<2r>

Of ye extraction of \Pure/ Square Cubick. Square-square & square-cubick rootes &c.

Let ye number whose roote is to bee extracted bee pointed bec makeing ye first point under ye {ut|n|ite} & comprizeing soe many numbers under each point as ye number hath dimensions as if ye number be square-cube tis thus pointed 5708.63524.10802.

Then then then t out of ye figures of ye first point next ye left hand extract ye \greatest/ roote proper to ye power of ye number & set yt downe in ye m{illeg}|Quo|tient wch is ye firt{illeg} {sic} side & is called A. (as ye roote quintuplicate of 5708 is (5) , & (5) quintuplicate is 3125 ) yn takeing yt roote duely multiplied out of ye number (as 3125 out of 5708 ) wth ye rest of ye numbers to ye next point. |seeke| {illeg}|y|e seacond side wch is found by divideing yt number by another number made out of ye first side (wch is called ye Divisor) & this second side I name E. (thus by divideing 258363524 by 5A/qq\\+10Ac+10Aq+5A / after such a mane {sic} yt 5 Aqq E + 10 Ac Eq + 10 Aq Ec + 5 A Eqq + Eqc may be conteined in ye number ye product of yt division shall be E =

<2v>

The extraction of ye sqare {sic} roote

The square to be resolved29.16.(54 The Product The square of  ye  first side250be taken away. The rest of  ye  sqare to be0416resolved. The divisor for finding  ye  seacond010sidie. which is  ye  first side doubled The rectangle by 2A & E04 The square of E0 00g 16q } to be substracted The sum¯e of  ye  rectangles40 16to be subducted 0 00The remainder

The extraction of ye cube roote

The cube to be resolved157. 464.(54 The cube to be subducted125 whose roote isA=5 The remainder foryefinding32 464of E The divisors foryefinding of (E)yeseacond side. { 0 7 0 0 5 0 15 0 3Aq 3A The sume ofyedivisors07 650 Sollids to be substracted { 0 30 0 2 0 0 0 0 0 40 0 064 3AqE 3AEq Ec The sume of those32 464sollids The remainder00 000

The extraction of ye square square roote

The square-square33. 1776.(24 The square-squ: to be subduc:16 =Aqq Remainder.17 1776 Divisors for findingye seacond side E. { 0 3 0 0 0 0 2 0 24 0 008 0 4Ac 6Aq 4A Theire sum¯e03 4480 Squ-Squares to be sub= =ducted { 12 3 0 0 8 84 5120 0256 4AcE 6AqEq 4AEc Eqq Theire Sum¯e17 1776

<3r>

The extraction of ye Square-Cube roote

The squ: cube to be resolved79. 62624.(24 Substract32 Aqc Remainde47 62624 Divisors { 5q 8 0 0 0 0 q 0 05q 80 Ac 0400 Aq 0010 A 5Aqq 10Ac 10Aq 5A The Sume ofyedivisors08 84100 Plano-Sollids to be substracted { q 32 q 12 q 02 q 00 q 00 0 q 80 q 5600 q 25600 q 01024 q 5AqqE 10AcEq 10AqEc 5AEqq Eqc. Theire Sum¯e47 62624 Remainder00 00000

Note yt ye 3d 4th 5th & other figures are found by ye same mann{illeg}|e|r yt ye seacond figure is found onely makeing all ye figures found to stand for A ye first side & {illeg} ye figure sought for e or ye 2d side

And if ye number {propounded} be {t} roote is found inexpressible in whole numbers yt|n| adding ciphers & pointing them f{illeg}rom ye unite towards ye right \{kind}/ as was before exp{illeg}|l|ained & soe hold on ye worke in decimalls.

As for ye Divisors they are easily found by ye 2d Table of Powers from a Binomial roote.

If ye {illeg} Number bee of 6.7.8.9.10 &c dimensions The roote may be extracted after ye same manner

<4r>

Of ye Exta{illeg}|c|tion {sic} of R{illeg}|o|otes in Affected powers.

The manner of ye extraction of rootes in pure & affected powers is verry {sic} much alike, \especially/ when ye affected powers are decently prepared, yt is, when theire affections are not over large & those altogether either affirmative or negative, & ye power affirmative, affirmations & negations so mixt yt there be noe ambiguity & all fractions & Asymmetry taken away

All ye figures in ye coeffents {sic} & affected power are to be pointed (af{illeg}|t|er ye manner before eplained {sic} in ye Analisis of pure powers) according to ye degree of theire dimensions & the worke onely differs from yt in pure powers yt in yt {illeg}|ye| coefficients enter into ye divisors

Let ye first side be called A. ye 2d be called E. ye Roote of ye equation B{L} ye coefficients B . Cq . Dc . Fqq . Gqc . Hcc &c ye Power P . Pq . Pc . Pqq &c & ye Operation follows

The analysis of Cubick Equations.

The equation supposed Lc* + 30L = 14356197 . Lc + CqL = Pc
0.The square coëfficient 0.The cube affected to be 0 0. 1 4. 0. 0 3 3 5 6. 0. 0. 0. 1 9 7. (243 Sollids to bee substracted { 0 8 0 0 0 0 0 0 0 6 0 0 0 0 0 0 =Ac =ACq Theire sum¯e 0 8 0 0 6 0 0 0 Rests 0 6 3 5 0 1 9 7 for findingye2dside The extraction ofyeseacond side 0.Coëfficient 0.The rest ofyecube to be 0. 0 0. 6 0 0. 0 3 5 0. 3 0. 0. 1 9 7. or superior divisor0. resolved0. The inferior divisors { 0 1 0 0 2 0 0 0 6 0 0 0 0 0 0 0 3Aq 3A Theire sum¯e 0 1 2 6 0 3 0 0 Sollids to be sub= stracted { 0 4 0 0 0 0 0 0 8 0 0 9 6 0 0 6 4 0 0 1 0 0 0 0 0 0 0 0 0 2 0 0 =3AqE =3AEq =Ec =ECq Theire sum¯e The superior part ofyedivisor 0 5 0 8 2 5 0 2 0 0 0

<4v>

The superior part ofyedivisor The remainder for finding 0 0 0 0 0 0 0 5 2 4 0 3 0. 9 9 7. oryesquare coefficient yethird side The inferior part ofye divisor { 0 0 1 7 2 0 0 0 8 0 0 7 2 0 3Aqthat is3×24×24 3Aor3×24. The sum¯eofyedivisor 0 0 1 7 3 5 5 0 Sollids to be taken away { 0 0 0 0 0 0 0 0 5 1 8 0 0 6 0 0 0 0 0 0 4 0 0 4 8 0 0 2 7 0 0 9 0 0 3AqE 3AEq Ec ECq Theire Sum¯e 0 0 5 2 4 9 9 7 Remaines 0 0 0 0 0 0 0 0

But ye Coëfficient maybe greater yn ye Power soe yt it cannot be substracted from it wch argues yt ye Cube {illeg} more propperly affects yn is affected. In this case ye coëfficient must descend towards ye unite soe many points untill it may be substracted, & soe many points as ye coëfficient is devolved soe many pricks {illeg}|mu|st be blotted out towards ye left hand in ye power affected. As ye Example shews
Lc+95400L = {illeg} / =1819459 . \

Sollids to be substracted { 0 9. 0 1. 5. 4 0 8 1 9. 0. 0. 0. 4 5 9. Coefficiens0. The Power0. Since9 is greateryn1make a devolution thus. 0.T 0.The Quote (19 0. 0 0. 1. 9. 5. 4. 8 1 9. 0 0. 0. 4 5 9. The Coefficient0. The affected power0. Sollids to be substracted { 0 0 0 0 9 5 4 0 0 1 0 0 0 0 0 0 ACq Ac Sum¯a 0 0 9 5 5 0 0 0 substrahenda Divisoru¯superior pars 0 0 0. 9 5 4 0 0. Coefficiens Planum 0 0 8 6 4 4 5 9 Potestas reliqua Divisoru¯pars inferior { 0 0 0 0 0 0 0 0 0 0 3 0 0 0 3 0 3Aq 3A Divisoru¯   sum¯a 0 0 0 9 5 7 3 0 Sollida ablativa { 0 0 0 0 0 0 0 0 8 5 8 0 0 2 0 0 2 0 0 0 6 0 0 7 0 0 4 3 0 7 2 9 ECq 3AqE 3AEq 3Ec Eoru¯  Summa 0 0 8 6 4 4 5 9 Restat 0 0 0 0 0 0 0 0

<5r>

To place ye unite of ye coefficient in its right place in respect of ye power make so many pricks above as there are under ye power begining at ye unit, & if ye coefficient be one dimension lesse yn ye power make a prick on every figure if 2 dimensions les yn every other figure of 3 dimensions lesse make it one each third figure &c

If there be many coefficients in ye equation each must be placed according to this rule.

Sometimes ye coefficient is under a negative sine as Lc10L=13584 & ye Analysis is as follows

Coëfficiens planum0 10.0sublaterale Cubus resolvendus+13. 584.(24 Solida ablativa { +08 0 0 20 Ac ACq Sum¯a+07 800 Restat+0.5 784.resolvendum Divisoru¯p_ssuperior+00 -10coëfficiens planum Divisoru¯p_sinferior { +01 +01 20 06 +3AA +3A Suma divisoru¯+01 250 Solida ablativa { +04 +0 +0 0 0 800 960 064 040 3AAE 3AEE EEE ECC Eoru¯  sum¯a+05 784

But sometimes ye square coëfficient hath more paires of figures yn ye cube to be analysed, \hath/ & yn \there is/ præfixing so many ciphers to ye cube as figures are wanting, ye first side will not much differ from ye square roote of ye coefficient. as Lc116620L=352947

1 1. 6 6 2. 0. 0. 0. Coefficiens planu¯ Cubus resolvendus + 0 0. 3 5 2. 9 4 7. (343 Sollida Ablativa { + 2 7 3 4 0 0 0 9 8 6 0 0 0 Ac ACq Restat auferendu¯ 0 7 9 8 6 0 0 0 Reliquum resolvendi 0 + 8 3 3 8 9 4 7 Cubi ________________________________________________________

<5v>

Divisoru¯p_ssuperior Coeff: 0 1. 1 6 6. 2 0. 0. planum. Reliquu¯  resolvendi cubi 0 + 8. 3 3 8. 9 4 7. negative affecti Divisoru¯p_sinferior { + + 2 + + 0 7 0 0 0 9 0 0 0 0 3AA. 3A. Sum¯a Divisorum { + * * + + 1. * * * 6 2 3. * * * 8 0. 0 ************ 3AA + 3A + Cq Sollida ablativa { + 1 0 0 + 1 0 + 0 0 4 8 0 0 4 4 0 0 6 4 6 6 4 0 0 0 0 0 0 0 0 0 8 0 0 3AAE 3AEE EEE CCE Eorum summa + + 7 6 3 9 2 0 0 Restat Resolvend¯ + + 0. 6 9 9. 7 4 7. pro3olatere Divisorum¯p_ssuperior ~ + 0. 1 1 6. 6 2 0. CC Divisoru¯p_sinferior { + + 0 + + 0 3 4 6 0 0 1 8 0 0 0 2 0 3AA 3A *********** * * * * * * * * * * Eorum Summa 0 0 0 2 3 1 2 0 0 =3AA+3A+CC Sollida ablativa { 0 + 1 0 + 0 0 + 0 0 0 0 4 0 0 0 9 0 0 0 3 4 9 4 0 0 1 8 0 0 2 7 8 6 0 3AAE 3AEE EEE ECC Eorum Summa 0 + 0 6 9 9 7 4 7 ________________________________________________________

Sometimes though there be as many 2 figures in the coefficient as 3 figures \in/ ye cube affected yet ye coëfficient may be so greate as to deceive an uwary {sic} Analist As in this Lc6400L=153000 . where ye roote of 64 is 8 wch cubed is 512 wch added to 153 makes 665 thē whose \roote/ ye number immediately greater is 9 wch make is ye first sid {sic} =A .

But if ye coefficient had beene affirmative, yn not ye aggregate of ye facts but ye difference must be taken as in this. Lc+64L=1024 .

Since ye roote of 64 is 8 . wch cubed is 512 . & 1024512= 512 . ye roote of wch is 8=A . The like is observable in equations of higher powers

If ye Cube be affected wth a negative sine as 13,104LLc=155,520 . Then ye Equation is expressible of 2 rootes: whereof is less ye other greater yn {i} ye square of one is <6r> lesse & the square of ye other is greater ther|n|e 131043 . & therefore one roote is lesse ye other greater then 15552013104 . & in this equation 27755LLqq=217944 are two rootes whereof one is greater ye other lesse then 21794427755 .

Suppose in ye former \cubick/ equation ye lesse roote be 12. yn 15552012=12960 . or else 1310412×12=12960 . & Lq+12L=12960 . where L=108 is ye greater roote.

And in ye latter equation if ye lesse |greater| roote be 27. & 21794427=8072 , c. or 27×27×27 = | + | 27755=8072 . {illeg} 27×27=729 . If there be 4 cubes continully {sic} proportionall whose greate extreame is 27c=19683 . & ye aggregate of ye 3 rest is 8072 & Lc ye lesse extreame, therefore Lc+27Lq+729 {illeg} | L | =8072 . ye roote of wch ./is\ 8 ye other roote of ye equation

Or haveing one roo{illeg}|t|e of an equation ye e|E|quation may be lessoned {sic} by division thus 13104llc {illeg}| = | 155520 or l3 {illeg} 13 {illeg} | 1 | 04l+155520=0 . & one roote is 12 . therefore divide this equation by l12 & the Quote is an equation conteing {sic} ye ot{illeg}|h|er roote viz: lq+12l=12960 .

<7v>

To find two meane proportionalls twixt BC & IK Figure

<8r>

Propositiones Geometricæ. Franc: Vietæ.

prop 1

Figure

abaccebd

pro{illeg}|p| 2

& if abacacbd: then acababce

prop 3. If ab×ac=bd×ce . then bdacacababce

prop 3. To find two mene {sic} proportionalls {twixt} Bc & IK . On ye center a wth the Rad ai describe Figure the circle ibck . inscribe b {illeg} | c | =cd . draw da through ye center & bg parallel to it. draw hk through A soe yt gh=ab=(=ai) . & ikhbhbhihibc . ∺

Figure

Prop: 4

I{illeg}f ad=db=cb . yn ye Angle {illeg} / c \ be is tripple to ye Angle abd .

Prop 5

{illeg}|I|f ab=bd=Rad. 3Ang:bad=cde

Figure Figure

Prop 6

Figure

If 3rpq=spq:recto . that is If 2qr=pr . then 3or×or=sp×sp+op×op+px×px

Prop 7

Figure

Iaf If ad=dc=ce=ef . then Figure ecf=efc=3dac=3dca & AC3=3AC×ad2+Cf×ad2 . Z3=aaz+b3.~ ~ ~

Figure

prop 8. If op=pr=qr=qs . yn, prt=3qsr &c and sr3=3sr×qr2or×qr2 . Z3=aazb3

Figure

prop 9 If{illeg} op=pr= ah=hb=bf=fd . & ch=2eh or ceh=3hce . then bg3 ac3=3ac×ah2db×ah2&cb3=3cb×ah2db×ah2 }Z3=aazb3.

<8v>

Prop 10

Figure

If de=ea & dbdaab×abdc×dc . yn be is a side of a 7 equall sided & angled figure. or 7eab=4right angles .

Figure

prop 10

If ac=ef . & aef a right angle & ab pas {sic} through ye center then cddededfdfdb . And if cddededfdfdb then ae is perpendicular to ef . 2hd is ye difference of ye extreames & 2do is ye difference of ye meanes. wch given ye proportionall lines may be found &c.

Figure

prop 11

\Pse{illeg}|u|domesolabium wherby/ To find 2 meane proportionalls. {illeg} If, aeececededeb . they be inscribed in the {illeg}|c|ircle acbd the diam : being ae+eb . If twixt g {illeg} | i | & {illeg} | i | h two meane proportionalls are sought on ye same center f wth ye Rad : gi+ih2 describe gkhl & inscribe a line kl parallel to cd cutting ab in ye point e | i | & gikikiililih .      Examine it.

[1] Figure

prop 12

If do=dh & ac bisected in b & bd bee drawne rd is i|ye| side of a pentagon wch may be inscribed in defcro

prop 13.

Figure

If rd be the side of a {octogondecagon} & pd ye side of an {hexagonoctogon} ye arch rp divided equally in o , od will be {illeg}|th|e side of an {heptagonenneagon} to be inscribed in ye circle ord & ye arch RP is bisected \rightly divided/ by Bisecting ye Line ac . /Examine it\

<9r>

Of Angular sections.

Figure

prop 14

If ead=cab . Then abab2cbeb×adae×dbacae×ad+eb×db or, abaq×abopeb×anae×nqaoae×an+eb×nq . But ye angles anq , aop are right ones and {illeg} | e | an=oap {illeg} | = | eabdab

prop 15

If \ye angle/ cab+dab=eab . or naq+oaq=eab . & anq , aop are right ang\les/ then Abab2Ebad×bc+ac×dbeaad×acdb×cb . or the triang: unequall. abap×aqeban×op+ao×nqeaan×aonq×op

prop 16.

In 2 rectang: triang: acb & aed , if ye first have an acute angle cab submultiple to the acute angle eab of ye 2d triang aeb ye si{illeg}|d|es of ye seacond have this proportion. Suppose ye Hypoten of ye first tri: be z . ye base b . ye Cathetus c .

If  ye  acute angle ofye  seacond triangle beto  ye  acute angle ofye  first triangle ina proportion{   Hypoten:BasePerpendicular Duple,Z2.B2.-C2.2BC. Triple,Z3.B3.-3DDC.3BBC.-C3. Quadruple,Z4.B4.-6B2C2.+C4.4B3C.-4BC3. Quintuple,Z5.B5.-10B3C2.+5BC4.5B4C.-10B2C3.+C5.

