<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 ab . ab+c . ab+2c . ab+3c &c {illeg} be the series: Multiply all their denominators together, & the product will bee b4+6b3c+11bbcc+6bc3 4321; each terme of wch being multiplyed by its dimensions of b, & the product againe multiplyed by ab, the result shall bee the Numerator of the desired aggregate {illeg} 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; wch multiplyed as before gives 5ab415abbcc+4ac4, ye Numerator of ye aggregate.

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

In numbers: If 12.13.14.15.16 is ye series, then putting b=2 yt ye series may bee 1b . 1b+1 . 1b+2 . 1b+3 . 1b+4 ; ye factus of their denominators will bee b5+10b4+35b3+50bb+24b the denominator, & consequently 5b4+40b3+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 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 Problems {sic} is by approximation. Suppose the number of termes in the propounded series of 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 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 .ab2c . abc . ab . ab+c . ab+2c. And making n the number of termes from ye said middle terme either way; as also {illeg} nn+n=m. 2n+1=p. mp3=r. 3mrr5=t. {illeg} 3mmr5t7=x. m3r2mmr3+t=z. &c, the desired Aggregate shall bee abinp+rccbb +tc4b4+xc6b6+zc8b8 &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 1e part of an unit. Then make a rude guesse how many times ncb bnc multiplyed into it selfe will bee about ye 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 c|t|o ye exactnesse of 18th pt 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 wch 612 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 abinp+rccbb: b {illeg} in the rest being of above 3 dimensions. And soe making 2n+1=7=p. & nn+n3p=28=r, the desired aggregate will bee {illeg} 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 abo{illeg}|v|e 1350 of an unit. Thus if the said series were continued to 21 termes 10000100 being ye 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 pt & 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 ye next nine, & lastly adding ye first terme to the other two aggregates. And this may bee done to about the 60th 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 ccbb.c4b4.c6b6.c8b8 &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 G {sic}, make AC=a Figure CF=b, & CD or CB=x. soe yt aba+x=DG & abax=BE. Then according to Mercator ye area GDCF is bxbxx2a+bx33aabx44a3 +bx55a4 &c. & ye 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, 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.

Your obliged friend & Servant

Is: Newton.

<1av>

These
To Mr John Collins his house in Bloomsbury next
doore to ye three Crowns
in
London

|Mr Newton about the Musicall Progression|

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Professor Rob Iliffe
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