Mathematical Notebook Isaac Newton c. 32,634 words Newton Project Brighton 2011 Newton Project, Sussex University

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 $5708$ is $\left(5\right)$, & $\left(5\right)$ 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 $258363524$ by $5A$$qq$$+10Ac+10Aq+5A$ after such a manemaner ytthat $5AqqE+10AcEq+10AqEc+5AEqq+Eqc$ may be conteined in yethe number yethe product of ytthat division shall be E =
The extraction of yethe sqaresquare roote The extraction of yethe cube roote $\begin{array}{cc}\hfill \text{The cube to be resolved}\phantom{\rule{1em}{0ex}}15\underset{\text{.}}{7}& 46\underset{\text{.}}{4}\phantom{\rule{1em}{0ex}}\text{(54}\hfill \\ \hfill \text{The cube to be subducted}\phantom{\rule{1em}{0ex}}125& \text{whose roote is}\phantom{\rule{1em}{0ex}}A=5\hfill \\ \hfill \text{The remainder for}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{finding}\phantom{\rule{1em}{0ex}}32& 464\phantom{\rule{1em}{0ex}}\text{of E}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{The divisors for}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{finding}\\ \text{of (E)}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{seacond side.}\end{array}& \left\{\begin{array}{cc}\phantom{0}& 7\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{cc}\begin{array}{cc}5& \phantom{0}\\ 15& \phantom{0}\end{array}& \begin{array}{c}3Aq\hfill \\ 3A\hfill \end{array}\end{array}\hfill \\ \hfill \text{The sume of}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{divisors}\phantom{\rule{1em}{0ex}}\phantom{0}7& 65\phantom{0}\phantom{\rule{1em}{0ex}}\text{}\hfill \\ \hfill \begin{array}{cc}\begin{array}{}\text{Sollids to be substracted}\end{array}& \left\{\begin{array}{cc}\phantom{0}& 30\\ \phantom{0}& 2\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{cc}\begin{array}{ccc}0& \phantom{0}& \phantom{0}\\ 40& \phantom{0}\\ \phantom{0}64\end{array}& \begin{array}{c}3AqE\hfill \\ 3AEq\hfill \\ Ec\hfill \end{array}\end{array}\hfill \\ \hfill \text{The sume of those}\phantom{\rule{1em}{0ex}}32& 464\phantom{\rule{1em}{0ex}}\text{sollids}\hfill \\ \hfill \text{The remainder}\phantom{\rule{1em}{0ex}}00& 000\phantom{\rule{1em}{0ex}}\text{}\hfill \end{array}$ The extraction of yethe square square roote $\begin{array}{cc}\hfill \text{The square-square}\phantom{\rule{1em}{0ex}}3\underset{\text{.}}{3}& 177\underset{\text{.}}{6}\phantom{\rule{1em}{0ex}}\text{(24}\hfill \\ \hfill \text{The square-squ: to be subduc:}\phantom{\rule{1em}{0ex}}16& =Aqq\\ \hfill \text{Remainder.}\phantom{\rule{1em}{0ex}}17& 1776\phantom{\rule{1em}{0ex}}\text{}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Divisors for finding}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\\ \text{seacond side E.}\phantom{\rule{0.5em}{0ex}}\end{array}& \left\{\begin{array}{cc}\phantom{0}& 3\\ \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{cc}\begin{array}{cc}2& \phantom{0}\\ 24& \phantom{0}\\ \phantom{0}\phantom{0}8& \phantom{0}\end{array}& \begin{array}{c}4Ac\hfill \\ 6Aq\hfill \\ 4A\hfill \end{array}\end{array}\hfill \\ \hfill \text{Theire su}\overline{\text{m}}\text{e}\phantom{\rule{1em}{0ex}}\phantom{0}3& 448\phantom{0}\phantom{\rule{1em}{0ex}}\text{}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Squ-Squares to be sub=}\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\\ \text{=ducted}\phantom{\rule{0.5em}{0ex}}\end{array}& \left\{\begin{array}{cc}& 12\\ & 3\\ & \phantom{0}\\ & \phantom{0}\end{array}\end{array}& \begin{array}{cc}\begin{array}{cc}8& \\ 84& \\ 512\phantom{0}& \\ \phantom{0}256& \end{array}& \begin{array}{c}4AcE\hfill \\ 6AqEq\hfill \\ 4AEc\hfill \\ Eqq\hfill \end{array}\end{array}\hfill \\ \hfill \text{Theire Su}\overline{\text{m}}\text{e}\phantom{\rule{1em}{0ex}}17& 1776\phantom{\rule{1em}{0ex}}\text{}\hfill \end{array}$ 3 The extraction of yethe Square-Cube roote $\begin{array}{cc}\hfill \text{The squ: cube to be resolved}\phantom{\rule{1em}{0ex}}7\underset{\text{.}}{9}& 6262\underset{\text{.}}{4}\phantom{\rule{1em}{0ex}}\text{(24}\hfill \\ \hfill \text{Substract}\phantom{\rule{1em}{0ex}}32& \phantom{\rule{1em}{0ex}}Aqc\hfill \\ \hfill \text{Remain}{\text{d}}^{\text{e}}\phantom{\rule{1em}{0ex}}47& 62624\phantom{\rule{1em}{0ex}}\text{}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Divisors}\phantom{\rule{0.5em}{0ex}}\end{array}& \left\{\begin{array}{cc}\phantom{5q}& 8\\ \phantom{\text{0}}& \phantom{0}\\ \phantom{\text{0}}& \phantom{0}\\ \phantom{\text{q}}& \phantom{0}\end{array}\end{array}& \begin{array}{cc}\begin{array}{cc}0\phantom{\text{5q}}& \\ 80& \phantom{\text{Ac}}\\ \phantom{0}40\phantom{0}& \phantom{\text{Aq}}\\ \phantom{0}\phantom{0}10& \phantom{\text{A}}\end{array}& \begin{array}{c}5Aqq\hfill \\ 10Ac\hfill \\ 10Aq\hfill \\ 5A\hfill \end{array}\end{array}\hfill \\ \hfill \text{The Sume of}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{divisors}\phantom{\rule{1em}{0ex}}\phantom{0}8& 8410\phantom{0}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Plano-Sollids to be}\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\\ \text{substracted}\phantom{\rule{0.5em}{0ex}}\end{array}& \left\{\begin{array}{cc}\phantom{\text{q}}& 32\\ \phantom{\text{q}}& 12\\ \phantom{\text{q}}& \phantom{0}2\\ \phantom{\text{q}}& \phantom{00}\\ \phantom{\text{q}}& \phantom{00}\end{array}\end{array}& \begin{array}{cc}\begin{array}{cc}0& \phantom{\text{q}}\\ 80& \phantom{\text{q}}\\ 560\phantom{0}& \phantom{\text{q}}\\ 2560\phantom{0}& \phantom{\text{q}}\\ \phantom{0}1024& \phantom{\text{q}}\end{array}& \begin{array}{c}5AqqE\hfill \\ 10AcEq\hfill \\ 10AqEc\hfill \\ 5AEqq\hfill \\ Eqc\text{.}\hfill \end{array}\end{array}\hfill \\ \hfill \text{Theire Su}\overline{\text{m}}\text{e}\phantom{\rule{1em}{0ex}}47& 62624\phantom{\rule{1em}{0ex}}\text{}\hfill \\ \hfill \text{Remainder}\phantom{\rule{1em}{0ex}}00& 00000\phantom{\rule{1em}{0ex}}\text{}\hfill \end{array}$ 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 $B.Cq.Dc.Fqq.Gqc.