Figure

Prop 17. I{illeg}|f| {\/} ab=bc=ce=eg &c: & ac=cd . & ae=ef &c then abacacadaeaf &c /& ed=ab & ac=gf &c. {nam} △ cde & cba , efg & eac &c: = & sim. \

Prop.18.

Figure

If bd=dg=gh=hk=pq=pw &c Then alak pepc eddo pdpc+do rggs rqqo qgqo+gs &c & if ( lf2 = ) from ʒ to ye center be drawne cʒ then alak didv iqqx dqdv+qx gzgx wzwx &c & Ergo acak ab +adadab+agagad+ahahag+ak &c.

<9v> Figure

Prop 19

If fa=ab=be=eh &c. &. af+ab+be+eh are greater y t |n| ye semiperiphery: then, Radge(=ec)(=ad) bd(=bc) cd(=cdde) de {illeg} dhbd & dh is ye greatest, db ye least line drawn from d to these points a , b , e , h . yn rad {illeg} | dh | db dade .

Prop 20

Out of ye 18th & 19th {illeg} Prop: To divide An angle into any number of pts in ye figure of ye 18th prop: al=diam= \ 2/ z . ah is ye greatest of ye inscribed lines =B : now z /{illeg}\ || Bah+ {illeg} | 2z. | therfore bb = ahinz + 2z2 . & bb 2 zz z =ah . And zB b22z2z b+ag . therfore b32zzb zz b = ag Likewise b44zzbb2+2z4z3 = ad . & B55zzB3+5z4B z4 = ab B6 6zzB4 + 9z4BB 2z6 zzzzz =
B77zzB5+14z4B37z6B z6 = to a seaventh line
B88zzb6 + 20z4b4 16z6bb + 2z8 z7 = to an eight {illeg} line 00 00 B99zzb7 + 27z4b5 30z6b6 + 9z8b z8 = a n nineth line 00 00 B10 10zzb8 + 35z4b6 50z6b4 + 25z8bb 2z10 z9 = {illeg}/tenth &\

Prop 21

out of ye 17th Theor.: in ye figure whereof if ab {\:/} ye least inscribed line =z . & ac ye next line bee B . then zB Bz+ae . & bb zz z = ae & B3 2z2B zz = ag & B4 3z2bb + z4 z3 = to a fift line. B5 4zzb3 + 3z4b z4 z4 = a sixt. & b6 5zzb4 + 6z4bb z6 = seaventh  {sic} b7 6zzb5 + 10z4b3 4z6b z6 = to an eight line
B8 7zzb6 + 15z4b4 10z6bb + z8 z7 = to a nineth line B9 + 8zzb7 + 21z4b5 20z6b3 + 5z8b z8 = to a tenth line B10 9zzb8 + 28z4b6 35z6b4 + 15z8bb z10 z9 = eleventh .

<10r> Figure

Prop 22.

If aq=ab=bd=dc=ch=hk=kl=lf . Then GK Rad kl kl el (=alah) And hl hm (=qhqd= akac) dl do (=qdqa=acab) lc cn (=qcqb= ahad &c ) Soe that Rad{illeg} ye Periph: divided into any number of pts. gl lk lk al ak lh ak ac lc ah ad dl ac ab &c. & gl lk ah lc lk ac ld lh ad lb lc ab al ah &c.

hence Prop 23.

In ye former scheame If al {illeg} =2x =hypotenusa . kl=b . xb b2xah & bb+2xx x = ah / xb bb+2xx x 2bxxb3 xx (=lc {illeg} | b | ) lc{}kl{=}dl{}\h/ therefore 3bxxb3 xx = lc . & 2x4 4bbxx + b4 x3 = ad ye base of ye 4th triang: &{c} 5bx4 5b3xx + b5 x4 = ye perpendicular ( {illeg} | bl | ) of ye 5t tri: &{c} 2x6 9x4bb + 6xxb4 b6 x5 = base of ye 6t triang. 7x6b 14x4b3 + 7x2b5 b7 x6 = perpendic of the 7th tri 2x8 16x6bb + 20x4b4 8xxb6 + b8 x7 = base of ye 8th tri. 9x8b 30x6b3 + 27x4b5 9xxb7 + b9 x8 = perp: of ye 9th tri:

<10v> Figure

Prop 24:

If {illeg}| bd | =dh=hk=kl= b {illeg} g &c: {illeg}|t|hen bh=gd & bk=gh & bl=hm &c: & then
xtbx acag ceeh eiem iook opon pqql qrqy rssx stsf baad therefore xtbx btdg+hm+kn+ly+xf . againe xtbt abbd accg cech eiim &c Therefore xtbt btbd+gh+mk+ml+yx+ft . & since, as xtbt bxdg+hm+kn+ly+xf Therefore xtbt bx+bt bd+dg+gh+hm+mk+kn+ +nl+ly+yx+xf+ft . And xtbt bx+bt+xt to all ye perpen dicular & transverse line + bt . that is
| ( {illeg} \5/ {illeg} ) | xtbt xt+bt+bx 2bd + 2bh + 2bk + 2bl + 2bx + 2bt .

Prop 24

Figure

If in ye circle cfgh Figure be inscribed ye helix bedc a|&| ac touch it in ye point c then ab = to ye circumference.

Prop 25

Figure

If apcr be les yn halfe ye circle. & vt=tp . & vo= to ye vrap : then rq×po 2 = 4 times ye section /rapc\

Prop 26

Figure

If ab=bd=ad & bh perpendicular to ad fro {sic} ye angle {illeg} | b |. ce=ed . yn aed=adi=3dae . & ed is ye side of a heptagon

Prop 27.

If a line be cut by extreame & meane proportion ye lesse {illeg}|s|egment almos{illeg}|t| is to ye whole line as ye diameter is to 56 of 5 times the periphery divided by 6 .

Prop 28

Si secetur linea per extremam & mediam proportionem er{illeg}|i|t proximè, ut tota line{illeg}a plus minori segmento ad {tot} bis totam lineam, ita quæ potest quadrato sesquialterum semidiametri, ad latus quadrati circulo equalis.
linea secta sit 100,000 . minus segmentū 38,197 . Semidiametrū 100,000 , quæ potest quadrato sesquialterū semidiametri paulo maior est quam 122,474 . Radix Peripheriæ, 177,245 .

<11r>

Prop 28.

Figure

If er=rh=or . & ao=fc= to ye side of a decagon; & fn parallell to cd yn en shall be almost equall to ye fourth te of a circle for af  {sic} is divided in extreame & meane propor in ye point c . & ecef hf ef512 Perimeter|hbk|fa hr524Perim 610ef(=de) 14Perim ; by ye 27th prop: & ecef ed en ( =14 Perim ) .

Prop 29.

Figure

If os=2cp . & co is divided by extreame & meane proportion in r . & od parallell to rp then db is ye side of a square = to o|t|he area of ye circle. for be|y| ye 28th prop: As br(=to line+less segm¯) bo(=twice  ye  line) bp ({illeg} ( =32of the square of  ye   semidiameter ) bd(=to  ye  roote of a square equall to  ye  area of a circle .

Prop 30

Figure

If ye line dc touch ye helix in ye line ag . & ye line hf toucheth ye beginning of it in ye center a & 4ac = af then 2ad shall bee equall to perim: asr . & ac being ye Diam: ye tri: acd area of ye triang acd = to ye area of ye circle asr

Prop 31

Figure

If bed be a square of one revolution of an helix & ye angle dbe=dba & through ye points a , d , in ye helix be drawne ye line adk & through ye points ed  {sic} in ye Helix be drawne edg . & ye angle kdg bisected by dh ; then dh shall almost touch ye helix in d . {illeg}|&| it shall be soe much ye nigher a touch line by how much ye angles
ebd dba are lesser.

<11v>

Prop 32

Figure

If many Polygons be inscribed in a circle ye number of theire sides increaseing in a double proportion. & theire apotomies, or ye t|b|ase \of/ a tri: whose cathetus is a leg of ye Polygon & hypotenusa is ye {illeg}|D|iam (as ye apotome of ye Polygō cgp is ce . of pacegi is ae &c) if ye Apotome of ye sides of ye first Polygō be called b . |of| ye 2d = c . |of| ye 3d = d . of ye 4th = f . of ye 5t = g of ye Sixt = h . & ye diameter be z then And ye first Pol{illeg}ygon be = p . ye 2d = q . ye 3d = r = abcdefghiopq . ye fourth = s . ye 5t = t ye sixt = v . ye {illeg} 7th = w &c then
pq bz . & p{illeg} pr bc zz . & ps bcd zzz . & pt bcdf z4 . & pv bcdfg z5 . & pw bcdfgh z6 &c



To know how many elections may bee made of \divers ways/ things, whereof some of ym are equall, \may bee ordered./ . as of . a.b.b.c.c.c.d.d. doe thus { a bb ccc dd 11 × 2×31×2 × 4×5×61×2×3 × 7×81×2 =112 } { 1a1 × 2b×3b1×2 × 4c×5c×6c1×2×3 × 7d×8d1×2 =112 } the number of changes, in order.

To know how many elections may bee made doe thus 2a × 3bb × 4ccc × 3dd × 5eeee = 360 = 2a × 2b×3b2 × 2c×3c×4c2×3 × 2d×3d2 × 2e×3e×4e×5e2×3×4 therefore there are 359=3601 elections in abbc3dde4 .

<12r>

Propositiones Geometricae Ex Schootenij
Sectionibus miscellaneis.

Sectio 1ma

To know how many changes 6 Bells, abcdef or how divers conjuctions ye 7 planets can make . or how many divisors abcde hath, or how man{{sic}} divers compositions ye 24 le{illeg}|t|ers {sic} can make &c the examples following show.
1. a1
2. b . ab3
4. c . cb . cab . ac . 7
8. d . da . db . dab . dc . dac . dcb . dcab . 15
16. e . ea . eb . eab . ec . eac . ecb . ecab . ed . eda . edb . edab . edc . edac . \ edcb . edcab . \ 31 / /
32. f . fa . fb . fab . fc . fcb . fac . fcab . fd . fda . fdb . fdab &c 63.
64. g . ga . gb . gab . gc . gcb . gac . gcab . gd . gda . gdb &c 127
wch shows yt in 7 letters 127 el{illeg}e|ec|tions may be made. yt 7 Planets may be conjoyned 120 divers ways. yt abcdefg . hath 128 divisors for an unite is one of {y}& 1×2×3×4×5×6.=720 ; are ye number of changss {sic} in six bells.

Sec 2

To know how many things & of wt sort they are wch may be chosen 15 ways. 15+1=16 . 162=8 . 82=4 . 42=2 . 22=1 . & {illeg}4{illeg} \ 21 . 21 . 21 . 21 . = 1 . 1 . 1 . 1 . = 4 . that 4 things all unequall/ m{illeg}|a|y be varyed 15 ways. also. 164=4 . 42=2 . 22=1 41 . 21 . 21 = 5 {illeg}|&|. 5 things whereof 3 are equall viz: a . a . a . b . c . {illeg}|&| 164=4 . 44=1 . 41 . + 41 = 3+3 = 6 . & 6 thins {sic} whereof 3 & 3 are eaquall {sic} as aaabbb . may be varied 15 ways. & 168=2 22=1 . 81 . 21 = 8 = 7+1 . & 8 things whereof 7 are = may be varyed 15 ways. as aaaaaaab . 1616=1 . 161 = 15 . 2 wherefore 15 alike things &c |as a \15/|. 2 wt things vary 23 ways. 23+1 = 24 & 24 admitts a {illeg}|7| fold divisor therefore ye multitude of things sought may be 7 fold but since 43 is a primary number (viz wch cannot bee divided) 42+1 = 43 . 4343=1 431 = 42 . therefore onely 42 like things can be varyed 42 ways as a42 .

<12v>

Sec 3

Every quantity hath one divisor more yt it hath aliquote pts (yt is pts greater yn an unite \of whole numbers./). How to find a quantity haveing a given multitude of divisors or ali{illeg}|q|uote{illeg} pts: suppose its aliq: pts must be 15. 15+1 = 16 & \soe/ by ye former section abcd . a3bc . a3b3 . a7b . a15 . may be varyed 15 ways. therefore they shall have 15 aliquote pts & 16 divisors. but since onely 42 like things (as a42 ) ca{illeg}|n| be varyed 42 ways therefore oenely {sic} a42 hath 42 aliquote pts & 43 divisors. &c

Sec 4

To find ye least numbers haveing a given multitude of divisors & aliquote pts instead of soe many letters in the former sec: put soe many p{illeg} least primary numbers & take ye least result from ym. as from ye former example: abcd . a3bc . a3b3 . a7b . a15 . that is 2 . 3 . 5 . 7 . or | 2 . | 2 . 2 . 3 . 5 . &c. {illeg} now. 2×3×5×7 . = 210 . & 2×2×2×3×5 = 120 . &c therefore 2×3×5×7 . \ =210 / are \is/ ye least number{s} haveing 16 divisors.

Sec: 5 conteines a table of Primary numbers.

Sec 6

To find progressions constituteing rectangular triangles wth sides rationall ye examples \following/ shew. take two numbers as 1 . 2 . yn 1×2 = 2 since ye product is eaved|n| double it viz: 2×2 = 4 . & 4 is ye numerato\r/ yn 1+2 = 3 & since 3 is od multiply it by the difference of ye termes: 1×3 = 3 & 3 is ye denominator. & ye first terme 43 . yn since (1) ye difference of ye termes is od multiply it by 4 . 4×1 = 4 & 4× per 2 majore terminum. 4×2 = 8 8+4 (the former numerato\{r}/) = 12 = numerator 2d. yn 3 (ye former denom) added to. {illeg}| 2 | (ye double square of ye diff: of ye termes because ye square (1) is o{illeg}|d|d) =5 ye 2d denominator. I ad anothe example take 1 . 3 . yn 1 × 3 = 3 = 1st numerator. yn 1+3 = 4 & since 4 is eaven halfe 4 {illeg} 1st denom. & ye 1st {illeg} \halfe/ is 4×22 (diff: of ye termes) =4 & ye first denom is 4. ye first terme{illeg} 34 . yn becaus ye diff of ye termes is eaven 2 2×2 = 4 & 4×3 = 12 & 12+3 = 15 . yn 2×2 = 4 . 4+4 = 8 . & 158 ye 2d terme & now {illeg}|t|er{illeg}|m|es may be had by Arithmeticall proportion. thus. 43 . 125 or 113 . 225 . 337 . 449 . 5511 . 6613 7715 8817 9919 101021 &c |&| 34 . 158 or 34 178 . 21112 . 31516 . 41920 . 52324 . 62728 . 73132 . 83536 &c thus may other progressions be obteined. For ye use take ye numerator for one leg & ye denom for another & ye Hypoten: will be rationall as in 225 5×2 = 10 . 10+2 = 12 or 125 144+25 = 169 = 13 . & in this 178 or 158 225+64 = 17 . Figure

<13r>

If ye {illeg}|s|uposed number{illeg}|s| be 2 . 5 . yn 2×5 = 10 . 10+10 = 20 . & 2+5 = 7 . 3×7 = 21 . so yt 2021 . yn {illeg}be 4×3 = 12 . 12×5 = 60 . {illeg} 60+20 = 80 . |&| 3×3 = 9 . 9doubled = 18 . 18+21 = 39 . & ye 2 first termes 2021 . 8039 or 2239 . Agai{illeg}|n|e, if ye numbers be 3 . 4 3×4 = 12 . 12×2 = 24 . & 3+4 = 7 . 1×7 = 7 . therefore 247 . yn 4×1 = 4 . 4×4 = 16 16+24 = 40 . & 1×1 = 1 . 2×1 = 2 . 7+2 = 9 therfore 409 is ye 2d & ye progres may be continued, as 2021 . 2239 . 3557 . 4875 . 51193 . & 337 . 449 . 5511 . 6613 &c.

Sec 7

To find a {number} wch divided by 7 leaves 2 . by 11 leaves 1 . by 13 leaves 9 . the least common divisor of 7 . 11 . 13 . is 7 × 11 × 13 × = {illeg} |1001 |. divide {illeg} |1001 | twice by each & consider ye remainder of ye {illeg}|s|eacond division thus.
2102105(1252 . because 1 is left 105 is ye multiplier.
210370(1323 . since 1 is left 70 is ye multiplier.
210542(258 since more yn 1 is left (viz: 2 ) multiply 2 till it divided by 5 leaves ( 1 ) 2×3 5 = 115 . therefore 42×3 = 126 is ye multiplyere. 2107307(427 since more yn 1 is left

1 Since more yn 1 is left (viz 3) multiply 3 till it divided by 7 leavs 1 . 5×37 = 217 therfore 5×143 = 715 ye multiplier | 10017 ( 1437 ( 2037 .

2 Since more yn 1 is left (viz: 3 ) 3×411 = 1111 therfore 4×91 = =364 ye multipl: | 100111 ( 9111 ( 8311 .

3 If but 1 had ben|e|ne left 77 had beene divisor but now 12×12 11 = 13111 . therfore 12×77 = 924 is multiplyer. 100113 ( 7713 ( 51213 . now the number sough {sic} is thus found.

< insertion from the center right of f 13r >

Divisor.Reliq:Multip. 70.2×715=1430. 11.1×364=0364. 13.9×924=8316. The Sum¯e10110.

< text from f 13r resumes >

Lastly did|v|{illeg}|i|de
by ye least com. divis: 101101001 ( 101001001 whe{illeg}|r|efore 100 ye number left is ye number sought.

Sec 8.

