Mathematical Notebook

2 Of yethe extraction of Pure Square Cubick. Square-square & square-cubick rootes &c.

Let yethe number whose roote is to bee extracted bee pointed ~~bec~~ makeing yethe first point under yethe utnite & comprizeing soe many numbers under each point as yethe number hath dimensions as if yethe number be square-cube tis thus pointed $570 \underset{.}{8} 6352 \underset{.}{4} 1080 \underset{.}{2}$

Then then then t out of yethe figures of yethe first point next yethe left hand extract yethe greatest roote proper to yethe power of yethe number & set ytthat downe in yethe mQuotient wchwhich is yethe firtfirst side & is called A. (as yethe roote quintuplicate of

5708

(5)

, &

(5)

quintuplicate is

3125

) ynthen takeing ytthat roote duely multiplied out of yethe number (as

3125

out of

5708

) wthwith yethe rest of yethe numbers to yethe next point. seeke yethe seacond side wchwhich is found by divideing ytthat number by another number made out of yethe first side (wchwhich is called yethe Divisor) & this second side I name E. (thus by divideing

258363524

5 A

qq

+ 10 A c + 10 A q + 5 A

after such a manemaner ytthat

5 A qq E + 10 A c E q + 10 A q E c + 5 A E qq + E qc

may be conteined in yethe number yethe product of ytthat division shall be E =

The extraction of yethe sqaresquare roote

\begin{matrix} The square to be resolved 2 \underset{.}{9} & 1 \underset{.}{6} (54 The Product \\ The square of y^{e} first side 25 & be taken away. \\ The rest of y^{e} sqare to be 4 & 16 resolved. \\ The divisor for finding y^{e} seacond 1 & 0 sidie. which is y^{e} first side doubled \\ \begin{matrix} The rectangle by 2A & E 4 \\ The square of E \end{matrix} & \begin{matrix} 0 \\ 16 \end{matrix}} to be substracted \\ The su \bar{m} e of y^{e} rectangles 0 & 16 to be subducted \\ 0 & 00 The remainder \end{matrix}

The extraction of yethe cube roote

\begin{matrix} The cube to be resolved 15 \underset{.}{7} & 46 \underset{.}{4} (54 \\ The cube to be subducted 125 & whose roote is A = 5 \\ The remainder for y^{e} finding 32 & 464 of E \\ \begin{matrix} \begin{matrix} The divisors for y^{e} finding \\ of (E) y^{e} seacond side. \end{matrix} & {\begin{matrix} 7 \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 5 \\ 15 \end{matrix} & \begin{matrix} 3 A q \\ 3 A \end{matrix} \end{matrix} \\ The sume of y^{e} divisors 7 & 65 \\ \begin{matrix} \begin{matrix} Sollids to be substracted \end{matrix} & {\begin{matrix} 30 \\ 2 \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ 40 \\ 64 \end{matrix} & \begin{matrix} 3 A q E \\ 3 A E q \\ E c \end{matrix} \end{matrix} \\ The sume of those 32 & 464 sollids \\ The remainder 00 & 000 \end{matrix}

The extraction of yethe square square roote

\begin{matrix} The square-square 3 \underset{.}{3} & 177 \underset{.}{6} (24 \\ The square-squ: to be subduc: 16 & = A q q \\ Remainder. 17 & 1776 \\ \begin{matrix} \begin{matrix} Divisors for finding y^{e} \\ seacond side E. \end{matrix} & {\begin{matrix} 3 \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 2 \\ 24 \\ 8 \end{matrix} & \begin{matrix} 4 A c \\ 6 A q \\ 4 A \end{matrix} \end{matrix} \\ Theire su \bar{m} e 3 & 448 \\ \begin{matrix} \begin{matrix} Squ-Squares to be sub= \\ =ducted \end{matrix} & {\begin{matrix} 12 \\ 3 \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 8 \\ 84 \\ 512 \\ 256 \end{matrix} & \begin{matrix} 4 A c E \\ 6 A q E q \\ 4 A E c \\ E qq \end{matrix} \end{matrix} \\ Theire Su \bar{m} e 17 & 1776 \end{matrix}

3 The extraction of yethe Square-Cube roote

\begin{matrix} The squ: cube to be resolved 7 \underset{.}{9} & 6262 \underset{.}{4} (24 \\ Substract 32 & A qc \\ Remain d^{e} 47 & 62624 \\ \begin{matrix} \begin{matrix} Divisors \end{matrix} & {\begin{matrix} 8 \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ 80 \\ 40 \\ 10 \end{matrix} & \begin{matrix} 5 A qq \\ 10 A c \\ 10 A q \\ 5 A \end{matrix} \end{matrix} \\ The Sume of y^{e} divisors 8 & 8410 \\ \begin{matrix} \begin{matrix} Plano-Sollids to be \\ substracted \end{matrix} & {\begin{matrix} 32 \\ 12 \\ 2 \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ 80 \\ 560 \\ 2560 \\ 1024 \end{matrix} & \begin{matrix} 5 A qq E \\ 10 A c E q \\ 10 A q E c \\ 5 A E qq \\ E qc . \end{matrix} \end{matrix} \\ Theire Su \bar{m} e 47 & 62624 \\ Remainder 00 & 00000 \end{matrix}

Note ytthat yethe 3d 4th 5th & other figures are found by yethe same manner ytthat yethe seacond figure is found onely makeing all yethe figures found to stand for A yethe first side & yethe figure sought for e or yethe 2d side And if yethe number propounded be t roote is found inexpressible in whole numbers ytnthen adding ciphers & pointing them from yethe unite towards yethe right kind as was before explained & soe hold on yethe worke in decimalls. As for yethe Divisors they are easily found by yethe 2d Table of Powers from a Binomial roote. If yethe Number bee of 6.7.8.9.10 &c dimensions The roote may be extracted after yethe same manner 4 Of yethe ExtactionExtraction of Rootes in Affected powers. The manner of yethe extraction of rootes in pure & affected powers is verryvery much alike, especially when yethe affected powers are decently prepared, ytthat is, when theire affections are not over large & those altogether either affirmative or negative, & yethe power affirmative, affirmations & negations so mixt ytthat there be noe ambiguity & all fractions & Asymmetry taken away All yethe figures in yethe coeffentscoefficients & affected power are to be pointed (after yethe manner before eplainedexplained in yethe Analisis of pure powers) according to yethe degree of theire dimensions & the worke onely differs from ytthat in pure powers ytthat in ytthat yethe coefficients enter into yethe divisors Let yethe first side be called A. yethe 2d be called E. yethe Roote of yethe equation BL yethe coefficients

B . C q . D c . F qq . G qc .

H cc

&c yethe Power

P . P q . P c . P qq

&c & yethe Operation follows The analysis of Cubick Equations. The equation supposed

L c^{*} + 30 L = 14356197

L c + C q L = P c

\begin{matrix} \begin{matrix} \begin{matrix} \underset{}{} & The square coëfficient \\ \underset{}{} & The cube affected to be \end{matrix} & \begin{matrix} \underset{}{} \\ 1 & \underset{.}{4} \end{matrix} \end{matrix} & \begin{matrix} \underset{}{} & 3 \\ 3 & 5 & \underset{.}{6} \end{matrix} & \begin{matrix} \begin{matrix} \underset{.}{0} & \underset{.}{} & \underset{.}{} \\ 1 & 9 & \underset{.}{7} \end{matrix} & \begin{matrix} (243 \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollids to bee substracted \end{matrix} & {\begin{matrix} 8 \end{matrix} \end{matrix} & \begin{matrix} 6 \end{matrix} & \begin{matrix} \begin{matrix} 0 \end{matrix} & \begin{matrix} = A c \\ = A C q \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Theire su \bar{m} e \end{matrix} & \begin{matrix} 8 \end{matrix} \end{matrix} & \begin{matrix} 0 & 0 & 6 \end{matrix} & \begin{matrix} \begin{matrix} 0 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Rests \end{matrix} & \begin{matrix} 6 \end{matrix} \end{matrix} & \begin{matrix} 3 & 5 & 0 \end{matrix} & \begin{matrix} \begin{matrix} 1 & 9 & 7 \end{matrix} & \begin{matrix} for finding y^{e} 2^{d} side \end{matrix} \end{matrix} \\ The extraction of & y^{e} seac & ond side \\ \begin{matrix} \begin{matrix} \underset{}{} & Coëfficient \\ \underset{}{} & The rest of y^{e} cube to be \end{matrix} & \begin{matrix} \underset{}{} \\ \underset{}{} & 6 \end{matrix} \end{matrix} & \begin{matrix} \underset{}{} \\ 3 & 5 & \underset{.}{0} \end{matrix} & \begin{matrix} \begin{matrix} 3 & \underset{.}{0} & \underset{.}{} \\ 1 & 9 & \underset{.}{7} \end{matrix} & \begin{matrix} or superior divisor & \underset{}{} \\ resolved & \underset{}{} \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} The inferior divisors \end{matrix} & {\begin{matrix} 1 \end{matrix} \end{matrix} & \begin{matrix} 2 \\ 6 \end{matrix} & \begin{matrix} \begin{matrix} 0 \end{matrix} & \begin{matrix} 3 A q \\ 3 A \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Theire su \bar{m} e \end{matrix} & \begin{matrix} 1 \end{matrix} \end{matrix} & \begin{matrix} 2 & 6 & 0 \end{matrix} & \begin{matrix} \begin{matrix} 3 & 0 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollids to be sub= \\ stracted \end{matrix} & {\begin{matrix} 4 \end{matrix} \end{matrix} & \begin{matrix} 8 \\ 9 & 6 \\ 6 & 4 \\ 1 \end{matrix} & \begin{matrix} \begin{matrix} 2 & 0 \end{matrix} & \begin{matrix} = 3 A q E \\ = 3 A E q \\ = E c \\ E C q \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Theire su \bar{m} e \end{matrix} & \begin{matrix} 5 \end{matrix} \end{matrix} & \begin{matrix} 8 & 2 & 5 \end{matrix} & \begin{matrix} \begin{matrix} 2 & 0 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \end{matrix}

\begin{matrix} \begin{matrix} \begin{matrix} The superior part of y^{e} divisor \\ The remainder for finding \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 5 & 2 & 4 \end{matrix} & \begin{matrix} \begin{matrix} 3 & \underset{.}{0} \\ 9 & 9 & \underset{.}{7} \end{matrix} & \begin{array}{l} or y^{e} square coefficient \\ y^{e} third side \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} The inferior part of y^{e} \\ divisor \end{matrix} & {\begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 1 & 7 & 2 \end{matrix} & \begin{matrix} \begin{matrix} 8 \\ 7 & 2 \end{matrix} & \begin{matrix} 3 A q that is 3 \times 24 \times 24 \\ 3 A or 3 \times 24 . \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} The su \bar{m} e of y^{e} divisor \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 1 & 7 & 3 \end{matrix} & \begin{matrix} \begin{matrix} 5 & 5 & 0 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollids to be taken \\ away \end{matrix} & {\begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 5 & 1 & 8 \\ 6 \end{matrix} & \begin{matrix} \begin{matrix} 4 \\ 4 & 8 \\ 2 & 7 \\ 9 & 0 \end{matrix} & \begin{matrix} 3 A q E \\ 3 A E q \\ E c \\ E C q \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Theire Su \bar{m} e \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 5 & 2 & 4 \end{matrix} & \begin{matrix} \begin{matrix} 9 & 9 & 7 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Remaines \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 0 & 0 & 0 \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 & 0 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \end{matrix}

But yethe Coëfficient maybe greater ynthan yethe Power soe ytthat it cannot be substracted from it wchwhich argues ytthat yethe Cube more propperly affects ynthan is affected. In this case yethe coëfficient must descend towards yethe unite soe many points untill it may be substracted, & soe many points as yethe coëfficient is devolved soe many pricks must be blotted out towards yethe left hand in yethe power affected. As yethe Example shews

L c + 95400 L

=

= 1819459

\begin{array}{l} \begin{matrix} \begin{matrix} \begin{matrix}  \end{matrix} & \begin{matrix} \underset{}{9} \\ \underset{.}{1} \end{matrix} \end{matrix} & \begin{matrix} \underset{}{5} & 4 & 0 \\ 8 & 1 & \underset{.}{9} \end{matrix} & \begin{matrix} \begin{matrix} \underset{.}{0} & \underset{.}{} & \underset{.}{} \\ 4 & 5 & \underset{.}{9} \end{matrix} & \begin{array}{l} Coefficiens & \underset{}{} \\ The Power & \underset{}{} \end{array} \end{matrix} \end{matrix} \\ Since 9 is greater y^{n} 1 make a devolution thus. \\ \begin{matrix} \begin{matrix} \begin{matrix} \underset{}{} \\ \underset{}{} & The Quote (19 \end{matrix} & \begin{matrix} \underset{}{} \\ \underset{}{} & \underset{}{1} \end{matrix} \end{matrix} & \begin{matrix} \underset{}{9} & \underset{}{5} & \underset{}{4} \\ 8 & 1 & \underset{.}{9} \end{matrix} & \begin{matrix} \begin{matrix} 0 & \underset{.}{0} & \underset{.}{} \\ 4 & 5 & \underset{.}{9} \end{matrix} & \begin{array}{l} The Coefficient & \underset{}{} \\ The affected power & \underset{}{} \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollids to be substracted \end{matrix} & {\begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 9 & 5 & 4 \\ 1 \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 \end{matrix} & \begin{array}{l} A C q \\ A c \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Su \bar{m} a \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 9 & 5 & 5 \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 \end{matrix} & \begin{matrix} substrahenda \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Divisor \bar{u} superior pars \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 9 & 5 \end{matrix} & \begin{matrix} \begin{matrix} 4 & 0 & \underset{.}{0} \end{matrix} & \begin{matrix} Coefficiens Planum \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix}  \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 8 & 6 & 4 \end{matrix} & \begin{matrix} \begin{matrix} 4 & 5 & 9 \end{matrix} & \begin{matrix} Potestas reliqua \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Divisor \bar{u} pars inferior \end{matrix} & {\begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix}  \end{matrix} & \begin{matrix} \begin{matrix} 3 \\ 3 \end{matrix} & \begin{array}{l} 3 A q \\ 3 A \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Divisor \bar{u} su \bar{m} a \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 9 & 5 \end{matrix} & \begin{matrix} \begin{matrix} 7 & 3 & 0 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollida ablativa \end{matrix} & {\begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 8 & 5 & 8 \\ 2 \\ 2 \end{matrix} & \begin{matrix} \begin{matrix} 6 & 0 & 0 \\ 7 \\ 4 & 3 \\ 7 & 2 & 9 \end{matrix} & \begin{array}{l} E C q \\ 3 A q E \\ 3 A E q \\ E c \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Eor \bar{u} Summa \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 8 & 6 & 4 \end{matrix} & \begin{matrix} \begin{matrix} 4 & 5 & 9 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \\ \begin{matrix} \begin{matrix} Restat \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 0 & 0 & 0 \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 & 0 \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} \end{matrix} \end{array}