$ $Hcc$ &c yethe Power $P.Pq.Pc.Pqq$ &c & yethe Operation follows The analysis of Cubick Equations. The equation supposed $L{c}^{*}+30L=14356197$. $Lc+CqL=Pc$ $\begin{array}{ccc}\hfill \begin{array}{cc}\begin{array}{cc}\underset{\phantom{.}}{\phantom{0}}& \text{The square coëfficient}\\ \underset{\phantom{.}}{\phantom{0}}& \text{The cube affected to be}\end{array}& \begin{array}{cc}\phantom{0}& \underset{\phantom{.}}{\phantom{0}}\\ 1& \underset{.}{4}\end{array}\end{array}& \begin{array}{ccc}\underset{\phantom{.}}{\phantom{0}}& \phantom{0}& 3\\ 3& 5& \underset{.}{6}\end{array}& \begin{array}{cc}\begin{array}{ccc}\underset{.}{0}& \underset{.}{\phantom{0}}& \underset{.}{\phantom{0}}\\ 1& 9& \underset{.}{7}\end{array}& \begin{array}{c}\phantom{\rule{1em}{0ex}}\left(243\end{array}\end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Sollids to bee substracted}\end{array}& \left\{\begin{array}{cc}\phantom{0}& 8\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}& 6\end{array}& \begin{array}{cc}\begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\\ 0& \phantom{0}& \phantom{0}\end{array}& \begin{array}{c}=Ac\\ =ACq\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Theire su}\overline{\text{m}}\text{e}\phantom{\rule{0.5em}{0ex}}\end{array}& \begin{array}{cc}\phantom{0}& 8\end{array}\end{array}& \begin{array}{ccc}0& 0& 6\end{array}& \begin{array}{cc}\begin{array}{ccc}0& \phantom{0}& \phantom{0}\end{array}& \begin{array}{}\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Rests}\phantom{\rule{0.5em}{0ex}}\end{array}& \begin{array}{cc}\phantom{0}& 6\end{array}\end{array}& \begin{array}{ccc}3& 5& 0\end{array}& \begin{array}{cc}\begin{array}{ccc}1& 9& 7\end{array}& \begin{array}{c}\text{for finding}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}{\text{2}}^{\text{d}}\phantom{\rule{0.5em}{0ex}}\text{side}\end{array}\hfill \end{array}\hfill \\ \hfill \text{The extraction of}& \hfill {\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{seac}& \text{ond side}\hfill \\ \hfill \begin{array}{cc}\begin{array}{cc}\underset{\phantom{.}}{\phantom{0}}& \text{Coëfficient}\\ \underset{\phantom{.}}{\phantom{0}}& \text{The rest of}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{cube to be}\end{array}& \begin{array}{cc}\underset{\phantom{.}}{\phantom{0}}& \phantom{0}\\ \underset{\phantom{.}}{\phantom{0}}& 6\end{array}\end{array}& \begin{array}{ccc}\phantom{0}& \underset{\phantom{.}}{\phantom{0}}& \phantom{0}\\ 3& 5& \underset{.}{0}\end{array}& \begin{array}{cc}\begin{array}{ccc}3& \underset{.}{0}& \underset{.}{\phantom{0}}\\ 1& 9& \underset{.}{7}\end{array}& \begin{array}{cc}\text{or superior divisor}& \underset{\phantom{.}}{\phantom{0}}\\ \text{resolved}& \underset{\phantom{.}}{\phantom{0}}\end{array}\end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{The inferior divisors}\end{array}& \left\{\begin{array}{cc}\phantom{0}& 1\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}2& \phantom{0}& \phantom{0}\\ \phantom{0}& 6& \phantom{0}\end{array}& \begin{array}{cc}\begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\\ 0& \phantom{0}& \phantom{0}\end{array}& \begin{array}{c}3Aq\\ 3A\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Theire su}\overline{\text{m}}\text{e}\phantom{\rule{0.5em}{0ex}}\end{array}& \begin{array}{cc}\phantom{0}& 1\end{array}\end{array}& \begin{array}{ccc}2& 6& 0\end{array}& \begin{array}{cc}\begin{array}{ccc}3& 0& \phantom{0}\end{array}& \begin{array}{}\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Sollids to be sub=}\\ \text{stracted}\end{array}& \left\{\begin{array}{cc}\phantom{0}& 4\\ \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}8& \phantom{0}& \phantom{0}\\ 9& 6& \phantom{0}\\ \phantom{0}& 6& 4\\ \phantom{0}& \phantom{0}& 1\end{array}& \begin{array}{cc}\begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}& \phantom{0}\\ 2\hfill & 0\hfill & \phantom{0}\end{array}& \begin{array}{c}=3AqE\\ =3AEq\\ =Ec\\ \phantom{=}ECq\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Theire su}\overline{\text{m}}\text{e}\phantom{\rule{0.5em}{0ex}}\\ \phantom{\text{The superior part of}}\phantom{\rule{0.5em}{0ex}}\phantom{{\text{y}}^{\text{e}}}\phantom{\rule{0.5em}{0ex}}\phantom{\text{divisor}}\end{array}& \begin{array}{cc}\phantom{0}& 5\\ \phantom{0}\end{array}\end{array}& \begin{array}{ccc}8& 2& 5\\ \phantom{0}\end{array}& \begin{array}{cc}\begin{array}{ccc}2& 0& \phantom{0}\\ \phantom{0}\end{array}& \begin{array}{}\end{array}\hfill \end{array}\hfill \end{array}$ $\begin{array}{ccc}\hfill \begin{array}{cc}\begin{array}{c}\text{The superior part of}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{divisor}\\ \text{The remainder for finding}\end{array}& \begin{array}{cc}\phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\\ 5& 2& 4\end{array}& \begin{array}{cc}\begin{array}{ccc}\phantom{0}& 3& \underset{.}{0}\\ 9& 9& \underset{.}{7}\end{array}& \begin{array}{l}\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{square coefficient}\\ \phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{third side}\end{array}\end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{The inferior part of}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\\ \text{divisor}\end{array}& \left\{\begin{array}{cc}& \phantom{0}\\ & \phantom{0}\end{array}\end{array}& \begin{array}{ccc}1& 7& 2\\ \phantom{0}& \phantom{0}& \phantom{0}\end{array}& \begin{array}{cc}\begin{array}{ccc}8& \phantom{0}& \phantom{0}\\ 7& 2& \phantom{0}\end{array}& \begin{array}{c}3Aq\phantom{\rule{0.5em}{0ex}}\text{that is}\phantom{\rule{0.5em}{0ex}}3×24×24\\ 3A\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}3×24\text{.