Touching ye Method of weights suppose a man have weigs {sic} of 1.2.3.4.5 1.2.4.8.16.32 pounds &c by ym all intermediate pounds may be thus weighed 123 1. 2 3. 4 5 6 7 8 9 10 11 12 13 14 1. 2 1+2 4 1+4. 2+4. 1+2+4. 8 8+1 8+2. 8+1+2. 8+4. 8+4+1. 8+4+2 &c or is|f| his wights {sic} be 1.3.9.27.81. all weights may be supplyed thus. 1. 2. 3. 4. 5 6 7 8 9 10 11 12 13 14 1 31. 3. 3+1. 913. 93. 9+13. 91. 9. 9+1. 9+31. 9+3. 9+3+1. 27931. &c Note yt weight marked wth − signifie ye wigh {sic} to be put in ye opposite scale ballance.

<13v>

{illeg}|S|ec. 9.

To find numeri amicabiles that is 2 numbers whose aliquote pts are mutually equall to theire wholes. take this Des-Cartes his rule

If (2), or any other number producc|e|d out of 2 {illeg} |a{illeg}|s|| 2×2. 2×2×2 &c (viz 2.4.8.16.32 &c .|)| bee such a number yt 1 taked|n| out of it triple there rests a primary number{,} & yt if {illeg} 1 taken f{illeg}|r|om it sextuple there rests a primary number, & if 1 taken from its square octodecuple a primary number rests: yn multiply this last prime number by ye assumed number doubled & ye product is one amicable number & ye aliquote pts of {illeg} it make ye other Example. if 2 be taken. 2×3 1 = 5 numero primario primo. 2×6 1 = 11 numero primario scdo. 2×2×18 1 = 71 numero primario tertio. 4×71 = 284 , one amicable number, & ye 2 former prime numbers \ × one another & ye product/ ×4 ye double of ye assumed number viz 5×11 = 55 55×4=220 . Thus from 8 . & 64 &c. may be deduced amicable numbers.

Sec 1{illeg}|0|

To find triangles whose sides, segments of theire bases, & Perpendiculars are expressible by rationall numbers
Figure 1st if ye perpendic: is without ye tri: let ac=z . bd=x cd=y . ad=z+y . {illeg} ad=y+b . xx + yy = yy + 2by + bb . y = xx bb 2b . & cd = z+y+a . xx + zz + 2yz + yy = zz + yy + aa + 2zy + 2za + 2ay . 2ay = xx aa 2za = axx abb b . bxx baa 2zab = axx abb . bxx baa axx + abb 2ab = z . puting any numbers for a , b , & x ; y & z may be found. then ad = z+y = = xx + bb 2b . cd = z+y+a = xx + aa 2a . wch reduced to ye common denominator 2ab ; & yt cast away. cd = bxx + baa . {illeg} ad = axx + abb . de = 2abx . ae = axx abb . ce = bxx baa . ac = bxx axx + abb aab .

Figure

In like manner if ye perpendicular fall wthin side. ab = bxx + baa . bd = 2abx . ad = bxx baa . dc = axx abb . bc = axx + abb . ac = bxx + axx abb baa .

Also by ye conjunction & disjunction of 2 triangles it may be found yt ab = bbx + aax ad = bbx aax . {illeg} ac = bbx aax aax + abb . {illeg} bc = axx + abb . {illeg} db = 2abx . dc = axx abb . For if bd=x dc = xx bb 2b . bc = xx + bb 2b . that is bd = 2bx . dc = xx bb . bc = xx + bb . Likewise bd = 2ab . ad = bb aa . ab = bb + aa . 2abx ye least quanti{illeg}|t|y divisible by 2bx & 2ab , being divided by ym, leaves a & x wch must {illeg} multiply ye bases & hy{illeg}|p|otenusas. If ye perpendic: fall wthout ye legs may be thus exprest cd = acc + ayy . da = yyc + aac <14r> ca = acc ayy + cyy aac . ae = yyc aac . ce = acc ayy . ad = 2acy .

Sec 11

To ma{illeg}|k|e yt two such tri: be of ye same base & altitude. Suppose an equation twixt ye bases & pependiculars {sic} of ye 2 last tri: as 2abx = 2acy . x = cy b . xx = ccyy bb . bbx aax axx + abb = acc ayy + yyc aac or bbcy aacy b accyy bb + abb = acc ayy + yyc aac & yy = + b3cy + aabbc aabc + ab4 abbcc bbc + acc abb . Suppose aabbc + ab4 = abbcc . {illeg}|o|r a = c bb c . let c=3 greater yn b=2 . a=53 . y=2261 . x=3361 & consequently Figure

Sec 1{illeg}|4| differs not from Cap 19: prob 18 Ou{illeg}|g|htred.

Sec: 15 Of P{illeg}|o|lygons or multangular num{illeg}|b|ers

\The sue of all ye tearms|e|s in/ an arithmet: progres: increasing from {illeg} \an/ unite by 1 compseth {sic} tra|i|{illeg} triangles. by 2 , composes □s. by 3 , composes pentangles. by 4 , hexang: &c {illeg}ke as 1. 2. 3. 4. 5. 6. compos {sic} ye triangles 1 3 6 10 15 &c likewise 1. 3. 5. 7. 9. compose 1 4 9 16 25 &c So 1. 4. 7. 10. 13. compose ye quintangles 1. 5. 12. 22. 35. 51. 70. &c. If a=1= ye first term{e} ye excess {illeg}|o|f the progression =x. The sume of ye termes =z= |to ye polygon| ye multitude of ye termes =t= to ye side of ye Polygon. {illeg} given to find {illeg} Suppos {sic} t given to find {illeg}z. z = 1tt+1t 2 or z = tt+t 2 in trigons. z=tt in 4gons. z = 3ttt 2 in 5gons. z = 2ttt in 6go z = 5tt3t 2 in {illeg}|7|{illeg}|g|ons. z = 3tt2t in 8gons. z = 7tt5t 2 in 9gons. &{c} {illeg}|&| z given t is found thus t = 1+1+8z 2 in tri. t = 0+16z 4 in 4go t = +1+1+24z 6 , in 5gons. { t = +2+4+32z } { t = +2+4+32z 8 } in 6gons &c. As {illeg} ye side 12 of a tri given. ye tri = z = 12×12+12 2 = 78 &c & if z=21 be octangled. t = +4+16+48z 12 = 4+16+48×21 12 t = 4+1024 12 = 3 .

<14v>

July 4th 1699. By consulting an accompt of my expenses at Cambridge in th{illeg}|e| {illeg}|y|ears 1663 & 1664 I find that in ye year 1664 a little before Christmas I being then Senior Sophister, I borrowed bought Schooten's Miscellanies & Carte's's Geometry (having read this Geometry & Oughtred's \Clavis/ above half a year before) & borrowed Wallis's works & by consequence made these Annotations out of Schooten & Wallis in winter between the years 1664 & 1665. At wch time I found the method of Infinite Is series. And in summer 169|6|5 being d|f|orced from Cambrid{illeg}|g|e by the Plague I computed ye area of ye Hyperbola a{illeg}|t| Boothby in Lincolnshire to I{illeg} two & fifty figures by the same method.                 Is. Newton

<15r>

Annotations out of Dr Wallis his Arithmetica infinitorum.

|1| A primanary series of {illeg}|q|uantys {sic} is arithmetically proportionall, as 0, 1, 2, 3, 4 . & its index is 1

|A| Secundanary series i|a|re those whose rootes are arithmetically proportionall; as 0, 1, 4, 9, 16 . & its index is 2

A Tertianary, quartanary, quintanary series of quantitys are those whose cube, square square, square cube rootes are Arithmetically Proportionall as 0, 1, 8, 27, 64 . / 0, 1, 16, 81, 156 . / 0, 1, 32, 243, 624 . &c Their indices being 3, 4, 5 &c.

3 Subsecunda\na/ry, sub{illeg}|t|ertianary, series &c: are those whose squares, cubes, &c are arithmetically proportionall, as 0, 1, 2, 3 . . c:0, c:1, c:2, c:3 &c. Theire indices being {illeg} 12, 13, &c.

2 Primary Secundanary, tertianare|y| series &c are said to bee reciprocally proportionall ({illeg} yt is to ye sam {sic} se{illeg} increasing) which continually decrease as. 10, 11, 12, 13, 14, . 10, 11, 14, 19, 116, . 10, 11, 18, 127, 164, . Their indices being negative as -1, -2, -3 .

4 The indices of compound |or mixt of rationall & irrati{illeg}{onall} |series|| series, by multiplying or dividing ye indices of ye simple series may bee found as in a subsecundanary progression cube{illeg}|d| 0, 1, 8, 27, 64 ye index is 12×3 = 32 . So in ye cube rootes of a secundanary progression, c:0, c:1, c:4, c:9 &c. ye index is 13×2 = 23 . so in irrationall reciprocal progressions
qq:10, qq:11, qq:12, qq:13 , ye index is -1×14 = -14 .

<15v> Figure

Now su{illeg}|p|pose ye line ac be divided into an infinite number of equall pts ad, de, ef, fg &c, from each of wch are drawne perpendiculars \parallels/ nd pe qf &c. wch wch ino|c|rease continually in so{illeg}|m|e of ye foregoeing prg|o|gressions or in some progresion {sic} compounded of ym, all those lines may be taken for ye surface bqnac , & to know wt proportion \all/ those lines have that superficies hath to ye superficies ambc yt is wt proportion all those lines have to soe may equal to ye greatest of ym, I say as ye index of ye progression increased by an unite is to an unite soe is ye square abcm to ye area of ye crooked line. As if abc is a parabola ye lines ad, pe, qf, &c ao|r|e a subsecundanary series (for y=rx) whos {sic} index is 12 wch added to an unite is 1+12 = 32 Therefore 32 1 3 2 so is ye square ambc to ye area of ye Parab. ({illeg}|ye| names of ye lines are (ad), {illeg} ae, af {illeg}|&|c =x . dn, pe, qf &c =y . ac=p. bc=q.) T{illeg}|h|e case is ye same if abc bee supposed a sollid, as suppose it{illeg}|s| {nature} t a Parabolicall conoides. yn since ye nature of it is rx = yy . yy designes ye squares nd, pe, qf &c: all wch taken together are equivalent to ye Sollid. & those □s increase in ye same proportion wch rx . or x doth. yt is they are a primanary series whose index i{illeg}|s| 1 to wch (according to ye rule I ad an unite & tis 2. Therefor 1 {illeg} 2 soe are all ye □s of ye Primary series to soe many □s equall to ye greatest of yt series. & soe is ye conoides to a cilinder of ye same altid|t|ude.

<16r>

Also if a superficies be compounded of 2 or more of these series, Their area is as easily found as if ye nature of ye line bee {illeg}, {illeg} y {illeg} = aa xx , or {illeg} y = a4 2aaxx + x4 or {illeg} y = a6 3a4xx + 3aax4 x6 . &c. Their areas will bee to ye parallelograms {ar|b|o{illeg}|u|t} them as {illeg}|2 | to {illeg}|3 |, as 8 to 15 15 , as 48 to 105 &c. but if I put in ye intermediate termes in these last named lines th{illeg}|e|ir order will bee y = aa xx , y = aa xx , y = aa xx aa xx ¯ , y = a4 2aaxx + x4 . y = a4 2aaxx + x4 aa xx ¯ . y = a6 3a4xx + 3aax4 x6 ; &c: & since these lines observe a geometricall progression their areas must observe some kind of progression. of wch every other terme is given viz 1. . 23. . 815. . 48105. . 384945. . 384010395 . T{b}{l}wixt wch {illeg}|t|ermes if ye intermediate termes . can bee found ye 2nd □ will give ye area of ye line y = aa xx , ye circle. Soe likwise {sic} in this progression of lines {illeg}| y | =1 . {illeg}| y | = ax xx . y = aa xx y = ax xx ax xx ¯ . y = aaxx 2ax3 + x4 &c: ye progression of their areas {illeg} is 1: : 16: : 130: : 1140: 1630: &c. ye 2nd if it can bee found givs {sic} ye area of ye ○ for as its denominator to its numerator so is ye □ of ye diameter to ye area of a semicircle. If th{illeg}|i|s last progresion {sic} bee multiplyed by ye respective termes in ye progress 1. 2. 3. 4 & it may bee diminished ye reslut {sic} being 1. 2. 12. 4. 16. 6. 120. 8.170 soe yt in this progression {illeg} 1, b, 12, c, 16, d, 120, e, 170, f, &c: if <16v> b can be found yn, {illeg} ye □ of ye diameter to ye area of ye circle is as ye denominator of b to its numerator. Likewise ye 1st series of areas may be diminished by multiplying each terme by its correspondent terme in this progression 1, 2, 3, 4, 5, 6, &c: & it will become, 1, a, 2, b, 83, c, 4815, d, 384105, e, 3840945 . &c. In wch if a can bee found yn as ye denom of a to its num: so ye □ of ye Radius is a semicircle, yt making ye radius =q . 2aq= . The same \kinds of/ changes may bee performed |by| other p any other progressions, as \by division/ by ye geometricall progression 1, 2, 4, 8, 16, & ye {illeg}|fi|rst series of areas becomes 1, g, 16, h, 130, k, 1140, l, 1630, &c viz ye same wth ye 2d series. Also these changes may be done by addition or substraction of mutuall termes in 2 proportions. Soe yt ye most convenient way \may be/ be chosen, {illeg}|w|he|r|by to reduce any series of proportions to ye most convent {sic} forme.

Now if it be propounded to find thes{illeg}e mi{illeg}|d|dle termes, f|i| |I|t w{illeg}|i|ll {illeg}|b|ee convenient to d|f|ind how the given proportion may bee deduced from an Arithmeticall, Geometricall, or some other familiar proportion, viz whose meane termes may be found, as this progression 1. 23. 815. 48105 deduceth its originall from this {illeg}| A | × 0×2×4×6×8 1×3×5×7×9 & in wch A is an infinine {sic} number =10.

It will also be convenient to find what relatiō all ye other meanes have to ye first soe yt if ye first bee had all ye other may bee deduced thence. As in this case suppose ye 1st meane to bee a . The progression will bee <17r> 12a: 1: a: 32: 4a3: 158: 8a5: 10548: 64a35: 945384: 640a315: deducing its originall from A × 0×2×4×6×8×10 1×3×5×7×9×11 {illeg}|&| from this A(=12) × 2a × 4a × 6a × 8a × 10a 1×3×5×7×9 . &c {+} (note yt ye proportions of th{illeg}|e|se te|m|eane termes to oneano {sic}ther, or to (a), are found f by finding ye proportion of ye circle y = aa xx to ye line y = aa xx aa xx ¯ &c).

In this case to find ye quantity a : it may be considered yt a 12a = 2 . {illeg} 3 1×2 = {illeg} 32 . 4a 3×a = 43 . Naming ye termes in ye progress: 12a: 1: a: 32: 4a3: 158: 8a5: 3516: bcdefghk. 1st observe yt db=2 . ec=32 . fd=43 . ge=54 . hf=65 &c ye proportions still {illeg}|de|creasing & therere|f|ore {sic} yt in cb . dc . ed . fe . gf . hg . kh &c: ye latter terme is lesse yn ye former; & therefore dd cc { is less  yn  dc×cb=db=2. greater  yn  dc×ed=ec=32. or / a= \ d is { less  yn  1×2=1+11 greater  yn  1×32=1+12 . Also
ff ee is { lesse  yn  fe×ed=fd=43 greater  y¯  fe×gf=ge=54 } = a: 8a9 = 2×4a3×3 . Therefore 3×34
Therefore a is { lesse  yn 3×3 2×4 43 greater  yn 3×3 2×4 54 . And So by ye /same reasoning.\ a is { less  yn 9×25×49×81×121×169 2×16×36×64×100×144×14 1413 . greater  yn 3×3×5×5×7×7×9×9×11×11×13×13 2×4×4×6×6×8×8×10×10×12×12×14 1514 . &c. Thus {illeg}|W|allis doth it. but it ma{illeg}|y| bee done thus. a is { greater  yn   1 . less then 32 . | 4a3 is { greater then 32 less then 158 Therefore <17v> a is { greater  yn   3×3 2×4 lesse then 3×3×5 2×4×4 . 8a5 is { greater  yn   158 lesse  yn   3516 yt is a is { greater then 3×3×5×5 2×4×4×6 = 3×5×5 2×4×4×2 lesse then 3×3×5×5×7 2×4×4×6×6 = 5×5×7 2×4×4×4 &c. By ye same reasoning
a is { greater  yn   32×34×54×56×76×78×98×910 = 5×49×81 2×16×4×64×2 . lesse  yn   32×34×54×56×76×78×98×910×1110 = 49×81×11 2×16×4×64×4 . Or
a is { greater  y¯   11×169×225×289×369×441 2×16×4×64×4×144×4×256×4×400×2 lesse  yn   169×225×289×369×441×23 2×16×4×64×4×144×4×256×4×400×4 . Note yt a is greater yn 12 these two summes.

