Letter from Newton to John Collins, dated 19 January 1669/70

<1r>

Trin: Coll: Cambridge.
Ian 1669

S^r

I received D^r Wallis his Mechanicks w^ch you sent to M^r Barrow for mee. I must needs acknowledg you more then ordinarily obliging, & my selfe puzzeled how I shall quit Courtesys.

The Problemes you proposed to mee I have considered & sent you here y^e best solutions of one of them that I can contrive; Namely how to find y^e aggregate of a series of fractions, whose numerato^rs are the same & their denominato^rs in arithmeticall progression. To doe this I shall propound two ways, The first by reduction to one common denominator as followeth.

If $\frac{\mathrm{a}}{\mathrm{b}}.\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+2\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+3\mathrm{c}}$ &c {illeg} be the series: Multiply all their denominators together, & the product will bee $\begin{array}{ccccccc}{\mathrm{b}}^4& +& 6{\mathrm{b}}^3\mathrm{c}& +& 11\mathrm{b}\mathrm{b}\mathrm{c}\mathrm{c}& +& 6\mathrm{b}{\mathrm{c}}^3\\ 4& & 3& & 2& & 1& \end{array}$; each terme of w^ch being multiplyed by its dimensions of b, & the product againe multiplyed by $\frac{\mathrm{a}}{\mathrm{b}}$, the result shall bee the Numerator of the desired aggregate {illeg} $\begin{array}{ccc}& \frac{4\mathrm{a}{\mathrm{b}}^3+18\mathrm{a}\mathrm{b}\mathrm{b}\mathrm{c}+22\mathrm{a}\mathrm{b}\mathrm{c}\mathrm{c}+6\mathrm{a}{\mathrm{c}}^3}{{\mathrm{b}}^3+6{\mathrm{b}}^3\mathrm{c}+11\mathrm{b}\mathrm{b}\mathrm{c}\mathrm{c}+6\mathrm{b}{\mathrm{c}}^3}& \end{array}$

If $\frac{\mathrm{a}}{\mathrm{b}-2\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}-\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}}.\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+2\mathrm{c}}$, be the series: The factus of their denominators is $\mathrm{b}in\overline{\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}}in\mathrm{b}\mathrm{b}-4\mathrm{c}\mathrm{c}$, or ${\mathrm{b}}^5-5{\mathrm{b}}^3\mathrm{c}\mathrm{c}+4\mathrm{b}{\mathrm{c}}^4$, the denominator; w^ch multiplyed as before gives $5\mathrm{a}{\mathrm{b}}^4-15\mathrm{a}\mathrm{b}\mathrm{b}\mathrm{c}\mathrm{c}+4\mathrm{a}{\mathrm{c}}^4$, y^e Numerator of y^e aggregate.

If $\frac{\mathrm{a}}{\mathrm{b}-\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+3\mathrm{c}}$ &c is y^e series. Then $\overline{\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}}\times \mathrm{b}+3\mathrm{c}$ or ${\mathrm{b}}^3+3\mathrm{b}\mathrm{b}\mathrm{c}-\mathrm{b}\mathrm{c}\mathrm{c}-3{\mathrm{c}}^3$ is y^e denominator and $3\mathrm{a}\mathrm{b}\mathrm{b}+6\mathrm{a}\mathrm{b}\mathrm{c}-\mathrm{a}\mathrm{c}\mathrm{c}$ y^e numerator of the Aggregate.

In numbers: If $\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}$ is y^e series, then putting $\mathrm{b}=2$ y^t y^e series may bee $\frac{1}{\mathrm{b}}.\frac{1}{\mathrm{b}+1}.\frac{1}{\mathrm{b}+2}.\frac{1}{\mathrm{b}+3}.\frac{1}{\mathrm{b}+4}$; y^e factus of their denominato^rs will bee ${\mathrm{b}}^5+10{\mathrm{b}}^4+35{\mathrm{b}}^3+50\mathrm{b}\mathrm{b}+24\mathrm{b}$ the denominator, & consequently $5{\mathrm{b}}^4+40{\mathrm{b}}^3+105\mathrm{b}\mathrm{b}+100\mathrm{b}+24$ the denomin|numer|ato^r of y^e aggregate.

But its {sic} better to put y^e {illeg}$\mathrm{b}=4$, that the series may bee $\frac{1}{\mathrm{b}-2}.\frac{1}{\mathrm{b}-1}.$ $\frac{1}{\mathrm{b}}.\frac{1}{\mathrm{b}+1}.\frac{1}{\mathrm{b}+2}.$ And so shall the aggregate bee $\frac{5{\mathrm{b}}^4-15\mathrm{b}\mathrm{b}+4}{{\mathrm{b}}^5-5{\mathrm{b}}^3+4\mathrm{b}}$.