}\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{The su}\overline{\text{m}}\text{e}\phantom{\rule{0.5em}{0ex}}\text{of}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{divisor}\end{array}& \begin{array}{cc}\phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}1& 7& 3\end{array}& \begin{array}{cc}\begin{array}{ccc}5& 5& 0\end{array}& \begin{array}{}\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Sollids to be taken}\\ \text{away}\end{array}& \left\{\begin{array}{cc}\phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}5& 1& 8\\ \phantom{0}& \phantom{0}& 6\\ \phantom{0}& \phantom{0}& \phantom{0}\\ \phantom{0}& \phantom{0}& \phantom{0}\end{array}& \begin{array}{cc}\begin{array}{ccc}4& \phantom{0}& \phantom{0}\\ 4& 8& \phantom{0}\\ \phantom{0}& 2& 7& \phantom{0}\\ \phantom{0}& 9& 0& \phantom{0}\end{array}& \begin{array}{c}3AqE\\ 3AEq\\ Ec\\ ECq\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Theire Su}\overline{\text{m}}\text{e}\phantom{\rule{0.5em}{0ex}}\end{array}& \begin{array}{cc}\phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}5& 2& 4\end{array}& \begin{array}{cc}\begin{array}{ccc}9& 9& 7\end{array}& \begin{array}{}\end{array}\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Remaines}\phantom{\rule{0.5em}{0ex}}\end{array}& \begin{array}{cc}\phantom{0}& \phantom{0}\end{array}\end{array}& \begin{array}{ccc}0& 0& 0\end{array}& \begin{array}{cc}\begin{array}{ccc}0& 0& 0\end{array}& \begin{array}{}\end{array}\hfill \end{array}\hfill \end{array}$ 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 $Lc+95400L$ $=$ $=1819459$. 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 $Lc-10L=13584$ & yethe Analysis is as follows 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 $Lc-116620L=352947$ $\begin{array}{ccc}\hfill \begin{array}{cc}\begin{array}{}\text{}\end{array}& \begin{array}{ccc}-& 1& \underset{\phantom{.}}{1}\end{array}\end{array}& \begin{array}{ccc}6& 6& \underset{\phantom{.}}{2}\end{array}& \begin{array}{cc}\begin{array}{ccc}\underset{.}{0}& \underset{.}{\phantom{0}}& \underset{.}{\phantom{0}}\end{array}& \begin{array}{l}\phantom{\rule{1em}{0ex}}\text{Coefficiens plan}\overline{\text{u}}\end{array}\end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{}\text{Cubus resolvendus}\end{array}& \begin{array}{ccc}\phantom{+}& 0& \underset{\phantom{.}}{0}\end{array}\end{array}& \begin{array}{ccc}3& 5& \underset{\phantom{.}}{2}\end{array}& \begin{array}{cc}\begin{array}{ccc}9& 4& \underset{.}{7}\end{array}& \begin{array}{l}\phantom{\rule{1em}{0ex}}\text{(343}\end{array}\end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{}\text{Sollida Ablativa}\end{array}& \left\{\begin{array}{ccc}+& 2& 7\\ -& 3& 4\end{array}\end{array}& \begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\\ 9& 8& 6\end{array}& \begin{array}{cc}\begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\end{array}& \begin{array}{l}\phantom{\rule{1em}{0ex}}Ac\\ \phantom{\rule{1em}{0ex}}ACq\end{array}\end{array}\hfill \\ \hfill \begin{array}{cc}\begin{array}{c}\text{Restat auferend}\overline{\text{u}}\end{array}& \begin{array}{ccc}\phantom{0}& -& 7\end{array}\end{array}& \begin{array}{ccc}9& 8& 6\end{array}& \begin{array}{cc}\begin{array}{ccc}\phantom{0}& \phantom{0}& \phantom{0}\end{array}& \begin{array}{l}\phantom{\rule{1em}{0ex}}\text{}\end{array}\end{array}\hfill \\ \hfill \begin{array}{rr}\begin{array}{}\text{Reliquum resolvendi}\end{array}& \begin{array}{ccc}\phantom{0}& +& 8\end{array}\end{array}& \begin{array}{ccc}3& 3& 8\end{array}& \begin{array}{cc}\begin{array}{ccc}9& 4& 7\end{array}& \begin{array}{l}\phantom{\rule{1em}{0ex}}\text{Cubi}\end{array}\end{array}\hfill \\ \phantom{_____________________________________________}& \phantom{__________}& \phantom{_}\end{array}$ 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 $Lc-6400L=153000$. where yethe roote of $64$ is $8$ 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. $Lc+64L=1024$. Since yethe roote of $64$ is $8$. 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 $13,104L-Lc=155,520$. 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 $27755L-Lqq=217944$ 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 $13104-12×12=12960$. & $Lq+12L=12960$. 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 $-27×27×27$ $=$ $+$ $27755=8072$. $27×27=729$. If there be 4 cubes continullycontinually proportionall whose greate extreame is $27c=19683$. & yethe aggregate of yethe 3 rest is 8072 & $Lc$ yethe lesse extreame, therefore $Lc+27Lq+729$ $L$ $=8072$. 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 $13104l-lc$ $=$ $155520$ or ${l}^{3}$ $-13$ $1$ $04l+155520=0$. & one roote is $12$ . therefore divide this equation by $l-12$ & the Quote is an equation conteingconteining yethe other roote viz: $lq+12l=12960$. To find two meane proportionalls twixt BC & IK 8 Propositiones Geometricæ. Franc: Vietæ. prop 1 $ab:ac\colon\colon ce:bd$ prop 2 & if $ab:ac\colon\colon ac:bd$: then $ac:ab\colon\colon ab:ce$ prop 3. If $ab×ac=bd×ce$. then $bd:ac\colon\colon ac:ab\colon\colon ab:ce$ $∺$ 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 $gh=ab=\text{(}=ai\text{)}$. & $ik:hb\colon\colon hb:hi\colon\colon hi:bc$. $∺$ Prop: 4 If $ad=db=cb$. ynthen yethe Angle $c$ $be$ is tripple to yethe Angle $abd$. Prop 5 If $ab=bd=Rad$. $3Ang:bad=cde$ Prop 6 If $3rpq=spq\text{:recto}$. that is If $2qr=pr$ . then $3or×or=sp×sp+op×op+px×px$ Prop 7 Iaf If $ad=dc=ce=ef$. then $ecf=efc=3dac=3dca$ & ${AC}^{3}=3AC×{ad}^{2}+Cf×{ad}^{2}$. $\overline{){Z}^{3}=aaz+{b}^{3}\text{.}\phantom{\rule{1em}{0ex}}\text{~ ~ ~}}$ prop 8. If $op=pr=qr=qs$. ynthen, $prt=3qsr$ &c and ${sr}^{3}=3sr×{qr}^{2}-or×{qr}^{2}$. $\overline{){Z}^{3}=aaz-{b}^{3}}$ prop 9 If $op=pr=$ $ah=hb=bf=fd$. & $ch=2eh$ or $ceh=3hce$. then $\begin{array}{}{bg}^{3}\\ \phantom{\rule{0ex}{1em}}\end{array}$ $\begin{array}{}\phantom{\rule{1em}{0ex}}{ac}^{3}=3ac×{ah}^{2}-db×{ah}^{2}\\ \text{&}{cb}^{3}=3cb×{ah}^{2}-db×{ah}^{2}\end{array}\phantom{\rule{1em}{0ex}}\right\}\phantom{\rule{1em}{0ex}}\overline{){Z}^{3}=aaz-{b}^{3}\text{.}}$ Prop 10 If $de=ea$ & $db:da\colon\colon ab×ab:dc×dc$. ynthen $be$ is a side of a $7$ equall sided & angled figure. or $7eab=4right angles$. prop 10 If $ac=ef$. & $aef$ a right angle & $ab$ paspasses through yethe center then $cd:de\colon\colon de:df\colon\colon df:db$. And if $cd:de\colon\colon de:df\colon\colon df:db$ then $ae$ is perpendicular to $ef$. $2hd$ is yethe difference of yethe extreames & $2do$ 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, $ae:ec\colon\colon ec:ed\colon\colon ed:eb$. 