<18r> Figure

Having ye signe of any angle to find ye angle or to find ye content of any segmnt of a circle

Suppose ye circle to be aec its |se{illeg}|m|i|diameter ap=pc=1 . ye given sine pq=x , viz: ye signe of ye angle epa . ye segment sought eapq . abcp the □ of its Radius. & yt, qi: qk: qg: qf: qe: qd: qh: ql: qr: &c are continally {sic} proportionall. Then is eq = 1 xx . fq = 1 xx . gq = {illeg} 1 xx in 1 xx ¯ . hq = 1 2xx + x4 . iq = 1 2x2 + x4 1 xx dq=1 . kq = 1 1 xx . lq = 1 1 xx . rq = 1 1 xx in 1 xx . &c & since all the{illeg} ordinately applyed lines in these figures abcq , aecq , afcq , agcq , &c are geometr{illeg}|i|cally proportionall their areas adqp , aeqp , afqp , agqp , ahpq &c will observe some proportion amongst one another. To find wch proportion, 1st adqp = 1×xx = x . 2dly afc is a parab: therefore afqp = x x34 . also since tis qh = 1 2xx + x4 , therefore ahqp = x 23x3 + 15x5 . |Also| vq = 1 3xx + 3x4 x6 , therefore avqp = {illeg}/ x x3 \ + 35x5 17x7 . & by ye same proceeding ye proportion may bee still continued after this manner <18v> x : x 13 x3 : x 23 x3 + 15 x5 : x 33 x3 + 35 x5 17 x7 : x 43 x3 + 65 x5 47 x7 + 19 x9 : x 53 x3 + 105 x5 107 x7 + 59 x9 111 x11 : x 63 x3 + 155 x5 207 x7 + 159 x9 611 x11 + 113 x3 : x 73 x3 + 215 x5 357 x7 + 359 x9 2111 x11 + 713 x13 115 x15 . &c.
And if ye meane termes be inserted it will bee
x : x : x 13 x3 : x 36 x3 + : x 23 x3 + 15 x5 : x 56 x3 + 25 x5 or x : x 16 x3 : x 13 x3 : x The first letters \ x / run in this {illeg} progression 1. 1. 1. 1. 1. &c. ye 2d | x3| in this -13. 03. 13. 23. 33. 43. 53 &c ye 3d x5 in this 0+1 . 1+2 . 3+3 . 3 . \ 6. 3. 1. 0. / 0+1=1 . 1+2=3 3+3=6 . 6+4=10 . 10+5=15 . ye 4th x7 th{illeg}|i|s 1st . +x×1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 2d . x33×0 . 0+1=1 . 1+1=2 . 2+1=3 . 3+1=4 . 4+1=50 . 6 . 7 . 8 . 9 . 10 . 3d . +x55×0 . 0+0=0 . 0+1=1 . 1+2=3 . 3+3=6 . 6+4=10 . 15 . 21 . 28 . 36 . 45 . 4th . x77×0 . 0+0=0 . 0+0=0 . 0+1=1 . 1+3=4 . 4+6=10 . 20 . 35 . 56 . 84 . 120 . 5th . +x99×0 . 0+0=0 . 0+0=0 . 0+0=0 . 0+1=1 . 1+4=50 . 15 . 35 . 70 . 126 . 210 . 6th . x1111×0 . 0+0=0 . 0+0=0 . 0+0=0 . 0+0=0 . 0+1=10 . 6 . 21 . 56 . 126 . 252 . 1st. .2d. .3d .4th .5th .6th 1 . 7 . 28 . 84 . 210 . | 7th . 1 . 8 . 36 . 120 . | 8th 1 . 9 . 45 . | 9th 1 . 10 . | 10th 1 . | 11th . Now if the meane ter{illeg}|m|es in these progressions can bee callculated ye first of ym gives ye area aeqp. Which is thus done 1st +x×1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 1 . 2d . x33×0 . 12 . 1 . 32 . 2 . 52 . 3 . 72 . 4 . 92 . 5 . 112 . 6 . 3d . +15x5×0 . 18 . 0 . 38 . 1 . 158 . 3 . 358 . 6 . 638 . 10 . 998 . 15 . 4 . 17x7×0 . +116+ . 0 . 116 . 0 . 516 . 1 . 3516 . 4 . 10516 . 10 . 23116 . 20 . 5 . +19x9×0 . 5128 . 0 . 3128 . 0 . 5128 . 0 . 35128 . 1 . 315128 . 5 . 1155128 . 15 . 6 . 111x11×0 . 7256 . 0 . 3256 . 0 . 3256 . 0 . 7256 . 0 . 63256 . 1 . 693256 . 6 . 7 . 113x13×0 . 211024 . 0 . 71024 . 0 . 51024 . 0 . 71024 . 0 . 211024 . 0 . 2311024 . 1 . 30031024 . Soe yt 1×x 12×13×x3 18×15×x5 116×17×x7 5128×19×x9 &c. is ye area, apqe yt is 00×x 00×12×13x3 00×12×14×15x5 00×12×14×36×17x7 1×3×5x9 2×4×6×8×9 &c: The progression may be deduced from hence 0× 1× -1× 3× -5× 7× -9× 11 0× 2× 4× 6× 8× 10× 12× 14 . &c <19r> Figure Soe yt if {illeg}|ye| given sine bee pq=te=x . & if ye Radius pc=1 . Then is ye superficies ape = x x1xx 16x3 140x5 x7112 5x91152 7x112816 &c: And ye area ade = x36 + x540 + x7112 + 5x91152 + 7x112816 + 21x1313312 + 11x1510240 + 429x17557056 + 715x191245184 + 2431x215505024 &c. By wch meanes ye angle ape is easily found for aecpa apc=90 ape ape .

The same may bee thus done.

adp = x2 . Or 2adp = x . 2afp = x+x33 . 2ahp = x+2x333x55 . And 2avp = x + x3 9x55 + 57x7 . &c. as in this order x . x + 13x3 . x + 23x3 35x5 . x + x3 95x5 + 5x77 . x + 4x33 18x55 + 20x77 79x9 . x + 53x3 305x5 + 50x77 35x99 + 9x1111 . x + 6x33 45x55 + 1007x7 105x99 + + 54x1111 11x1313 . x + 7x33 63x55 + 175x77 245x99 + 18911x11 77x1313 + 13x1515 . &c Which progression wth their intermediate termes {illeg}|may| bee thus exhibited. By{illeg} wch it may appeare yt if pe=1. pq=x. yn aep = 12x + x312 + 3x580 + 5x7224 + 35x92304 &c. And ye area aep given gives ye angle ape for apce apc=90d ape ape Likewise ye angle ape given its sign ma{illeg}|y| bee found hereby |&c| +911x11in0 2×adpa= 2×aep= 2afp= 2agp= 2ahp= 2aip= 2avp= +xin1 . 1 . 1 . 1 . 1 . 1 . 1 . x33in0 . 12 . 1 . 32 . 2 . 52 . 3 . 35x5in0 . 18 . 0 . 38 . 1 . 158 . 3 . +57x7in0 . 116 . 0 . 116 . 0 . 516 . 1 . 79x9in0 . 5128 . 0 . 3128 . 0 . 5128 . 0 . +911x11in0 . 7256 . 0 . 3256 . 0 . 3256 . 0 . 2×adpa= . 2×aep= . 2afp= . 2agp= . 2ahp= . 2aip= . 2avp= . Note yt 1xx = x2 2x36 5x580 7x7224 45x92304 77x115632 &c that is 1xx = x2 x33 x516 x732 5x9256 7x11512 21x132048 {illeg}|&|c. According to this progression 12 × 12 × 14 × 3×5×7×9×11×13×15×17 6×8×10×12×14×16×18×20 &c. Note also yt ye segment ae = x312 + 3x580 + 5x7224 + 35x92304 . &c. aep = x2 + x312 + 3x580 + 5x7224 + 35x93304 + 63x115632 + 231x1326624 + 143x1520480 + 6435x171114112 + 12155x192490368 + 46189x2111010048 .

<19v> Figure

If pq=a. qd=x. pc=1=pb. db= 1 aa 2ax xx yn ye areas of ye lines in this progression. 1 1 aa 2ax xx . 1 aa 2ax xx . 1 aa 2ax xx ¯ }32 . 1 2aa 4ax 2xx + a4 + 4a3x + 6aaxx + 4ax3 + x4 . . 1 3aa + 3a4 + 8a3x + 8aaxx & (supposeing also b 2ax xx _ }32 . bb 4abx 2bxx + 4ax3 + x4 + 4aa . b3 6abbx 3bbxx + 8abx3 + 3bx4 6ax5 x6 + 12aabx2 8a3x3 12aax4 + 4abx3 . &c

<20r>

To squar{illeg}|e| ye Hyperbola.

Epitome Geometriæ

Figure

So if nadm is an Hyperbola. & cp=1=pa. pq=x . qd , qe , qf , qg &c =y . & 11+x = y = dq . 1 = y = eq . 1+x = y = qf . 1+2xxx = y = qg . 1+3x+3xx+x3. 1+4x+6x2+4x3+x4. &c. There {sic} squares are. . x . x + xx2 . x + 2xx2 + x33 . x + 3xx2 + 3x33 + x44 . x + 4xx2 + 6x33 + 4x44 + x55 . x + 5xx2 + 10x33 + 10x44 + 5x55 + x66 . &c As in ye following table. By whose first terme is represented ye square of ye Hyperbola, viz yt it is -0.0 0×-=. =. =. =. =. =. =. =. =. 0 -0.0 x×a. a. a. a. 1. 1. 1. 1. 1. -0.0 x×p. p. p. p. 1. 4. 10. 20. 35. -0.0 x×q. q. q. q. 0. 1. 5. 15. 35. -0.0 x×d. e. f. g. 0. 0. 1. 6. 21. =. =. =. =. 0. 0. 1. 6. 21. -0.0 x×1. 1. 1. 1. 1. 1. 1. 1. 1. -0.0 x22×- 1. 0. 1. 2. 3. 4. 5. 6. 7. -0.0 x33×1. 0. 0. 1. 3. 6. 10. 15. 21. -0.0 x44×- 1. 0. 0. 0. 1. 4. 10. 20. 35. -0.0 x55×1. 0. 0. 0. 0. 1. 5. 15. 35. -0.0 x66×- 1. 0. 0. 0. 0. 0. 1. 6. 21. -0.0 x77×1. 0. 0. 0. 0. 0. 0. 1. 7. -0.0 0×-=. =. =. =. =. =. =. =. =. -0.0 0000. 00000. 00000. -0.0 3333. 33333. 33333 . 33333. 33333. 3 -0.0 0000. 00000. 00000 . 00000. 00000. 0 -0.0 8571. 42857. 14285 . 71428. 57142. 8 -0.0 1111. 11111. 11111 . 11111. 11111. 1 -0.0 9090. 90909. 09090 . 90909. 09090. 9 0 0 xx22+x33x44+x55x66+ +x77x88+x99x1010. &c. 0 As ifx=110. orcq=1110=1,1 . yn isdapq=1101200+13000 140000+150000016000000. &c yt0is0apqd=0,10000.00000.000000. 13x3=+0,00033.33333.333333. 15x5=+0,00000.20000.000000. 17x7=+0,00000.00142.857142. 19x9=+0,00000.00001.111111. 111x11=+0,00000.00000.009090. 113x13=+0,00000.00000.000076. 000115x15= +0,00000.00000.000000.¯ 0,10033.53477.310755. Summa000000000 . -0.0 9230. 76923. 07692 . 30769. 23076. 9230769230. -0.0 6666. 66666. 66666 . 66666. 66666. 6666666666 . -0.0 0058. 82352. 94117 . 64705. 88235. 2941176470 . -0.0 0000. 52631. 57947 . 36842. 10526. 3157947368. -0.0 0000. 00476. 19047 . 61904. 76190. 4761904761 . -0.0 0000. 00004. 34782 . 60869. 56521. 7391304347 . -0.0 0000. 00000. 04000 . 00000. 00000. 0000000000 . 0000000000. 00000000 -0.0 0000. 00000. 000 ==. =====. =====. 0000000000. 0000000000. 00000000 -0.0 0000. 00000. 000 37. 03703. 70370. 3703703703 . 0000000000. 00000000 -0.0 0000. 00000. 00000. 34482. 75862. 0689655172. 4137931034. 48275862 -0.0 0000. 00000. 00000. 00 ===. =====. = 000000000. 0000000000. 00000000 -0.0 0000. 00000. 00000. 00322. 58064. 5161290322. 5806451612. 90322580 -0.0 0000. 00000. 00000. 00003. 03030. 3030303030. 3030303030. 30303030 -0.0 0000. 00000. 00000. 00000. 02857. 1428571428. 5714285714. 28571428 -0.0 0000. 00000. 00000. 00000. 00027. 0270270270. 2702702702. 70270270 -0.0 0000. 00000. 00000. 00000. 00000. 2564010256. 4010256401. 02564010 -0.0 0000. 00000. 00000. 00000. 00000. 0024390243. 9024390243. 90243902 -=. 8063. 57265. 52007. 40736. 63159. 415063 &c = summæ -0.0 0000. 00000. 00000. 00000. 00000. 0000000000. 0000000000. 00000000 -0. 00500. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 0000 2. 50000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 00000. 0 1666. 66666. 66666. 66666. 66666. 66666. 66666. 66666. 66666. 6. -0. 00000. 000 12. 50000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 00000. 00000 10000. 00000. 00000. 00000. 00000. 00000. 00000. 00000. 0. -0. 00000. 00000 000 83. 33333. 33333. 33333. 33333. 33333. 33333. 33333. 3. -0. 00000. 00000 00000 71428. 57142. 85714. 28571. 42857. 14285. 71428. 5. -0. 00000. 00000 00000 00 630. 55555. 55555. 55555. 55555. 55555. 55555. 5. -0. 00000. 00000 00000 00000 0 5045. 45454. 54545. 45454. 54545. 45454. 5. -0. 00000. 00000 00000 00000 00000 41666. 66666. 66666. 66666. 66666. 6. -0. 00000. 00000 00000 00000 00000 00 384. 61538. 46153. 84615. 38461. 5. -0. 00000. 00000 00000 00000 00000 0000 3. 57142. 85714. 28571. 42857. 1. -0. 00000. 00000 00000 00000 00000 00000 0 3364. 58333. 33333. 33333. 3. -0. 00000. 00000 00000 00000 00000 00000 00000 29411. 76470. 58823. 5. -0. 00502. 51679. 26750. 72059. 17744. 28779. 27385. 30147. 14044. 12586. 0.

<20v>

cui addendum -277.77777.77777.7 00-2.63157.89421.0 000.-02523.80952.4 000.00000.-22727.3 000.00000.00-217.4 that is 000.00000.0000-2.1 -280.43459.71098.0 that is And so ye summe will bee +0.10033.53477.31075.58063.57265.52007.40736.63159.41506.3 -0.00502.51679.26750.72059.17144.28779.27385.30427.57503.8 -0.09530.01798.04324.86004.40121.23228.13351.32731.84002.5 wch is ye quantity of ye area adpq . If cpab=1. & cp= ab=10pq & qde||ap||bc=ap . In like manner if I make x= 1100=pq . The opperacon followeth. +0,01000.03333.33333.33333.33333.33333.33333.33333.33333.30 +15x5+17x7= 33333.20001.42857.14285.71428.57142.85714.28571.40 19x9+111x11= ==.=====.=====.=====.=====.=0202.00 19x9+111x11= 11.11202.02020.20202.02020.20202.00 113x13+115x15= 769.23076.92307.69230.70 117x17= 6666.66666.66666.60 119x19= 58823.52941.10 121x21= 5.26315.70 47.60 +0,01000.03333.53334.76201.58821.07551.40422.38870.97309.30 -0,00005.00025.00166.66666.66666.66666.66666.66666.66666.6. 18x8110x10112x12= -1250.10000.83333.33333.33333.33333.3. 114x14= -7.14285.71428.57142.8. 116x16118x18= -62.50555.55555.5. =====.=====.=. 120x20= -5000.4. -0,00005.00025.00166.67916.76667.50007.14348.21984.17699.0. +0,00995.03308.53168.08284.82153.57544.26074.16886.79610.0.
wch is ye quantity of ye area apqd if 100p=cp . and abcp=1

<21r>

y=db . x=ba aay=x3 . a |b | + y | z| =y {illeg} z=bc aab+aa y |z| =x3. a |b|y |z|=y= {illeg} z=bf. aabaa y |z| =x3. y |z|a |b|=y= {illeg}z=bg aa y |z||a| ab=x3. {illeg} x=d+ξ {illeg} ξ=ah. {illeg} aa{illeg}|y|d33ddξ3dξξξ3=0 ξ aab+aaz aabaaz aazaab } = d3+3ddξ+3dξξ+ξ3 . x=bξ . aq=ξ aay aab±aaz aazaab } = d33ddξ+3dξξξ3 . x=ξb . ak=x aay aab±aaz aazaab } = ξ33dξξ3ddξd3 .

na=ξ . nd=z . |Or.| ξ3a2z+bbξ aban ab . &. abnb ac then ξ3 = b3az bbcaξ . or if abn is \a/ right angle. & || | ξ=d+ζ {illeg} ζ=mn | d3+3d2ζ / +bbcζ \ + 3dζζ + ζ3 b3az + bbcda

<21v> <22r>

db=x . ba=y . aax=y3 {illeg}. x=b+z. z=bc. y=c+ξ. ξ=ah aab + aaz = ( x=bz.z=bf. ) aab aaz = ( x=zb.z=bg ) aaz aab = } ξ3 + 3cξ2 + 3ccξ + ξ3 . y=cξ . ξ=aq . c3 3ccξ + 3cξ2 ξ2 . y=ξc . ξ=ak . ξ3 3cξ2 + 3ξc2 c3 . na=y . d {illeg}| n | =x . aban ab . abnb ac . & y3 = b3ax bbcay . Or d2x = εεy + y3 . ( x=z+o.z=nc ) ξ3 d2z + d2o = ξ3 ( x=zo.z=gn ) ξ3 ddz ddo = ξ3 ( x=oz.z=nf ) ξ3 ddo ddz = ξ3 ddz } εεy + y3 . ( y=n+ξ.ξ=na ) ξ2n + εεξ + n3 + 3nnξ + 3nξ2 + ξ3 . ( y=nξ.ξ=as ) εεn εεξ + n3 3nnξ + 3nξ2 ξ3 . ( y=ξn.ξ=av. ) εεξ εεn + ξ3 3nξ2 + 3ξn2 n3 . ddz al=y . dl=x . ab {illeg} \ al / ab &. {illeg} albl bc . whence y3 = xb3 + yb2c a or y3 = yb2c a = {illeg} / y3 εεy = ddx \ &c: as before onely varying ye signes at εεn & εεξ . ao=y. do=x. ab bddo . bc doob . a3bx = y3 3cxyyb + 3ccxxybb c3x3 bb .