5 To place yethe unite of yethe coefficient in its right place in respect of yethe power make so many pricks above as there are under yethe power begining at yethe unit, & if yethe coefficient be one dimension lesse ynthan yethe power make a prick on every figure if 2 dimensions les ynthan every other figure of 3 dimensions lesse make it one each third figure &c If there be many coefficients in yethe equation each must be placed according to this rule. Sometimes yethe coefficient is under a negative sine as

L c - 10 L = 13584

& yethe Analysis is as follows

\begin{matrix} Coëfficiens planum - & 1 \underset{.}{0} sublaterale \\ Cubus resolvendus + 1 \underset{.}{3} & 58 \underset{.}{4} (24 \\ \begin{matrix} \begin{matrix} Solida ablativa \end{matrix} & {\begin{matrix} + 8 \\ - \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 20 \end{matrix} & \begin{matrix} A c \\ A C q \end{matrix} \end{matrix} \\ Su \bar{m} a + 7 & 80 \\ Restat + 5 & 78 \underset{.}{4} resolvendum \\ Divisor \bar{u} {\underline{p}}^{s} superior & - 10 coëfficiens planum \\ \begin{matrix} \begin{matrix} Divisor \bar{u} {\underline{p}}^{s} inferior \end{matrix} & {\begin{matrix} + 1 \\ + \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 2 \\ 6 \end{matrix} & \begin{matrix} + 3 A A \\ + 3 A \end{matrix} \end{matrix} \\ Suma divisor \bar{u} + 1 & 25 \\ \begin{matrix} \begin{matrix} Solida ablativa \end{matrix} & {\begin{matrix} + 4 \\ + \\ + \\ - \end{matrix} \end{matrix} & \begin{matrix} \begin{matrix} 8 \\ 96 \\ 64 \\ 40 \end{matrix} & \begin{matrix} 3 A A E \\ 3 A E E \\ E E E \\ E C C \end{matrix} \end{matrix} \\ Eor \bar{u} su \bar{m} a + 5 & 784 \end{matrix}

But sometimes yethe square coëfficient hath more paires of figures ynthan yethe cube to be analysed, hath & ynthen there is præfixing so many ciphers to yethe cube as figures are wanting, yethe first side will not much differ from yethe square roote of yethe coefficient. as

L c - 116620 L = 352947

\begin{matrix} \begin{matrix} \begin{matrix}  \end{matrix} & \begin{matrix} - & 1 & \underset{}{1} \end{matrix} \end{matrix} & \begin{matrix} 6 & 6 & \underset{}{2} \end{matrix} & \begin{matrix} \begin{matrix} \underset{.}{0} & \underset{.}{} & \underset{.}{} \end{matrix} & \begin{array}{l} Coefficiens plan \bar{u} \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Cubus resolvendus \end{matrix} & \begin{matrix} 0 & \underset{}{0} \end{matrix} \end{matrix} & \begin{matrix} 3 & 5 & \underset{}{2} \end{matrix} & \begin{matrix} \begin{matrix} 9 & 4 & \underset{.}{7} \end{matrix} & \begin{array}{l} (343 \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollida Ablativa \end{matrix} & {\begin{matrix} + & 2 & 7 \\ - & 3 & 4 \end{matrix} \end{matrix} & \begin{matrix} 9 & 8 & 6 \end{matrix} & \begin{matrix} \begin{matrix}  \end{matrix} & \begin{array}{l} A c \\ A C q \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Restat auferend \bar{u} \end{matrix} & \begin{matrix} - & 7 \end{matrix} \end{matrix} & \begin{matrix} 9 & 8 & 6 \end{matrix} & \begin{matrix} \begin{matrix}  \end{matrix} & \begin{array}{l}  \end{array} \end{matrix} \\ \begin{array}{r} \begin{matrix} Reliquum resolvendi \end{matrix} & \begin{matrix} + & 8 \end{matrix} \end{array} & \begin{matrix} 3 & 3 & 8 \end{matrix} & \begin{matrix} \begin{matrix} 9 & 4 & 7 \end{matrix} & \begin{array}{l} Cubi \end{array} \end{matrix} \end{matrix}

\begin{matrix} \begin{matrix} \begin{matrix} Divisor \bar{u} {\underline{p}}^{s} superior Coeff: \end{matrix} & \begin{matrix} - & \underset{}{1} \end{matrix} \end{matrix} & \begin{matrix} 1 & 6 & \underset{}{6} \end{matrix} & \begin{matrix} \begin{matrix} 2 & \underset{.}{0} & \underset{.}{} \end{matrix} & \begin{array}{l} planum. \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Reliqu \bar{u} resolvendi cubi \end{matrix} & \begin{matrix} + & \underset{}{8} \end{matrix} \end{matrix} & \begin{matrix} 3 & 3 & \underset{.}{8} \end{matrix} & \begin{matrix} \begin{matrix} 9 & 4 & \underset{.}{7} \end{matrix} & \begin{array}{l} negative affecti \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Divisor \bar{u} {\underline{p}}^{s} inferior \end{matrix} & {\begin{matrix} + & 2 \\ + \end{matrix} \end{matrix} & \begin{matrix} 7 \\ 9 \end{matrix} & \begin{matrix} \begin{matrix}  \end{matrix} & \begin{array}{l} 3 A A . \\ 3 A . \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Su \bar{m} a Divisorum \end{matrix} & \begin{matrix} * & * \\ + & \underset{}{1} \end{matrix} \end{matrix} & \begin{matrix} * & * & * \\ 6 & 2 & \underset{}{3} \end{matrix} & \begin{matrix} \begin{matrix} * & * & * \\ 8 & \underset{.}{} \end{matrix} & \begin{array}{l} ************ \\ 3 A A + 3 A + C q \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollida ablativa \end{matrix} & {\begin{matrix} + & 1 & 0 \\ + & 1 \\ + \\ - & 4 \end{matrix} \end{matrix} & \begin{matrix} 8 \\ 4 & 4 \\ 6 & 4 \\ 6 & 6 & 4 \end{matrix} & \begin{matrix} \begin{matrix} 8 \end{matrix} & \begin{array}{l} 3 A A E \\ 3 A E E \\ E E E \\ C C E \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Eorum summa \end{matrix} & \begin{matrix} + & 7 \end{matrix} \end{matrix} & \begin{matrix} 6 & 3 & 9 \end{matrix} & \begin{matrix} \begin{matrix} 2 \end{matrix} & \begin{array}{l}  \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Restat Resolven \bar{d} \end{matrix} & \begin{matrix} + \end{matrix} \end{matrix} & \begin{matrix} 6 & 9 & \underset{}{9} \end{matrix} & \begin{matrix} \begin{matrix} 7 & 4 & \underset{.}{7} \end{matrix} & \begin{array}{l} pro 3^{o} latere \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Divisoru \bar{m} {\underline{p}}^{s} superior ~ \end{matrix} & \begin{matrix} - \end{matrix} \end{matrix} & \begin{matrix} 1 & 1 & \underset{}{6} \end{matrix} & \begin{matrix} \begin{matrix} 6 & 2 & \underset{.}{0} \end{matrix} & \begin{array}{l} C C \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Diviso \bar{ru} {\underline{p}}^{s} inferior \end{matrix} & {\begin{matrix} + \\ + \end{matrix} \end{matrix} & \begin{matrix} 3 & 4 & 6 \\ 1 \end{matrix} & \begin{matrix} \begin{matrix} 8 \\ 0 & 2 \end{matrix} & \begin{array}{l} 3 A A \\ 3 A \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} *********** \end{matrix} & \begin{matrix} * & * & * \end{matrix} \end{matrix} & \begin{matrix} * & * & * \end{matrix} & \begin{matrix} \begin{matrix} * & * & * \end{matrix} & \begin{array}{l} * \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Eorum Summa \end{matrix} & \begin{matrix}  \end{matrix} \end{matrix} & \begin{matrix} 2 & 3 & 1 \end{matrix} & \begin{matrix} \begin{matrix} 2 & 0 & 0 \end{matrix} & \begin{array}{l} = 3 A A + 3 A + C C \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Sollida ablativa \end{matrix} & {\begin{matrix} + & 1 \\ + \\ + \\ - \end{matrix} \end{matrix} & \begin{matrix} 0 & 4 & 0 \\ 9 \\ 3 & 4 & 9 \end{matrix} & \begin{matrix} \begin{matrix} 4 \\ 1 & 8 \\ 2 & 7 \\ 8 & 6 & 0 \end{matrix} & \begin{array}{l} 3 A A E \\ 3 A E E \\ E E E \\ E C C \end{array} \end{matrix} \\ \begin{matrix} \begin{matrix} Eorum Summa \end{matrix} & \begin{matrix} + \end{matrix} \end{matrix} & \begin{matrix} 6 & 9 & 9 \end{matrix} & \begin{matrix} \begin{matrix} 7 & 4 & 7 \end{matrix} & \begin{array}{l}  \end{array} \end{matrix} \end{matrix}

Sometimes though there be as many 2 figures in the coefficient as 3 figures in yethe cube affected yet yethe coëfficient may be so greate as to deceive an uwaryunwary Analist As in this

L c - 6400 L = 153000

. where yethe roote of

64

8

wchwhich cubed is

512

wchwhich added to

153

makes

665

thēen whose roote yethe number immediately greater is

9

wchwhich make is yethe first sidside

= A

. But if yethe coefficient had beene affirmative, ynthen not yethe aggregate of yethe facts but yethe difference must be taken as in this.

L c + 64 L = 1024

. Since yethe roote of

64

8

. wchwhich cubed is

512

. &

1024 - 512 =

512

. yethe roote of wchwhich is

8 = A

. The like is observable in equations of higher powers If yethe Cube be affected wthwith a negative sine as

13,104 L - L c = 155,520

. Then yethe Equation is expressible of 2 rootes: whereof is less yethe other greater ynthan i yethe square of one is 6 lesse & the square of yethe other is greater therne

\frac{13104}{3}

. & therefore one roote is lesse yethe other greater then

\frac{155520}{13104}

. & in this equation

27755 L - L qq = 217944

are two rootes whereof one is greater yethe other lesse then

\frac{217944}{27755}

. ☞ Suppose in yethe former cubick equation yethe lesse roote be 12. ynthen

\frac{155520}{12} = 12960

. or else

13104 - 12 \times 12 = 12960

. &

L q + 12 L = 12960

. where

L = 108

is yethe greater roote. And in yethe latter equation if yethe lesse greater roote be 27. &

\frac{217944}{27} = 8072

, c. or

- 27 \times 27 \times 27

=

+

27755 = 8072

27 \times 27 = 729

. If there be 4 cubes continullycontinually proportionall whose greate extreame is

27 c = 19683

. & yethe aggregate of yethe 3 rest is 8072 &

L c

yethe lesse extreame, therefore

L c + 27 L q + 729

L

= 8072

. yethe roote of wchwhich .is 8 yethe other roote of yethe equation ☞ Or haveing one roote of an equation yethe eEquation may be lessonedlessened by division thus

13104 l - l c

=

155520

l^{3}

- 13

1

04 l + 155520 = 0

. & one roote is

12

. therefore divide this equation by

l - 12

& the Quote is an equation conteingconteining yethe other roote viz:

l q + 12 l = 12960

. To find two meane proportionalls twixt BC & IK 8 Propositiones Geometricæ. Franc: Vietæ. prop 1

ab ∶ ac ∷ ce ∶ bd

prop 2 & if

ab ∶ ac ∷ ac ∶ bd

: then

ac ∶ ab ∷ ab ∶ ce

prop 3. If

ab \times ac = bd \times ce

. then

bd ∶ ac ∷ ac ∶ ab ∷ ab ∶ ce

∺

prop 3. To find two menemeane proportionalls twixt

Bc

IK

. On yethe center

a

wthwith the Radradius

ai

describe the circle

ibck

. inscribe

b

c

= cd

. draw

da

through yethe center &

bg

parallel to it. draw

hk

through

A

soe ytthat

gh = ab = (= ai)

. &

ik ∶ hb ∷ hb ∶ hi ∷ hi ∶ bc

∺

Prop: 4 If

ad = db = cb

. ynthen yethe Angle

c

be

is tripple to yethe Angle

abd

. Prop 5 If

ab = bd = Rad

3 Ang: bad = cde

Prop 6 If

3 rpq = spq :recto

. that is If

2 qr = pr

. then

3 or \times or = sp \times sp + op \times op + px \times px

Prop 7 Iaf If

ad = dc = ce = ef

. then

ecf = efc = 3 dac = 3 dca

{AC}^{3} = 3 AC \times {ad}^{2} + Cf \times {ad}^{2}

Z^{3} = a a z + b^{3} . ~ ~ ~

prop 8. If

op = pr = qr = qs

. ynthen,

prt = 3 qsr

&c and

{sr}^{3} = 3 sr \times {qr}^{2} - or \times {qr}^{2}

Z^{3} = a a z - b^{3}

prop 9 If

op = pr =

ah = hb = bf = fd

. &

ch = 2 eh

ceh = 3 hce

. then

\begin{matrix} {bg}^{3} \end{matrix}

\begin{matrix} {ac}^{3} & = & 3 ac & \times & {ah}^{2} & - & db \times {ah}^{2} \\ & & {cb}^{3} & = & 3 cb & \times & {ah}^{2} & - & db \times {ah}^{2} \end{matrix}} Z^{3} = a a z - b^{3} .