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 $e$ $i$ & $gi:ki\colon\colon ki:il\colon\colon il:ih$. 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 $\left\{\begin{array}{}\text{octogon}\\ \text{decagon}\end{array}\right\}$ & $pd$ yethe side of an $\left\{\begin{array}{}\text{hexagon}\\ \text{octogon}\end{array}\right\}$ yethe arch $rp$ divided equally in $o$, $od$ will be the side of an $\left\{\begin{array}{}\text{heptagon}\\ \text{enneagon}\end{array}\right\}$ 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 $ab:{ab}^{2}\colon\colon cb:eb×ad-ae×db\colon\colon ac:ae×ad+eb×db$ or, $ab:aq×ab\colon\colon op:eb×an-ae×nq\colon\colon ao:ae×an+eb×nq$. 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 $Ab:{ab}^{2}\colon\colon Eb:ad×bc+ac×db\colon\colon ea:ad×ac-db×cb$. or the triang: unequall. $ab:ap×aq\colon\colon eb:an×op+ao×nq\colon\colon ea:an×ao-nq×op$ 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 $c$. Prop 17. If $ab=bc=ce=eg$ &c: & $ac=cd$. & $ae=ef$ &c then $ab:ac\colon\colon ac:ad\colon\colon ae:af$ &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 $al:ak\colon\colon pe:pc\colon\colon ed:do\colon\colon pd:pc+do\colon\colon$ $rg:gs\colon\colon rq:qo\colon\colon qg:qo+gs\colon\colon$ &c & if $\left(\frac{lf}{2}=lʒ\right)$ from ʒ to yethe center be drawne $cʒ$ then $al:ak\colon\colon di:dv\colon\colon iq:qx\colon\colon dq:dv+qx\colon\colon gz:gx\colon\colon wz:wx$ &c & Ergo $ac:ak\colon\colon ab\colon\colon a\delta +ad:ad:ab+ag:ag:ad+ah:ah:ag+ak$ &c. Prop 19 If $fa=ab=be=eh$ &c. &. $af+ab+be+eh$ are greater y t n than yethe semiperiphery: then, $Rad:ge\left(=ec\right)\left(=ad\right)\colon\colon bd\left(=bc\right)$ $:cd\left(=cd-de\right)\colon\colon de:$ $dh-bd:$ & $dh$ is yethe greatest, $db$ yethe least line drawn from $d$ to these points $a$, $b$, $e$, $h$. ynthen $rad:$ $dh$ $\colon\colon db:da-de$. 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$ $\colon\colon$ $:$ $B\colon\colon ah+$ $2z$. therfore $bb=ahinz+2{z}^{2}$. & $\frac{bb-2zz}{z}=ah$. And $z:B\colon\colon \frac{{b}^{2}-2{z}^{2}}{z}:b+ag$ . therfore $\frac{{b}^{3}-2zzb}{zz}-b=ag$ Likewise $\frac{{b}^{4}-4zzb{b}^{2}+2{z}^{4}}{{z}^{3}}=ad$. & $\frac{{B}^{5}-5zz{B}^{3}+5{z}^{4}B}{{z}^{4}}=ab$ $\frac{{B}^{6}-6zz{B}^{4}+9{z}^{4}BB-2{z}^{6}}{zzzzz}=a\delta$ $\frac{{B}^{7}-7zz{B}^{5}+14{z}^{4}{B}^{3}-7{z}^{6}B}{{z}^{6}}=\text{to a seaventh line}$ $\frac{{B}^{8}-8zz{b}^{6}+20{z}^{4}{b}^{4}-16{z}^{6}bb+2{z}^{8}}{{z}^{7}}=\text{to an eight}$ $\text{line}\genfrac{}{}{0}{}{\phantom{{0}^{0}}}{\phantom{{0}^{0}}}$ $\frac{{B}^{9}-9zz{b}^{7}+27{z}^{4}{b}^{5}-30{z}^{6}{b}^{6}+9{z}^{8}b}{{z}^{8}}=\text{a}$ n $\text{nineth line}\genfrac{}{}{0}{}{\phantom{{0}^{0}}}{\phantom{{0}^{0}}}$ $\frac{{B}^{10}-10zz{b}^{8}+35{z}^{4}{b}^{6}-50{z}^{6}{b}^{4}+25{z}^{8}bb-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\colon\colon B:z+ae$. & $\frac{bb-zz}{z}=ae$ & $\frac{{B}^{3}-2{z}^{2}B}{zz}=ag$ & $\frac{{B}^{4}-3{z}^{2}bb+{z}^{4}}{{z}^{3}}=\text{to a fift line.}$ $\frac{{B}^{5}-4zz{b}^{3}+3{z}^{4}b{z}^{4}}{{z}^{4}}=\text{a sixt.}$ & ${b}^{6}-5zz{b}^{4}+6{z}^{4}bb-{z}^{6}=\text{seaventh}$ $\frac{{b}^{6}-5zz{b}^{4}+6{z}^{4}bb-{z}^{6}}{{z}^{5}}=\text{seaventh}$ $\frac{{b}^{7}-6zz{b}^{5}+10{z}^{4}{b}^{3}-4{z}^{6}b}{{z}^{6}}=\text{to an eight line}$ $\frac{{B}^{8}-7zz{b}^{6}+15{z}^{4}{b}^{4}-10{z}^{6}bb+{z}^{8}}{{z}^{7}}=\text{to a nineth line}$ $\frac{{B}^{9}+8zz{b}^{7}+21{z}^{4}{b}^{5}-20{z}^{6}{b}^{3}+5{z}^{8}b}{{z}^{8}}=\text{to a tenth line}$ $\frac{{B}^{10}-9zz{b}^{8}+28{z}^{4}{b}^{6}-35{z}^{6}{b}^{4}+15{z}^{8}bb-{z}^{10}}{{z}^{9}}=\text{eleventh}$. 10 Prop 22. If $aq=ab=bd=dc=ch=hk=kl=lf$. Then $GK Rad:kl\colon\colon kl:el\left(=al-ah\right)\colon\colon$ And $hl:hm\left(=qh-qd=$ $ak-ac\right)\colon\colon dl:do\left(=qd-qa=ac-ab\right)\colon\colon lc:cn\left(=qc-qb=$ $ah-ad$ &c $\right)$ Soe that Rad yethe Periph: divided into any number of ptspoints. $gl:lk\colon\colon lk:al-ak\colon\colon lh:ak-ac\colon\colon lc:ah-ad\colon\colon dl:ac-ab$ &c. & $gl:lk\colon\colon ah:lc-lk\colon\colon ac:ld-lh\colon\colon ad:lb-lc\colon\colon ab:al-ah$ &c. hence Prop 23.In yethe former scheame If $al$ $=2x=\text{hypotenusa}$. $kl=b$. $x:b\colon\colon b:2x-ah$ & $\frac{-bb+2xx}{x}=ah$ / $x:b\colon\colon \frac{-bb+2xx}{x}:\frac{2bxx-{b}^{3}}{xx}\left(=lc-$ $b$ $\right)$ $lc$$kl$=$dl$h therefore $\frac{3bxx-{b}^{3}}{xx}=lc$. & $\frac{2{x}^{4}-4bbxx+{b}^{4}}{{x}^{3}}=ad$ yethe base of yethe 4th triang: &c $\frac{5b{x}^{4}-5{b}^{3}xx+{b}^{5}}{{x}^{4}}=$ yethe perpendicular $\left($ $bl$ $\right)$ of yethe 5t tri:triangle &c $\frac{2{x}^{6}-9{x}^{4}bb+6xx{b}^{4}-{b}^{6}}{{x}^{5}}=$ 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 $\frac{2{x}^{8}-16{x}^{6}bb+20{x}^{4}{b}^{4}-8xx{b}^{6}+{b}^{8}}{{x}^{7}}=$ base of yethe 8th tri. $\frac{9{x}^{8}b-30{x}^{6}{b}^{3}+27{x}^{4}{b}^{5}-9xx{b}^{7}+{b}^{9}}{{x}^{8}}=$ perp: of yethe 9th tri: Prop 24: If $bd$ $=dh=hk=kl=$ $b$ $g$ &c: then $bh=gd$ & $bk=gh$ & $bl=hm$ &c: & then $xt:bx\colon\colon ac:ag\colon\colon ce:eh\colon\colon ei:em\colon\colon io:ok\colon\colon$ $op:on\colon\colon pq:ql\colon\colon qr:qy\colon\colon rs:sx\colon\colon st:sf\colon\colon ba:ad$ therefore $xt:bx\colon\colon bt:dg+hm+kn+ly+xf$. againe $xt:bt\colon\colon ab:bd\colon\colon ac:cg\colon\colon ce:ch\colon\colon ei:im$ &c Therefore $xt:bt\colon\colon bt:bd+gh+mk+ml+yx+ft$. & since, as $xt:bt\colon\colon bx:dg+hm+kn+ly+xf$ Therefore $xt:bt\colon\colon bx+bt:bd+dg+gh+hm+mk+kn+$ $+nl+ly+yx+xf+ft$. And $xt:bt\colon\colon bx+bt+xt:$ to all yethe perpen dicular & transverse line $+bt$. that is ( 5 ) $xt:bt\colon\colon xt+bt+bx:2bd+2bh+2bk+2bl+2bx+2bt$. Prop 24 If in yethe circle $cfgh$ be inscribed yethe helix $bedc$ a& $ac$ 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 $\frac{rq×po}{2}=$ 4 times yethe section $rapc$ Prop 26 If $ab=bd=ad$ & $bh$ perpendicular to $ad$ frofrom yethe angle $b$ . $ce=ed$. ynthen $aed=adi=3dae$. & $ed$ 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 $\frac{5}{6}$ of $5$ 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 teparte of a circle for $af$ $ef$ is divided in extreame & meane propor in yethe point $c$. & $ec:ef\colon\colon$ hf $ef:\frac{5}{12}$Perimmetereterhbkfa $\colon\colon hr:\frac{5}{24}Perim\colon\colon \frac{6}{10}ef\left(=de\right)\colon\colon \frac{1}{4}Perim$; by yethe 27th prop: & $ec:ef\colon\colon ed:en\left(=\frac{1}{4}Perim\right)$. Prop 29. If $os=2cp$. & $co$ 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 ( . Prop 30 If yethe line $dc$ touch yethe helix in yethe line $ag$. & yethe line $hf$ toucheth yethe beginning of it in yethe center $a$ & $4ac=af$ then $2ad$ 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 $cgp$ is $ce$. of $pacegi$ is $ae$ &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 $p:q\colon\colon b:z$. & p $p:r\colon\colon bc:zz$. & $p:s\colon\colon bcd:zzz$. & $p:t\colon\colon bcdf:{z}^{4}$. & $p:v\colon\colon bcdfg:{z}^{5}$. & $p:w\colon\colon bcdfgh:{z}^{6}$ &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 $\begin{array}{ccccccc}a& & bb& & ccc& & dd\\ \frac{1}{1}& ×& \frac{2×3}{1×2}& ×& \frac{4×5×6}{1×2×3}& ×& \frac{7×8}{1×2}& =112\end{array}$ $\frac{\stackrel{a}{1}}{1}×\frac{\stackrel{b}{2}×\stackrel{b}{3}}{1×2}×\frac{\stackrel{c}{4}×\stackrel{c}{5}×\stackrel{c}{6}}{1×2×3}×\frac{\stackrel{d}{7}×\stackrel{d}{8}}{1×2}=112$ the number of changes, in order. To know how many elections may bee made doe thus $\stackrel{a}{2}×\stackrel{bb}{3}×\stackrel{ccc}{4}×\stackrel{dd}{3}×\stackrel{eeee}{5}=360=\stackrel{a}{2}×\frac{\stackrel{b}{2}×\stackrel{b}{3}}{2}×\frac{\stackrel{c}{2}×\stackrel{c}{3}×\stackrel{c}{4}}{2×3}×\frac{\stackrel{d}{2}×\stackrel{d}{3}}{2}×\frac{\stackrel{e}{2}×\stackrel{e}{3}×\stackrel{e}{4}×\stackrel{e}{5}}{2×3×4}$ therefore there are $359=360-1$ elections in $abb{c}^{3}dd{e}^{4}$. 12 Propositiones Geometricae Ex Schootenij Sectionibus miscellaneis. Sectio 1ma To know how many changes 6 Bells, $abcdef$ or how divers conjuctions yethe 7 planets can make . or how many divisors $abcde$ hath, or how maney divers compositions yethe 24 letersletters can make &c the examples following show. $1.$ $a\phantom{\rule{1em}{0ex}}1$ $2.$ $b.ab\phantom{\rule{1em}{0ex}}3$ $4.$ $c.cb.cab.ac.\phantom{\rule{1em}{0ex}}7$ $8.$ $d.da.db.dab.dc.dac.dcb.dcab.\phantom{\rule{1em}{0ex}}15$ $16.$ $e.ea.eb.eab.ec.eac.ecb.ecab.ed.eda.edb.edab.edc.edac.$ $edcb.edcab.$ $31$ $32.$ $f.fa.fb.fab.fc.fcb.fac.fcab.fd.fda.fdb.fdab\phantom{\rule{1em}{0ex}}\text{&c}\phantom{\rule{1em}{0ex}}63.$ $64.$ $g.ga.gb.gab.gc.gcb.gac.gcab.gd.gda.gdb\phantom{\rule{1em}{0ex}}\text{&c}\phantom{\rule{1em}{0ex}}127$ wchwhich shows ytthat in 7 letters 127 eleections may be made. ytthat 7 Planets may be conjoyned 120 divers ways. ytthat $abcdefg$. hath 128 divisors for an unite is one of ythem& $1×2×3×4×5×6.=720$; 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 $aaabbb$. 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 $aaaaaaab$. $\frac{16}{16}=1$. $16-1=15$. 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 $abcd.{a}^{3}bc.{a}^{3}{b}^{3}.{a}^{7}b.{a}^{15}.$ 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: $abcd.{a}^{3}bc.{a}^{3}{b}^{3}.{a}^{7}b.{a}^{15}.$ that is $2.3.5.7.$ or $2.$ $2.2.3.5.$ &c. now. $2×3×5×7.=210$. & $2×2×2×3×5=120$. &c therefore $2×3×5×7.$ $=210$ 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 $1×2=2$ since yethe product is eavedn double it viz: $2×2=4$. & $4$ is yethe numerator ynthen $1+2=3$ & since $3$ is od multiply it by the difference of yethe termes: $1×3=3$ & $3$ is yethe denominator. & yethe first terme $\frac{4}{3}$. ynthen since (1) yethe difference of yethe termes is od multiply it by $4$. $4×1=4$ & $4×$ per 2 majoreem terminum. $4×2=8$ $8+4$ (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 $1×3=3=$ 1st numerator. ynthen $1+3=4$ & since $4$ is eaven halfe $4$ 1st denom. & yethe 1st halfe is $\frac{4×2}{2}$ (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 $2$ $2×2=4$ & $4×3=12$ & $12+3=15$. ynthen $2×2=4.4+4=8$. & $\frac{15}{8}$ yethe 2d terme & now termes may be had by Arithmeticall proportion. thus. $\frac{4}{3}.\frac{12}{5}$ or $1\frac{1}{3}.2\frac{2}{5}.3\frac{3}{7}.4\frac{4}{9}.5\frac{5}{11}.6\frac{6}{13}\phantom{\rule{1em}{0ex}}7\frac{7}{15}\phantom{\rule{1em}{0ex}}8\frac{8}{17}\phantom{\rule{1em}{0ex}}9\frac{9}{19}\phantom{\rule{1em}{0ex}}10\frac{10}{21}\phantom{\rule{1em}{0ex}}$ &c & $\frac{3}{4}.\frac{15}{8}$ or $\frac{3}{4}\phantom{\rule{1em}{0ex}}1\frac{7}{8}.2\frac{11}{12}.3\frac{15}{16}.4\frac{19}{20}.5\frac{23}{24}.6\frac{27}{28}.7\frac{31}{32}.8\frac{35}{36}$ &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 $2\frac{2}{5}$ $5×2=10.10+2=12$ or $\frac{12}{5}$ $\sqrt{144+25}=\sqrt{169}=13$. & in this $1\frac{7}{8}$ or $\frac{15}{8}$ $\sqrt{225+64}=17$. 13 If yethe suposed numbers be $2.5$. ynthen $2×5=10$. $10+10=20$. & $2+5=7$. $3×7=21$. so ytthat $\frac{20}{21}$. ynthen be $4×3=12$. $12×5=60$. $60+20=80$. & $3×3=9$. $9\phantom{\rule{0.5em}{0ex}}\text{doubled}=18$. $18+21=39$. & yethe 2 first termes $\frac{20}{21}.\frac{80}{39}$ or $2\frac{2}{39}$. Againe, if yethe numbers be $3.4$ $3×4=12$. $12×2=24$. & $3+4=7$. $1×7=7$. therefore $\frac{24}{7}$. ynthen $4×1=4$. $4×4=16$ $16+24=40$. & $1×1=1$. $2×1=2$. $7+2=9$ 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 $7.11.13.$ is $7×11×13×=$ $1001$. divide $1001$ twice by each & consider yethe remainder of yethe seacond division thus. $\begin{array}{ccc}\frac{210}{2}& \frac{{105}^{\left(1}}{2}& 52\end{array}$. because $1$ is left $105$ is yethe multiplier. $\begin{array}{ccc}\frac{210}{3}& \frac{{70}^{\left(1}}{3}& 23\end{array}$. since $1$ is left $70$ is yethe multiplier. $\begin{array}{ccc}\frac{210}{5}& \frac{{42}^{\left(2}}{5}& 8\end{array}$ since more ynthan $1$ is left (viz: $2$ ) multiply $2$ till it divided by 5 leaves ( $1$ ) $\frac{2×3}{5}=1\frac{1}{5}$. therefore $42×3=126$ is yethe multiplyere. $\begin{array}{ccc}\frac{210}{7}& \frac{30}{7}& \left(4\frac{2}{7}\end{array}$ since more ynthan $1$ is left 1 Since more ynthan $1$ is left (viz $3$) multiply $3$ till it divided by $7$ leavs $1$ . $\frac{5×3}{7}=2\frac{1}{7}$ therfore $5×143=715$ yethe multiplier $|\frac{1001}{7}\left(\frac{143}{7}\left(20\frac{3}{7}$. 