# Letter from Michael Dary to Newton, dated 15 October 1674

Tower the 15 of October 1674

Sir

Although I sent you three papers yesterday, I Cannot refrain from sending you this: I have had fresh Thoughts this morning about those two sorts of Equations which wee have latly Bandied about, and I have attained an Universall series for any Equation of two Cossique notes. Truly it pleaseth me well. But yet I doe hereby submitt it to your Censure.

To Extract the root of an Equation Consisting

of Two severall powers (or potestates) and

an absolute number; per an approximation

easily performed by logarithmes.

As for example $+{\mathrm{z}}^{\mathrm{p}}=\mathrm{a}{\mathrm{z}}^{\mathrm{q}},\mathrm{n}$

In which example and in all such like Equations, you must observe that z is the unknown symbol or root sought, $\mathrm{p}=$ the index (or power note) of the highest power, $\mathrm{a}=$ the known number or Coefficient of the midle term, $\mathrm{q}=$ the index of the inferior power; $\mathrm{n}=$ the absolute number or resolvend.

The rule is thus: First guess at the root as nearly as you can, the nearer the better (not for necessity but for accomodaccomodation) and suppose that guess to be z.

Then observing the following series, you shall approach (from this supposed z) toward the true z which is sought: Because every term in this series brings you nearer and nearer, for if your supposition be too great, every term in this series makes it less and less; or if your supposition be too litle, every term in this series makes it greater and greater. So when you are pleased to make a Ceasation, the last term is that which you seek. The series followes:

$\sqrt[\mathrm{p}]{:+\mathrm{a}{\mathrm{z}}^{\mathrm{q}}+\mathrm{n}:}=\mathrm{b}$ $\sqrt[\mathrm{p}]{:+\mathrm{a}{\mathrm{b}}^{\mathrm{q}}+\mathrm{n}:}=\mathrm{c}$ $\sqrt[\mathrm{p}]{:+\mathrm{a}{\mathrm{c}}^{\mathrm{q}}+\mathrm{n}:}=\mathrm{d}$.

$\sqrt[\mathrm{p}]{:+\mathrm{a}{\mathrm{d}}^{\mathrm{q}}+\mathrm{n}:}=\mathrm{e}$ $\sqrt[\mathrm{p}]{:+\mathrm{a}{\mathrm{e}}^{\mathrm{q}}+\mathrm{n}:}=\mathrm{f}$ $\sqrt[\mathrm{p}]{:+\mathrm{a}{\mathrm{f}}^{\mathrm{q}}+\mathrm{n}:}=\mathrm{g}$, &c:

In which series you must note that + only intimates the retaining of the proper signes whether they be + or −.

Sir pray doe not count mee troublesome for I could not forbear but send this, by Stiles the Carrier, who is paid for the Carriage. pray Remember mee about the series of Logarithmes.

Your most humble and oblidged servant

Mich: Dary

A true Coppie of that which I sent to M^{r} Newton this morning.