Prop 10 If

de = ea

db ∶ da ∷ ab \times ab ∶ dc \times dc

. ynthen

be

is a side of a

7

equall sided & angled figure. or

7 eab = 4 right angles

. prop 10 If

ac = ef

. &

aef

a right angle &

ab

paspasses through yethe center then

cd ∶ de ∷ de ∶ df ∷ df ∶ db

. And if

cd ∶ de ∷ de ∶ df ∷ df ∶ db

then

ae

is perpendicular to

ef

2 hd

is yethe difference of yethe extreames &

2 do

is yethe difference of yethe meanes. wchwhich given yethe proportionall lines may be found &c. prop 11 Pseudomesolabium wherby To find 2 meane proportionalls. If,

ae ∶ ec ∷ ec ∶ ed ∷ ed ∶ eb

. they be inscribed in the circle

acbd

the

diam

: being

ae + eb

. If twixt

g

i

i

h

two meane proportionalls are sought on yethe same center

f

wthwith yethe

Rad

\frac{gi + ih}{2}

describe

gkhl

& inscribe a line

kl

parallel to

cd

cutting

ab

in yethe point

e

i

gi ∶ ki ∷ ki ∶ il ∷ il ∶ ih

. Examine it. prop 12. 13 & I think 11 are trew onely mechanically. prop 12 If

do = dh

ac

bisected in

b

bd

bee drawne

rd

is iyethe side of a pentagon wchwhich may be inscribed in

defcro

prop 13. If

rd

be the side of a

{\begin{matrix} octogon \\ decagon \end{matrix}}

pd

yethe side of an

{\begin{matrix} hexagon \\ octogon \end{matrix}}

yethe arch

rp

divided equally in

o

od

will be the side of an

{\begin{matrix} heptagon \\ enneagon \end{matrix}}

to be inscribed in yethe circle

ord

& yethe arch

RP

is bisected rightly divided by Bisecting yethe Line

ac

. Examine it 9 Of Angular sections. prop 14 If

ead = cab

. Then

ab ∶ {ab}^{2} ∷ cb ∶ eb \times ad - ae \times db ∷ ac ∶ ae \times ad + eb \times db

or,

ab ∶ aq \times ab ∷ op ∶ eb \times an - ae \times nq ∷ ao ∶ ae \times an + eb \times nq

. But yethe angles

anq

aop

are right ones and

e

an = oap

=

eab - dab

prop 15 If yethe angle

cab + dab = eab

. or

naq + oaq = eab

. &

anq

aop

are right angles then

Ab ∶ {ab}^{2} ∷ Eb ∶ ad \times bc + ac \times db ∷ ea ∶ ad \times ac - db \times cb

. or the triang: unequall.

ab ∶ ap \times aq ∷ eb ∶ an \times op + ao \times nq ∷ ea ∶ an \times ao - nq \times op

prop 16. In 2 rectang: triang:

acb

aed

, if yethe first have an acute angle

cab

submultiple to the acute angle

eab

of yethe 2d triang

aeb

yethe sides of yethe seacond have this proportion. Suppose yethe Hypoten of yethe first tri: be

z

. yethe base

b

. yethe Cathetus

c

\begin{matrix} \begin{matrix} If y^{e} acute angle of \\ y^{e} seacond triangle be \\ to y^{e} acute angle of \\ y^{e} first triangle in \\ a proportion \end{matrix} & {\begin{matrix} Hypoten: & Base & Perpendicular \\ \begin{array}{r} Duple, Z^{2} . \end{array} & \begin{matrix} B^{2} . \\ {-C}^{2} . \end{matrix} & \begin{matrix} 2 B C . \end{matrix} \\ \begin{array}{r} Triple, Z^{3} . \end{array} & \begin{matrix} B^{3} . \\ -3 D D C . \end{matrix} & \begin{matrix} 3 B B C . \\ {-C}^{3} . \end{matrix} \\ \begin{array}{r} Quadruple, Z^{4} . \end{array} & \begin{matrix} B^{4} . \\ -6 B^{2} C^{2} . \\ {+C}^{4} . \end{matrix} & \begin{matrix} 4 B^{3} C . \\ -4 B C^{3} . \end{matrix} \\ \begin{array}{r} Quintuple, Z^{5} . \end{array} & \begin{matrix} B^{5} . \\ -10 B^{3} C^{2} . \\ +5 B C^{4} . \end{matrix} & \begin{matrix} 5 B^{4} C . \\ -10 B^{2} C^{3} . \\ {+C}^{5} . \end{matrix} \end{matrix} \end{matrix}

Prop 17. If ∠

ab = bc = ce = eg

&c: &

ac = cd

. &

ae = ef

&c then

ab ∶ ac ∷ ac ∶ ad ∷ ae ∶ af

&c &

ed = ab

ac = gf

&c. nam △triangle

cde

cba

efg

eac

&c: = & sim. Prop.18. If

bd = dg = gh = hk = pq = pw

&c Then

al ∶ ak ∷ pe ∶ pc ∷ ed ∶ do ∷ pd ∶ pc + do ∷

rg ∶ gs ∷ rq ∶ qo ∷ qg ∶ qo + gs ∷

&c & if

(\frac{lf}{2} = lʒ)

from ʒ to yethe center be drawne

cʒ

then

al ∶ ak ∷ di ∶ dv ∷ iq ∶ qx ∷ dq ∶ dv + qx ∷ gz ∶ gx ∷ wz ∶ wx

&c & Ergo

ac ∶ ak ∷ ab ∷ aδ + ad ∶ ad ∶ ab + ag ∶ ag ∶ ad + ah ∶ ah ∶ ag + ak

&c. Prop 19 If

fa = ab = be = eh

&c. &.

af + ab + be + eh

are greater y t n than yethe semiperiphery: then,

Rad ∶ ge (= ec) (= ad) ∷ bd (= bc)

∶ cd (= cd - de) ∷ de ∶

dh - bd ∶

dh

is yethe greatest,

db

yethe least line drawn from

d

to these points

a

b

e

h

. ynthen

rad ∶

dh

∷ db ∶ da - de

. Prop 20 Out of yethe 18th & 19th Prop:Propositions To divide An angle into any number of ptspoints in yethe figure of yethe 18th prop:

al = diam =

2

z

ah

is yethe greatest of yethe inscribed lines

= B

: now

z

∷

∶

B ∷ ah +

2 z

. therfore

b b = ah in z + 2 z^{2}

. &

\frac{b b - 2 z z}{z} = ah

. And

z ∶ B ∷ \frac{b^{2} - 2 z^{2}}{z} ∶ b + ag

. therfore

\frac{b^{3} - 2 z z b}{z z} - b = ag

Likewise

\frac{b^{4} - 4 z z b b^{2} + 2 z^{4}}{z^{3}} = ad

. &

\frac{B^{5} - 5 z z B^{3} + 5 z^{4} B}{z^{4}} = ab

\frac{B^{6} - 6 z z B^{4} + 9 z^{4} B B - 2 z^{6}}{z z z z z} = aδ

\frac{B^{7} - 7 z z B^{5} + 14 z^{4} B^{3} - 7 z^{6} B}{z^{6}} = to a seaventh line

\frac{B^{8} - 8 z z b^{6} + 20 z^{4} b^{4} - 16 z^{6} b b + 2 z^{8}}{z^{7}} = to an eight

line \binom{}{}

\frac{B^{9} - 9 z z b^{7} + 27 z^{4} b^{5} - 30 z^{6} b^{6} + 9 z^{8} b}{z^{8}} = a

nineth line \binom{}{}

\frac{B^{10} - 10 z z b^{8} + 35 z^{4} b^{6} - 50 z^{6} b^{4} + 25 z^{8} b b - 2 z^{10}}{z^{9}} =

tenth & Prop 21 out of yethe 17th Theor.: in yethe figure whereof if

ab

: yethe least inscribed line

= z

. &

ac

yethe next line bee

B

. then

z ∶ B ∷ B ∶ z + ae

. &

\frac{b b - z z}{z} = ae

\frac{B^{3} - 2 z^{2} B}{z z} = ag

\frac{B^{4} - 3 z^{2} b b + z^{4}}{z^{3}} = to a fift line.

\frac{B^{5} - 4 z z b^{3} + 3 z^{4} b z^{4}}{z^{4}} = a sixt.

b^{6} - 5 z z b^{4} + 6 z^{4} b b - z^{6} = seaventh

\frac{b^{6} - 5 z z b^{4} + 6 z^{4} b b - z^{6}}{z^{5}} = seaventh

\frac{b^{7} - 6 z z b^{5} + 10 z^{4} b^{3} - 4 z^{6} b}{z^{6}} = to an eight line

\frac{B^{8} - 7 z z b^{6} + 15 z^{4} b^{4} - 10 z^{6} b b + z^{8}}{z^{7}} = to a nineth line

\frac{B^{9} + 8 z z b^{7} + 21 z^{4} b^{5} - 20 z^{6} b^{3} + 5 z^{8} b}{z^{8}} = to a tenth line

\frac{B^{10} - 9 z z b^{8} + 28 z^{4} b^{6} - 35 z^{6} b^{4} + 15 z^{8} b b - z^{10}}{z^{9}} = eleventh

. 10 Prop 22. If

aq = ab = bd = dc = ch = hk = kl = lf

. Then

GK Rad ∶ kl ∷ kl ∶ el (= al - ah) ∷

And

hl ∶ hm (= qh - qd =

ak - ac) ∷ dl ∶ do (= qd - qa = ac - ab) ∷ lc ∶ cn (= qc - qb =

ah - ad

)

Soe that Rad yethe Periph: divided into any number of ptspoints.

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 yethe former scheame If

al

= 2 x = hypotenusa

kl = b

x ∶ b ∷ b ∶ 2 x - ah

\frac{- b b + 2 x x}{x} = ah

x ∶ b ∷ \frac{- b b + 2 x x}{x} ∶ \frac{2 b x x - b^{3}}{x x} (= lc -

b

)

lc

−

kl

dl

−h therefore

\frac{3 b x x - b^{3}}{x x} = lc

. &

\frac{2 x^{4} - 4 b b x x + b^{4}}{x^{3}} = ad

yethe base of yethe 4th triang: &c

\frac{5 b x^{4} - 5 b^{3} x x + b^{5}}{x^{4}} =

yethe perpendicular

(

bl

)

of yethe 5t tri:triangle &c

\frac{2 x^{6} - 9 x^{4} b b + 6 x x b^{4} - b^{6}}{x^{5}} =

base of yethe 6t triang.

\frac{7 x^{6} b - 14 x^{4} b^{3} + 7 x^{2} b^{5} - b^{7}}{x^{6}} =

perpendicpendicular of the 7th tritriangle

\frac{2 x^{8} - 16 x^{6} b b + 20 x^{4} b^{4} - 8 x x b^{6} + b^{8}}{x^{7}} =

base of yethe 8th tri.

\frac{9 x^{8} b - 30 x^{6} b^{3} + 27 x^{4} b^{5} - 9 x x b^{7} + b^{9}}{x^{8}} =

perp: of yethe 9th tri: Prop 24: If

bd

= dh = hk = kl =

b

g

&c: then

bh = gd

bk = gh

bl = hm

&c: & then

xt ∶ bx ∷ ac ∶ ag ∷ ce ∶ eh ∷ ei ∶ em ∷ io ∶ ok ∷

op ∶ on ∷ pq ∶ ql ∷ qr ∶ qy ∷ rs ∶ sx ∷ st ∶ sf ∷ ba ∶ ad

therefore

xt ∶ bx ∷ bt ∶ dg + hm + kn + ly + xf

. againe

xt ∶ bt ∷ ab ∶ bd ∷ ac ∶ cg ∷ ce ∶ ch ∷ ei ∶ im

&c Therefore

xt ∶ bt ∷ bt ∶ bd + gh + mk + ml + yx + ft

. & since, as

xt ∶ bt ∷ bx ∶ dg + hm + kn + ly + xf

Therefore

xt ∶ bt ∷ bx + bt ∶ bd + dg + gh + hm + mk + kn +

+ nl + ly + yx + xf + ft

. And

xt ∶ bt ∷ bx + bt + xt ∶

to all yethe perpen dicular & transverse line

+ bt

. that is ( 5 )

xt ∶ bt ∷ xt + bt + bx ∶ 2 bd + 2 bh + 2 bk + 2 bl + 2 bx + 2 bt

. Prop 24 If in yethe circle

cfgh

be inscribed yethe helix

bedc

ac

touch it in yethe point

c

then

ab =

to yethe circumference. Prop 25 If

apcr

be les ynthan halfe yethe circle. &

vt = tp

. &

vo =

to yethe

vrap

: then

\frac{rq \times po}{2} =

4 times yethe section

rapc

Prop 26 If

ab = bd = ad

bh

perpendicular to

ad

frofrom yethe angle

b

ce = ed

. ynthen

aed = adi = 3 dae

. &

ed

is yethe side of a heptagon Prop 27. If a line be cut by extreame & meane proportion yethe lesse segment almost is to yethe whole line as yethe diameter is to

\frac{5}{6}

5

times the periphery divided by

6

. Prop 28 Si secetur linea per extremam & mediam proportionem erit proximè, ut tota linea 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 segmentūum

38,197

. Semidiametrūum

100,000

, quæ potest quadrato sesquialterūum semidiametri paulo maior est quam

122,474

. Radix Peripheriæ,

177,245

. 11 Prop 28. If

er = rh = or

. &

ao = fc =

to yethe side of a decagon; &

fn

parallell to

cd

ynthen

en

shall be almost equall to yethe fourth ꝑteparte of a circle for

af

ef

is divided in extreame & meane propor in yethe point

c

. &

ec ∶ ef ∷

ef ∶ \frac{5}{12}

Perimmetereterhbkfa

∷ hr ∶ \frac{5}{24} Perim ∷ \frac{6}{10} ef (= de) ∷ \frac{1}{4} Perim

; by yethe 27th prop: &

ec ∶ ef ∷ ed ∶ en (= \frac{1}{4} Perim)

. Prop 29. If

os = 2 cp

. &

co

is divided by extreame & meane proportion in

r

. &

od

parallell to

rp

then

db

is yethe side of a square = to othe area of yethe circle. for bey yethe 28th prop: As

br (= to line + less s \bar{egm}) ∶ bo (= twice y^{e} line) ∷ bp

(

(= \sqrt{\frac{3}{2} of the square of y^{e} semidiameter}) ∶ bd (= to y^{e} roote

of a square equall to y^{e} area of a circle

. Prop 30 If yethe line

dc

touch yethe helix in yethe line

ag

. & yethe line

hf

toucheth yethe beginning of it in yethe center

a

4 ac = af

then

2 ad

shall bee equall to perim:

asr

. &

ac

being yethe Diammeter: yethe tri:

acd

area of yethe triang

acd

= to yethe area of yethe circle

asr

Prop 31 If

bed

be a square of one revolution of an helix & yethe angle

dbe = dba

& through yethe points

a

d

, in yethe helix be drawne yethe line

adk

& through yethe points

ed

e

d

in yethe Helix be drawne

edg

. & yethe angle

kdg

bisected by

dh

; then

dh

shall almost touch yethe helix in

d

. & it shall be soe much yethe nigher a touch line by how much yethe angles

ebd

dba

are lesser. Prop 32 If many Polygons be inscribed in a circle yethe number of theire sides increaseing in a double proportion. & theire apotomies, or yethe tbase of a tri: whose cathetus is a leg of yethe Polygon & hypotenusa is yethe Diammeter (as yethe apotome of yethe Polygōon

cgp

ce

. of

pacegi

ae

&c) if yethe Apotome of yethe sides of yethe first Polygōon be called

b

. of yethe 2d

= c

. of yethe 3d

= d

. of yethe 4th

= f

. of yethe 5t

= g

of yethe Sixt

= h

. & yethe diameter be z then And yethe first Polygon be

= p

. yethe 2d

= q

. yethe 3d

= r = abcdefghiopq

. yethe fourth

= s

. yethe 5t

= t

yethe sixt

= v

. yethe 7th

= w

&c then

p ∶ q ∷ b ∶ z

. & p

p ∶ r ∷ b c ∶ z z

. &

p ∶ s ∷ b c d ∶ z z z

. &

p ∶ t ∷ b c d f ∶ z^{4}

. &

p ∶ v ∷ b c d f g ∶ z^{5}

. &

p ∶ w ∷ b c d f g h ∶ z^{6}

&c To know how many elections may bee made of divers ways things, whereof some of ymthem are equall, may bee ordered. . as of .

