Trinity College Cambridge.
Ian 1669


I received Dr Wallis his Mechanicks which 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 the best solutions of one of them that I can contrive; Namely how to find the 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 ab . ab+c . ab+2c . ab+3c &c be the series: Multiply all their denominators together, & the product will bee b4+6b3c+11bbcc+6bc3 4321; each terme of which being multiplyed by its dimensions of b, & the product againe multiplyed by ab, the result shall bee the Numerator of the desired aggregate 4ab3+18abbc+22abcc+6ac3 b3+6b3c+11bbcc+6bc3

If ab2c . abc . ab . ab+c . ab+2c, be the series: The factus of their denominators is binbbccinbb4cc, or b55b3cc+4bc4, the denominator; which multiplyed as before gives 5ab415abbcc+4ac4, the Numerator of the aggregate.

If abc . ab+c . ab+3c &c is the series. Then bbcc×b+3c or b3+3bbcbcc3c3 is the denominator and 3abb+6abcacc the numerator of the Aggregate.

In numbers: If is the series, then putting b=2 that the series may bee 1b . 1b+1 . 1b+2 . 1b+3 . 1b+4 ; the factus of their denominators will bee b5+10b4+35b3+50bb+24b the denominator, & consequently 5b4+40b3+105bb+100b+24 the numerator of the aggregate.

But it is better to put b=4, that the series may bee 1b2 . 1b1 . 1b . 1b+1 . 1b+2 . And so shall the aggregate bee 5b415bb+4 b55b3+4b .

The annexed table willmuch faciliateye multiplicationof denominators together. +11 +15+4 +114+4936 +130+273820+576 +155+10237645+5127614400 b11. b9cc. b7c4. b5c6. b3c8. bc10 +155+10237645+5127614400 for 3 terms for 5 termes for 7 termes for 9 termes for 11 termes

This rule holds good though the differences of the denominators bee not equall: as if ab+c . ab+d . abe are to bee added the factus of their denominators is b3+bbc+bbdbbe+bcdbcebdecde the denominator, which multiplyed by the dimensions of b & againe by ab produces 3abb+2abc+2abd2abe+acdace bde the numerator of the desired summ.

The other way of resolving this Probleme is by approximation. Suppose the number of termes in the propounded series bee p. And make ppp2=q. 2pqq3=r. qq=s. 6qrr5=t. 4qss3=v. 12rs5t7=x. 2ssv=y. rvrs3+t=z. &c. Now if the propounded series bee ab . ab+c . ab+2c . ab+3c &c their Aggregate shall bee abinpqcb+rccbbsc3b3tc4b4 &c: In which progression the farther you proceede, the nearer you approach to truth.

But it is better to put b for the Denominator of the middle terme of the propounded series, thus .ab2c . abc . ab . ab+c . ab+2c. And making n the number of termes from the said middle terme either way; as also nn+n=m. 2n+1=p. mp3=r. 3mrr5=t. 3mmr5t7=x. m3r2mmr3+t=z. &c, the desired Aggregate shall bee abinp+rccbb +tc4b4+xc6b6+zc8b8 &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 <1v> you would have their aggregate; suppose not erring from truth above 1e part of an unit. Then make a rude guesse how many times bnc multiplyed into it selfe will bee about the bignesse of 5ae2b 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 10000100+10000106+10000112+10000118+10000124+10000130+10000136 bee desired to the exactnesse of 18th part of an unite Then is a=10000. b=118. c=6. n=3. e=8. bnc=11818 or about 612. 5ae2b=200000118 or about 1700; to which 612 square-squared or multiplyed 3 times into it selfe is about equall. Therefore I take only the two first termes of the rule abinp+rccbb: b in the rest being of above 3 dimensions. And soe making 2n+1=7=p. & nn+n3p=28=r, the desired aggregate will bee ab×7+28ccbb or 70000118in1406813924, 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 1350 of an unit. Thus if the said series were continued to 21 termes 10000100 being the first, 10000160 the middle, & 10000220 the last terme: three termes of the rule would give an aggregate too little by about 12 of an unit, 4 termes by about 112 or 115 part & 5 termes by about 1100th 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 the next nine, & lastly adding the first terme to the other two aggregates. And this may bee done to about the 60th part of an unit by using onely the three first termes of the rule.

From these instances may bee guessed what is to bee done in other cases. But it may bee further noted that it will much expedite the work to subduct the Logarithm of b from that of c and multiply the remainder by &c which products shall bee the Logarithms of ccbb.c4b4.c6b6.c8b8 &c. whose computation in propper numbers would bee troublesom.

This Probleme much resembles the squaring of the Hyperbola: That being only to find the aggregate of a series of fractions infinite in number & littlenesse, with one common numerator to denominators whose differences are equall & infinitely little. And as I referred all the series to the middle terme, the like may bee done conveniently in the Hyperbola. If AC AH are its rectangled Asymptot{es} & the area BDGE is desired: bisect BD in C=b, make AC=a Figure CF=b, & CD or CB=x. soe that aba+x=DG & abax=BE. Then according to Mercator the area GDCF is bxbxx2a+bx33aabx44a3 +bx55a4 &c. & the area BCFE is bx+bxx2a+bx33aa+bx44a3+bx55a4 &c. And the summ of these two make the whole area BDGE= 2bx+2bx33aa+2bx55a5 &c. Where each other terme is wanting, & x is lesse by half then it would otherwise have beene, which makes the series more converging towar{d} the truth.

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

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

Your obliged friend & Servant

Is: Newton.



To Mr John Collins his house in Bloomsbury next

doore to the three Crowns



Mr Newton about the Musicall Progression

© 2024 The Newton Project

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

Privacy Statement

  • University of Oxford
  • Arts and Humanities Research Council
  • JISC