<22v>

Dr Wi|a|llis in a letter to {illeg}|S|r Kenelme Digby promiseth ye squareing of ye Hyperbola by finding a meane propor{illeg}|t|ion twixt 1 , & 56 in the progression 1, 56, 3130, 209140, 1471630, 106252772 &c.

<23r>

The resolution of cubick equations out of Dr Wallis in his dedication before Meibomius confuted

suppose x=ae. yn x3 = a3 3aae 3aee e3 . or x3 = \+ /{illeg} 3aex a3 e3 . that is ma{illeg}|k|ing a3 + e3 = q {illeg}. yn x3 = 3aex c3 . & 3ae = p . yn x3× |=| + px q . Againe suppose {illeg}| y | = b | a | O {illeg}| e | Then {illeg}| y3 | Againe suppose x = a e . yn x3 = a3 3aae + 3aee e3 . yt is making a3 e3 = q , & 3ae = p , yn x3 = px q .

Then in the first o{illeg}|f| th|e|se p = 3ae . or p3e = a . or p327e3 = a3 = qe3 . Therefore e6 = qe3 p327 . {illeg}|&| e3 = 12q 14qq p327 . & by y{illeg}|e| \same/ reason a3 = 12q O 14qq p327 where ye irrationall quantitys have. divers signes other{illeg}|w|ise a3 + e3 = q would bee false. Soe that
x = a e = c: 12q O 14qq 127p3 c: 12q 14qq 127p3 . is a rule for resolving ye equation x3 p O q = 0 , when it hath but one roote yt is when it may be generated according to the supposition x = a e . &c. By ye same reason x3 + px O q . may be resolved by this rule x = a e = c: 12q 14qq + 127p3 c: 12q O 14qq + 127p3 .

But here of|b|serve yt Dr Wallis would h|A|rgue yt since in the first of these two cases fsometimetimes (viz when ye equation hath 3 \reall/ rootes) {sic} first rule faile|th| as it were imposs|i|ble {sic} for ye equation to have rootes when yet it hath, therefore ye fault is in Algebra. & therefore when Anal{illeg}|i|ses Analysis leads us to an imposs|i|bility {sic} wee ought not to conclude ye thing |absolutely| imposible, untill wee have tryed all ye ways yt may bee.

But let me answer yt ye fault is {illeg}i{illeg}|not| in ye Analysis in this example, but \in/ his opperation. for {illeg} when ye equation x3 + p x q = 0 , hath 3 root{illeg}|s| hee supposeth it to have but one roote viz x = a e . but sin{illeg}|c|e ye Equation cannot be then generated according to yt supposition it is impose|i|ble {sic} it should be relolved {sic} by it.

<23v>

In like manner hee s{illeg}|a|yeth yt Algebra representehteth a thing possible when tis not so as in this examl|p|le, in ye △ abc, make ab=1 . bc=2 ac=4. FigureThen to find dc=x, worke |thus,| {illeg} ad = 4x . bd×bd = 116+8xx2 = 4x2 {illeg}|t|herefore 8x = 19 . or x = 198 . In wch oppera{illeg}|c|on all things proceede as possible though they are not soe for ac is greater yn ab+bc .

yet I answer yt if ye opperation & conclusion be compared together ye absurdity will appeare. for in ye equation bd×bd = 4xx = 436164 = 25636164 or bd×bd = -1058 . but it is impossible yt a □ number should be negative.

Thus x=-b is impossible. square it & tis xx=-b. Againe, & tis x4 = bb . Extract ye roote & tis xx = b or x=b. wch is possible. The reason of this proceeding Event is yt x4bb=0 hath two possible rootes viz x=b . x=-b. & two impossible viz: x=-b. x=--b.

Thus ye valors of x8 a8 = 0 are x=a , -a , -aa, --aa , 4:-a4, - 4:-a4, --a4, ---a4.

<24r>

Dr Wallis in a letter to Sr Kenelme Digby teacheth how to find ye center of gra{illeg}|v|ity in divers lines first when their position is as in Figure this figure.

Suppose ad ye Axis, a their vertex & as 1 to ye series of ye progresion cons. Then saying, as 1 to ye index of ye line increased by an unite (vide pag {illeg}|2|da) so cd to ca Then c is their center of gravity.

The Demonstracon.

Let p bee ye index of ye series according to wch ye odinately aplyed lines (parallell to db) increase, yn {illeg} {illeg} 1p+1 area of ye line ∶ to nmbq . ye distances of those ordinate lines from ye vertex a are equall to ye intercepted diameters & therefore a primanary series |(whos index is 1. | & since supposing a ye center of ye ballance ye whole weight of ye surface or figure is composed of its magnitude & distance from ye center and therefore ye index of all its moments or whole weight is p+1, viz: ye aggregate of ye other two. Therefore as all its mom moments (or ye \weight of the/ figure in its site in respect of ye center a are to soe many of ye greatest (or to ye weight of ye ▭ nmbq hung on ye point d ) soe is 1, to p+2. and if apad 1p+2 , then nmbq hung on ye point q shall counterballance ye figure in its site &c therefore if {illeg} accd p+11 , {illeg}|c | shall be ye center of gravity of those figures.

Figure

Also as the figure is now put extending infinitely towards δ if -2p + 1 - {p } +1 amac . {illeg}|m | being ye center of qnbd yn c shall bee ye center of gravity of ye whole figure qndbδ .

Demonstration

<24v>

sincce {sic} ye lines parallell to aδ increase in series reciprocl|a|lly proportionall their index is -p & since ye halfe{illeg}|s| of those lines increas {sic} in ye same proportion their index is -p. whose extremitys or middle points of ye whole lines (suposing a ye center of ye ballance) are theire centers of gravity, their distances from a being proportionall to ye lines whose centers they are & consequently their index is -p & since all ye moments (or whole weight of ye figure) increase in a proportion compounded of ye proportion of ye magnitudes & distances of ye lines from ye center a, they will be in a du{illeg}|p|licate proportion of ye lines magnitudes that is a series reciprocall series whose index is -2p. Therefore ye figure is t{illeg}|o| ye inscribed parallelogram as 1 to 1p. & all its moments or w{illeg}|h|ole weight of the Para in this its site to {illeg}|t|he weight of ye pgr as 1 to 12p. Therefore if, amap 12p 1 , the paralelogra {sic} hanging on ye point p s{illeg}|h|all counterballanc{e} ye whole figure in its site &c: whence ye point c may be found easily, {illeg} viz amac 12p 1p .

<26r>

Of Refractions.

1 If ye {illeg} ray ac bee refracted at the center Figureof ye circle acdg towards d & abbegced . Then suppose abed de . {illeg}|S|ee Cartes Dioptricks

2 If there be an hyperbola Figurewhose \the/ distance of whose foci bd are to its transverse axis hf as d to e . Then ye ray acbd {illeg}|i|s refracted to ye exterior focus (d). See C: Dioptr

3 Having ye proportion of d to e, or. bdhf. The Hyperbola{illeg} may bee thus described.

1 Upon ye centers a, b Figure let ye instrument adbtec bee moved in wch instrumnt observe yt ad {illeg}de c {illeg} et & yt the beame cet is not in ye same plane wth adbe but intersects it at ye angle tev soe yt if tvev, then de ettv . Or de Radsine of ∠ tev . Also make de=q2, i.e half ye t{illeg}|r|ansverse diamet{er.} Then place \the fiduciall side of/ plate chm in the same plaine wth ab . & moving ye instrument adbect to & fro its edge cet shall cut or weare its {illeg} into ye shape of ye desired Parabola. Or the plate chm may bee filed away untill ye edg {sic} cet exactly touch it everywhere.

2 By the same proceeding Des=Cartes concave Hyperbolicall wheele may bee described by beeing turned wth a chissell {illeg}| d |tec whose edge is a streight line inclined to the edge \axis/ of the mandrill by ye ∠ tev wch angle is found by making de ettv Radsine of etv .

3 By the same reason a wheele ma{illeg}|y| be turned Hyperbolically concave ye Hyperbola being convex. Or a Plate may bee turned Hyperbolically concave

<26v>

Figure Also Des=Cartes his Convex wheele \ B / may be turned {illeg}|o|r {ground} trew a concave wheele \ A / being made use of instead Figure of a patterne

5 In turning the concave wheele A it will \perhaps/ bee best to weare it wth a stone p & let {illeg}|t|he streight edged chissell d serve for a patterne. And it may bee convenient to grind ye stone (or iron &c {illeg}) p into ye fas{illeg}|h|ion Figure of a cone S That it may fit ye hollow of the wheele A. The angle of wch conce {sic} being a right one or something greater it will almost grind the wheele to a {illeg}tre{illeg} figur Same done by helpe of a Cone.

6 Figure Draw ab=q; ac & cb of any length or intersecting one another at any angles. to make up the triangl{illeg}{e} abc . [Suppose yt ac bee called b, & that e=q= {illeg} then is d the distance of ye foci]. produc{e} ac to d soe that ad = ddeeb . Then draw bk through ye point d, & draw eh parallell to bc , lastly wth the sides he , ek & angle hek describe the cone heklm . Then produce ba to g &c \indefinitely/ & ag being ye axis of a section mal shall be ye sought Par Hyperbola

7 Since the proportion of cb to ab & ∠ cba is not deter{illeg}|m|ined it will be most convenient to make <27r> cb=ab=e=q . & cbabah. And then there will bee little danger of error at ye vertex of the {illeg}|Hy|perbola. And ye calculation is readier for drawing bpca, Then is cp=bp=ap=e2=ee2 & pd=dde2 . Soe that eedd cp=bppd Radtangen RadiusTangent of pbd soe yt ye ∠ cbk=hek is {illeg} easily found.

9 Halving such a cone smo|o|thly pollished wthin & wthout, by the helpe of a square set ye plate perpendicular {illeg}|t|o one side hae the fiduciall edge being distant from ye vertex the length of ae = edde3 dd+ee & if ye edge of ye plaine every where tou{illeg}ch the cone, tis trew.

10 The exact distance \(ae)/ of ye plate from the vertex of ye cone neede not bee much regarded for that changeth onely the shape \bigness/ not ye shape of ye figure.

[By ye broken lookinglasse I find in glasse refraction, yt de 4328 1000651+ 1536 {illeg} 1000. These are insensib almost insensibly different from truth de 2013 10065 1531000 . Or de 23{illeg} /2315\ 100652+ de 6643 100651,5151+ . Or de 100651

For ye Ellipsid|s| ddee d x + eedd dd xx = yy

<27v>

The former {descriptionspropositions} demonstrated.

Lemma. If in ye Opposite Hyperbolas one of abc Figure edf (one of wch are to bee described) supposing bd=d. hf=e. gh=x gc=y. gcghd & {illeg} point & gc terminated by ye hyperbola Then is ddee ee xx + ddee e x = yy . b d | h | =de2 . dh=d+e2 . d |b| g=2xd+e2. gd=2x+d+e2. dc2= 4xx + 4dx + 4ex + dd + 2ed + ee + 4yy 4=gd×gd +gc×gc. bc2= 4xx 4dx + 4ex + dd 2ed + ee + 4yy 4=gb2 +gc2. And since af dc=bc+hf. /Or dc2 = bc2 + 2bc×hf + hf2 \ Therefore 2dx + ed ee = e 4xx 4dx + 4ex + dd 2ed + ee + 4yy . Both ts of wch □ed & ordered ye result is 4ddxx 4eexx + 4eddx 4e3x 4eeyy = 0 . That is ddee e x + ddee ee xx = yy .

Desciption {sic} ye 1st demonstrated \Synthetically/. See yt Scheame

Naming ye quantitys ed=e2=dh {illeg} de{illeg} ∶ e{illeg} tv Nameing ye quantitys \{illeg}/ ed=dh=e2. gh=x. gc=be=y. dg=x+e2=nc. cgdhg. cd2 = x2 + ex + ee4 + y ce2 = xx + ex + yy . eg2 = xx + {illeg} ex . de Also de ettv ceeg , therefore ddxx + ddex = eex2 + e3x + e2y That is ddee e x + ddee ee xx = yy . As in ye lea

The Same demonstrated synthe \Analy/tically.

Nameing ye quantitys, de=dh={illeg}|a |. gh=x . gc=y. dg=a+x dc2 = aa + 2ax + xx + yy . ce2 = 2ax + x2 + yy . eg2 = xx + ex . Supose yt {illeg} bc ettv ceeg .

<28r>

Then is bbxx + bbex = 2ccax + ccxx + ccyy . That is bbcccc xx + bbe2ccacc x = yy . Therefore ye line chm is a Conick Section & since (bb) is greater yn (cc) tis an Hyperbola, wch yt it may bee {illeg} ye same wth yt in ye lemma, Their correspondnt termes are to bee compared together & soe I find yt bbcccc xx = ddeeee xx . & bbe2ccacc x = ddeee x by ye 1st =tion bb = ccdd ee . Or b=cde. yt is bc de . by ye 2nd ccee ccdd + bbee = 2ccea . And by substituting ccdd ee into the place of bb {illeg}|A|nd ordering it tis {illeg}2{illeg} ccee = 2ccea . Or e2 = a . Therefore if I take e2 = a = de . & de bc cttv . then shall chm bee ye Para|Hyper|bola desired Q:E:D.

The 2d 3d 4th & 5th Propositions are manifest from this

The 6th Description demonstrated Syntheticaly The quantity named are ab=e. Figure

Instead of ye 6th & 7th Descriptions wch are false use these

6 Draw 2 concentrick circles (na & cd) wth ye Radij e & d. Then from ye comon center b draw 2 lines bc & {babd} at the given angle {bae=abcaed=cbd} of {sectionyeCone} then draw a line cad thr from c throug by ye end of ye Rad {babd} & to ye intersection of yt line wth ye circle {cdna} draw {bdba} & so the ∠ of {yeCone,hek=cbdsection,eab=abc} is found.

<28v>

Or wch is the same {illeg} make ab=e. bd=d & then if yt cone is sought make cd=2r the ∠ cba being given, make ac=a. Then is cd = dd ee + aa a . & soe ye ∠ cbc=aed is knowne & also ae = ed = d3dee dd ee + aa , & ad= dd ee a . Buf But if ye ∠ bae=abc of ye section is sought y{illeg}|e| cone b|e|ing given yn make cd=2b. And it will bee ac=b+eedd+bb . & soe ∠ abc=bae is given also ad=beedd+bb . & ae = dbdeedd+bb 2b

{illeg}|I|n generall observe yt in any cone cut any ways bd = be+ea = d . & {illeg}|b|a=e.

7. DesCartes his wheele \thus described/ cut by any plaine produceth one of ye Conick=Sections.

Description ye 6th Demonstrated. Synthetically.

Call, {illeg} bd=d. ba=e. {illeg} cp=pd=a. bp=ddaa. ag=x ap=eedd+aa. ac=a+eedd+aa. ad=aeedd+aa. baac aggh= ax + xeedd+aa e . baad bggk= ea+ax e exe eedd+aa . \ gk×gh=gm2=y2 ./ Therefore ddee ee xx + ddee e x = yy . by ordering ye result of /gk×gh. \ wch is like yt in the lemma.

The 7th Proposition may be easyly demonstrated {illeg} after the same manner.

If the two \equall/ cones bad bcd intersect the Figure one the other soe yt ab=bc their intersection (bf) shall bee one of ye Conick sections as they had each beene intersected by the plane bf .

<29r>

To describe ye Parabola (& other figures after ye same manner) pretty exactly.

Figure

Thake a squire cbe , soe yt cb=r2 (for then the{illeg} circle described by (bc) will bee as crooked as ye Parabola at the vertex d ). Divide ye other leg (be) of ye Squire into any number of pts, Then get a plate of Br{illeg}|a|sse &c: lkfd streight & eaven. And taking one point d for ye vertex of it & another point c for ye Squire to move{illeg}n soe yt cd= cb=r2 , & wearein{illeg}|g| away ye edge of the plate untill (ye {illeg}|S|quire being erected) ab= qd. the squire touching ye plate at a . thus shall ye edge adf become Parabolicall. wth ye Rad: ab describe a circle adg & by that {meanes} it may bee knowne when ab=a. \Instead of ye leg be a/ Demonstraco\circle may be used/ Supose aq=y. cd=cb=r2 then {illeg}. Then is {illeg}rr4+ {illeg} zz {illeg} cq {illeg} ad=xx+yy=ab. & ac = rr4 + xx + yy . {illeg}|A|nd cq=rr4+ xx . & dq=x . Demonstracon. qd=x. cd=r2=cb. cq=r2x. aq=rx=y. ac2=rr4+xx ab2=rr4+xxrr4 & ab=x. Q.E.D.

Another description of ye Parabola y|w|e {sic} ye compasses. Make ab= bc=r4. Make ce=cd Figure & cebd. Make af=ae, & bf=bd then shall f be a point in ye Parabola.

Another. {illeg}|M|ake ab = r+x2 = ac . | eb=xce | & ye point c shall bee in ye parabola. This like ye first by calculation may bee made use of in other lines.

<29v>

The manner whereby any kind of \little/ lines may {illeg}be described very accurately. And that the same Instrument serve for all lines (though never so small) differing in quantity but not{illeg} in qual{illeg}ity.

Figure Make ye plate d of ye figure requ{illeg}|ir|ed (by some of ye former meanes) the larger the better. Then hold the streight \steele/ staffe b again{illeg}|st| the center a & {roule}{route} i{illeg}|t| to {illeg}|&| fro it shall grind c into ye same figure but soe much lesse as ac is lesse yn ad.