a . b . b . c . c . c . d . d .

doe thus

\begin{matrix} a & b b & c c c & d d \\ \frac{1}{1} & \times & \frac{2 \times 3}{1 \times 2} & \times & \frac{4 \times 5 \times 6}{1 \times 2 \times 3} & \times & \frac{7 \times 8}{1 \times 2} & = 112 \end{matrix}

\frac{\overset{a}{1}}{1} \times \frac{\overset{b}{2} \times \overset{b}{3}}{1 \times 2} \times \frac{\overset{c}{4} \times \overset{c}{5} \times \overset{c}{6}}{1 \times 2 \times 3} \times \frac{\overset{d}{7} \times \overset{d}{8}}{1 \times 2} = 112

the number of changes, in order. To know how many elections may bee made doe thus

\overset{a}{2} \times \overset{bb}{3} \times \overset{ccc}{4} \times \overset{dd}{3} \times \overset{eeee}{5} = 360 = \overset{a}{2} \times \frac{\overset{b}{2} \times \overset{b}{3}}{2} \times \frac{\overset{c}{2} \times \overset{c}{3} \times \overset{c}{4}}{2 \times 3} \times \frac{\overset{d}{2} \times \overset{d}{3}}{2} \times \frac{\overset{e}{2} \times \overset{e}{3} \times \overset{e}{4} \times \overset{e}{5}}{2 \times 3 \times 4}

therefore there are

359 = 360 - 1

elections in

a b b c^{3} d d e^{4}

. 12 Propositiones Geometricae Ex Schootenij Sectionibus miscellaneis. Sectio 1ma To know how many changes 6 Bells,

a b c d e f

or how divers conjuctions yethe 7 planets can make ♄ ♃ ♂ ☉ ♀ ☿ ☾. or how many divisors

a b c d e

hath, or how maney divers compositions yethe 24 letersletters can make &c the examples following show.

1 .

a 1

2 .

b . a b 3

4 .

c . c b . c a b . a c . 7

8 .

d . d a . d b . d a b . d c . d a c . d c b . d c a b . 15

16 .

e . e a . e b . e a b . e c . e a c . e c b . e c a b . e d . e d a . e d b . e d a b . e d c . e d a c .

e d c b . e d c a b .

31

32 .

f . f a . f b . f a b . f c . f c b . f a c . f c a b . f d . f d a . f d b . f d a b &c 63 .

64 .

g . g a . g b . g a b . g c . g c b . g a c . g c a b . g d . g d a . g d b &c 127

wchwhich shows ytthat in 7 letters 127 eleections may be made. ytthat 7 Planets may be conjoyned 120 divers ways. ytthat

a b c d e f g

. hath 128 divisors for an unite is one of ythem&

1 \times 2 \times 3 \times 4 \times 5 \times 6 . = 720

; are yethe number of changsschanges in six bells. Sec 2 To know how many things & of wtwhat sort they are wchwhich may be chosen 15 ways.

15 + 1 = 16 . \frac{16}{2} = 8 . \frac{8}{2} = 4 . \frac{4}{2} = 2 . \frac{2}{2} = 1

. & 4

2 - 1 . 2 - 1 . 2 - 1 . 2 - 1 . = 1.1.1.1 . = 4

. that 4 things all unequall may be varyed 15 ways. also.

\frac{16}{4} = 4 . \frac{4}{2} = 2 . \frac{2}{2} = 1

4 - 1 . 2 - 1 . 2 - 1 = 5

&. 5 things whereof 3 are equall viz:

a . a . a . b . c .

\frac{16}{4} = 4 . \frac{4}{4} = 1

4 - 1 . + 4 - 1 = 3 + 3 = 6

. & 6 thinsthings whereof 3 & 3 are eaquallequall as

a a a b b b

. may be varied 15 ways. &

\frac{16}{8} = 2

\frac{2}{2} = 1

8 - 1 . 2 - 1 = 8 = 7 + 1

. & 8 things whereof 7 are = may be varyed 15 ways. as

a a a a a a a b

\frac{16}{16} = 1

16 - 1 = 15

. 2 wherefore 15 alike things &c as a 15. 2 wtwhat things vary 23 ways.

23 + 1 = 24

& 24 admitts a 7 fold divisor therefore yethe multitude of things sought may be 7 fold but since 43 is a primary number (viz wchwhich cannot bee divided)

42 + 1 = 43

\frac{43}{43} = 1

43 - 1 = 42

. therefore onely 42 like things can be varyed 42 ways as

a^{42}

. Sec 3 Every quantity hath one divisor more ytthat it hath aliquote ptsparts (ytthat is ptsparts greater ynthan an unite of whole numbers.). How to find a quantity haveing a given multitude of divisors or aliquote ptsparts: suppose its aliq: ptsparts must be 15.

15 + 1 = 16

& soe by yethe former section

a b c d . a^{3} b c . a^{3} b^{3} . a^{7} b . a^{15} .

may be varyed 15 ways. therefore they shall have 15 aliquote ptsparts & 16 divisors. but since onely 42 like things (as

a^{42}

) can be varyed 42 ways therefore oenelyonely

a^{42}

hath 42 aliquote ptsparts & 43 divisors. &c Sec 4 To find yethe least numbers haveing a given multitude of divisors & aliquote ptsparts instead of soe many letters in the former sec: put soe many p least primary numbers & take yethe least result from ymthem. as from yethe former example:

a b c d . a^{3} b c . a^{3} b^{3} . a^{7} b . a^{15} .

that is

2.3.5.7 .

2 .

2.2.3.5 .

&c. now.

2 \times 3 \times 5 \times 7 . = 210

. &

2 \times 2 \times 2 \times 3 \times 5 = 120

. &c therefore

2 \times 3 \times 5 \times 7 .

= 210

are is yethe least numbers haveing 16 divisors. Sec: 5 conteines a table of Primary numbers. Sec 6 To find progressions constituteing rectangular triangles wthwith sides rationall yethe examples following shew. take two numbers as

1.2

. ynthen

1 \times 2 = 2

since yethe product is eavedn double it viz:

2 \times 2 = 4

. &

4

is yethe numerator ynthen

1 + 2 = 3

& since

3

is od multiply it by the difference of yethe termes:

1 \times 3 = 3

3

is yethe denominator. & yethe first terme

\frac{4}{3}

. ynthen since (1) yethe difference of yethe termes is od multiply it by

4

4 \times 1 = 4

4 \times

per 2 majoreem terminum.

4 \times 2 = 8

8 + 4

(the former numerator)

= 12 =

numerator 2d. ynthen

3

(yethe former denom) added to.

2

(yethe double square of yethe diff: of yethe termes because yethe square (1) is odd)

= 5

yethe 2d denominator. I ad anotheanother example take

1.3 .

ynthen

1 \times 3 = 3 =

1st numerator. ynthen

1 + 3 = 4

& since

4

is eaven halfe

4

1st denom. & yethe 1st halfe is

\frac{4 \times 2}{2}

(diff: of yethe termes)

= 4

& yethe first denom is

4

. yethe first terme

\frac{3}{4}

. ynthen becaus yethe diff of yethe termes is eaven

2

2 \times 2 = 4

4 \times 3 = 12

12 + 3 = 15

. ynthen

2 \times 2 = 4 . 4 + 4 = 8

. &

\frac{15}{8}

yethe 2d terme & now termes may be had by Arithmeticall proportion. thus.

\frac{4}{3} . \frac{12}{5}

1 \frac{1}{3} . 2 \frac{2}{5} . 3 \frac{3}{7} . 4 \frac{4}{9} . 5 \frac{5}{11} . 6 \frac{6}{13} 7 \frac{7}{15} 8 \frac{8}{17} 9 \frac{9}{19} 10 \frac{10}{21}

&c &

\frac{3}{4} . \frac{15}{8}

\frac{3}{4} 1 \frac{7}{8} . 2 \frac{11}{12} . 3 \frac{15}{16} . 4 \frac{19}{20} . 5 \frac{23}{24} . 6 \frac{27}{28} . 7 \frac{31}{32} . 8 \frac{35}{36}

&c thus may other progressions be obteined. For yethe use take yethe numerator for one leg & yethe denom for another & yethe Hypoten: will be rationall as in

2 \frac{2}{5}

5 \times 2 = 10 . 10 + 2 = 12

\frac{12}{5}

\sqrt{144 + 25} = \sqrt{169} = 13

. & in this

1 \frac{7}{8}

\frac{15}{8}

\sqrt{225 + 64} = 17

. 13 If yethe suposed numbers be

2.5

. ynthen

2 \times 5 = 10

10 + 10 = 20

. &

2 + 5 = 7

3 \times 7 = 21

. so ytthat

\frac{20}{21}

. ynthen be

4 \times 3 = 12

12 \times 5 = 60

60 + 20 = 80

. &

3 \times 3 = 9

9 doubled = 18

18 + 21 = 39

. & yethe 2 first termes

\frac{20}{21} . \frac{80}{39}

2 \frac{2}{39}

. Againe, if yethe numbers be

3.4

3 \times 4 = 12

12 \times 2 = 24

. &

3 + 4 = 7

1 \times 7 = 7

. therefore

\frac{24}{7}

. ynthen

4 \times 1 = 4

4 \times 4 = 16

16 + 24 = 40

. &

1 \times 1 = 1

2 \times 1 = 2

7 + 2 = 9

therfore

\frac{40}{9}

is yethe 2d & yethe progres may be continued, as

\frac{20}{21} . 2 \frac{2}{39} . 3 \frac{5}{57} . 4 \frac{8}{75} . 5 \frac{11}{93}

. &

3 \frac{3}{7} . 4 \frac{4}{9} . 5 \frac{5}{11} . 6 \frac{6}{13}

&c. Sec 7 To find a number wchwhich divided by

7

leaves

2

. by

11

leaves

1

. by

13

leaves

9

. the least common divisor of

7.11.13 .

7 \times 11 \times 13 \times =

1001

. divide

1001

twice by each & consider yethe remainder of yethe seacond division thus.

\begin{matrix} \frac{210}{2} & \frac{105^{(1}}{2} & 52 \end{matrix}

. because

1

is left

105

is yethe multiplier.

\begin{matrix} \frac{210}{3} & \frac{70^{(1}}{3} & 23 \end{matrix}

. since

1

is left

70

is yethe multiplier.

\begin{matrix} \frac{210}{5} & \frac{42^{(2}}{5} & 8 \end{matrix}

since more ynthan

1

is left (viz:

2

) multiply

2

till it divided by 5 leaves (

1

)

\frac{2 \times 3}{5} = 1 \frac{1}{5}

. therefore

42 \times 3 = 126

is yethe multiplyere.

\begin{matrix} \frac{210}{7} & \frac{30}{7} & (4 \frac{2}{7} \end{matrix}

since more ynthan

1

is left 1 Since more ynthan

1

is left (viz

3

) multiply

3

till it divided by

7

leavs

1

\frac{5 \times 3}{7} = 2 \frac{1}{7}

therfore

5 \times 143 = 715

yethe multiplier

| \frac{1001}{7} (\frac{143}{7} (20 \frac{3}{7}

. 2 Since more ynthan

1

is left (viz:

3

)

\frac{3 \times 4}{11} = 1 \frac{1}{11}

therfore

4 \times 91 =

= 364

yethe multipl:

| \frac{1001}{11} (\frac{91}{11} (8 \frac{3}{11}

. 3 If but

1

had benene left

77

had beene divisor but now

\frac{12 \times 12}{11} = 13 \frac{1}{11}

. therfore

12 \times 77 = 924

is multiplyer.

\frac{1001}{13} (\frac{77}{13} (5 \frac{12}{13}

. now the number soughsought is thus found.

\begin{matrix} \begin{matrix} Divisor. & Reliq: & Multip. \\ 7 . & 2 & \times & 715 & = & 1430 . \\ 11 . & 1 & \times & 364 & = & 364 . \\ 13 . & 9 & \times & 924 & = & 8316 . \end{matrix} \\ The Su \bar{m} e 10110 . \end{matrix}

Lastly didvide by yethe least com. divis:

\frac{10110}{1001} (10 \frac{100}{1001}

wherefore

100

yethe number left is yethe number sought. Sec 8. Touching yethe Method of weights suppose a man have weigsweights of

1.2.3.4.5

1.2.4.8.16.32

pounds &c by ymthem all intermediate pounds may be thus weighed

\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}

\begin{matrix} 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 \end{matrix}

&c or isf his wightsweights be

1.3.9.27.81 .

all weights may be supplyed thus.

\begin{matrix} 1 . & 2 . & 3 . & 4 . & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ 1 & 3 - 1 . & 3 . & 3 + 1 . & 9 - 1 - 3 . & 9 - 3 . & 9 + 1 - 3 . & 9 - 1 . & 9 . & 9 + 1 . & 9 + 3 - 1 . & 9 + 3 . & 9 + 3 + 1 . & 27 - 9 - 3 - 1 . \end{matrix}

&c Note ytthat weight marked wthwith

-

signifie yethe wighweight to be put in yethe opposite scale ballance. Sec. 9. To find numeri amicabiles that is 2 numbers whose aliquote ptsparts are mutually equall to theire wholes. take this Des-Cartes his rule If

(2)

, or any other number producced out of

2

2 \times 2 . 2 \times 2 \times 2

&c (viz

2.4.8.16.32

&c .) bee such a number ytthat

1

takedn out of it triple there rests a primary number, & ytthat if

1

taken from it sextuple there rests a primary number, & if

1

taken from its square octodecuple a primary number rests: ynthen multiply this last prime number by yethe assumed number doubled & yethe product is one amicable number & yethe aliquote ptspoints of it make yethe other Example. if

2

be taken.

2 \times 3 - 1 = 5

numero primario primo.

2 \times 6 - 1 = 11

numero primario scdosecundo.

2 \times 2 \times 18 - 1 = 71

numero primario tertio.