$\begin{array}{cc}\begin{array}{c}\text{The annexed table will}\\ \text{much faciliate}{\text{y}}^{\text{e}}\text{multiplication}\\ \text{of denominators together.}\end{array}& \begin{array}{cc}\begin{array}{c}\begin{array}{rrrrrr}& & & & +1& -1\\ & & & +1& -5& +4\\ & & +1& -14& +49& -36\\ & +1& -30& +273& -820& +576\\ +1& -55& +1023& -7645& +51276& -14400\end{array}\\ \begin{array}{cccccc}{\mathrm{b}}^{11}.& {\mathrm{b}}^9\mathrm{c}\mathrm{c}.& {\mathrm{b}}^7{\mathrm{c}}^4.& {\mathrm{b}}^5{\mathrm{c}}^6.& {\mathrm{b}}^3{\mathrm{c}}^8.& \mathrm{b}{\mathrm{c}}^{10}\\ \phantom{+1}& \phantom{-55}& \phantom{+1023}& \phantom{-7645}& \phantom{+51276}& \phantom{-14400}\end{array}\end{array}& \begin{array}{cc}& \begin{array}{l}\text{for 3 terms}\\ \text{for 5 termes}\\ \text{for 7 termes}\\ \text{for 9 termes}\\ \text{for 11 termes}\end{array}\end{array}\end{array}\end{array}$

This rule holds good though the differences of the denominato^rs bee not equall: as if $\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+\mathrm{d}}.\frac{\mathrm{a}}{\mathrm{b}-\mathrm{e}}$ are to bee added the factus of their denominators is ${\mathrm{b}}^3+\mathrm{b}\mathrm{b}\mathrm{c}+\mathrm{b}\mathrm{b}\mathrm{d}-\mathrm{b}\mathrm{b}\mathrm{e}+\mathrm{b}\mathrm{c}\mathrm{d}-\mathrm{b}\mathrm{c}\mathrm{e}-\mathrm{b}\mathrm{d}\mathrm{e}-\mathrm{c}\mathrm{d}\mathrm{e}$ the denominator, which multiplyed by the dimensions of b & againe by $\frac{\mathrm{a}}{\mathrm{b}}$ produces $3\mathrm{a}\mathrm{b}\mathrm{b}+2\mathrm{a}\mathrm{b}\mathrm{c}+2\mathrm{a}\mathrm{b}\mathrm{d}-2\mathrm{a}\mathrm{b}\mathrm{e}+\mathrm{a}\mathrm{c}\mathrm{d}-\mathrm{a}\mathrm{c}\mathrm{e}$$-\mathrm{b}\mathrm{d}\mathrm{e}$ the numerator of the desired summ.

The other way of resolving this Problems {sic} is by approximation. Suppose the number of termes in the propounded series of bee p. And make $\frac{\mathrm{p}\mathrm{p}-\mathrm{p}}{2}=\mathrm{q}$. $\frac{2\mathrm{p}\mathrm{q}-\mathrm{q}}{3}=\mathrm{r}$. $\mathrm{q}\mathrm{q}=\mathrm{s}$. $\frac{6\mathrm{q}\mathrm{r}-\mathrm{r}}{5}=\mathrm{t}$. $\frac{4\mathrm{q}\mathrm{s}-\mathrm{s}}{3}=\mathrm{v}$. $\frac{12\mathrm{r}\mathrm{s}-5\mathrm{t}}{7}=\mathrm{x}$. $2\mathrm{s}\mathrm{s}-\mathrm{v}=\mathrm{y}$. $\frac{\mathrm{r}\mathrm{v}-\mathrm{r}\mathrm{s}}{3}+\mathrm{t}=\mathrm{z}$. &c. Now if the propounded series \bee/ $\frac{\mathrm{a}}{\mathrm{b}}.\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+2\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+3\mathrm{c}}$ &c their Aggregate shall bee $\frac{\mathrm{a}}{\mathrm{b}}in\mathrm{p}-\mathrm{q}\frac{\mathrm{c}}{\mathrm{b}}+\mathrm{r}\frac{\mathrm{c}\mathrm{c}}{\mathrm{b}\mathrm{b}}-\mathrm{s}\frac{{\mathrm{c}}^3}{{\mathrm{b}}^3}-\mathrm{t}\frac{{\mathrm{c}}^4}{{\mathrm{b}}^4}$ &c: In w^ch progression the farther you proceede, the nearer you approach to truth.