2 Since more ynthan $1$ is left (viz: $3$ ) $\frac{3×4}{11}=1\frac{1}{11}$ therfore $4×91=$ $=364$ yethe multipl: $|\frac{1001}{11}\left(\frac{91}{11}\left(8\frac{3}{11}$. 3 If but $1$ had benene left $77$ had beene divisor but now $\frac{12×12}{11}=13\frac{1}{11}$. therfore $12×77=924$ is multiplyer. $\frac{1001}{13}\left(\frac{77}{13}\left(5\frac{12}{13}$. now the number soughsought is thus found. $\begin{array}{c}\begin{array}{ccccc}\text{Divisor.}& \text{Reliq:}& & \text{Multip.}& \\ 7\phantom{0}.& \hfill 2& ×& 715& =& 1430\text{.}\\ 11.& \hfill 1& ×& 364& =& \phantom{0}364\text{.}\\ 13.& \hfill 9& ×& 924& =& 8316\text{.}\end{array}\text{The Su}\overline{\text{m}}\text{e}\phantom{\rule{3em}{0ex}}10110\text{.}\end{array}$ Lastly didvide by yethe least com. divis: $\frac{10110}{1001}\left(10\frac{100}{1001}$ 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 $\begin{array}{}1\\ 2\\ 3\end{array}$ $\begin{array}{cccccccccccccc}1.& 2& 3.& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14\\ 1.& 2& 1+2& 4& 1+4.& 2+4.& 1+2+4.& 8& 8+1& 8+2.& 8+1+2.& 8+4.& 8+4+1.& 8+4+2\end{array}$ &c or isf his wightsweights be $1.3.9.27.81.$ all weights may be supplyed thus. $\begin{array}{cccccccccccccc}1.& 2.& 3.& 4.& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14\\ 1& 3-1.& 3.& 3+1.& 9-1-3.& 9-3.& 9+1-3.& 9-1.& 9.& 9+1.& 9+3-1.& 9+3.& 9+3+1.& 27-9-3-1.\end{array}$ &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 $\left(2\right)$, or any other number producced out of $2$ as $2×2.2×2×2$ &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. $2×3-1=5$ numero primario primo. $2×6-1=11$ numero primario scdosecundo. $2×2×18-1=71$ numero primario tertio. $4×71=284$, one amicable number, & yethe 2 former prime numbers $×$ one another & yethe product $×4$ yethe double of yethe assumed number viz $5×11=5555×4=220$. 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 $ac=z$. $bd=x$ $cd=y$. $ad=z+y$. $ad=y+b$. $xx+yy=yy+2by+bb$. $y=\frac{xx-bb}{2b}$. & $cd=z+y+a$. $xx+zz+2yz+yy=zz+$ $yy+aa+2zy+2za+2ay$. $2ay=xx-aa-2za=\frac{axx-abb}{b}$. $bxx-baa-2zab=axx-abb$. $\frac{bxx-baa-axx+abb}{2ab}=z$. puting any numbers for $a$ , $b$ , & $x$ ; $y$ & $z$ may be found. then $ad=z+y=$ $=\frac{xx+bb}{2b}$. $cd=z+y+a=\frac{xx+aa}{2a}$. wchwhich reduced to yethe common denominator $2ab$; & ytthat cast away. $cd=bxx+baa$. $ad=axx+abb$. $de=2abx$. $ae=axx-abb$. $ce=bxx-baa$. $ac=bxx-axx+abb$ $-aab$. In like manner if yethe perpendicular fall wthwithin side. $ab=bxx+baa$. $bd=2abx$. $ad=bxx-baa$. $dc=axx-abb$. $bc=axx+abb$. $ac=bxx+axx-abb-baa$. Also by yethe conjunction & disjunction of 2 triangles it may be found ytthat $ab=bbx+aax$ $ad=bbx-aax$. $ac=bbx-aax-aax$ $+abb$. $bc=axx+abb$. $db=2abx$. $dc=axx-abb$. For if $bd=x$ $dc=\frac{xx-bb}{2b}$. $bc=\frac{xx+bb}{2b}$. that is $bd=2bx$. $dc=xx-bb$. $bc=xx+bb$. Likewise $bd=2ab$. $ad=bb-aa$. $ab=bb+aa$. $2abx$ yethe least quantity divisible by $2bx$ & $2ab$, being divided by ymthem, leaves $a$ & $x$ wchwhich must multiply yethe bases & hypotenusas. If yethe perpendic: fall wthwithout yethe legs may be thus exprest $cd=acc+ayy$. $da=yyc+aac$ 14 $ca=acc-ayy+cyy-aac$. $ae=yyc-aac$. $ce=acc-ayy$. $ad=2acy$. 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 $2abx=2acy$. $x=\frac{cy}{b}$. $xx=\frac{ccyy}{bb}$. $bbx-aax-axx+abb=acc-ayy+yyc-aac$ or $\frac{bbcy-aacy}{b}-\frac{accyy}{bb}+abb=acc-ayy+yyc-aac$ & $yy=\frac{+{b}^{3}cy+aabbc-aabc+a{b}^{4}-abbcc}{bbc+acc-abb}$. Suppose $aabbc+a{b}^{4}=abbcc$. or $a=c-\frac{bb}{c}$. 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 summe 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 $\begin{array}{ccccc}1\begin{array}{}●\end{array}& \phantom{\rule{1em}{0ex}}3\begin{array}{}●\\ ●●\end{array}& \phantom{\rule{1em}{0ex}}6\begin{array}{}●\\ ●●\\ ●●●\end{array}& \phantom{\rule{1em}{0ex}}10\begin{array}{}●\\ ●●\\ ●●●\\ ●●●●\end{array}& \phantom{\rule{1em}{0ex}}15\begin{array}{}●\\ ●●\\ ●●●\\ ●●●●\\ ●●●●●\end{array}\end{array}$ &c likewise $1.3.5.7.9.$ compose $\begin{array}{ccccc}1\begin{array}{}●\end{array}& \phantom{\rule{1em}{0ex}}4\begin{array}{c}●●\\ ●●\end{array}& \phantom{\rule{1em}{0ex}}9\begin{array}{c}●●●\\ ●●●\\ ●●●\end{array}& \phantom{\rule{1em}{0ex}}16\begin{array}{c}●●●●\\ ●●●●\\ ●●●●\\ ●●●●\end{array}& \phantom{\rule{1em}{0ex}}25\begin{array}{c}●●●●●\\ ●●●●●\\ ●●●●●\\ ●●●●●\\ ●●●●●\end{array}\end{array}$ &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 $z$. $z=\frac{1tt+1t}{2}$ or $z=\frac{tt+t}{2}$ in trigons. $z=tt$ in 4gons. $z=\frac{3tt-t}{2}$ in 5gons. $z=2tt-t$ in 6goons $z=\frac{5tt-3t}{2}$ in 7gons. $z=3tt-2t$ in 8gons. $z=\frac{7tt-5t}{2}$ in 9gons. &c & $z$ given $t$ is found thus $t=\frac{-1+\sqrt{1+8z}}{2}$ in tri. $t=\frac{\sqrt{0+16z}}{4}$ in 4goons $t=\frac{+1+\sqrt{1+24z}}{6}$, in 5gons. $t=+2+\sqrt{4+32z}$ $t=\frac{+2+\sqrt{4+32z}}{8}$ in 6gons &c. As yethe side $12$ of a tri given. yethe $tri=z=\frac{12×12+12}{2}=78$ &c & if $z=21$ be octangled. $t=\frac{+4+\sqrt{16+48z}}{12}=\frac{4+\sqrt{16+48×21}}{12}$ $t=\frac{4+\sqrt{1024}}{12}=3$. 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 $\frac{1}{2}×3=\frac{3}{2}$. 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 $\frac{1}{3}×2=\frac{2}{3}$. 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 $-1×\frac{1}{4}=-\frac{1}{4}$. (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 $ndpe$ $qf$ &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 $y=\sqrt{rx}$) whoswhose index is $\frac{1}{2}$ wchwhich added to an unite is $1+\frac{1}{2}=\frac{3}{2}$ Therefore $\frac{3}{2}:1\colon\colon 3:2$ so is yethe square $ambc$ to yethe area of yethe Parab. (yethe names of yethe lines are $\left(ad\right),$ $ae,af$ &c $=x$ . $dn,pe,qf$ &c $=y$ . $ac=p$. $bc=q$.) 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 $rx=yy$. $yy$ designes yethe squares $nd,pe,qf$ &c: all wchwhich taken together are equivalent to yethe Sollid. & those □ssquares increase in yethe same proportion wchwhich $rx$. 