4 \times 71 = 284

, one amicable number, & yethe 2 former prime numbers

\times

one another & yethe product

\times 4

yethe double of yethe assumed number viz

5 \times 11 = 55 55 \times 4 = 220

. Thus from

8

. &

64

&c. may be deduced amicable numbers. Sec 10 To find triangles whose sides, segments of theire bases, & Perpendiculars are expressible by rationall numbers 1st if yethe perpendic: is without yethe tri: let

ac = z

bd = x

cd = y

ad = z + y

ad = y + b

x x + y y = y y + 2 b y + b b

y = \frac{x x - b b}{2 b}

. &

cd = z + y + a

x x + z z + 2 y z + y y = z z +

y y + a a + 2 z y + 2 z a + 2 a y

2 a y = x x - a a - 2 z a = \frac{a x x - a b b}{b}

b x x - b a a - 2 z a b = a x x - a b b

\frac{b x x - b a a - a x x + a b b}{2 a b} = z

. puting any numbers for

a

b

, &

x

;

y

z

may be found. then

ad = z + y =

= \frac{x x + b b}{2 b}

cd = z + y + a = \frac{x x + a a}{2 a}

. wchwhich reduced to yethe common denominator

2 a b

; & ytthat cast away.

cd = b x x + b a a

ad = a x x + a b b

de = 2 a b x

ae = a x x - a b b

ce = b x x - b a a

ac = b x x - a x x + a b b

- a a b

. In like manner if yethe perpendicular fall wthwithin side.

ab = b x x + b a a

bd = 2 a b x

ad = b x x - b a a

dc = a x x - a b b

bc = a x x + a b b

ac = b x x + a x x - a b b - b a a

. Also by yethe conjunction & disjunction of 2 triangles it may be found ytthat

ab = b b x + a a x

ad = b b x - a a x

ac = b b x - a a x - a a x

+ a b b

bc = a x x + a b b

db = 2 a b x

dc = a x x - a b b

. For if

bd = x

dc = \frac{x x - b b}{2 b}

bc = \frac{x x + b b}{2 b}

. that is

bd = 2 b x

dc = x x - b b

bc = x x + b b

. Likewise

bd = 2 a b

ad = b b - a a

ab = b b + a a

2 a b x

yethe least quantity divisible by

2 b x

2 a b

, being divided by ymthem, leaves

a

x

wchwhich must multiply yethe bases & hypotenusas. If yethe perpendic: fall wthwithout yethe legs may be thus exprest

cd = a c c + a y y

da = y y c + a a c

ca = a c c - a y y + c y y - a a c

ae = y y c - a a c

ce = a c c - a y y

ad = 2 a c y

. Sec 11 To make ytthat two such tri: be of yethe same base & altitude. Suppose an equation twixt yethe bases & pependicularsperpendiculars of yethe 2 last tri: as

2 a b x = 2 a c y

x = \frac{c y}{b}

x x = \frac{c c y y}{b b}

b b x - a a x - a x x + a b b = a c c - a y y + y y c - a a c

\frac{b b c y - a a c y}{b} - \frac{a c c y y}{b b} + a b b = a c c - a y y + y y c - a a c

y y = \frac{+ b^{3} c y + a a b b c - a a b c + a b^{4} - a b b c c}{b b c + a c c - a b b}

. Suppose

a a b b c + a b^{4} = a b b c c

. or

a = c - \frac{b b}{c}

. let

c = 3

greater ynthan

b = 2

a = \frac{5}{3}

y = \frac{22}{61}

x = \frac{33}{61}

& consequently Sec 14 differs not from Cap 19: prob 18 Oughtred. Sec: 15 Of Polygons or multangular numbers The sum̄mme of all yethe tearmses in an arithmet: progres: increasing from an unite by

1

compsethcomposeth trai triangles. by

2

, composes □ssquares. by

3

, composes pentangles. by

4

, hexang: &c ke as

1.2.3.4.5.6 .

composcompose yethe triangles

\begin{matrix} 1 \begin{matrix} ● \end{matrix} & 3 \begin{matrix} ● \\ ● ● \end{matrix} & 6 \begin{matrix} ● \\ ● ● \\ ● ● ● \end{matrix} & 10 \begin{matrix} ● \\ ● ● \\ ● ● ● \\ ● ● ● ● \end{matrix} & 15 \begin{matrix} ● \\ ● ● \\ ● ● ● \\ ● ● ● ● \\ ● ● ● ● ● \end{matrix} \end{matrix}

&c likewise

1.3.5.7.9 .

compose

\begin{matrix} 1 \begin{matrix} ● \end{matrix} & 4 \begin{matrix} ● ● \\ ● ● \end{matrix} & 9 \begin{matrix} ● ● ● \\ ● ● ● \\ ● ● ● \end{matrix} & 16 \begin{matrix} ● ● ● ● \\ ● ● ● ● \\ ● ● ● ● \\ ● ● ● ● \end{matrix} & 25 \begin{matrix} ● ● ● ● ● \\ ● ● ● ● ● \\ ● ● ● ● ● \\ ● ● ● ● ● \\ ● ● ● ● ● \end{matrix} \end{matrix}

&c So

1.4.7.10.13 .

compose yethe quintangles

1.5.12.22.35.51.70 .

&c. If

a = 1 =

yethe first terme yethe excess of the progression

= x

. The summme of yethe termes

= z =

to yethe polygon yethe multitude of yethe termes

= t =

to yethe side of yethe Polygon. given to find SupposSuppose

t

given to find

z

z = \frac{1 t t + 1 t}{2}

z = \frac{t t + t}{2}

in trigons.

z = t t

in 4gons.

z = \frac{3 t t - t}{2}

in 5gons.

z = 2 t t - t

in 6goons

z = \frac{5 t t - 3 t}{2}

in 7gons.

z = 3 t t - 2 t

in 8gons.

z = \frac{7 t t - 5 t}{2}

in 9gons. &c &

z

given

t

is found thus

t = \frac{- 1 + \sqrt{1 + 8 z}}{2}

in tri.

t = \frac{\sqrt{0 + 16 z}}{4}

in 4goons

t = \frac{+ 1 + \sqrt{1 + 24 z}}{6}

, in 5gons.

t = + 2 + \sqrt{4 + 32 z}

t = \frac{+ 2 + \sqrt{4 + 32 z}}{8}

in 6gons &c. As yethe side

12

of a tri given. yethe

tri = z = \frac{12 \times 12 + 12}{2} = 78

&c & if

z = 21

be octangled.

t = \frac{+ 4 + \sqrt{16 + 48 z}}{12} = \frac{4 + \sqrt{16 + 48 \times 21}}{12}

t = \frac{4 + \sqrt{1024}}{12} = 3

. July 4th 1699. By consulting an accompt of my expenses at Cambridge in the years 1663 & 1664 I find that in yethe 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 wchwhich time I found the method of Infinite Is series. And in summer 16965 being dforced from Cambridge by the Plague I computed yethe area of yethe Hyperbola at Boothby in Lincolnshire to I two & fifty figures by the same method. Is. Newton 15 (1 Annotations out of Dr Wallis his Arithmetica infinitorum. 1 A primanary series of quantysquantitys is arithmetically proportionall, as

0, 1, 2, 3, 4

. & its index is

1

A Secundanary series iare 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 Subsecundanary, subtertianary, series &c: are those whose squares, cubes, &c are arithmetically proportionall, as

\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}

. \sqrt{c : 0}, \sqrt{c : 1}, \sqrt{c : 2}, \sqrt{c : 3}

&c. Theire indices being

\frac{1}{2}, \frac{1}{3},

&c. 2 Primary Secundanary, tertianarey series &c are said to bee reciprocally proportionall ( ytthat is to yethe samsame se increasing) which continually decrease as.

\frac{1}{0}, \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},

\frac{1}{0}, \frac{1}{1}, \frac{1}{4}, \frac{1}{9}, \frac{1}{16},

\frac{1}{0}, \frac{1}{1}, \frac{1}{8}, \frac{1}{27}, \frac{1}{64},

. Their indices being negative as

- 1, - 2, - 3

. 4 The indices of compound or mixt of rationall & irrationall series series, by multiplying or dividing yethe indices of yethe simple series may bee found as in a subsecundanary progression cubed

\sqrt{0}, \sqrt{1}, \sqrt{8}, \sqrt{27}, \sqrt{64}

yethe index is

\frac{1}{2} \times 3 = \frac{3}{2}

. So in yethe cube rootes of a secundanary progression,

\sqrt{c : 0}, \sqrt{c : 1}, \sqrt{c : 4}, \sqrt{c : 9}

&c. yethe index is

\frac{1}{3} \times 2 = \frac{2}{3}

. so in irrationall reciprocal progressions

\sqrt{qq : \frac{1}{0}}, \sqrt{qq : \frac{1}{1}}, \sqrt{qq : \frac{1}{2}}, \sqrt{qq : \frac{1}{3}}

, yethe index is

- 1 \times \frac{1}{4} = - \frac{1}{4}

. (2 Now suppose yethe line

ac

be divided into an infinite number of equall ptsparts

ad, de, ef, fg

&c, from each of wchwhich are drawne perpendiculars parallels

nd pe

qf

&c. wchwhich wchwhich inocrease continually in some of yethe foregoeing prgogressions or in some progresionprogression compounded of ymthem, all those lines may be taken for yethe surface

bqnac

, & to know wtwhat proportion all those lines have that superficies hath to yethe superficies

ambc

ytthat is wtwhat proportion all those lines have to soe may equal to yethe greatest of ymthem, I say as yethe index of yethe progression increased by an unite is to an unite soe is yethe square

abcm

to yethe area of yethe crooked line. As if

abc

is a parabola yethe lines

ad, pe, qf,

&c aore a subsecundanary series (for

y = \sqrt{r x}

) whoswhose index is

\frac{1}{2}

wchwhich added to an unite is

1 + \frac{1}{2} = \frac{3}{2}

Therefore

\frac{3}{2} ∶ 1 ∷ 3 ∶ 2

so is yethe square

ambc

to yethe area of yethe Parab. (yethe names of yethe lines are

(ad),

ae, af

= x

dn, pe, qf

= y

ac = p

bc = q

.) The case is yethe same if

abc

bee supposed a sollid, as suppose its nature t a Parabolicall conoides. ynthen since yethe nature of it is

r x = y y

y y

designes yethe squares

nd, pe, qf

&c: all wchwhich taken together are equivalent to yethe Sollid. & those □ssquares increase in yethe same proportion wchwhich

r x

. or

x

doth. ytthat is they are a primanary series whose index is

1

to wchwhich (according to yethe rule I ad an unite & tis

2

. Therefor

1

∶ 2 ∷

soe are all yethe □ssquares of yethe Primary series to soe many □ssquares equall to yethe greatest of ytthat series. & soe is yethe conoides to a cilinder of yethe same altidtude. 16 (3 Also if a superficies be compounded of 2 or more of these series, Their area is as easily found as if yethe nature of yethe line bee ,

y

= a a - x x

, or

y = a^{4} - 2 a a x x + x^{4}

y = a^{6} - 3 a^{4} x x + 3 a a x^{4} - x^{6}

. &c. Their areas will bee to yethe parallelograms arbout them as

2

3

, as

8

\frac{1}{5}

15

, as

48

105

&c. but if I put in yethe intermediate termes in these last named lines their order will bee

y = \sqrt{a a - x x}

y = a a - x x

y = \bar{a a - x x \sqrt{a a - x x}}

y = a^{4} - 2 a a x x + x^{4}

y = \bar{a^{4} - 2 a a x x + x^{4} \sqrt{a a - x x}}

y = a^{6} - 3 a^{4} x x + 3 a a x^{4} - x^{6}

; &c: & since these lines observe a geometricall progression their areas must observe some kind of progression. of wchwhich every other terme is given viz

1 . □ . \frac{2}{3} . ∗ . \frac{8}{15} . ∗ . \frac{48}{105} . ∗ . \frac{384}{945} . ∗ . \frac{3840}{10395}

. Tblwixt wchwhich termes if yethe intermediate termes

□ . ∗

can bee found yethe 2nd □square will give yethe area of yethe line

y = \sqrt{a a - x x}

, yethe circle. Soe likwiselikewise in this progression of lines

y

= 1

y

= \sqrt{a x - x x}

y = a a - x x

y = \bar{a x - x x \sqrt{a x - x x}}

y = a a x x - 2 a x^{3} + x^{4}

&c: yethe progression of their areas is

1 : □ : \frac{1}{6} : ∗ :

\frac{1}{30} : ∗ : \frac{1}{140} : ∗ \frac{1}{630} :

&c. yethe 2nd if it can bee found givsgives yethe area of yethe ○circle for as its denominator to its numerator so is yethe □square of yethe diameter to yethe area of a semicircle. If this last progresionprogression bee multiplyed by yethe respective termes in yethe progressprogression

1.2.3.4

& it may bee diminished yethe reslutresult being

1 . 2 □ . \frac{1}{2} . 4 ∗ . \frac{1}{6} . 6 ∗ . \frac{1}{20} . 8 ∗ . \frac{1}{70}

soe ytthat in this progression

1, b, \frac{1}{2}, c, \frac{1}{6}, d, \frac{1}{20}, e, \frac{1}{70}, f,

&c: if

b

can be found ynthen, yethe □square of yethe diameter to yethe area of yethe circle is as yethe denominator of

b

to its numerator. Likewise yethe 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, \frac{8}{3}, c, \frac{48}{15},

d, \frac{384}{105}, e, \frac{3840}{945}

. &c. In wchwhich if

a

can bee found ynthen as yethe denomdenominator of

a

to its numnumerator: so yethe □square of yethe Radius is a semicircle, ytthat making yethe radius

= q

2 a q = ○

. The same kinds of changes may bee performed by other p any other progressions, as by division by yethe geometricall progression

1, 2, 4, 8, 16,

& yethe first series of areas becomes

1, g, \frac{1}{6}, h, \frac{1}{30}, k, \frac{1}{140}, l, \frac{1}{630},

&c viz yethe same wthwith yethe 2d series. Also these changes may be done by addition or substraction of mutuall termes in 2 proportions. Soe ytthat yethe most convenient way may be be chosen, wherby to reduce any series of proportions to yethe most conventconvenient forme. Now if it be propounded to find these middle termes, fi It will bee convenient to dfind 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 . \frac{2}{3} . \frac{8}{15} . \frac{48}{105}

deduceth its originall from this

A

\times \frac{0 \times 2 \times 4 \times 6 \times 8}{1 \times 3 \times 5 \times 7 \times 9}

& in wchwhich

A

is an infinineinfinite number

= \frac{1}{0}

. It will also be convenient to find what relatiōon all yethe other meanes have to yethe first soe ytthat if yethe first bee had all yethe other may bee deduced thence. As in this case suppose yethe 1st meane to bee

a

. The progression will bee 17 (5

\frac{1}{2} a : 1 : a : \frac{3}{2} : \frac{4 a}{3} : \frac{15}{8} : \frac{8 a}{5} : \frac{105}{48} : \frac{64 a}{35} : \frac{945}{384} : \frac{640 a}{315} :

deducing its originall from

A \times \frac{0 \times 2 \times 4 \times 6 \times 8 \times 10}{1 \times 3 \times 5 \times 7 \times 9 \times 11}