Figure Soe if ye glass c bee fastened upon ye mandrill f, it may be ground acording to ye sollid f{illeg} figure d by ye helpe of a stick of steele (as a cilind cone) {illeg} whose cuspis is in ye hole a upon wch it is moved as on a center. when ye stick cone b leanes uppon ye vertices of d & c it must be perpendicular to the mandrill f. Perhaps it may be convenient to cause ye cone b to turne about its axis. Or it may bee {illeg} better {illeg} instead of ye nutt at a wth a hole in it to make a sharpe pointed nutt, & instead of ye cilinder /cone\ b to make use of a broad plate to cover a, c & d & move every way upon them

<30r>

Another way to describe lines on plates

Figure

Suppose ye plate bee abc, whose edg {sic} boc is to be made into ye fashion of a given crooked line suppoes {sic} (o) is its vertex & yt a circle described wth ye {illeg}|R|adius eo would bee as crooked as ye given line at its vertex. A{illeg}gaine suppose two streight rulers mn & pq to bee very trew & steddyly fastened together Figure wch must a \very/ little incline ye one to ye other, soe as that being produced they would meete at ar. Then are ye lines pn=a {illeg}, & pr=b given.

Suppose yn ye point d in ye crooked line is to bee found yn is dc given by supposition, & {illeg} consequently (supposing dk to bee a tangent) dg=y. gc=x. fg=v. fd=s ec=c. fk=v+yyv {illeg}. {illeg} ef=v+xc. ek=cx+yyv. & (if ehdkdf) then is eh= cv xv + yy vv + yy = d . (eh) being thus foun{illeg}|d|, supposing yt pn=a=ec , then I take re=bda. that is pe=bbda . {illeg} haveing thus found ye point e lay ye plate twixt two rule the two rulers so yt ye point of it, fall upon ye point e yn should ye line mn touch ye plate in d. But note yt pnmn.

In both telescopes & microscopes tis most convenient to have a convex glasse next ye eye for by that meanes ye angle of vision will bee much greater yn it will bee wth a concave one (though both doe magnifie alike). If ye convex glasse be Hyperbolicall (&c) make it soe bigg yt ye penecilli may crosse in ye pupill; yt is, ye exterior focus will be as far distant from ye vertex as ye eye is. let ye glass bee as thinn yt ye as may bee yt ye eye bee not to{illeg}|o| farr {sic} from ye vertex yet {sic} it should bee about as thick as ye distance of ye interior focus from ye vertex.

And by this meanes also, (ye focus of ye objectglasse being within ye telescope twixt ye glasses) there may bee placed at {illeg}|t|hat {illeg}|f|ocus ye edge of <30v> a steele ruler accurately divided into equall parts (to measure ye diameters or distances of starrs &c) wch should bee soe made yt by a pinne or handle it may be placed i{illeg}|n| any posture & in any parte of ye focus, wthout otherwise altering ye Telescope in observations.

Note that were not ye glasses faulty they would not onely magnify objects but render vision more distinct; each of the penicilli passing through (perhaps but) the 10th, 20th or 100th part{illeg}|e| of ye pupill must bee more exactly refracted to one point {illeg} of ye Tunica Retina yn in ordinary visio in wch each of ye penicill{illeg}|i| spreads over all the pupill.

Note also that Figureyt ye glasse a may be ground Parabolic \Hyperbolicall/ by ye line cb, if it turne on ye mandrill e whilst c{illeg}|b | turnes on ye axis rd being inclined to it as was shewed before. If the edge (cb) bee not durable enough, inough instead thereof use a long small cilinder: wch I conceive to bee the best way, of{illeg} all. For a Cilinder of all sollids is most easily made exact (being Figure turned, as in the figure, by a gage untill its thicknesse bee every where equall). 2 the Cilinder may bee made {illeg} to slip up & downe & turne round whereby it will not onely grinde ye glase crosse wise to take of all hubbes, but also ye glasse & cilinder will grinde ye one ye other truer & truer. All ye difficulty is in placing ye axis rd perpendicular to the Mandrill ae \& vertex to vertex/, wch yet may bee done exactly severall ways. & untill yn the glasse & Cilinder will not fit. & should ye axis not intersect ye glasse would bee still Hyperbolicall except a point at the vertex of it. The same instrument may also serve for severall glasses onely making df longer or shorter. Let the Cilinder han{g} over the glasse.

<31r>

To Grinde Sphæricall optick Glasses

If ye glasse (bc) is to bee ground sphærically Figure hollow: naile a steele plate to ye beame (fg), on ye upper side: In wch make a center hole for ye steele point (f) of ye shaft (def): to wch shaft fasten a plugg (a) of stone or leade or leather &c: (wth wch you intend to grinde ye glasse (bc)): wch shaft & plugg being swung to & fro upon ye center f will grind ye glasse bc sphærically hollow.

The manner whereby Figure glasses may bee ground sf|p|hærically convex may appeare by ye annexed {illeg} figure (being ye former way inverted). Also ye plugg (a), in ye firstsecond figure, is ground sphærically convex.concave.

But if this way bee not exact enough yet hereby may bee {grownd}{ground} plates of mettall well nigh sphæricall, And by those plates may bee ground glasses after ye usual manner; If a circular hoope of steele \(abc)/ bee put about ye edge of ye glasse (d) to keepe it Figure from grinding away at ye edges faster yn in ye middle.

But the best way of all will bee to ti{illeg}|u|rne ye glass circularly upon a mandrill whilest ye plate is steadily rubbed upon it or else <31v> to turne ye plate upon a mandrill whilest ye glasse is rubbed upon it \or let sometimes ye one, sometimes ye other bee turned./: & by this meanes they will either of them weare the other to a truely sphericall forme. but however let there bee a hoope or of some mettall wch {illeg}|w|eares more difficultly then glasse to defend ye glasse from wearing more at its edges then in ye middle. Perhaps it may doe well \first/ to weare ye plate sphæricall by ye hoope alone wthout the glasse.

The same meanes may bee used for gr{illeg}|i|nding plaine glasses.

Let not an object glasse bee ground sphærically convex on both sides, but sphaerically convex on one side & concave or plaine \plane or but a little convex ~/ on ye other, & turne ye convexest side towards ye object.

<32r>

If the Glasses of a Telescope bee not truely ground, how to find where the fault is, & consequently to rectify it.

Figure Take two plates abfh, dgkh of {illeg} wood or brasse i{illeg}|at| ye midst of ye sides of wch {boare} two very small holes \c & e/ (viz whose diameters are about ye {illeg}|2|0th or {illeg}|3|0th pte of an inch, that they may transmit but soe much light as may serve to see {illeg}|ye| edge of ye sunne or a starre of ye first magnitude)

Also make two Figure other plates rs, & tv like ye former but ye holes in ym must be as small as can bee (viz \about/ ye 100th pte of an inch in diameter or lesse). For ye small end of ye tube

Also make an\o/ther plate wzxy wth a hole in \the midst of/ it about ye 5th or sixt pte of an inch in diameter (viz equall to ye diameter of ye pupill of ye eye or eyeglasse). Make ye like \plate A/ (but wth a very small hole) for ye eyeglaess {sic})

First cover ye object glasse wth ye plates af, & gh distant about ye ye holes in ym being distant about ye sixt parte of an inch; & placed neare ye center of ye object glasse. Also cover ye eye glasse wth ye plate A soe yt its hole exactly respect ye center of ye eye glasse then turne ye tube to a Starre wch will appeare like two starrs if ye tube bee two {sic} long or short, wch bee shortned or lengthned untill there appeare but one, And then is ye Tube of a good distance length for ye vertices of ye Glasses.

Secondly remove those plates, and instead thereof cover ye object glasse wth ye plate wz, its hole exactly respecting ye center of ye glasse. (Fig 2)

<32v> Figure

If ye Glasses of a Telescope bee not truely ground The{illeg}|ir|e errors may bee may bee thus found.

Because an error is much more easily discernable in ye object glasse yn in ye eye {sic} glasse let us first suppose ye eye glasse to bee ground true towards its center, (tis exact enough if it be sphericall, & not Hyperbolicall), & so wee may find & rectifie ye errors of ye object glasse.

First make a thin plate (A) of brasse & in the center of it a Small hole (whose diameter perhaps may bee about ye 50th or 100dth parte of an inch. {illeg}|W|ith wch plate cover ye eye glass ye center of it respecting ye center of ye glasse.

Secondly make two other plates the one B wth two holes \as/ neare to its edge as may bee their{e}{} distance being about ye 5t {sic} pte of an inch or lesse, & ye other C wth one hole close to ye midst of its edge. Let ye diameters of these 3 holes bee about a 20th pte of an inch or lesse. And theire edges must bee true that they may slide one upon another, & yet not let ye suns rays passe through, to wch purpose make ym oblique. Wth these two plates cover ye object glase {sic} (first stopping ye hole of C \ye holes of ye other plate respecting ye center of ye glasse/ & looke at a stare (or ye edge of ye sunne &c) & if ye object appeare double (like two starrs &c) make ye Tube longer or shorter untill it appeare single. Then open ye hole of C , & ye plate B being fixed, slide ye plate C up & downe still looking at ye starre, When then appeares <33r> but one starre yt part of ye glasse under ye hole of C is truely ground in respect of ye 2 parts of ye glasse under ye two holes of B. But {no} when ye starre appeares double. And ye position of ye starre caused by ye hole of {illeg} C in repect {sic} of ye starre caused by ye holes of B, shews ye error of ye incli wch way ye glasse under ye hole of C is erroneously inclined; the distance of ye two starres giving ye quantity of yt inclinati error.

Thus ye errors of ye of object glasse bein{g} found in every place of it they may bee all rectified, & found againe, & againe rectified, untill they almost or altogether vanish.

Then may ye eye=glasse bee rectified much after ye same manner, in every parte of it, & if it bee necessary ye object glasse may bee aganie {sic} rectified & againe ye eye=glasse untill ye Teles{illeg}|c|ope bee as perfect as ye workeman can make: Whome perhaps experience & other may teach by this & ye former rules to make telescopes as perfect as {illeg} {illeg}men can \hope to/ make them.

These glasses may also bee rectified whilst on ye Mandrill by observing ye images mab|d|e by reflection from ye vertex & all other ts of ye glasse wt proportion they have one to another & how much they are longer yn broader in one place then another. &c.

<33v>

BLANKThe sines measuring refractions are in Aere42 water56 Glasse65 christall70 The proportions of the motions of theextreamely heterogeneous rays are in} 39,4.40,4. 7038.7138. 095110.096110 11013.11113 The proportions ofyesines of refraction of the extreamely hetero-00 geneous rays into aire out of Their common sine of incidence00 Which substracted the difference is00 000Ty000 c00 000gf000 00 00 000Ty000 9023.9123. 000Ty000 6813 2213.2313. 000Ty000 068.0006900 000Ty000 4414 02334.02434 000Ty000 06145.06245. 000Ty000 3645 024.00025.00 The like proportions for refrac-tions made into water out of 27545.27645 23825 03725.03825 19613.19713 15749 03919.04019

of the method of infinite series

I Newton

<35r>

Theoremata varia. Circa angulorum æqualitates.

Figure

si ang DAB = |&| DAE bisecentur a rectis FH et IG et ducatur quævis KLMN . Erit
1. AK.AM KL.LM \{illeg} KN.MN / Euclid 6 3
2. AK×AM = ALq + KL×LM = A {illeg}+ KL×LM ALq . Scho{illeg}|{o}|te de {concis} /{æqu{is}}\
3. AM+AK . PK AQ.AL posito AP=AM .

Figure Si in angulo quovis PAQ inseribantur æquales AB, BC, CD, DE, EF, FG, GH &c anguli BA{illeg} P erit angulus CBQ dul|p|lus {illeg} \ DCP/ tripl, ED P|Q| quadr FEQ quint, GFQ sext, HGP sept. IHQ oct &c. Horu vero angulorum positi|o| ra{illeg}|di|o AB {illeg} sinus erunt Bβ , Cχ &c cosinus AB , Bχ , Cδ &c. Ergo si AB=r, & AB=x erit \ AC=2x / =2xxr. =2xx4xr . AD = (2AB) = 4xxrrr . = 4x3rrxrr &c

<163v>

In Generall.

mm+8mn+15nn¯ × ddx2n1 mm2mn¯ d eex1 in dxm+n+exm = y . 1=v . &, 2m+6n¯ × ddx2n + 2ndexn 2m4n¯ × ee in dxm+n+exm = {illeg} z.

<40v>

VII. 14

{illeg} Geomet. {illeg}. Frn Vieta Schooten {illeg} July 4th 1699 {illeg}

<41r>

To find the sume of ye squares cu{bes} &c. of ye rootes of an equation

If a , bg , c , d , e , f &c be the rootes of ye equation x6 + px5 + qx4 + rx3 + sxx + tx + v = 0 . yn is {illeg} a + b + c + d + e + f = p(=g) a2 + b2 + c2 + d2 + e2 + f2 = pp2q . (= pg 2q {illeg} =h) a3 + b3 + c3 + d3 + e3 + f3 = p3 3pq + 3r . (= ph qg + 3r = k ) a4 + b4 &c = p4 4ppq + 4pr + 2qq 4s . (= pk qh + rg 4s = l ) a5 &c = p5 5p3q + 5pqq + 5ppr 5ps 5qr + 5t . (= pl qk + rh sg + 5t {illeg} = { m) } a6 &c: = p6 6p4q + 9ppqq + 6p3r 12pqr 6pps + 6pt 2q3 + 3rr + 6qs 6v .

<80r> Figure

ag=a . ab=x . bh=dxc . bc=y . bg=xxaa gh = ddxxeexx+eeaaee . dg=b. ce = ydx ddxxeexx+eeaa = fg . cf=dx+eydxxxaa . df = b ydx ddxxeexx+eeaa . z2=dc2={ xxaa+2eyxd2eyaadx+bb+yy 2bydx ddxxeexx+eeaa 2bydx ddxxeexx+eeaa = xx+yyzz+bbaa+2eyxx2eyaadx . 4bbddxx ddx6 + 4deyx5 + 2ddyyx4 4deaayx3 4bbddyyxx 4dey3aax + 4e2a4y2 2ddzzx4 + 4dey3 + 4bbeeyy + 4deyz2a2 + 2ddbb 4deyzz 8aaeeyy 4deybba2 2ddaa + 4deybb + 2ddy4 + 4deya4 4bbe2a2y2 + 4eeyy 4deyaa

Ad constructionem Canonis angularis.

90gr 905gr = 18gr . 185gr = 3gr + 36 . Et 603gr = 20gr . 203gr = 6gr + 40 6gr + 40 2 = 3gr + 20 . 3gr + 36 3gr 20 = 16 . 162=8 . 82=4 . 42=2 22=1 .

If r= radius. Then
yesine of 0 0 0 78degr is, r5r+r30+65 8 . 66degr is, r5r+r3065 8 . 42degr is, 5rr+r+30rr+6rr5 8 . 06degr is, 30rr6rr55rrr 8 .

<80v>

Suppose gh=x . nh=r . Then gh=x. unisectio ab×r=2rrxx. bisectio hb2×r2=3rrxx3. trisectio ab3×r3=2r44rrxx+x4.quadriseco. hb4×r4=5r4x5rrx3+x5.quintusecto. ab5×r5=2r69r4x2+6rrx4x6. hb×r6=7r6x14r4x3+7rrx5x7. ab×r7=2r816r6x2+20r4x48rrx6+x8. hb×r8=9r8x30r6x3+27r4x59rrx7+x9. ab×r9=2r1025r8x2+50r6x435r4x6+10rrx8x10. hb×r10=11r10x55r8x3+77r6x544r4x7+11rrx9x11.

< insertion from the top right of f 80v >

As on ye other leafe excepting some signes here changed.

< text from f 80v resumes >

If gh=x . bh=y .Then y=bh. duplicatio angulihag yyxx=x×hb2. triplicatio angulihag. y32xxy=xx×hb3. quadruplicatio. y43xxyy+x4=x3×b4h. quinto y54xxy3+3x4y=x4×hb5. sexto y65xxy4+6x4yyx6=x5×hb. septo y76xxy5+10x4y34x6y=x6×hb. octo y87xxy6+15x4y410x6yyx8=x7×hb. nonc y98xxy7+21x4y520x6y3+5x8y=x8×hb. dec y109xxy8+28x4y635x6y4+15x8y2x10=x9×hb. und, y1110xxy9+36x4y756x6y5+35x8y36x10y=x10×hb. duod

<81r> Figure

Of Angular sections

Suppose ab=q . ah2=r . & ag=x. & yt ye arches {illeg} hg, gb, bb are equall. By ye following Equations an angle \bah/ may bee divided into any number of partes. x=q. unisectio. x22rr=rq. bisectio. x33rrx=rrq. trisectio. x44rrxx+2r4=r3q. quadrisectio. x55rrx3+5r4x=r4q. quintusectio. x66rrx4+9r4xx2r6=r5q. sextusectio. x77rrx5+14r4x37r6x=r6q. septusectio. x88rrx6+20r4x416r6x2+2r8=r7q. x99rrx7+27r4x530r6x3+9r8x=r8q. x1010rrx8+35r4x650r6x4+25r8x22r10=r9q. x1111rrx9+44r4x777r6x5+55r8x311r10x=r10q. x1212rrx10+54r4x8112r6x6+105r8x436r10x2+2r12=r11q. x1313rrx11+65r4x9156r6x7+182r8x591r10x3+13r12x=r12q. x1414rrx12+77r4x10210r6x8+294r8x6196r10x4+49r12x2&c x1515rrx13+90r4x11275r6x9+450r8x7318r10x5+140r12x3&c x1616rrx14+104r4x12352r6x10+660r8x8672r10x6+336r12x4&c x1717rrx15+119r4x13442r6x11+935r8x91122r10x7+714r12x5&c x1818rrx16+135r4x14546r6x12+1287r8x101782r10x8+1386r12x6&c x1919rrx17+152r4x15665r6x13+1729r8x112717r10x9+2508r12x7&c x2020rrx18+170r4x16800r6x14+2275r8x124604r10x10+4290r12x8&c

This scheame is ye forme\r/ inversed. Figure

<81v>

Suppose ye perifery {illeg} \ bgh / to bee a & ye whole perifery to bee p. The line bh subtends these arches. a. pa. p+a. 2pa. 2p+a. 3pa. 3p+a. 4pa. 4p+a. 5pa. 5p+a. 6pa. 6p+a. &c: All wch are bisected, trisected, quadrisected, quintuse{illeg}|c|ted &c after same manner. |As for example|

The rootes of ye equation hb2 × rr = 3rrx x3 . are 3. The first whereof subtends ye arches a3 . 3pa3. 3p+a3. 6pa3. 6p+a3. 9pa3. 9p+a3 &c. The second subtends ye arches pa3. 2p+a3. 4pa3. 5p+a3. 7pa3. &c. The 3d p+a3. 2pa3. 4p+a3. 5pa3. 7p+a3 &c.