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 $1$ $:2\colon\colon$ 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 , $y$ $=aa-xx$, or $y={a}^{4}-2aaxx+{x}^{4}$ or $y={a}^{6}-3{a}^{4}xx+3aa{x}^{4}-{x}^{6}$. &c. Their areas will bee to yethe parallelograms arbout them as $2$ to $3$ , as $8$ to $\frac{1}{5}$ $15$ , as $48$ to $105$ &c. but if I put in yethe intermediate termes in these last named lines their order will bee $y=\sqrt{aa-xx}$, $y=aa-xx$, $y=\overline{aa-xx\sqrt{aa-xx}}$, $y={a}^{4}-2aaxx+{x}^{4}$. $y=\overline{{a}^{4}-2aaxx+{x}^{4}\sqrt{aa-xx}}$. $y={a}^{6}-3{a}^{4}xx+3aa{x}^{4}-{x}^{6}$; &c: & since these lines observe a geometricall progression their areas must observe some kind of progression. of wchwhich every other terme is given viz $1.\square .\frac{2}{3}.\ast .\frac{8}{15}.\ast .\frac{48}{105}.\ast .\frac{384}{945}.\ast .\frac{3840}{10395}$. Tblwixt wchwhich termes if yethe intermediate termes $\square .\ast$ can bee found yethe 2nd square will give yethe area of yethe line $y=\sqrt{aa-xx}$, yethe circle. Soe likwiselikewise in this progression of lines $y$ $=1$ . $y$ $=\sqrt{ax-xx}$ . $y=aa-xx$ $y=\overline{ax-xx\sqrt{ax-xx}}$. $y=aaxx-2a{x}^{3}+{x}^{4}$ &c: yethe progression of their areas is $1:\square :\frac{1}{6}:\ast :$ $\frac{1}{30}:\ast :\frac{1}{140}:\ast \frac{1}{630}:$ &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 $1.2\square .\frac{1}{2}.4\ast .\frac{1}{6}.6\ast .\frac{1}{20}.8\ast .\frac{1}{70}$ 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 $=q$ . $2aq=○$. 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 $A$ $×\frac{0×2×4×6×8}{1×3×5×7×9}$ & 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 $\frac{1}{2}a:1:a:\frac{3}{2}:\frac{4a}{3}:\frac{15}{8}:\frac{8a}{5}:\frac{105}{48}:\frac{64a}{35}:\frac{945}{384}:\frac{640a}{315}:$ deducing its originall from $A×\frac{0×2×4×6×8×10}{1×3×5×7×9×11}$ & from this $A\left(=\frac{1}{2}\right)×\frac{2a×4a×6a×8a×10a}{1×3×5×7×9}$. &c + (note ytthat yethe proportions of these temeane termes to oneanoone another, or to $\left(a\right)$, are found f by finding yethe proportion of yethe circle $y=\sqrt{aa-xx}$ to yethe line $y=\overline{aa-xx\sqrt{aa-xx}}$ &c). In this case to find yethe quantity $a$ : it may be considered ytthat $\frac{a}{\frac{1}{2}a}=2$. $\frac{3}{1×2}=$ $\frac{3}{2}$. $\frac{4a}{3×a}=\frac{4}{3}$. Naming yethe termes in yethe progress: $\begin{array}{}\frac{1}{2}a:1:a:\frac{3}{2}:\frac{4a}{3}:\frac{15}{8}:\frac{8a}{5}:\frac{35}{16}:\\ b& c& d& e& f& g& h& k\text{.}\end{array}$ 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 or $a=$ . Also . Therefore $3×3\sqrt{\frac{4}{}}$ Therefore . And So by yethe same reasoning. &c. Thus Wallis doth it. but it may bee done thus. $\frac{4a}{3}\phantom{\rule{1em}{0ex}}\text{is}\left\{\begin{array}{}\text{greater then}\frac{3}{2}\\ \text{less then}\frac{15}{8}\end{array}$ Therefore . ytthat is $a\phantom{\rule{1em}{0ex}}\text{is}\left\{\begin{array}{}\text{greater then}\frac{3×3×5×5}{2×4×4×6}=\frac{3×5×5}{2×4×4×2}\\ \text{lesse then}\frac{3×3×5×5×7}{2×4×4×6×6}=\frac{5×5×7}{2×4×4×4}\end{array}$ &c. By yethe same reasoning Or . 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 $eq=\sqrt{1-xx}$. $fq=1-xx$. $gq=$ $\overline{1-xxin\sqrt{1-xx}}$. $hq=1-2xx+{x}^{4}$. $iq=1-2{x}^{2}+{x}^{4}\sqrt{1-xx}$ $dq=1$. $kq=\frac{1}{\sqrt{1-xx}}$. $lq=\frac{1}{1-xx}$. $rq=\frac{1}{1-xx\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}\sqrt{1-xx}}$. &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 $adqp=1×xx=x$. 2dly $afc$ is a parab: therefore $afqp=x-\frac{{x}^{3}}{4}$. also since tis $qh=1-2xx+{x}^{4}$, therefore $ahqp=x-\frac{2}{3}{x}^{3}+\frac{1}{5}{x}^{5}$. Also $vq=1-3xx+3{x}^{4}-{x}^{6}$, 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 $x:x-:x-\frac{1}{3}{x}^{3}:x-\frac{3}{6}{x}^{3}+:x-\frac{2}{3}{x}^{3}+\frac{1}{5}{x}^{5}:x-\frac{5}{6}{x}^{3}+\frac{2}{5}{x}^{5}-$ or $x:x-\frac{1}{6}{x}^{3}:x-\frac{1}{3}{x}^{3}:x$ 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{array}{}\\ \begin{array}{ccccccccccccccccccccccccc}{\text{1}}^{\text{st}}\hfill & .& +x×1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& \\ {\text{2}}^{\text{d}}\hfill & .& -\frac{{x}^{3}}{3}×0& .& 0+1=1& .& 1+1=2& .& 2+1=3& .& 3+1=4& .& 4+1=5\phantom{0}& .& 6& .& 7& .& 8& .& 9& .& 10& .& \\ {\text{3}}^{\text{d}}\hfill & .& +\frac{{x}^{5}}{5}×0& .& 0+0=0& .& 0+1=1& .& 1+2=3& .& 3+3=6& .& 6+4=10& .& 15& .& 21& .& 28& .& 36& .& 45& .\\ {\text{4}}^{\text{th}}\hfill & .& -\frac{{x}^{7}}{7}×0& .& 0+0=0& .& 0+0=0& .& 0+1=1& .& 1+3=4& .& 4+6=10& .& 20& .& 35& .& 56& .& 84& .& 120& .\\ {\text{5}}^{\phantom{\text{th}}}\hfill & .& +\frac{{x}^{9}}{9}×0& .& 0+0=0& .& 0+0=0& .& 0+0=0& .& 0+1=1& .& 1+4=5\phantom{0}& .& 15& .& 35& .& 70& .& 126& .& 210& .\\ {\text{6}}^{\phantom{\text{th}}}\hfill & .& -\frac{{x}^{11}}{11}×0& .& 0+0=0& .& 0+0=0& .& 0+0=0& .& 0+0=0& .& 0+1=1\phantom{0}& .& 6& .& 21& .& 56& .& 126& .& 252& .\\ \hfill & & {\text{1}}^{\text{st}}\text{.}& & {\phantom{\text{.}}}^{\text{∗}}\phantom{\rule{2em}{0ex}}{\text{2}}^{\text{d}}\text{.}\hfill & & {\phantom{\text{.}}}^{\text{∗}}\phantom{\rule{2em}{0ex}}{\text{3}}^{\text{d}}& & {\phantom{\text{.}}}^{\text{∗}}\phantom{\rule{2em}{0ex}}{\text{4}}^{\text{th}}& & {\phantom{\text{.}}}^{\text{∗}}\phantom{\rule{2em}{0ex}}{\text{5}}^{\text{th}}& & {\phantom{\text{.}}}^{\text{∗}}\phantom{\rule{2em}{0ex}}{\text{6}}^{\text{th}}& & 1& .& 7& .& 28& .& 84& .& 210& .\\ \hfill & & & & & & & & & & & & & {\phantom{\text{|}}}^{\text{∗}}& {\text{7}}^{\text{th}}& .& 1& .& 8& .& 36& .& 120& .\\ \hfill & & & & & & & & & & & & & & & {\phantom{\text{|}}}^{\text{∗}}& {\text{8}}^{\text{th}}& & 1& .& 9& .& 45& .\\ \hfill & & & & & & & & & & & & & & & & & {\phantom{\text{|}}}^{\text{∗}}& {\text{9}}^{\text{th}}& & 1& .& 10& .\\ \hfill & & & & & & & & & & & & & & & & & & & {\phantom{\text{|}}}^{\text{∗}}& {\text{10}}^{\text{th}}& & 1& .\\ \hfill & & & & & & & & & & & & & & & & & & & & & {\phantom{\text{|}}}^{\text{∗}}& {\text{11}}^{\text{th}}& .