& from this

A (= \frac{1}{2}) \times \frac{2 a \times 4 a \times 6 a \times 8 a \times 10 a}{1 \times 3 \times 5 \times 7 \times 9}

. &c + (note ytthat yethe proportions of these temeane termes to oneanoone another, or to

(a)

, are found f by finding yethe proportion of yethe circle

y = \sqrt{a a - x x}

to yethe line

y = \bar{a a - x x \sqrt{a a - x x}}

&c). In this case to find yethe quantity

a

: it may be considered ytthat

\frac{a}{\frac{1}{2} a} = 2

\frac{3}{1 \times 2} =

\frac{3}{2}

\frac{4 a}{3 \times a} = \frac{4}{3}

. Naming yethe termes in yethe progress:

\begin{matrix} \frac{1}{2} a & : & 1 & : & a & : & \frac{3}{2} & : & \frac{4 a}{3} & : & \frac{15}{8} & : & \frac{8 a}{5} & : & \frac{35}{16} & : \\ b & c & d & e & f & g & h & k & . \end{matrix}

1st observe ytthat

\frac{d}{b} = 2 . \frac{e}{c} = \frac{3}{2} . \frac{f}{d} = \frac{4}{3} . \frac{g}{e} = \frac{5}{4} . \frac{h}{f} = \frac{6}{5}

&c yethe proportions still decreasing & therereforetherefore ytthat in

\frac{c}{b} . \frac{d}{c} . \frac{e}{d} . \frac{f}{e} . \frac{g}{f} . \frac{h}{g} . \frac{k}{h}

&c: yethe latter terme is lesse ynthan yethe former; & therefore

\frac{d d}{c c} {is \begin{matrix} less y^{n} \frac{d}{c} \times \frac{c}{b} = \frac{d}{b} = 2 . \\ greater y^{n} \frac{d}{c} \times \frac{e}{d} = \frac{e}{c} = \frac{3}{2} . \end{matrix}

a =

d is {\begin{matrix} less y^{n} 1 \times \sqrt{2} = \sqrt{1 + \frac{1}{1}} \\ greater y^{n} 1 \times \sqrt{\frac{3}{2}} = \sqrt{1 + \frac{1}{2}} \end{matrix}

. Also

\frac{f f}{e e} is {\begin{matrix} lesse y^{n} \frac{f}{e} \times \frac{e}{d} = \frac{f}{d} = \frac{4}{3} \\ greater & \bar{y} & \frac{f}{e} \times \frac{g}{f} = \frac{g}{e} = \frac{5}{4} \end{matrix}} = a : \frac{8 a}{9} = \frac{2 \times 4 a}{3 \times 3}

. Therefore

3 \times 3 \sqrt{\frac{4}{}}

Therefore

a is {\begin{matrix} lesse y^{n} & \frac{3 \times 3}{2 \times 4} \sqrt{\frac{4}{3}} \\ greater y^{n} & \frac{3 \times 3}{2 \times 4} \sqrt{\frac{5}{4}} \end{matrix}

. And So by yethe same reasoning.

a is {\begin{matrix} less y^{n} & \frac{9 \times 25 \times 49 \times 81 \times 121 \times 169}{2 \times 16 \times 36 \times 64 \times 100 \times 144 \times 14} \sqrt{\frac{14}{13}} & . \\ greater y^{n} & \frac{3 \times 3 \times 5 \times 5 \times 7 \times 7 \times 9 \times 9 \times 11 \times 11 \times 13 \times 13}{2 \times 4 \times 4 \times 6 \times 6 \times 8 \times 8 \times 10 \times 10 \times 12 \times 12 \times 14} \sqrt{\frac{15}{14}} & . \end{matrix}

&c. Thus Wallis doth it. but it may bee done thus.

a is {\begin{matrix} greater & y^{n} & 1 & . \\ less then & \frac{3}{2} & . \end{matrix} |

\frac{4 a}{3} is {\begin{matrix} greater then & \frac{3}{2} \\ less then & \frac{15}{8} \end{matrix}

Therefore

a is {\begin{matrix} greater y^{n} \frac{3 \times 3}{2 \times 4} \\ lesse then \frac{3 \times 3 \times 5}{2 \times 4 \times 4} \end{matrix}

\frac{8 a}{5} is {\begin{matrix} greater y^{n} \frac{15}{8} \\ lesse y^{n} \frac{35}{16} \end{matrix}

ytthat is

a is {\begin{matrix} greater then & \frac{3 \times 3 \times 5 \times 5}{2 \times 4 \times 4 \times 6} = \frac{3 \times 5 \times 5}{2 \times 4 \times 4 \times 2} \\ lesse then & \frac{3 \times 3 \times 5 \times 5 \times 7}{2 \times 4 \times 4 \times 6 \times 6} = \frac{5 \times 5 \times 7}{2 \times 4 \times 4 \times 4} \end{matrix}

&c. By yethe same reasoning

a is {\begin{matrix} greater y^{n} \frac{3}{2} \times \frac{3}{4} \times \frac{5}{4} \times \frac{5}{6} \times \frac{7}{6} \times \frac{7}{8} \times \frac{9}{8} \times \frac{9}{10} = \frac{5 \times 49 \times 81}{2 \times 16 \times 4 \times 64 \times 2} . \\ lesse y^{n} \frac{3}{2} \times \frac{3}{4} \times \frac{5}{4} \times \frac{5}{6} \times \frac{7}{6} \times \frac{7}{8} \times \frac{9}{8} \times \frac{9}{10} \times \frac{11}{10} = \frac{49 \times 81 \times 11}{2 \times 16 \times 4 \times 64 \times 4} . \end{matrix}

a is {\begin{matrix} greater \bar{y} \frac{11 \times 169 \times 225 \times 289 \times 369 \times 441}{2 \times 16 \times 4 \times 64 \times 4 \times 144 \times 4 \times 256 \times 4 \times 400 \times 2} \\ lesse y^{n} \frac{169 \times 225 \times 289 \times 369 \times 441 \times 23}{2 \times 16 \times 4 \times 64 \times 4 \times 144 \times 4 \times 256 \times 4 \times 400 \times 4} \end{matrix}

. Note ytthat

a

is greater ynthan

\frac{1}{2}

these two summes. 18 (7 Having yethe signe of any angle to find yethe angle or to find yethe content of any segmntsegment of a circle Suppose yethe circle to be

aec

its semidiameter

ap = pc = 1

. yethe given sine

pq = x

, viz: yethe signe of yethe angle

epa

. yethe segment sought

eapq

abcp

the □square of its Radius. & ytthat,

qi : qk : qg : qf : qe : qd : qh : ql : qr :

&c are continallytinually proportionall. Then is

eq = \sqrt{1 - x x}

fq = 1 - x x

gq =

\bar{1 - x x in \sqrt{1 - x x}}

hq = 1 - 2 x x + x^{4}

iq = 1 - 2 x^{2} + x^{4} \sqrt{1 - x x}

dq = 1

kq = \frac{1}{\sqrt{1 - x x}}

lq = \frac{1}{1 - x x}

rq = \frac{1}{1 - x x in \sqrt{1 - x x}}

. &c & since all the ordinately applyed lines in these figures

abcq, aecq, afcq, agcq,

&c are geometrically proportionall their areas

adqp, aeqp, afqp, agqp, ahpq

&c will observe some proportion amongst one another. To find wchwhich proportion, 1st

adqp = 1 \times x x = x

. 2dly

afc

is a parab: therefore

afqp = x - \frac{x^{3}}{4}

. also since tis

qh = 1 - 2 x x + x^{4}

, therefore

ahqp = x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5}

. Also

vq = 1 - 3 x x + 3 x^{4} - x^{6}

, therefore

avqp =

x - x^{3}

+ \frac{3}{5} x^{5} - \frac{1}{7} x^{7}

. & by yethe same proceeding yethe proportion may bee still continued after this manner

x : x - \frac{1}{3} x^{3} : x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} : x - \frac{3}{3} x^{3} + \frac{3}{5} x^{5} - \frac{1}{7} x^{7} :

x - \frac{4}{3} x^{3} + \frac{6}{5} x^{5} - \frac{4}{7} x^{7} + \frac{1}{9} x^{9} : x - \frac{5}{3} x^{3} + \frac{10}{5} x^{5} - \frac{10}{7} x^{7} + \frac{5}{9} x^{9} - \frac{1}{11} x^{11} :

x - \frac{6}{3} x^{3} + \frac{15}{5} x^{5} - \frac{20}{7} x^{7} + \frac{15}{9} x^{9} - \frac{6}{11} x^{11} + \frac{1}{13} x^{3} : x - \frac{7}{3} x^{3} + \frac{21}{5} x^{5} - \frac{35}{7} x^{7} + \frac{35}{9} x^{9} - \frac{21}{11} x^{11}

+ \frac{7}{13} x^{13} - \frac{1}{15} x^{15}

. &c. And if yethe meane termes be inserted it will bee

x : x - : x - \frac{1}{3} x^{3} : x - \frac{3}{6} x^{3} + : x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} : x - \frac{5}{6} x^{3} + \frac{2}{5} x^{5} -

x : x - \frac{1}{6} x^{3} : x - \frac{1}{3} x^{3} : x

The first letters

x

run in this progression

1.1.1.1.1 .

&c. yethe 2d

x^{3}

in this

\frac{- 1}{3} . \frac{0}{3} . \frac{1}{3} . \frac{2}{3} . \frac{3}{3} . \frac{4}{3} . \frac{5}{3}

&c yethe 3d

x^{5}

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

. yethe 4th

x^{7}

this

\begin{matrix} \begin{matrix} 1^{st} & . & + x \times 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . \\ 2^{d} & . & - \frac{x^{3}}{3} \times 0 & . & 0 + 1 = 1 & . & 1 + 1 = 2 & . & 2 + 1 = 3 & . & 3 + 1 = 4 & . & 4 + 1 = 5 & . & 6 & . & 7 & . & 8 & . & 9 & . & 10 & . \\ 3^{d} & . & + \frac{x^{5}}{5} \times 0 & . & 0 + 0 = 0 & . & 0 + 1 = 1 & . & 1 + 2 = 3 & . & 3 + 3 = 6 & . & 6 + 4 = 10 & . & 15 & . & 21 & . & 28 & . & 36 & . & 45 & . \\ 4^{th} & . & - \frac{x^{7}}{7} \times 0 & . & 0 + 0 = 0 & . & 0 + 0 = 0 & . & 0 + 1 = 1 & . & 1 + 3 = 4 & . & 4 + 6 = 10 & . & 20 & . & 35 & . & 56 & . & 84 & . & 120 & . \\ 5^{} & . & + \frac{x^{9}}{9} \times 0 & . & 0 + 0 = 0 & . & 0 + 0 = 0 & . & 0 + 0 = 0 & . & 0 + 1 = 1 & . & 1 + 4 = 5 & . & 15 & . & 35 & . & 70 & . & 126 & . & 210 & . \\ 6^{} & . & - \frac{x^{11}}{11} \times 0 & . & 0 + 0 = 0 & . & 0 + 0 = 0 & . & 0 + 0 = 0 & . & 0 + 0 = 0 & . & 0 + 1 = 1 & . & 6 & . & 21 & . & 56 & . & 126 & . & 252 & . \\ 1^{st} . & ^{∗} 2^{d} . & ^{∗} 3^{d} & ^{∗} 4^{th} & ^{∗} 5^{th} & ^{∗} 6^{th} & 1 & . & 7 & . & 28 & . & 84 & . & 210 & . \\ ^{∗} & 7^{th} & . & 1 & . & 8 & . & 36 & . & 120 & . \\ ^{∗} & 8^{th} & 1 & . & 9 & . & 45 & . \\ ^{∗} & 9^{th} & 1 & . & 10 & . \\ ^{∗} & 10^{th} & 1 & . \\ ^{∗} & 11^{th} & . \end{matrix} \end{matrix}

Now if the meane termes in these progressions can bee callculated yethe first of ymthem gives yethe area

aeqp

. Which is thus done

\begin{matrix} \begin{array}{c} 1^{st} & + x \times 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . \\ 2^{d} & . & - \frac{x^{3}}{3} \times 0 & . & \frac{1}{2} & . & 1 & . & \frac{3}{2} & . & 2 & . & \frac{5}{2} & . & 3 & . & \frac{7}{2} & . & 4 & . & \frac{9}{2} & . & 5 & . & \frac{11}{2} & . & 6 & . \\ 3^{d} & . & + \frac{1}{5} x^{5} \times 0 & . & - \frac{1}{8} & . & 0 & . & \frac{3}{8} & . & 1 & . & \frac{15}{8} & . & 3 & . & \frac{35}{8} & . & 6 & . & \frac{63}{8} & . & 10 & . & \frac{99}{8} & . & 15 & . \\ 4 & . & - \frac{1}{7} x^{7} \times 0 & . & + \frac{1}{16} & . & 0 & . & - \frac{1}{16} & . & 0 & . & \frac{5}{16} & . & 1 & . & \frac{35}{16} & . & 4 & . & \frac{105}{16} & . & 10 & . & \frac{231}{16} & . & 20 & . \\ 5 & . & + \frac{1}{9} x^{9} \times 0 & . & - \frac{5}{128} & . & 0 & . & \frac{3}{128} & . & 0 & . & - \frac{5}{128} & . & 0 & . & \frac{35}{128} & . & 1 & . & \frac{315}{128} & . & 5 & . & \frac{1155}{128} & . & 15 & . \\ 6 & . & - \frac{1}{11} x^{11} \times 0 & . & \frac{7}{256} & . & 0 & . & - \frac{3}{256} & . & 0 & . & \frac{3}{256} & . & 0 & . & - \frac{7}{256} & . & 0 & . & \frac{63}{256} & . & 1 & . & \frac{693}{256} & . & 6 & . \\ 7 & . & \frac{1}{13} x^{13} \times 0 & . & - \frac{21}{1024} & . & 0 & . & \frac{7}{1024} & . & 0 & . & - \frac{5}{1024} & . & 0 & . & \frac{7}{1024} & . & 0 & . & - \frac{21}{1024} & . & 0 & . & \frac{231}{1024} & . & 1 & . & \frac{3003}{1024} & . \end{array} \end{matrix}

Soe ytthat

1 \times x - \frac{1}{2} \times \frac{1}{3} \times x^{3} - \frac{1}{8} \times \frac{1}{5} \times x^{5} - \frac{1}{16} \times \frac{1}{7} \times x^{7} - \frac{5}{128} \times \frac{1}{9} \times x^{9}