Soe ye rootes of ye equation ~ hb×r4=5r4x5rrx3+x5 , doe ye first subtend ye arches a5 . 5pa5. 5p+a5 &c: ye 2d pa5. 4p+a5. 6pa5. ye 3d p+a5. 4pa5. 6p+a5 &c. ye 4th 2pa5. 3p+a5. &c ye 5t 2p+a5. 3pa5. 7p+a5. &c.

Hence may appeare ye reason of ye number of rootes in these equations & yt ye points of ye circuference {sic} \to wch they are extended/ æquidistant. & by ye lower scheme may bee known wch rootes are affirmative & wch negative.

The numerall cöefficients of ye afforesaid equations may bee deduced from this progression (if 1n .) 1 × 0+n × 1+n 1 × 1n × n2 × n3 2 × 2n × n4 × n5 3 × 3n × n6 × n7 4 × 4n × n8 × n9 5 × 5n × n10 × n11 6 × 6n &c. As if x n=10. ye ression 1 × 10 × 72 × 107 × 12 × 225 × 0 . And ye coefficients 1 . 10 . +35 . 50 . +25 . 2 .

<82r>

1663 /4 January.

All ye parallell lines wch can be understoode to bee drawne uppon any superficies are equiv{illeg}|a|lent to it, as Figure all ye lines drawne from (ao) to (co) may be used instead of ye superficies (aco.)

If all ye parallell lines drawne uppon any superficies be multiplied by another line they produce a Sollid like yt wch relults {sic} from ye superficies drawne into ye lame {sic} line Figure as if either al{illeg}|l| ye lines in ye superficies (oac) or if ye superficies oac be drawne into ye line (b) they both produce ye same lollid {sic} (d) whence All ye parallell superficies wch {illeg} can bee understoode to bee in any sollid are equivalent to yt Sollid. And If all ye lines in any triangle, wch are parallell to one of ye sides, be squ\a/red there results a Pyramid. if those in a square, there results a cube. If those in a crookelined figure there resuts {sic} a sollid wth 4 sides {illeg} terminated & bended after according to ye fasshion {sic} of ye crookelined figure{.}

If each line in one superficies bee drawne into each correspondent line in another superficies as in aebk, & omnc Figure if ae×dh. bk×cn. qv×wx. &c. they produce a sollid whos {sic} opposite sides are fashioned by one of ye superfic as {illeg} figure|Sollid| fpsrg. where all ye lines drawne from fr {illeg} to ps are equall to all equall to all the correspondent lines drawne from ow to mx. & those drawne from fg to fr are equall to ye correspondent lines drawne from qz to vz.

<82v>

Theorema. 1

If in the Circle abcdeP there be Figure inscribed any Poligon abcde wth an odd number of sides, & from{illeg} any point in ye circumference P there bee drawne lines Pe, Pa, Pb, Pc, Pd to e{illeg}|v|ery corner of ye Pol{illeg}ygon: ye summ of every other line {illeg} is equall to ye summ of ye rest, Pa+Pb+Pc{illeg} =Pd+Pe. & soe are their cubes Pa3+Pb3+Pc3=Pd3+Pe3 . unless ye figure be a Tri/gon\

Theor 2

If from{illeg} ye points of ye Polygon Figure then {illeg}|b|ee drawne perpendicular ap, br, ct, ds, e{illeg}q to any Diameter pt: ye summe of ye Pe{illeg}|r|pendiculars on one side ye Diameter is {equall}{equal} to their summe on ye other ap+br+ct=eq+ds . & soe is ye summe of their cubes (unlesse wn ye figure is a Trigon), ap3+br3+ct3= eq3+ds3 . & of theire square cubes (except wn ye figure is a Trigon or Pentagon. &c.

Theor {illeg}|3|

If ye 2 circles (fig 1 & 2) be equall wth like Poligo{illeg}{ns} inscribed, & Pa in fig {illeg}|1| be assumed double to pa in fig 2. then are all ye ye other corresponding lines in fig 1 double to those in fig 2 viz Pb=2rb, Pc=2tc, Pd=2sd, Pe=2qe.

<83r>

To square ye Parabola

In ye Parabola cae suppose ye Figure Parameter ab=r. ad=y. dc=x. & ry=xx or xxr=y. Now suppose every ye lines called x doe increase in arithmeticall proportion all ye x's taken together ye supp make ye superficies dch wch is halfe a square let every line drawne from cd to hd be square & they produce {illeg}|a| Pyramid equall to every xx=x33. wch if divided by {illeg}y r {illeg} there remaines every x33r= =yx3 equall to every xxr {illeg}|e|quall to every (y) or all ye lines drawne from ag to accc equall to ye superficies ag {illeg}| c| equall to a 3d pte of ye superficies adcg & ye superficie acd=2yx3.

Otherwise. suppose ce=b. co=x. to=y. & ry = bxxx ye lines x increasing in arithmeticall proportion every x is equall to \4 times/ ye superficies cdh=bb2 wch drawne into b produceth ye sollid b32 but if every x be squarered {sic} {illeg}|t|hey poduce {sic} a {illeg}|p|yramid equall to b33. wherefore every bxxx=b36 equall to every ry equall to ye superfies|ci|es adce drawne into r & b36r= to cade as before.

<83v> Figure

* = pd = q+y5 & qy+yy5 = xx

<84r>

To Square ye Hyperbola

In ye Hyperbola eqaw. suppose ef=a. fa=b. ap=rq=y \{=d}/ pq=ar=x. ad=q=\5/oa=\5/ac=\5/ d= |r=|. & da+ar ar arrq . xx = dy+ {illeg} yx . |In wch equation| Every x \taken together/ is equall to ye triangle aβb \equall to aa2/ & eve{illeg}|r|y xx {illeg}|ta|k{illeg}en together is a pyramid = a33 . Every y taken together is equall to ye superficies eba=mkt If yn gh=lm=ns==d . every dy is equall to ye solid nglmhs. If ye angle mhk is a right one & if mh=gl=ba=ef=a {illeg}f=kh, then all ye lin that is if ye triangle mhk=abβ. every {illeg}|y |x will be equall to ye sollid mhstk Joyne these two sollids together as in lmtng=a33 . *

⊛ Againe Suppose every x taken together to be equall to ye superficies aef , {illeg} ye line squared is 4xx {illeg} every 4xx composeth a Sollid like (♊♉♌) a|n| quarter \eighth pte/ whereof (wch is equall to every xx2) being like \♊♉♓o=/ xyzv; xv will be equall to ♊ {illeg}|| =ef{illeg} =a=st=xz=o♓=hm. & vz=♊♓ ={illeg}km . whence ye {illeg} covexe {sic} superficies xyv of ye figure xyvz will fitly joyne wth ye concave superficies mst of ye figure shmkt . If every x is equall to ye superficies aef , every y shall be equall to ye triangle afπ=bb2. every yy=bbb3 every qy=qbb2 & therefor ye Sollid yxzv=b330+qbb20=2b3+3qbb60. Joyne ye Sollid shmkt to yxvz & there resulteth shmztk= =aab2 from wch againe substract xvzy=2b3+3qbb60 & there remaines ye sollid mhstk=30aab2b33qbb60 wch substract from ye sollid ntmlg=a33 & there remaines nglmhs=20a3+2b3+3qbb30aab60 b wch being divided by θ=5rr4. there remaines 40a3+4b3+6qbb60aabr5= to ye superficies abe

<85r>

The squareing of severall croked {sic} lines of ye Seacond kind.

Figure In a \any two/ crooked lines I call ye Parameter or right side \of ye greater./ (r). \but of ye lesse (s)/ Transverse side (q). ye right axis as cf (x) \or ef=v . / y Transverse axis as fe {illeg} y, \or/ fd z.

Suppose in{illeg} ye Parab: ddc: ac=r. & in eec: bc=s rx=zz=df2. sx=yy=fe2. rxsx=de= =p. rx=sx+pp+2psx. rxsxpp=2psx rrxx2rsxx+ssxx2pprx2ppsx+p4=4ppsx. Or p42rxpp6sxpp+rrxx2rsxx+ssxx=0. if p=y. xx=+2ryy+6syyxy4rr2rs+ss. make cf=a. fd=b. fe=c. ceeff=2ac3. & cddeff=2ab3 therefore 2ab2ac3=cddee ye square of ye crooked line cdd (when ye line cee is p|s|upposed to {sic} close wth ye line cf ) whose nature is exprest by ye foregoing Equation.

Figure 2 lk=b. li=x. qi=y. in=z. {illeg} +bx\xx/ =ry {illeg}bx \xx/ =sz. xx+bxr=y. xx+bxs=z qn=sxxs+bsx+rxxrbrxrs=v={illeg}y or, {illeg} \r/rx{illeg}sxx+bb\s/xbb\r/xrsy=0. Or {illeg} xxbxrsyrs=0

Figure 3 tg=x. dg=z. gp=y. rxrz=dp2. {illeg} rxrz+zz=y2. zz=rzrx+yy
z = r2 14rrrx+yy . ra xrrz z . 12rr r 14rrrx+yy = a rxrz . 12r4 + rryy r3x {illeg} a a r x + r z a a = r20 12r = r 14rrrx+yy

<86r>

14r6 + r4yy + rry4 r5x = 2r3yyx + r4xx r2aax 2ryyaax + 2rraaxx + a4xx ra rxrz zz . zz = arxarz . zz = az + ax . z {illeg} = 12a + 14aa+ax . axaz = rz rx + yy Or yy + rx + ax = 12aa 12ar + ar14aa+ax . 18631 35459 944

y4 2rx yy + rrxx + 14a4 + 2aarx 2ax + 2arxx + 12a3 + arrx aa + aaxx + 14aarr + a3x ar = 14a4+a3x+14rraa+rrax 0 y4 2rx yy + rrxx = 0 2ax + 2arxx aa + aaxx ar + 12a3r + 2aarx . . xx + 2aar x + y4 2ryy aayy 2ayy aryy + 12a3r rr+2ar+aa =0 .

17548875.(5849625 51)358(7,019608. 00)357¯(0,000-58. 00)00100¯(0,0-00. 00)000490(0,-00. 00)000459¯(0,-00. 00)0000310(,-00. 00)0000306¯(,-00. 00)000000400(,-. 0000000 51)197(3,862745 00)153¯(0,00+392 000)440(0,0+000 000)408¯(0,0+000 0000)320(0,0+00 0000)306¯(0,0+00 00000)140(0,0+0 00000)102¯(0,0+0 000000)380(0,0+ 000000)357¯(0,0+ 0000000)230(0,+ 0000000)204¯(0,+ 00000000)26(0,+ 00)93¯(0000000. 53)372(7018868. 00)371¯(000+682. 0000)100¯(000+0. 00000)470(00+0.00000)424¯(00+0. 000000)460(00+.000000)424¯(00+. 0000000)360(0+. 0000000)318¯(0+. 0000000)420(+. 000 212)819¯(3863207 00)636(0000070 00)1830(000000 00)1696¯(000000 000)1340(00000 000)1272¯(00000 000000)680(0000 000000)636(0000 0000000)44(0000 0000000)424¯(000 00000000)160(00

<86v>

0·0·9·7·5·4·5·5·5·9·7·7·7·0·0·3·0·0· 0·8·9·0·0·0·2·0·0·2·0·0·

<87r>

4 In ye Hyperbola Parabola Figure cb=a . be=x . 2aa 2ax {illeg} aa + 2ax xx = ed2 aa xx = ed2 = yy . aa xx r=y . cp=cb eb×df = fg \{ ×c }/ = {illeg} zc. aax x3 r + xx =zc x3 rxx aax + rcz = 0 . Since all eb×df = 18 all co2 = 14ab×ab×r . ab=b. all eb2 = a33 therefore bgpf 14bbr + a33 .

<87v> Figure <88r>

ab= {illeg}| b |e=y. bd=x. bq=dg=b. nb=c. ybxyc = bxc = ef Then shall bq&c: be the axis of gravity in feb&c & bqgd.

<88v> Figure <89r>

In ye 1st figure.

gccd cfedckhg = cd×cfed gc . ac=gc. xz zaxy . {illeg} zza x = ckhg . {illeg} or xxy z = cdef . Suppose cdca acbc ye swiftnesse of de ∶ to ye swiftnesse of gh . de×its swiftnesgh×its swiftness gccd . de×cd gh×ca de×ac gh×bc gc×cd .

Fig 2d {sic}. 3d.

c d | θ | ca acbc nmam swiftnessθeswiftnessegh . de× its {illeg}|s|wiftnesse gh× it {sic} swifnes {sic}c d | k|×degh c | × | ac de×ac gh×bc de× c |k | gh×ac de×ac gh×bc {illeg} \/ de×nm gh×am gccd . de×ck de×ck×cd = gh×ac×gc . de×cd = gh×bc . de×nm = gh×gc .

Fig 4

ckca motion of ye point a from c ∶ motion of ye point a from m ∷ ckca increasing of ac=gc∶increasing of cd ∷motion of gh ∶ motion of de. &c as before.

These are to find such figures cghk, cfed, as doe equiponderate in respect of ye axis acfk.

<89v> Figure
<90r>

Reasonings concerning chance.

If by one of ye equall chances a, I gaine p, by ye chance the equall chances \ a , b , c , &c/ are such yt one of ym must necessarily happen, & yt if one of ye chances a happen I gaine p thereby, or q by one of ye chances b , or r by one of ye chances c . My chance or expectation is worth pa+qb+rca+b+c

1 If p is ye number of chances by \one of/ wch I may gaine a, & q those by \one of/ wch I may gaine b, & r those by one of which I may gaine c; soe yt those chances are \all/ equall & one of them must necessarily happen: My hopes or chance is worth pa+qb+rcp+q+r=A. |The same is true if p, q, r signify |an|y|y|{e} proportion of chances for a, b, c.|

2. If I bargaine for more yn one cha{illeg}c|n|ce (viz: yt after I have taken ye gaines by my first chance, from the stake a+b+c; I will venter another chance at ye remaining stake &c) my second lott is worth pa+qb+rca+b+c×p+q+r {illeg} A A AAa+b+c = A AAa+b+c = B . My third lot is worth A AA AB a+b+c = C . My Fourth lot is worth A AA AB AC a+b+c = D . My Fift lot is worth A AA AB AC AD a+b+c = E . My sixt lot is worth A A × A+B+C+D+E a+b+c . &c

As if 6 me{illeg}|n| ( 1. 2. 3. 4. 5. 6. ) cast a die soe yt he gaines a who throws a cise first: since there is but one point \chance/ to gaine a & 5 to gaine nothing at each cast, I make b=0=c=r. p=1 & q=5. Therefore by the <90v> The first mans lot is worth a6 The 2ds is worth a6 a36 = 5a36 . The thirds is worth 5a36 5a216 = 25a216 . The 4d|t|hs is 25a216 25a1296 = 125a1296 The fifts lot is worth 125a1296 25a7776 = 625a7776 . The Sixts lot is 625a7776 625a46656 = 3125a46656 . &c. Soe yt their lots are as 7776: 6480: 5400: 4 {illeg}| 5 | 00 : 4 | 3 | 950:3 3 | 1 | 25 .

Soe yt if I cast a die two or more times tis 1. to 5 yt I cast a c{illeg}|i|s{illeg}|e| at ye first cast & 11 to 25 yt I throw it at two casts, & 91 to 125 yt I cast it at thrice, & 671 to 625 yt I cast it once in 4 trialls, & 4651 to 3125 yt I cast it once in 4|5| times. &c

3. If I bargaine to cast severall sorts of of {sic} lots successively at ye same stake ye valor of each lot is thus found viz: The first prop: gives ye valor of ye f{illeg}|ir|st lot; wch valor being destructed from ye stake, ye remainder is ye stake of ye 2d lot wch therefore may bee also found by ye first prop: &c.

As if {illeg}|I| gaine a by throwing 12 at ye first cast, or 11 at ye 2d or 10 at ye 3d &c wth two dice. Since at ye first cast there is but one chance for a (viz 12 ) & 35 for nothing Therefore its valor is a36 (by Prop 1). & ye stake{illeg} for ye 2d cast is a a36 = 35a36 . Now since there is|ar|e two chances for i{illeg}|t| (viz: ⚅⚄ & ⚄⚅) at ye 2d cast & 34 for 0 at ye 2d cast therefore its valor is 2×35a 36×36 = 35a 648 . as ye stake for ye 3d chance lot is 595a 648 f{illeg}|o|r wch there are 3 chances (f{illeg}|vi|z ⚄⚄, ⚅⚃, ⚃⚅{sic}) & 33 for nothing Therefore its valor is 595a7776.

<91r>

4 If I bargaine wth one or two more to cast lots in order untill one of us by an assigned lott shall win ye stake a: Since ye chances may succede infinitly I onely consider ye first revolution of them The valor of each mans \whole/ expectation being in such proportion one to another as ye valors of their lots in one revolution. & ye valors of each mans f{illeg}|ir|st lot being to ye valor of his whole expectation as ye summe of ye valors of their f{illeg}|irs|t lots to ye stake a.

