<1r>

Trin: Coll: Cambridge.
Ian 1669

Sr

I received Dr Wallis his Mechanicks wch you sent to Mr 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 ye best solutions of one of them that I can contrive; Namely how to find ye aggregate of a series of fractions, whose numerators are the same & their denominators in arithmeticall progression. To doe this I shall propound two ways, The first by reduction to one common denominator as followeth.

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

If $\frac{a}{b-2c}.\frac{a}{b-c}.\frac{a}{b}.\frac{a}{b+c}.\frac{a}{b+2c}$, be the series: The factus of their denominators is $bin\stackrel{‾}{bb-cc}inbb-4cc$, or ${b}^{5}-5{b}^{3}cc+4b{c}^{4}$, the denominator; wch multiplyed as before gives $5a{b}^{4}-15abbcc+4a{c}^{4}$, ye Numerator of ye aggregate.

If $\frac{a}{b-c}.\frac{a}{b+c}.\frac{a}{b+3c}$ &c is ye series. Then $\stackrel{‾}{bb-cc}×b+3c$ or ${b}^{3}+3bbc-bcc-3{c}^{3}$ is ye denominator and $3abb+6abc-acc$ ye numerator of the Aggregate.

In numbers: If $\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}$ is ye series, then putting $b=2$ yt ye series may bee $\frac{1}{b}.\frac{1}{b+1}.\frac{1}{b+2}.\frac{1}{b+3}.\frac{1}{b+4}$; ye factus of their denominators will bee ${b}^{5}+10{b}^{4}+35{b}^{3}+50bb+24b$ the denominator, & consequently $5{b}^{4}+40{b}^{3}+105bb+100b+24$ the denomin|numer|ator of ye aggregate.

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

$\begin{array}{cc}\begin{array}{c}\text{The annexed table will}\\ \text{much faciliate}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{multiplication}\\ \text{of denominators together.}\end{array}& \begin{array}{cc}\begin{array}{c}\begin{array}{rrrrrr}& & & & +1& -1\\ & & & +1& -5& +4\\ & & +1& -14& +49& -36\\ & +1& -30& +273& -820& +576\\ +1& -55& +1023& -7645& +51276& -14400\end{array}\\ \begin{array}{cccccc}{b}^{11}.& {b}^{9}cc.& {b}^{7}{c}^{4}.& {b}^{5}{c}^{6}.& {b}^{3}{c}^{8}.& b{c}^{10}\\ \phantom{+1}& \phantom{-55}& \phantom{+1023}& \phantom{-7645}& \phantom{+51276}& \phantom{-14400}\end{array}\end{array}& \begin{array}{cc}& \begin{array}{l}\text{for 3 terms}\\ \text{for 5 termes}\\ \text{for 7 termes}\\ \text{for 9 termes}\\ \text{for 11 termes}\end{array}\end{array}\end{array}\end{array}$

This rule holds good though the differences of the denominators bee not equall: as if $\frac{a}{b+c}.\frac{a}{b+d}.\frac{a}{b-e}$ are to bee added the factus of their denominators is ${b}^{3}+bbc+bbd-bbe+bcd-bce-bde-cde$ the denominator, which multiplyed by the dimensions of b & againe by $\frac{a}{b}$ produces $3abb+2abc+2abd-2abe+acd-ace$$-bde$ 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 $\frac{pp-p}{2}=q$. $\frac{2pq-q}{3}=r$. $qq=s$. $\frac{6qr-r}{5}=t$. $\frac{4qs-s}{3}=v$. $\frac{12rs-5t}{7}=x$. $2ss-v=y$. $\frac{rv-rs}{3}+t=z$. &c. Now if the propounded series \bee/ $\frac{a}{b}.\frac{a}{b+c}.\frac{a}{b+2c}.\frac{a}{b+3c}$ &c their Aggregate shall bee $\frac{a}{b}inp-q\frac{c}{b}+r\frac{cc}{bb}-s\frac{{c}^{3}}{{b}^{3}}-t\frac{{c}^{4}}{{b}^{4}}$ &c: In wch progression the farther you proceede, the nearer you approach to truth.

But its {sic} better to put b for the Denominator of ye middle termes of ye propounded series, thus $.\frac{a}{b-2c}.\frac{a}{b-c}.\frac{a}{b}.\frac{a}{b+c}.\frac{a}{b+2c}$. And making n the number of termes from ye said middle terme either way; as also {illeg} $nn+n=m$. $2n+1=p$. $\frac{mp}{3}=r$. $\frac{3mr-r}{5}=t$. {illeg} $\frac{3mmr-5t}{7}=x$. $\frac{{m}^{3}r-2mmr}{3}+t=z$. &c, the desired Aggregate shall bee $\frac{a}{b}inp+r\frac{cc}{bb}$$+t\frac{{c}^{4}}{{b}^{4}}+x\frac{{c}^{6}}{{b}^{6}}+z\frac{{c}^{8}}{{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 $\frac{1}{e}$ part of an unit. Then make a rude guesse how many times $\frac{nc}{b}$ $\frac{b}{nc}$ multiplyed into it selfe will bee about ye bignesse of $\frac{5ae}{2b}$ 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 ye exactnesse of $\frac{1}{8}\text{th}$ pt of an unite Then is $a=10000$. $b=118$. $c=6$. $n=3$. $e=8$. $\frac{b}{nc}=\frac{118}{18}$ or about $6\frac{1}{2}$. $\frac{5ae}{2b}=\frac{200000}{118}$ or about 1700; to wch $6\frac{1}{2}$ square-squared or multiplyed 3 times into it selfe is ab{illeg}|o|ut equall. Therefore I take only ye two first termes of the rule $\frac{a}{b}inp+r\frac{cc}{bb}$: b {illeg} in the rest being of above 3 dimensions. And soe making $2n+1\left(=7\right)=p$. & $\frac{nn+n}{3}p\left(=28\right)=r$, the desired aggregate will bee {illeg} $\frac{a}{b}×7+28\frac{cc}{bb}$ 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 ye 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 ye next nine, & lastly adding ye 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 ye three first termes of the rule.

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

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

As to yor other Problem about ye resolution of Equations by tables. There may bee such Tables made for Cubick equations; & consequen{t}ly wch 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 ye ordinary tables of Logarithms, & ye 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 ye drudgery of making it.

Yor Kinck-Huysons Algebra I have made some notes upon. I suppose you are not much in hast of it, wch makes me doe yt onely at my leisure.

Is: Newton.

<1av>

## London

|Mr Newton about the Musicall Progression|

Professor Rob Iliffe
Director, AHRC Newton Papers Project

Scott Mandelbrote,
Fellow & Perne librarian, Peterhouse, Cambridge

Faculty of History, George Street, Oxford, OX1 2RL - newtonproject@history.ox.ac.uk

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