&c. is yethe area,

apqe

ytthat is

\frac{0}{0} \times x - \frac{0}{0} \times \frac{1}{2} \times \frac{1}{3} x^{3} - \frac{0}{0} \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{5} x^{5} - \frac{0}{0} \times \frac{1}{2} \times \frac{1}{4} \times \frac{3}{6} \times \frac{1}{7} x^{7} - \frac{1 \times 3 \times 5 x^{9}}{2 \times 4 \times 6 \times 8 \times 9}

&c: The progression may be deduced from hence

\frac{0 \times 1 \times - 1 \times 3 \times - 5 \times 7 \times - 9 \times 11}{0 \times 2 \times 4 \times 6 \times 8 \times 10 \times 12 \times 14}

. &c 19 (9 Soe ytthat if yethe given sine bee

pq = te = x

. & if yethe Radius

pc = 1

. Then is yethe superficies

ape = x - x \sqrt{1 - x x} - \frac{1}{6} x^{3} - \frac{1}{40} x^{5} - \frac{x^{7}}{112} - \frac{5 x^{9}}{1152} - \frac{7 x^{11}}{2816}

&c: And yethe area

ade = \frac{x^{3}}{6} + \frac{x^{5}}{40} + \frac{x^{7}}{112} + \frac{5 x^{9}}{1152} + \frac{7 x^{11}}{2816} + \frac{21 x^{13}}{13312}

+ \frac{11 x^{15}}{10240} + \frac{429 x^{17}}{557056} + \frac{715 x^{19}}{1245184} + \frac{2431 x^{21}}{5505024}

&c. By wchwhich meanes yethe angle

ape

is easily found for

aecpa ∶ ∠ apc = 90 ∷ ape ∶ ∠ ape

. The same may bee thus done.

adp = \frac{x}{2}

. Or

2 adp = x

2 afp = x + \frac{x^{3}}{3}

2 ahp = x + \frac{2 x^{3}}{3} - \frac{3 x^{5}}{5}

. And

2 avp = x + x^{3} - \frac{9 x^{5}}{5} + \frac{5}{7} x^{7}

. &c. as in this order

x . x + \frac{1}{3} x^{3} . x + \frac{2}{3} x^{3} - \frac{3}{5} x^{5} . x + x^{3} - \frac{9}{5} x^{5} + \frac{5 x^{7}}{7} . x + \frac{4 x^{3}}{3} - \frac{18 x^{5}}{5} + \frac{20 x^{7}}{7} -

- \frac{7}{9} x^{9} . x + \frac{5}{3} x^{3} - \frac{30}{5} x^{5} + \frac{50 x^{7}}{7} - \frac{35 x^{9}}{9} + \frac{9 x^{11}}{11} . x + \frac{6 x^{3}}{3} - \frac{45 x^{5}}{5} + \frac{100}{7} x^{7} - \frac{105 x^{9}}{9} +

+ \frac{54 x^{11}}{11} - \frac{11 x^{13}}{13} . x + \frac{7 x^{3}}{3} - \frac{63 x^{5}}{5} + \frac{175 x^{7}}{7} - \frac{245 x^{9}}{9} + \frac{189}{11} x^{11} - \frac{77 x^{13}}{13} + \frac{13 x^{15}}{15}

. &c Which progression wthwith their intermediate termes may bee thus exhibited. By wchwhich it may appeare ytthat if

pe = 1

pq = x

. ynthen

aep = \frac{1}{2} x + \frac{x^{3}}{12} + \frac{3 x^{5}}{80} + \frac{5 x^{7}}{224} + \frac{35 x^{9}}{2304}

&c. And yethe area

aep

given gives yethe angle

ape

for

apce ∶ ∠ apc = 90^{d} ∷ ape ∶ ∠ ape

Likewise yethe angle

ape

given its sign may bee found hereby &c

\begin{matrix} \begin{matrix} 2 \times adpa = & 2 \times aep = & 2 afp = & 2 agp = & 2 ahp = & 2 aip = & 2 avp = \end{matrix} \\ \begin{array}{c} + x in 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . & 1 & . \\ \frac{x^{3}}{3} in 0 & . & \frac{1}{2} & . & 1 & . & \frac{3}{2} & . & 2 & . & \frac{5}{2} & . & 3 & . \\ - \frac{3}{5} x^{5} in 0 & . & - \frac{1}{8} & . & 0 & . & \frac{3}{8} & . & 1 & . & \frac{15}{8} & . & 3 & . \\ + \frac{5}{7} x^{7} in 0 & . & \frac{1}{16} & . & 0 & . & - \frac{1}{16} & . & 0 & . & \frac{5}{16} & . & 1 & . \\ - \frac{7}{9} x^{9} in 0 & . & - \frac{5}{128} & . & 0 & . & \frac{3}{128} & . & 0 & . & - \frac{5}{128} & . & 0 & . \\ + \frac{9}{11} x^{11} in 0 & . & \frac{7}{256} & . & 0 & . & - \frac{3}{256} & . & 0 & . & \frac{3}{256} & . & 0 & . \end{array} \end{matrix}

Note ytthat

\sqrt{1 - x x} = \frac{x}{2} - \frac{2 x^{3}}{6} - \frac{5 x^{5}}{80} - \frac{7 x^{7}}{224} - \frac{45 x^{9}}{2304} - \frac{77 x^{11}}{5632}

&c that is

\sqrt{1 - x x} = \frac{x}{2} - \frac{x^{3}}{3} - \frac{x^{5}}{16} - \frac{x^{7}}{32} - \frac{5 x^{9}}{256} - \frac{7 x^{11}}{512} - \frac{21 x^{13}}{2048}

&c. According to this progression

\frac{1}{2} \times \frac{1}{2} \times \frac{1}{4} \times \frac{3 \times 5 \times 7 \times 9 \times 11 \times 13 \times 15 \times 17}{6 \times 8 \times 10 \times 12 \times 14 \times 16 \times 18 \times 20}

&c. Note also ytthat yethe segment

ae = \frac{x^{3}}{12} + \frac{3 x^{5}}{80} + \frac{5 x^{7}}{224} + \frac{35 x^{9}}{2304}

. &c.

aep = \frac{x}{2} + \frac{x^{3}}{12} + \frac{3 x^{5}}{80} + \frac{5 x^{7}}{224} + \frac{35 x^{9}}{3304} + \frac{63 x^{11}}{5632} + \frac{231 x^{13}}{26624} + \frac{143 x^{15}}{20480} + \frac{6435 x^{17}}{1114112} + \frac{12155 x^{19}}{2490368} + \frac{46189 x^{21}}{11010048}

. If

pq = a

qd = x

pc = 1 = pb

db = \sqrt{1 - a a - 2 a x - x x}

ynthen yethe areas of yethe lines in this progression.

1 - \sqrt{1 - a a - 2 a x - x x} . 1 - a a - 2 a x - x x . \bar{1 - a a - 2 a x - x x}} \frac{3}{2} .

1 - 2 a a - 4 a x - 2 x x + a^{4} + 4 a^{3} x + 6 a a x x + 4 a x^{3} + x^{4} . * .

1 - 3 a a + 3 a^{4} + 8 a^{3} x + 8 a a x x

& (supposeing also

\overline{b - 2 a x - x x}} \frac{3}{2} . \begin{matrix} b b - 4 a b x - 2 b x x + 4 a x^{3} + x^{4} \\ + 4 a a \end{matrix} . *

\begin{matrix} b^{3} - 6 a b b x - 3 b b x x + 8 a b x^{3} + 3 b x^{4} - 6 a x^{5} - x^{6} \\ + 12 a a b x^{2} - 8 a^{3} x^{3} - 12 a a x^{4} \\ + 4 a b x^{3} \end{matrix}

. &c 20 (11 To square yethe Hyperbola. Epitome Geometriæ So if

nadm

is an Hyperbola. &

cp = 1 = pa

pq = x

qd, qe, qf, qg

= y

. &

\frac{1}{1 + x} = y = dq

1 = y = eq

1 + x = y = qf

1 + 2 x - x x = y = qg

1 + 3 x + 3 x x + x^{3}

1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}

. &c. ThereTheir squares are.

* . x . x + \frac{x x}{2} . x + \frac{2 x x}{2} + \frac{x^{3}}{3} . x + \frac{3 x x}{2} + \frac{3 x^{3}}{3} + \frac{x^{4}}{4} .

x + \frac{4 x x}{2} + \frac{6 x^{3}}{3} + \frac{4 x^{4}}{4} + \frac{x^{5}}{5} . x + \frac{5 x x}{2} + \frac{10 x^{3}}{3} + \frac{10 x^{4}}{4} + \frac{5 x^{5}}{5} + \frac{x^{6}}{6}

. &c As in yethe following table. By whose first terme is represented yethe square of yethe Hyperbola, viz ytthat it is

\begin{array}{l} \begin{matrix} \begin{matrix} \begin{array}{l} \begin{array}{l} a & a & a & a \\ p & p & p & p \\ q & q & q & q \\ d & e & f & g \\ = & = & = & = \end{array} \\ \begin{array}{l} x \times 1 . & 1 . & 1 . & 1 . & 1 . & 1 . & 1 . & 1 . & 1 . \\ \frac{x^{2}}{2} \times - 1 . & 0 . & 1 . & 2 . & 3 . & 4 . & 5 . & 6 . & 7 . \\ \frac{x^{3}}{3} \times 1 . & 0 . & 0 . & 1 . & 3 . & 6 . & 10 . & 15 . & 21 . \\ \frac{x^{4}}{4} \times - 1 . & 0 . & 0 . & 0 . & 1 . & 4 . & 10 . & 20 . & 35 . \\ \frac{x^{5}}{5} \times 1 . & 0 . & 0 . & 0 . & 0 . & 1 . & 5 . & 15 . & 35 . \\ \frac{x^{6}}{6} \times - 1 . & 0 . & 0 . & 0 . & 0 . & 0 . & 1 . & 6 . & 21 . \\ \frac{x^{7}}{7} \times 1 . & 0 . & 0 . & 0 . & 0 . & 0 . & 0 . & 1 . & 7 . \end{array} \\ \begin{array}{l} 0000.00000.00000 . \\ 3333.33333.33333.33333.33333.3 \\ 0000.00000.00000.00000.00000.0 \\ 8571.42857.14285.71428.57142.8 \\ 1111.11111.11111.11111.11111.1 \\ 9090.90909.09090.90909.09090.9 \end{array} \end{array} & \begin{matrix} \begin{matrix} x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} - \frac{x^{6}}{6} + \\ + \frac{x^{7}}{7} - \frac{x^{8}}{8} + \frac{x^{9}}{9} - \frac{x^{10}}{10} . &c. \end{matrix} \\ \begin{matrix} \begin{array}{r} \begin{array}{r} \begin{matrix} As if x = \frac{1}{10} . 
or cq = \frac{11}{10} = 1,1 . \\ y^{n} is dapq = \frac{1}{10} - \frac{1}{200} + \frac{1}{3000} - \\ - \frac{1}{40000} + \frac{1}{500000} - \frac{1}{6000000} . &c \end{matrix} \end{array} \end{array} \\ \begin{array}{r} y^{t} is apqd = 0,10000.00000.000000 \\ \frac{1}{3} x^{3} = + 0,00033.33333.333333 \\ \frac{1}{5} x^{5} = + 0,00000.20000.000000 \\ \frac{1}{7} x^{7} = + 0,00000.00142.857142 \\ \frac{1}{9} x^{9} = + 0,00000.00001.111111 \\ \frac{1}{11} x^{11} = + 0,00000.00000.009090 \\ \frac{1}{13} x^{13} = + 0,00000.00000.000076 \\ \underline{\frac{1}{15} x^{15} = + 0,00000.00000.000000} \\ 0,10033.53477.310755 . \\ Summa \end{array} \end{matrix} \end{matrix} \end{matrix} \\ \begin{array}{l} \begin{array}{l} 9230.76923.07692.30769.23076.9230769230 \\ 6666.66666.66666.66666.66666.6666666666 . \\ 0058.82352.94117.64705.88235.2941176470 . \\ 0000.52631.57947.36842.10526.3157947368 \\ 0000.00476.19047.61904.76190.4761904761 . \\ 0000.00004.34782.60869.56521.7391304347 . \end{array} \end{array} \end{matrix} \\ \begin{array}{l} \begin{array}{l} 0000.00000.04000.00000.00000.0000000000 \\ 0000.00000.000 == ===== ===== . \\ 37.03703.70370.3703703703 \\ 0000.00000.00000.34482.75862.0689655172.4137931034.48275862 \\ 0000.00000.00000.00 === ===== = \\ 322.58064.5161290322.5806451612.90322580 \\ 0000.00000.00000.00003.03030.3030303030.3030303030.30303030 \\ 0 02857.1428571428.5714285714.28571428 \\ 0000.00000.00000.00000 00027.0270270270.2702702702.70270270 \\ 0000.00000.00000.00000 00000.2564010256.4010256401.02564010 \\ 0000.00000.00000.00000 00000.0024390243.9024390243.90243902 \end{array} \\ \begin{matrix} = 8063.57265.52007.40736.63159.415063 &c = summæ \end{matrix} \\ \begin{array}{r} - 0.00500.00000.00000.00000.00000.00000.00000.00000.00000.00000.0 \\ 2.50000.00000.00000.00000.00000.00000.00000.00000.00000.0 \\ 1666.66666.66666.66666.66666.66666.66666.66666.66666.6 \\ 12.50000.00000.00000.00000.00000.00000.00000.00000.0 \\ 10000.00000.00000.00000.00000.00000.00000.00000.0 \\ 83.33333.33333.33333.33333.33333.33333.33333.3 \\ 71428.57142.85714.28571.42857.14285.71428.5 \\ 630.55555.55555.55555.55555.55555.55555.5 \\ 5045.45454.54545.45454.54545.45454.5 . \\ 41666.66666.66666.66666.66666.6 \\ 384.61538.46153.84615.38461.5 \\ 3.57142.85714.28571.42857.1 \\ 3364.58333.33333.33333.3 \\ 29411.76470.58823.5 \end{array} \\ - 0.00502.51679.26750.72059.17744.28779.27385.30147.14044.12586 . \end{array} \end{array}

cui addendum

\begin{matrix} \begin{array}{r} - 277.77777.77777.7 \\ - 2.63157.89421.0 \\ - 02523.80952.4 \\ - 22727.3 \\ - 217.4 & that is \\ 2.1 \end{array} \\ \begin{matrix} - 280.43459.71098.0 \end{matrix} \\ \begin{matrix}  \end{matrix} \end{matrix}

