorem +x=+t—m turns into the second case, or —x=—t—m, and accordingly it converges by descending.

By the same reasoning, it will be clear to anyone looking into the nature of equations that all cases reduce to the first two, and accordingly converge; that is, all those that have (+m) converge to the first, and those that have (-m) converge to the second. The last four are not possible unless, with the signs of (x) changed, they are converted into the first four; which will easily be clear to anyone considering it.