As if I contend wth another yt who first throws 12 wth 2 dice shall h{illeg}|a|ve a, I haveing ye dice. {illeg}|M|y first lot is worth a36 (by second prop 1), The 2d \his first/ lot is worth 35a36×36. And a3635a36×36 3635 my expectationto his. for ye two first lots make one revolution because {illeg} I have ye same lot If I throw a 2d time yt I had at ye first. Therefore ( 36+35=71 a 3636a71 ) 36a71 is my interest in ye stake.

If or bargaine bee soe yt there is some lott at ye beginning of or play wch returnes not in ye {illeg} after revolutions, detract ye valor of those irregular lotts from ye stake & ye rest shall bee ye stake of ye regular lots wch follow & revolve successively. As if I contend wth another yt who first casts 11 must have a , onely I have {ye} first cast for 12. My first lot is worth a36. & ye stake for or after throws is 35a36. his firts lot being 35a648. & my next lot 595a11664. soe yt his share in ye stake 35a36 is to mine as 35a648 595a11664 1817 . Soe yt my share in it is 17a36. To wch adding {illeg} ye valor of my f{illeg}|i|rst lot viz: a36, ye summe is 18a36 = a2 , my interest in ye stake a at ye begining.

5 If ye Proportion of {illeg}|t|he chances for any stake bee irrationall {illeg}|t|he interest in the stake may bee found after ye same manner. As if ye Radij ab , ac , divide ye \horizontall/ circle \ bcd / into two pts <91v> abec & abdc in such proportion as 1.|2 | to Figure 5 . And if a ball falling perpendicularly upon ye center a doth tumble into ye portion abec I winn (a): {illeg}|b|ut if into ye other portion, I win b . my hopes is worth 2a+b5 2+5 .

Soe if a die bee not a Regular body but a Parallelipipi|e|don or otherwise unequall sided, it may bee found how much one side is cast is more easily gotten then another.

6 Soe yt ye facility of ye chances & ye stake belonging to each chance being knowne ye worth of of {sic} the lott may bee ever found by ye precedent precepts. And if they bee not both immediatly found known they must bee sought before ye valor of ye lott can bee found.

As if I want two games at Irish & my adversary three to win a , & I would know ye value of my interest in ye stake (a.) my first lot {illeg} can gaine me nothing but ye advantage of another lot, & therefore to know its vallue I must first find ye value of yt other lot &c. First therefore if wee each wanted one lot to win a or interest in it would bee equall viz my lot worth a2. \2dly/ If I want thr{illeg} lo two {illeg} \one/ games & my adversary one|two|, {illeg} & I gaine ye next game yn I gaine a but if I loose it I onely gaine an{illeg} equall lot for a at ye {illeg}|n|ext game wch is worth 12a, Therefore my interest in ye stake is a+12a 2 = 3a 4 . 3dly If I want one game & my adversary {illeg}|t|hree & I gaine ye next game I get a; but if I loose it, then I want one game & my adversary but two, yt is I get 3a4: Therefore (there being one chance for a & one for 3a4) my interest in ye stake is a+3a4 2 = 7a 8 . 4thly If I want 2 games & my adversary 3; & I win I get 7a8. but if I loose I get 12a for or chances <92r> will then bee equall; Therefore my interest in ye stake is 11a16. Soe if I want {illeg}| 1 | games & my adversary 4 my interest in a is 15a16. If I want two and hee 4 , it is 13a16. If I want 3 and hee 4 it is 21a32. If I 1 and hee 5 , it is: 31a32. If I 2 and hee 5 it is 57a64. If I 3 and hee 5 it is 99128a. If I 4 and hee 5 , it is: 163256a. |(|The like may bee done if 3 or more play together. |(|as if one wants one game, another 3 a third 4: Their lots are as 616:82:31 . &c.|)| As also if their lots bee of divers sorts.|)|

By this meanes also some of ye precedent questions may bee resolved. as if I have two throws for a cise to win a , wth one die; If I have missed my first lot alredy {sic}, I have at my second cast five chances for nothing. & one for a . therefore yt cast is worth a6. Soe yt \in/ my first cast I had five chances for 16a & one for a , wch therefore is worth (wth my 2d cast) is worth 1136a. That is tis {illeg} to 11 to 25 whith {sic} yt I cast a cise at once in two throws. as before

By this meanes also my lot may bee known if I am to draw 4 cards of severall sorts out of 40 cards {illeg}| 1 | 0 of each s{illeg}i|or|t.

Or if out of two white & 3 black stones I am blindfold to chose a white & a black one.

<92v>

Figure Symbol (equal subscript u) in text Equation

An equation given; & \if/ either \both/ x , & y , be of a /have\ divers quan dimensions, try if ye roote \of one of y / may be extracted: {illeg} whereby ye Symbol (equal subscript u) in text may often be diminished riduced to finer termes & sometimes to finer dimensions if it can {illeg} \{illeg}/ ye line & {illeg} If a quantity wherein x | y | is not is divided by {illeg}| x | in ye {illeg}q{illeg} {illeg}g line equall to x . yt crooked cannot be squared.

<93r>

The line cdf is a Parab. \4 /ac=\2 /ad=r=\2 /na=\2 /ge. ce=x. ef=y. rx=yy. g{illeg} e ef 12r rx eoap . eo=ʒ. {illeg} eo=z. ap=a. 12ra = z rx . 14rraa = rzzx . raa4 = zzx . or supposeing ea= x|y|. {illeg} x = y+14r & raa4 = zzy + |14 |zzr. {illeg} wch shews ye nature of ye crooked line po. now if dt=ap. yn drst=eoap. for supposeing eo moves uniformely from ap , & rs moves from dt wth motion decreaseing in ye portio yt ye line eo doth shorten. Suppos aq=ap=r2=a & eq= x | y |. x= y14r . then r316 = zzy 14zzr . suppose z=a+y. yn r3=aax+2ayx+yyx. Or aax + 2ayx + yyx + 14aar + 18ayr + 14yyr = r316 . Or aax + 2ayx + yyx 14aar + 18ayr 14yyr = r316 . Or suppose z=ya . yn r316 = aax 2ayx + yyx . Or aax + /\ 2ayx + yyx + 14aar 18ayr + 14yyr = |r316 | Or aax 2ayx + yyx 14aar + 18ayr 14yyr = r316 . Or, if x=a y | r |. r316 = zza zz y | x |. &
a3 + \ 2 / aay + ayy {illeg}{illeg} aax 2ayx a |y | yx = r316 . Or a3 2aay + aay aax + 2ayx yyx = r3 16 . 2= \2/z2. mp={illeg}y x |ʒ |. mq = a + x | ʒ |=. mo = y = a + x | ʒ | 2z2 {illeg} pq(=a) aq(=12r) ƅq(=x+z) = y + 2zz ay+a2zz = 12rx + 12rz . 2ay+2z2aa r 12z = x = r3+4zzr 16zz . 32zzay + 32z3a2 8z3r r4 4zzrr = 0 mv = ξ = a + ʒ r z2. ξaʒ2 = z . ξ22ξa2ξʒ+aa+2aʒ+ʒ2 2 = zz . ξ33ξ2a3ξ2ʒ+ 3aaξ+6aʒξ+3ʒʒξa33aaʒ3aʒʒʒ3 8 = z3 a2+ 322aa 8r = c cz3+32aξzzr4 4rr = 0 <93v> Figure <94r> cξ3 3caξ2 + 3caaξ caʒ + 32aξ3 3cʒξ2 + 6caʒξ 3ca2ʒ + 3cʒʒξ 3caʒ2 64aaξ2 + 32a3ξ cʒ3 64aʒξ2 + 64aaʒξ 4rraa 4rrξ2 + 32aʒʒξ 8rraʒ + 8rraξ 4rrʒʒ + 8rrʒξ r4 = 0 Or
ξ3 deʒξ2 + ggξ fʒ d + hʒ ʒm2 + igʒʒ nʒ2 onʒ3 = 0 .

Let ed=a. ae=x ab=y a {illeg} y {illeg}/{illeg}\ se=v. sb=s. ss = y2 + vv + xx 2vx . y3 = vvy + ssy + 2vxy xxy = a2 {illeg} a {illeg} 2 {illeg} v {illeg} ss {illeg} xv{illeg} v4x2 aax + a3 = yyx y2=ssvv+2vxxx aax + a3 = ssx + /\ {illeg} vvx + 2vx2 x3 x3 2vx2 +vv ss +aa x + a3 = 0 2 1 0 1 2x3 2vx2 + a3 = 0 . 2x3a3 2xx = v

aax = yyx + yya . y2 + = ss vv + 2vx {illeg} xx aax = ssx vvx + 2vxx x3 + ssa + 2avx {illeg} axx avv . x3 2v +a x2 ss + vv 2av + aa x ssa + avv = 0 . x22ex+ee×x+f x3 2e +f x2 +eefx +2eex +eef =0 2v+a = f 2e . f = a+2e {illeg} 2v. eef=avvass eea 2e3 + 2eev + avv a = ss . vvx2vvx+x3 vva2vax+ax2 +aax = a x } 2eev + aav 2e3 eea a

<94v>

To square those lines in wch is y onely

If y is in but one terme onely of ye Equation (as xx=ay. or, a3=xxy) resolve ye Eq: into ye proport ya (as ya xxaa . or, ya aaxx .) If ye line hath Assymptotes
x3=aay . v = 3x5 a4 + x .

<95r>

+a +x a x vv 2vx.2 2vax 2vee. 2xaeev +x.3 +ax.2 +aa.x +2e.3 +ee.a +2xae.3 +xee. =0 v= 4x3 + 2axx + aa.x + 2x4a 4xx + 2ax + 2x3a 0 v= 4ax2 + 2aax + a3 + 2x3 4ax + 2aa + 2xx 0 sa= +4ax2 + 2x3 + 2aax + a3 4ax2 2x3 2aax 4ax + 2aa + 2xx 0 sa= a3 2x2 + 4ax + 2aa 2a4x x+a 2a6 4xx + 8ax + 4aa {illeg} 2aax 2a6 8x3 + 24ax + 24aax + 8a3 \2 / x3 + 2ax2 ad. 2a4x2+2a5x 8x3& = a4x 4xx + 8ax + 4aa a00aa x = yy aayy x2 = y4a2 a4+2a2y2+y4 a5yy 4aax2 + 8a3x + 4a4 4yy 8yya 4yyaa 2x3+4ax2+2aaxa3=0 oe=p ayy aaxa+x a32x2+4ax+2aa pz z2aax a+x = a6pp 4x4+16ax3+8aaxx+16a3x+4a4 +16aa a5pp+a4xpp=4zzx5+16zzax4+24aazzx3+16zza3x2+4a4z2x divided by x+a it produceth. 4zzx4+12zzax3+12zzaax2+4zza3x + /\ a4pp=0

<95v> Figure

By ye Squares of ye simplest lines to square lines more compound. 1s {sic} those whein y.

find ye valor of y. If ye number of ye termes in ye denom: thereof be neither 1.3.6.10.15.21.28. &c. ye line cannot be squared If it have but one terme tis squared by finding ye square of |{illeg} each| particular terme in ye valor of y & yn adding all those squares together. Example 1st. 3x4+a4=yaxx . & y=3x4+a4axx. Then makeing y equall to each particular terme. 3xxa=y. a3xx=y or 3xx=ay whose □ is x3a. & a3=xxy. whose □ is a3x Ad {sic} these 2 □s together & they (viz: x4+a4ax) are the □ of ye line 3x4+a4=ayxx. Againe 2a7 + \/ 2 a | b |x6+x7=a3x3y. Or y = 2a72bx6+x7 a3x3 . yn disjoynting ye valor of y. y=2a4x3 . y=x4a3. y=2bx3aaa Or x3y=2a4 , whose □ is a4xx . ya3=x4 , whose □ x55a3. y{illeg}| a3 |=2bx3, whose □ x4b2a3. which 3 □s (viz 10a7 + 2x7 5bx6 10a3xx ) is ye{illeg} taken together are ye square sought for. And these lines may bee ever squared unless in ye valor of y there bee found aax, abx, cc+dex, &c. for ye Squareing of yt line depends on ye squareing of ye Hyperbola. As in ye line {illeg}| x | xxy = {illeg}xx /x4 \ +a3x+a4 .

<96r>

aax=y3 . cd=a. ce=x. eh=y. oc=v. ok=s. ss = y2 + x2 2vx + v2 . x2 = 2vx + ss y2 v2 . x = v + ss y2 . aav aa ss y2 = y3 . a4v2 2aavy3 + y6 a4ss + a4y2 = 0 0 3 6 0 2 6y6 6aavy3 + \2 / a4yy = 0 v = a4+3y43y4 3aay . y a4+3y4 3aay y3 aa a4y3+3y7 3a4yy = a4y+3y5 3a4 vx = a4+3y43y4 3aay vx = a4 3aay = aa 3y . y aa 3y y3aa aay3 3aayy = y3 . make md=dv that is aa=3yy . & a3 = y = md = mv . x = dc = a 27 = x ds = dc = aa27 caay aa 3y aa27 z=en aax a6 3aax a3 2727 z3 aax2z3 = a7 8127 xxz3 = a5 8127 wch expresseth ye nature of ye crooked line ns . & vl=ds . vlfr=dens.

2dly. I{illeg}|f| it have 2|3| termes. bee 6,10,15,21, termes See if it may be reduced to {one orfewer} dimensions by addin{illeg}|g| or subtracting a{illeg} knowne quantity to or from x . Example. 2bax + axx = bby + 2bxy + xxy . wch (makeing x+b = z ) i{illeg}|s| thus red{illeg}|u|ced
zzy= {illeg}{illeg} bba+azz . Or bba+azzzz=y

<97r>

aax+bby=y3 . x=vssyy . aavaassy2+bby= \y3 / aav+bbyy3=aassy2 .
a4v2+2aavbby2aavy3+b4y22bby4+y6 a4ss +a4y2 013246 6y58bby3+2b4y+2a4y 2aabb+6aayy =v vx= 00 00 2b4y+6bby36y5 +2b4y+2bby3+6y5 +2a4y8bby3 vx=2a4y6aay22aabb vx=aay3y2bb . bby+ aa | y aay 3y2 bb y3bby aa & ad. aay4aabbyy 3aay3aabby y3bby 3y2bb .

{illeg} y3y2bb bb {illeg} y aay 3y2bb = y3bby aa. a4 = 3y4 4bbyy + b4
{illeg} y4 = 4bbyyb4+a4 3 y = bb 3 4b4 / a=b\ y2 = 4bb 3 . y = 2b 3 = dm = dv . 8b3 33 {illeg} 2bbb 3 aax . 2b 33 = dc = ds . y aay 3y2bb 2b33 z . yz = 2aaby 9y233bb3 9yyz3bb {illeg} z=2aab3 An Equation expressing ye nature of ye line ns .

<98r>

aax+byx=y3 . x=vssyy.
aav+byvy3= aa by ssyy
a4v2+2aav2by2aavy3+bbvvy22bvy4+y6 a4ss +a4yy+bb bbss 21+1024 01+3246 ss = 2a4vv + 2aavvby + aavy3 + 4bvy4 2bby4 4y6 2a4 12a4y6 16a4bvy4 + 4a4bby4 + 4bba4vvy4 12a6vy3 + 4a6vvby +4a8yy 12a4y6 16a4bvy4 +4a6vvby 8b3vy6 + 8a4bby4 12a6vy3 + 4a8yy +4b4y62aabbvy54aavvb3 } 0 +8y8bb¯ +4a6by +4aab3y3 { vv= 0 0 12a6y3av8a4bby4 2aab2y58y8bb 16a4by44b4y6 8b3y612a4y6 4a6by 4aab3y3 v= 6a6y2 + a2b2y4 + 8a4by3 + 4b3y5 4a6b 4aab3y2 36a12y4+8aab5y9 + 12a8b2y6+64a8b8y6 + 16a10by5+16bby10b4 + 24a8b3y7 + a4b4y8 b5 16a12b232a8b4y2 +16a4b6y4 . 8a4bby3b0 8y7bbb0 4b4y5 + 12a4y5b3 + a4b4y8 b5 + 4a6b 4aab3y2

<99r>

a3x=y4 . x=vssyy . a3vy4=a3ssyy abv22a3vy4+y8+a6yy abss 0482 . v=4y6+a64a3y2 vx=4y6+a64y64a3y2=a34y2 .
y · a34y2 y4a3 a3y44a3y3 = y4

<100r>

aaxaay=y3 . aavaayy3=aassyy a4v22a4vy2aavy3+2a4y2+2aay4+y6 a4ss 013246 v=3y5+4aay3+2a4ya4+3aayy
vx = 3y.5 + 4aa.y3 + 2a4y aa.y3 3y.5 ya.4 3aa.y3 a4 + 3aayy vx = aay aa+3yy . y aay a2+3y2 a2y+y3 aa a4y2+a2y4 a4y+3a2y3 = aay+y3 aa + 3yy y3+aay aa = aay aa+3yy yyaa + 3y4 + \ 3 / aayy + a4 = y {illeg}yy{illeg}aay

<101r>

aax = by2+y3 . aav by2 y3 = aa ss yy a4v2 2aabyyv 2aavy3 + bvy4 + 2by5 + y6 a4ss + a4 023456 v = 3y4+5by3+2bby2+a4yy 2aab+3aay vx = 3y.4+5by.3+2bby.22bbyy3by.32by33y.4+a4 2aab+3aay {illeg} vx = a4 2aab+3aay y by3+bby2 aab + 3aay by2+y3 aa ad 2bby4+by5+b3y3 a4+3aay a=b . a3 + 3aay = byy + bby aa+3ayyyay = 0 yy = 2ay + aa y = a + aa + aa . y = a + a2 = dm = vd . a3 + 3a32 + 6a3 + 2a32 a3 + 2a32 + 2a3 aa = 10a + 7a2 = dc y y3+b