And so yethe summe will bee

\begin{matrix} \begin{array}{r} + 0.10033.53477.31075.58063.57265.52007.40736.63159.41506.3 \\ - 0.00502.51679.26750.72059.17144.28779.27385.30427.57503.8 \end{array} \\ \begin{matrix} 0.09530.01798.04324.86004.40121.23228.13351.32731.84002.5 \end{matrix} \\ \begin{matrix}  \end{matrix} \end{matrix}

wchwhich is yethe quantity of yethe area

adpq

. If

cpab = 1

. &

cp =

ab = 10 pq

qde || ap || bc = ap

. In like manner if I make

x = \frac{1}{100} = pq

. The opperacotion followeth.

\begin{matrix} \begin{array}{r} 0, 01000.03333.33333.33333.33333.33333.33333.33333.33333.3 \\ + \frac{1}{5} x^{5} + \frac{1}{7} x^{7} = 33333.20001.42857.14285.71428.57142.85714.28571.40 \\ == ===== ===== ===== ===== = \\ \frac{1}{9} x^{9} + \frac{1}{11} x^{11} = 11.11202.02020.20202.02020.20202.0 \\ \frac{1}{13} x^{13} + \frac{1}{15} x^{15} = 769.23076.92307.69230.7 \\ \frac{1}{17} x^{17} = 6666.66666.66666.6 \\ \frac{1}{19} x^{19} = 58823.52941.1 \\ \frac{1}{21} x^{21} = 5.26315.7 \\ 47.6 \end{array} \\ \begin{array}{r} + 0, 01000.03333.53334.76201.58821.07551.40422.38870.97309 . \end{array} \\ \begin{array}{r} - 0, 00005.00025.00166.66666.66666.66666.66666.66666.66666.6 . \\ - \frac{1}{8} x^{8} - \frac{1}{10} x^{10} - \frac{1}{12} x^{12} = - 1250.10000.83333.33333.33333.33333.3 \\ - \frac{1}{14} x^{14} = - 7.14285.71428.57142.8 . \\ - \frac{1}{16} x^{16} - \frac{1}{18} x^{18} = - 62.50555.55555.5 \\ ===== ===== = \\ - \frac{1}{20} x^{20} = - 5000.4 . \end{array} \\ \begin{array}{r} - 0, 00005.00025.00166.67916.76667.50007.14348.21984.17699 . \end{array} \\ \begin{array}{r} + 0, 00995.03308.53168.08284.82153.57544.26074.16886.79610 \end{array} \\ \begin{matrix}  \end{matrix} \end{matrix}

wchwhich is yethe quantity of yethe area

apqd

100 p = cp

. and

abcp = 1

21 (13

y = db

x = ba

a a y = x^{3}

a

b

+

y

z

= y

z = bc

a a b + a a

y

z

= x^{3}

a

b

-

y

z

= y =

z = bf

a a b - a a

y

z

= x^{3}

y

z

-

a

b

= y =

z = bg

a a

y

z

-

a

a b = x^{3}

x = d + ξ

ξ = ah

a a

y

- d^{3} - 3 d d ξ - 3 d ξ ξ - ξ^{3} = 0

\begin{matrix} a a b + a a z \\ a a b - a a z \\ a a z - a a b \end{matrix}} = \begin{matrix} d^{3} + 3 d d ξ + 3 d ξ ξ + ξ^{3} \end{matrix}

x = b - ξ

aq = ξ

\begin{matrix} a a y \\ a a b \pm a a z \\ a a z - a a b \end{matrix}} = \begin{matrix} d^{3} - 3 d d ξ + 3 d ξ ξ - ξ^{3} \end{matrix}

x = ξ - b

ak = x

\begin{matrix} a a y \\ a a b \pm a a z \\ a a z - a a b \end{matrix}} = \begin{matrix} ξ^{3} - 3 d ξ ξ - 3 d d ξ - d^{3} \end{matrix}

na = ξ

nd = z

. Or.

ξ^{3} - a^{2} z + b b ξ

ab ∶ an ∷ a ∶ b

. &.

ab ∶ nb ∷ a ∶ c

then

ξ^{3} = \frac{b^{3}}{a} z - \frac{b b c}{a} ξ

. or if

abn

is a right angle. &

ξ = d + ζ

ζ = mn

d^{3} + 3 d^{2} ζ

+ b b c ζ

+ 3 d ζ ζ + ζ^{3} - \frac{b^{3}}{a} z + \frac{b b c d}{a}

22 (15

db = x

ba = y

a a x = y^{3}

x = b + z

z = bc

y = c + ξ

ξ = ah

\begin{matrix} a a b + a a z = \\ (x = b - z . z = bf .) \\ a a b - a a z = \\ (x = z - b . z = bg) \\ a a z - a a b = \end{matrix}} \begin{array}{l} ξ^{3} + 3 c ξ^{2} + 3 c c ξ + ξ^{3} . y = c - ξ . ξ = aq \\ c^{3} - 3 c c ξ + 3 c ξ^{2} - ξ^{2} . y = ξ - c . ξ = ak . \\ ξ^{3} - 3 c ξ^{2} + 3 ξ c^{2} - c^{3} . \end{array}

na = y

d

n

= x

ab ∶ an ∷ a ∶ b

ab ∶ nb ∷ a ∶ c

. &

y^{3} = \frac{b^{3}}{a} x - \frac{b b c}{a} y

. Or

d^{2} x = ε ε y + y^{3}

\begin{matrix} (x = z + o . z = nc) \\ d^{2} z + d^{2} o = \\ (x = z - o . z = gn) \\ d d z - d d o = \\ (x = o - z . z = nf) \\ d d o - d d z = \\ d d z \end{matrix}} \begin{array}{l} ε ε y + y^{3} . (y = n + ξ . ξ = na) \\ ξ^{2} n + ε ε ξ + n^{3} + 3 n n ξ + 3 n ξ^{2} + ξ^{3} . \\ (y = n - ξ . ξ = as) \\ ε ε n - ε ε ξ + n^{3} - 3 n n ξ + 3 n ξ^{2} - ξ^{3} . \\ (y = ξ - n . ξ = av .) \\ ε ε ξ - ε ε n + ξ^{3} - 3 n ξ^{2} + 3 ξ n^{2} - n^{3} . \end{array}

al = y

dl = x

ab ∶

al

∷ a ∶ b

al ∶ bl ∷ b ∶ c

. whence

y^{3} = \frac{x b^{3} + y b^{2} c}{a}

y^{3} = \frac{y b^{2} c}{a} =

y^{3} - ε ε y = d d x

&c: as before onely varying yethe signes at

εεn

εεξ

ao = y

do = x

a ∶ b ∷ bd ∶ do

b ∶ c ∷ do ∶ ob

\frac{a^{3}}{b} x = y^{3} - \frac{3 c x y y}{b} + \frac{3 c c x x y}{b b} - \frac{c^{3} x^{3}}{b b}

. Dr Wiallis in a letter to Sr Kenelme Digby promiseth yethe squareing of yethe Hyperbola by finding a meane proportion twixt

1

, &

\frac{5}{6}

in the progression

1, \frac{5}{6}, \frac{31}{30}, \frac{209}{140}, \frac{1471}{630}, \frac{10625}{2772}

&c. 23 (17 The resolution of cubick equations out of Dr Wallis in his dedication before Meibomius confuted suppose

x = ♉ a ♉ e

. ynthen

x^{3} = ♉ a^{3} ♉ 3 a a e ♉ 3 a e e ♉ e^{3}

. or

x^{3} =

+

3 a e x ♉ a^{3} ♉ e^{3}

. that is making

a^{3} + e^{3} = q

. ynthen

x^{3} = 3 a e x ♉ c^{3}

. &

3 a e = p

. ynthen

x^{3}

\times

=

+ p x ♉ q

. Againe suppose

y

= ♉

b

a

\underset{\cap}{O}

e

Then

y^{3}

Againe suppose

x = a - e

. ynthen

x^{3} = a^{3} - 3 a a e + 3 a e e - e^{3}

. ytthat is making

a^{3} - e^{3} = ♉ q

, &

3 a e = p

, ynthen

x^{3} = - p x ♉ q

. Then in the first of these

p = 3 a e

. or

\frac{p}{3 e} = a

. or

\frac{p^{3}}{27 e^{3}} = a^{3} = q - e^{3}

. Therefore

e^{6} = q e^{3} - \frac{p^{3}}{27}

. &

e^{3} = \frac{1}{2} q ♉ \sqrt{\frac{1}{4} q q - \frac{p^{3}}{27}}

. & by yethe same reason

a^{3} = \frac{1}{2} q \underset{\cap}{O} \sqrt{\frac{1}{4} q q - \frac{p^{3}}{27}}

where yethe irrationall quantitys have. divers signes otherwise

a^{3} + e^{3} = q

would bee false. Soe that

x = ♉ a ♉ e = ♉ \sqrt{c : \frac{1}{2} q \underset{\cap}{O} \sqrt{\frac{1}{4} q q - \frac{1}{27} p^{3}}} ♉ \sqrt{c : \frac{1}{2} q ♉ \sqrt{\frac{1}{4} q q - \frac{1}{27} p^{3}}}

. is a rule for resolving yethe equation

x^{3} * - p \underset{\cap}{O} q = 0

, when it hath but one roote ytthat is when it may be generated according to the supposition

x = ♉ a ♉ e

. &c. By yethe same reason

x^{3} * + p x \underset{\cap}{O} q

. may be resolved by this rule

x = a - e = \sqrt{c : \frac{1}{2} q ♉ \sqrt{\frac{1}{4} q q + \frac{1}{27} p^{3}}} - \sqrt{c : \frac{1}{2} q \underset{\cap}{O} \sqrt{\frac{1}{4} q q + \frac{1}{27} p^{3}}}

. But here ofbserve ytthat Dr Wallis would hArgue ytthat since in the first of these two cases fsometimetimes (viz when yethe equation hath 3 reall rootes) yyethe first rule faileth as it were impossibleimpossible for yethe equation to have rootes when yet it hath, therefore yethe fault is in Algebra. & therefore when Analises Analysis leads us to an impossibilityimpossibility wee ought not to conclude yethe thing absolutely imposible, untill wee have tryed all yethe ways ytthat may bee. But let me answer ytthat yethe fault is inot in yethe Analysis in this example, but in his opperation. for when yethe equation

x^{3} * + p x ♉ q = 0

, hath 3 roots hee supposeth it to have but one roote viz

x = ♉ a ♉ e

. but since yethe Equation cannot be then generated according to ytthat supposition it is imposeibleimpossible it should be relolvedresolved by it. In like manner hee sayeth ytthat Algebra representehteth a thing possible when tis not so as in this examlple, in yethe △triangle

abc

, make

ab = 1

bc = 2

ac = 4

. Then to find

dc = x

, worke thus,

ad = 4 - x

bd \times bd = 1 - 16 + 8 x - x^{2} = 4 - x^{2}

therefore

8 x = 19

. or

x = \frac{19}{8}

. In wchwhich opperacon all things proceede as possible though they are not soe for

ac

is greater ynthan

ab + bc

. yet I answer ytthat if yethe opperation & conclusion be compared together yethe absurdity will appeare. for in yethe equation

bd \times bd = 4 - x x = 4 - \frac{361}{64} = \frac{256 - 361}{64}

bd \times bd = \frac{- 105}{8}

. but it is impossible ytthat a □square number should be negative. Thus

x = \sqrt{- b}

is impossible. square it & tis

x x = - b

. Againe, & tis

x^{4} = b b

. Extract yethe roote & tis

x x = b

x = \sqrt{b}

. wchwhich is possible. The reason of this proceeding Event is ytthat

x^{4} - b b = 0

hath two possible rootes viz

x = \sqrt{b}

x = - \sqrt{b}

. & two impossible viz:

x = \sqrt{- b}

x = - \sqrt{- b}

. Thus yethe valors of

x^{8} - a^{8} = 0

are

x = a

- a

\sqrt{- a a}

- \sqrt{- a a}

\sqrt{4 : - a^{4}}

- \sqrt{4 : - a^{4}}

\sqrt{- \sqrt{- a^{4}}}

- \sqrt{- \sqrt{- a^{4}}}

. 24 19 Dr Wallis in a letter to Sr Kenelme Digby teacheth how to find yethe center of gravity in divers lines first when their position is as in this figure. Suppose

ad

yethe Axis,

a

their vertex & as

1

to yethe series of yethe progresion cons. Then saying, as

1

to yethe index of yethe line increased by an unite (vide pag 2daam) so

cd

ca

Then

c

is their center of gravity. The Demonstracoation. Let

p

bee yethe index of yethe series according to wchwhich yethe odinately aplyed lines (parallell to

db

) increase, ynthen

1 ∶ p + 1 ∷

area of yethe line

∶

nmbq

. yethe distances of those ordinate lines from yethe vertex

a

are equall to yethe intercepted diameters & therefore a primanary series (whos index is

1

. & since supposing

a

yethe center of yethe ballance yethe whole weight of yethe surface or figure is composed of its magnitude & distance from yethe center and therefore yethe index of all its moments or whole weight is

p + 1

, viz: yethe aggregate of yethe other two. Therefore as all its mom moments (or yethe weight of the figure in its site in respect of yethe center

a

are to soe many of yethe greatest (or to yethe weight of yethe ▭rectangle

nmbq

hung on yethe point

d

) soe is

1

, to

p + 2

. and if

ap ∶ ad ∷ 1 ∶ p + 2

, then

nmbq

hung on yethe point

q

shall counterballance yethe figure in its site &c therefore if

ac ∶ cd ∷ p + 1 ∶ 1

c

shall be yethe center of gravity of those figures. Also as the figure is now put extending infinitely towards

δ

- 2 p + 1 ∶ -

p

+ 1 ∷ am ∶ ac

m

being yethe center of

qnbd

ynthen

c

shall bee yethe center of gravity of yethe whole figure

qndbδ

. Demonstration sinccesince yethe lines parallell to

aδ

increase in series reciproclally proportionall their index is

- p

& since yethe halfes of those lines increasincrease in yethe same proportion their index is

- p

. whose extremitys or middle points of yethe whole lines (suposing

a

yethe center of yethe ballance) are theire centers of gravity, their distances from

a

being proportionall to yethe lines whose centers they are & consequently their index is

- p

& since all yethe moments (or whole weight of yethe figure) increase in a proportion compounded of yethe proportion of yethe magnitudes & distances of yethe lines from yethe center

a

, they will be in a duplicate proportion of yethe lines magnitudes that is a series reciprocall series whose index is

- 2 p

. Therefore yethe figure is to yethe inscribed parallelogram as

1

1 - p

. & all its moments or whole weight of the Para in this its site to the weight of yethe pgrparallelogram as

1

1 - 2 p

. Therefore if,

am ∶ ap ∷ 1 - 2 p ∶ 1

, the paralelograamparallelogram hanging on yethe point

p

shall counterballance yethe whole figure in its site &c: whence yethe point

c

may be found easily, viz

am ∶ ac ∷ 1 - 2 p ∶ 1 - p