But its {sic} better to put b for the Denominator of y^e middle termes of y^e propounded series, thus $.\frac{\mathrm{a}}{\mathrm{b}-2\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}-\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}}.\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{a}}{\mathrm{b}+2\mathrm{c}}$. And making n the number of termes from y^e said middle terme either way; as also {illeg} $\mathrm{n}\mathrm{n}+\mathrm{n}=\mathrm{m}$. $2\mathrm{n}+1=\mathrm{p}$. $\frac{\mathrm{m}\mathrm{p}}{3}=\mathrm{r}$. $\frac{3\mathrm{m}\mathrm{r}-\mathrm{r}}{5}=\mathrm{t}$. {illeg} $\frac{3\mathrm{m}\mathrm{m}\mathrm{r}-5\mathrm{t}}{7}=\mathrm{x}$. $\frac{{\mathrm{m}}^3\mathrm{r}-2\mathrm{m}\mathrm{m}\mathrm{r}}{3}+\mathrm{t}=\mathrm{z}$. &c, the desired Aggregate shall bee $\frac{\mathrm{a}}{\mathrm{b}}in\mathrm{p}+\mathrm{r}\frac{\mathrm{c}\mathrm{c}}{\mathrm{b}\mathrm{b}}$$+\mathrm{t}\frac{{\mathrm{c}}^4}{{\mathrm{b}}^4}+\mathrm{x}\frac{{\mathrm{c}}^6}{{\mathrm{b}}^6}+\mathrm{z}\frac{{\mathrm{c}}^8}{{\mathrm{b}}^8}$ &c: {illeg}|a| progression wanting each other terme & also converging much more towards the truth then the former.

Now a series of fractions being propounded: first consider how exact <1v> you would have their aggregate; suppose t|n|ot erring from truth above $\frac{1}{\mathrm{e}}$ part of an unit. Then make a rude guesse how many times $\frac{\mathrm{n}\mathrm{c}}{\mathrm{b}}$ $\frac{\mathrm{b}}{\mathrm{n}\mathrm{c}}$ multiplyed into it selfe will bee about y^e bignesse of $\frac{5\mathrm{a}\mathrm{e}}{2\mathrm{b}}$ more or lesse. And omit all those termes of the progression where b is of more then soe many dimensions. For example if the aggregate of $\frac{10000}{100}+\frac{10000}{106}+\frac{10000}{112}+\frac{10000}{118}+\frac{10000}{124}+\frac{10000}{130}+\frac{10000}{136}$ bee desired c|t|o y^e exactnesse of $\frac{1}{8}\text{th}$ pt of an unite Then is $\mathrm{a}=10000$. $\mathrm{b}=118$. $\mathrm{c}=6$. $\mathrm{n}=3$. $\mathrm{e}=8$. $\frac{\mathrm{b}}{\mathrm{n}\mathrm{c}}=\frac{118}{18}$ or about $6\frac{1}{2}$. $\frac{5\mathrm{a}\mathrm{e}}{2\mathrm{b}}=\frac{200000}{118}$ or about 1700; to w^ch $6\frac{1}{2}$ square-squared or multiplyed 3 times into it selfe is ab{illeg}|o|ut equall. Therefore I take only y^e two first termes of the rule $\frac{\mathrm{a}}{\mathrm{b}}in\mathrm{p}+\mathrm{r}\frac{\mathrm{c}\mathrm{c}}{\mathrm{b}\mathrm{b}}$: b {illeg} in the rest being of above 3 dimensions. And soe making $2\mathrm{n}+1\left(=7\right)=\mathrm{p}$. & $\frac{\mathrm{n}\mathrm{n}+\mathrm{n}}{3}\mathrm{p}\left(=28\right)=\mathrm{r}$, the desired aggregate will bee {illeg} $\frac{\mathrm{a}}{\mathrm{b}}\times 7+28\frac{\mathrm{c}\mathrm{c}}{\mathrm{b}\mathrm{b}}$ or $\frac{70000}{118}in\frac{14068}{13924}$, wanting about an eighth part of an unit. But if an exacter aggregate bee desired, take another terme of the rule & the error will not bee abo{illeg}|v|e $\frac{1}{350}$ of an unit. Thus if the said series were continued to 21 termes $\frac{10000}{100}$ being y^e first, $\frac{10000}{160}$ the middle, & $\frac{10000}{220}$ the last terme: three termes of the rule would give an aggregate too little by about $\frac{1}{2}$ of an unit, 4 termes by about $\frac{1}{12}$ or $\frac{1}{15}$ pt & 5 termes by about ${\frac{1}{100}}^{\text{th}}$ part, or lesse. But perhaps it might bee more convenient to resolve this at twice, first finding the aggregate of the last eleven termes, & then of y^e next nine, & lastly adding y^e first terme to the other two aggregates. And this may bee done to about the ${60}^{\text{th}}$ part of an unit by using onely y^e three first termes of the rule.

