44.
Trinity Coll:College Aug: 27th, 1675.

SrSir,

In the Theorems ytthat I sent youu I perceive I com̄mmitted a mistake in transcribeing them from the papers where I had computed them. They should have been $\begin{array}{l}\text{2)}\phantom{\rule{0.5em}{0ex}}\mathrm{B}+\frac{\mathrm{A}}{\mathrm{B}}=\sqrt{}\mathrm{A}\text{.}\\ \text{3)}\phantom{\rule{0.5em}{0ex}}2\mathrm{B}+\frac{\mathrm{A}}{\mathrm{B}\mathrm{B}}=\sqrt{}c:\phantom{\rule{0.5em}{0ex}}\mathrm{A}\text{.}\\ \text{4)}\phantom{\rule{0.5em}{0ex}}3\mathrm{B}+\frac{\mathrm{A}}{{\mathrm{B}}^{c:}}=\sqrt{}qq.\phantom{\rule{0.5em}{0ex}}\mathrm{A}\text{.}\end{array}$

In words at length: To finde the cube root of $\mathrm{A}$ to 11 decimal places: seek the Root by Logarithms to 5 decimal places, and suppose it $\mathrm{B}$. Then square $\mathrm{B}$, not by Logarithms, but by com̄mmon Arithmetick, ytthat youu may have its exact square to 10 decimal places, and by this square Divide $\mathrm{A}$ to 11 decimal places, and to the Quotient add $2\mathrm{B}$: The third part of the Quotient shall be the root cubical of $\mathrm{A}$: to 11 Decimal places. yoryour surest way will be to finde first the whole series of yethe Roots, $\mathrm{B.}$ by Logarithms, & try whether it be Regular by Differencing it: Then square those Roots by Nepeirs bones, and lastly Divide each NumbrNumber $\mathrm{A.}$ by the correspondent square, and add $2\mathrm{B}$ to each Quotient, and try the Resulting series againe by differencing it, whether it be Regular. If it be regular, I suppose youu know the differences will at last come to be equal: what is said of Cubes is easily applyable to Square=Squares: I would have given youu examples in numbers; but that I have lent my Bookes of Logarithms to a ꝑperson, who is out of Towne.
yoryour humble SirvantServant
Is: Newton
I thank youu for yoryour intended present.