Letter from Newton to John Collins, dated 19 January 1669/70 Isaac Newton c. 1,213 words The Newton Project Falmer 2012 Newton Project, University of Sussex

2 ff.

Published in H.W. Turnbull (ed), The Correspondence of Isaac Newton, vol. 1 (Cambridge: 1959), pp. 16-21

Letter from Newton to John Collins, dated 6 February 1669/70 [MS Add. 9597/2/18/2]
UKCambridgeCambridge University Library Macclesfield Collection MS Add. 9597/2/18/1
19 January 1669/70 England English Holograph Unknown Cataloguer Unknown Hand MathematicsCorrespondence Daniele Cassisa started tagged transcription Catalogue information compiled from CUL Janus Catalogue by Michael Hawkins Proofed by Robert Iliffe Code audit by Michael Hawkins
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Trin:Trinity Coll:College Cambridge. Ian 1669

SrSir

I received Dr Wallis his Mechanicks wchwhich 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 yethe best solutions of one of them that I can contrive; Namely how to find yethe 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 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 wchwhich 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 $\begin{array}{ccc}& \frac{4a{b}^{3}+18abbc+22abcc+6a{c}^{3}}{{b}^{3}+6{b}^{3}c+11bbcc+6b{c}^{3}}& \end{array}$

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; wchwhich multiplyed as before gives $5a{b}^{4}-15abbcc+4a{c}^{4}$, yethe Numerator of yethe aggregate. If $\frac{a}{b-c}.\frac{a}{b+c}.\frac{a}{b+3c}$ &c is yethe series. Then $\stackrel{‾}{bb-cc}×b+3c$ or ${b}^{3}+3bbc-bcc-3{c}^{3}$ is yethe denominator and $3abb+6abc-acc$ yethe numerator of the Aggregate. In numbers: If $\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}$ is yethe series, then putting $b=2$ ytthat yethe series may bee $\frac{1}{b}.\frac{1}{b+1}.\frac{1}{b+2}.\frac{1}{b+3}.\frac{1}{b+4}$; yethe 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 denominnumerator of yethe aggregate. But itsit is better to put yethe $b=4$, 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$$-bde$ the numerator of the desired summ. The other way of resolving this ProblemsProbleme is by approximation. Suppose the number of termes in the propounded series of bee $p$. And make $\frac{pp-p}{2}=q$. $\frac{2pq-q}{3}=r$. $qq=s$. $\frac{6qr-r}{5}=t$. $\frac{4qs-s}{3}=v$. $\frac{12rs-5t}{7}=x$. $2ss-v=y$. $\frac{rv-rs}{3}+t=z$. &c. Now if the propounded series bee $\frac{a}{b}.\frac{a}{b+c}.\frac{a}{b+2c}.\frac{a}{b+3c}$ &c their Aggregate shall bee $\frac{a}{b}inp-q\frac{c}{b}+r\frac{cc}{bb}-s\frac{{c}^{3}}{{b}^{3}}-t\frac{{c}^{4}}{{b}^{4}}$ &c: In wchwhich progression the farther you proceede, the nearer you approach to truth. But itsit is better to put $b$ for the Denominator of yethe middle termes of yethe propounded series, thus $.\frac{a}{b-2c}.\frac{a}{b-c}.\frac{a}{b}.\frac{a}{b+c}.\frac{a}{b+2c}$. And making $n$ the number of termes from yethe said middle terme either way; as also $nn+n=m$. $2n+1=p$. $\frac{mp}{3}=r$. $\frac{3mr-r}{5}=t$. $\frac{3mmr-5t}{7}=x$. $\frac{{m}^{3}r-2mmr}{3}+t=z$. &c, the desired Aggregate shall bee $\frac{a}{b}inp+r\frac{cc}{bb}$$+t\frac{{c}^{4}}{{b}^{4}}+x\frac{{c}^{6}}{{b}^{6}}+z\frac{{c}^{8}}{{b}^{8}}$ &c: 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 you would have their aggregate; suppose tnot 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 yethe 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 cto yethe exactnesse of $\frac{1}{8}\text{th}$ ptpart 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 wchwhich $6\frac{1}{2}$ square-squared or multiplyed 3 times into it selfe is about equall. Therefore I take only yethe two first termes of the rule $\frac{a}{b}inp+r\frac{cc}{bb}$: $b$ 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 $\frac{a}{b}×7+28\frac{cc}{bb}$ 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 above $\frac{1}{350}$ of an unit. Thus if the said series were continued to 21 termes $\frac{10000}{100}$ being yethe 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 $\frac{1}{15}$ ptpart & 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 yethe next nine, & lastly adding yethe 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 yethe three first termes of the rule. From these instances may bee guessed wtwhat is to bee done in other cases. But it may bee further noted ytthat it will much expedite the work to subduct yethe Logarithm of $b$ from ytthat of $c$ and multiply yethe remainder by $2.4.6.8$ &c wchwhich products shall bee yethe 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 yethe squaring of the Hyperbola: That being only to find yethe aggregate of a series of fractions infinite in number & littlenesse, wthwith one common numerator to denominators whose differences are equall & infinitely little. And as I referred all the series to yethe middle terme, the like may bee done conveniently in yethe Hyperbola. If $AC$ $AH$ are its rectangled Asymptotes & yethe area $BDGE$ is desired: bisect $BD$ in $\mathrm{G}$$C=b$, make $AC=a$ $CF=b$, & $CD$ or $CB=x$. soe ytthat $\frac{ab}{a+x}=DG$ & $\frac{ab}{a-x}=BE$. Then according to Mercator yethe area $GDCF$ is $bx-\frac{bxx}{2a}+\frac{b{x}^{3}}{3aa}-\frac{b{x}^{4}}{4{a}^{3}}$$+\frac{b{x}^{5}}{5{a}^{4}}$ &c. & yethe 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, wchwhich makes yethe series more converging toward yethe truth. As to yoryour other Problem about yethe resolution of Equations by tables. There may bee such Tables made for Cubick equations; & consequently wchwhich 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 column of figures bee added to yethe ordinary tables of Logarithms, & yethe 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 yethe drudgery of making it. YorYour Kinck-Huysons Algebra I have made some notes upon. I suppose you are not much in hast of it, wchwhich makes me doe ytthat onely at my leisure. Your obliged friend & Servant Is: Newton.
These To Mr John Collins his house in Bloomsbury next doore to yethe three Crowns in London Mr Newton about the Musicall Progression