From these instances may bee guessed w^t is to bee done in other cases. But it may bee further noted y^t it will much expedite the work to subduct y^e Logarithm of b from y^t of c and multiply y^e remainder by $2.4.6.8$ &c w^ch products shall bee y^e Logarithms of $\frac{\mathrm{c}\mathrm{c}}{\mathrm{b}\mathrm{b}}.\frac{{\mathrm{c}}^4}{{\mathrm{b}}^4}.\frac{{\mathrm{c}}^6}{{\mathrm{b}}^6}.\frac{{\mathrm{c}}^8}{{\mathrm{b}}^8}$ &c. whose computation in propper numbers would bee troublesom.

This Probleme much resembles y^e squaring of the Hyperbola: That being only to find y^e aggregate of a series of fractions infinite in number & littlenesse, w^th one common numerator to denominators whose differences are equall & infinitely little. And as I referred all the series to y^e middle terme, the like may bee done conveniently in y^e Hyperbola. If AC AH are its rectangled Asymptot{es} & y^e area BDGE is desired: bisect BD in $\mathrm{G}$ {sic}, make $\mathrm{AC}=\mathrm{a}$ $\mathrm{CF}=\mathrm{b}$, & CD or $\mathrm{CB}=\mathrm{x}$. soe y^t $\frac{\mathrm{a}\mathrm{b}}{\mathrm{a}+\mathrm{x}}=\mathrm{DG}$ & $\frac{\mathrm{a}\mathrm{b}}{\mathrm{a}-\mathrm{x}}=\mathrm{BE}$. Then according to Mercator y^e area GDCF is $\mathrm{b}\mathrm{x}-\frac{\mathrm{b}\mathrm{x}\mathrm{x}}{2\mathrm{a}}+\frac{\mathrm{b}{\mathrm{x}}^3}{3\mathrm{a}\mathrm{a}}-\frac{\mathrm{b}{\mathrm{x}}^4}{4{\mathrm{a}}^3}$$+\frac{\mathrm{b}{\mathrm{x}}^5}{5{\mathrm{a}}^4}$ &c. & y^e area BCFE is $\mathrm{b}\mathrm{x}+\frac{\mathrm{b}\mathrm{x}\mathrm{x}}{2\mathrm{a}}+\frac{\mathrm{b}{\mathrm{x}}^3}{3\mathrm{a}\mathrm{a}}+\frac{\mathrm{b}{\mathrm{x}}^4}{4{\mathrm{a}}^3}$$+\frac{\mathrm{b}{\mathrm{x}}^5}{5{\mathrm{a}}^4}$ &c. And the summ of these two make the whole area $\mathrm{BDGE}=$$2\mathrm{b}\mathrm{x}+\frac{2\mathrm{b}{\mathrm{x}}^3}{3\mathrm{a}\mathrm{a}}+\frac{2\mathrm{b}{\mathrm{x}}^5}{5{\mathrm{a}}^5}$ &c. Where each other terme is wanting, & x is lesse by half then it would otherwise have beene, w^ch makes y^e series more converging towar{d} y^e truth.

As to yo^r other Problem about y^e resolution of Equations by tables. There may bee such Tables made for Cubick equations; & consequen{t}ly w^ch shall serve for those of foure dimensions too: But scarcely for any others. Indeed could all Equations bee reduced to three termes only, tables might bee made for all: but that's beyond my skill to doe it, & beleife that it can bee done. For those of three dimensions there needs but one colum{n} of figures bee added to y^e ordinary tables of Logarithms, & y^e construction of it is pretty easy & obvious enough. If you please I will some time send you a specimen of its composition & use, but I cannot perswade my selfe to undertake y^e drudgery of making it.

Yo^r Kinck-Huysons Algebra I have made some notes upon. I suppose you are not much in hast of it, w^ch makes me doe y^t onely at my leisure.

Your obliged friend & Servant

Is: Newton.

<1av>

These

To M^r John Collins his house in Bloomsbury next

doore to y^e three Crowns

in

London

|M^r Newton about the Musicall Progression|