\end{array}\end{array}$ Now if the meane termes in these progressions can bee callculated yethe first of ymthem gives yethe area $aeqp$. Which is thus done $\begin{array}{}\begin{array}{cccccccccccccccccccccccccccc}{\text{1}}^{\text{st}}\hfill & & \hfill +x×1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .\\ {\text{2}}^{\text{d}}\hfill & .& \hfill -\frac{{x}^{3}}{3}×0& .& \frac{1}{2}& .& 1& .& \frac{3}{2}& .& 2& .& \frac{5}{2}& .& 3& .& \frac{7}{2}& .& 4& .& \frac{9}{2}& .& 5& .& \frac{11}{2}& .& 6& .\\ {\text{3}}^{\text{d}}\hfill & .& \hfill +\frac{1}{5}{x}^{5}×0& .& -\frac{1}{8}\phantom{-}& .& 0& .& \frac{3}{8}& .& 1& .& \frac{15}{8}& .& 3& .& \frac{35}{8}& .& 6& .& \frac{63}{8}& .& 10& .& \frac{99}{8}& .& 15& .\\ \text{4}\hfill & .& \hfill -\frac{1}{7}{x}^{7}×0& .& +\frac{1}{16}\phantom{+}& .& 0& .& -\frac{1}{16}\phantom{-}& .& 0& .& \frac{5}{16}& .& 1& .& \frac{35}{16}& .& 4& .& \frac{105}{16}& .& 10& .& \frac{231}{16}& .& 20& .\\ \text{5}\hfill & .& \hfill +\frac{1}{9}{x}^{9}×0& .& -\frac{5}{128}\phantom{-}& .& 0& .& \frac{3}{128}& .& 0& .& -\frac{5}{128}\phantom{-}& .& 0& .& \frac{35}{128}& .& 1& .& \frac{315}{128}& .& 5& .& \frac{1155}{128}& .& 15& .\\ \text{6}\hfill & .& \hfill -\frac{1}{11}{x}^{11}×0& .& \frac{7}{256}& .& 0& .& -\frac{3}{256}\phantom{-}& .& 0& .& \frac{3}{256}& .& 0& .& -\frac{7}{256}\phantom{-}& .& 0& .& \frac{63}{256}& .& 1& .& \frac{693}{256}& .& 6& .\\ \text{7}\hfill & .& \hfill \frac{1}{13}{x}^{13}×0& .& -\frac{21}{1024}\phantom{-}& .& 0& .& \frac{7}{1024}& .& 0& .& -\frac{5}{1024}\phantom{-}& .& 0& .& \frac{7}{1024}& .& 0& .& -\frac{21}{1024}\phantom{-}& .& 0& .& \frac{231}{1024}& .& 1& .& \frac{3003}{1024}& .\end{array}\\ \end{array}$ Soe ytthat $1×x-\frac{1}{2}×\frac{1}{3}×{x}^{3}-\frac{1}{8}×\frac{1}{5}×{x}^{5}-\frac{1}{16}×\frac{1}{7}×{x}^{7}-\frac{5}{128}×\frac{1}{9}×{x}^{9}$ &c. is yethe area, $apqe$ ytthat is $\frac{0}{0}×x-\frac{0}{0}×\frac{1}{2}×\frac{1}{3}{x}^{3}-\frac{0}{0}×\frac{1}{2}×\frac{1}{4}×\frac{1}{5}{x}^{5}-\frac{0}{0}×\frac{1}{2}×\frac{1}{4}×\frac{3}{6}×\frac{1}{7}{x}^{7}-\frac{1×3×5\phantom{\rule{1em}{0ex}}{x}^{9}}{2×4×6×8×9}$ &c: The progression may be deduced from hence $\frac{0\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}1\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}-1\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}3\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}-5\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}7\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}-9\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}11}{0\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}2\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}4\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}6\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}8\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}10\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}12\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}14}$. &c 19 (9 Soe ytthat if yethe given sine bee $pq=te=x$. & if yethe Radius $pc=1$. Then is yethe superficies $ape=x-x\sqrt{1-xx}-\frac{1}{6}{x}^{3}-\frac{1}{40}{x}^{5}-\frac{{x}^{7}}{112}-\frac{5{x}^{9}}{1152}-\frac{7{x}^{11}}{2816}$ &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 $aecpa:\angle apc=90\colon\colon ape:\angle ape$. The same may bee thus done. $adp=\frac{x}{2}$. Or $2adp=x$. $2afp=x+\frac{{x}^{3}}{3}$. $2ahp=x+\frac{2{x}^{3}}{3}-\frac{3{x}^{5}}{5}$. And $2avp=x+{x}^{3}-\frac{9{x}^{5}}{5}+\frac{5}{7}{x}^{7}$. &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 $apce:\angle apc={90}^{\text{d}}\colon\colon ape:\angle ape$ Likewise yethe angle $ape$ given its sign may bee found hereby &c $\begin{array}{}\begin{array}{cccccccccccccc}\phantom{\hfill +\frac{9}{11}{x}^{11}in0}\hfill 2×adpa=& & 2×aep=& & 2afp=& & 2agp=& & 2ahp=& & 2aip=& & 2avp=& \end{array}\begin{array}{cccccccccccccc}\hfill +xin1& .& 1& .& 1& .& 1& .& 1& .& 1& .& 1& .\\ \hfill \frac{{x}^{3}}{3}in0& .& \frac{1}{2}& .& 1& .& \frac{3}{2}& .& 2& .& \frac{5}{2}& .& 3& .\\ \hfill -\frac{3}{5}{x}^{5}in0& .& -\frac{1}{8}\phantom{-}& .& 0& .& \frac{3}{8}& .& 1& .& \frac{15}{8}& .& 3& .\\ \hfill +\frac{5}{7}{x}^{7}in0& .& \frac{1}{16}& .& 0& .& -\frac{1}{16}\phantom{-}& .& 0& .& \frac{5}{16}& .& 1& .\\ \hfill -\frac{7}{9}{x}^{9}in0& .& -\frac{5}{128}\phantom{-}& .& 0& .& \frac{3}{128}& .& 0& .& -\frac{5}{128}\phantom{-}& .& 0& .\\ \hfill +\frac{9}{11}{x}^{11}in0& .& \frac{7}{256}& .& 0& .& -\frac{3}{256}\phantom{-}& .& 0& .& \frac{3}{256}& .& 0& .\\ \phantom{\hfill 2×adpa=}\phantom{.}\phantom{2×aep=}\phantom{.}\phantom{2afp=}\phantom{.}\phantom{2agp=}\phantom{.}\phantom{2ahp=}\phantom{.}\phantom{2aip=}\phantom{.}\phantom{2avp=}\phantom{.}\end{array}\end{array}$ Note ytthat $\sqrt{1-xx}=\frac{x}{2}-\frac{2{x}^{3}}{6}-\frac{5{x}^{5}}{80}-\frac{7{x}^{7}}{224}-\frac{45{x}^{9}}{2304}-\frac{77{x}^{11}}{5632}$ &c that is $\sqrt{1-xx}=\frac{x}{2}-\frac{{x}^{3}}{3}-\frac{{x}^{5}}{16}-\frac{{x}^{7}}{32}-\frac{5{x}^{9}}{256}-\frac{7{x}^{11}}{512}-\frac{21{x}^{13}}{2048}$ &c. According to this progression $\frac{1}{2}×\frac{1}{2}×\frac{1}{4}×\frac{3×5×7×9×11×13×15×17}{6×8×10×12×14×16×18×20}$ &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 $pq=a$. $qd=x$. $pc=1=pb$. $db=\sqrt{1-aa-2ax-xx}$ ynthen yethe areas of yethe lines in this progression. $1-\sqrt{1-aa-2ax-xx}.1-aa-2ax-xx.\overline{1-aa-2ax-xx}\right\}\frac{3}{2}.$ $1-2aa-4ax-2xx+{a}^{4}+4{a}^{3}x+6aaxx+4a{x}^{3}+{x}^{4}.\ast .$ $1-3aa+3{a}^{4}+8{a}^{3}x+8aaxx$ & (supposeing also $\stackrel{_}{b-2ax-xx}\right\}\frac{3}{2}.\begin{array}{c}bb-4abx-2bxx+4a{x}^{3}+{x}^{4}\\ +4aa\end{array}.\ast$ $\begin{array}{c}{b}^{3}-6abbx-3bbxx+8ab{x}^{3}+3b{x}^{4}-6a{x}^{5}-{x}^{6}\\ +12aab{x}^{2}-8{a}^{3}{x}^{3}-12aa{x}^{4}\\ +4ab{x}^{3}\end{array}$. &c 20 (11 To square yethe Hyperbola. Epitome Geometriæ So if $nadm$ is an Hyperbola. & $cp=1=pa$. $pq=x$ . $qd,qe,qf,qg$ &c $=y$ . & $\frac{1}{1+x}=y=dq$. $1=y=eq$. $1+x=y=qf$. $1+2x-xx=y=qg$. $1+3x+3xx+{x}^{3}$. $1+4x+6{x}^{2}+4{x}^{3}+{x}^{4}$. &c. ThereTheir squares are. $\ast .x.x+\frac{xx}{2}.x+\frac{2xx}{2}+\frac{{x}^{3}}{3}.x+\frac{3xx}{2}+\frac{3{x}^{3}}{3}+\frac{{x}^{4}}{4}.$ $x+\frac{4xx}{2}+\frac{6{x}^{3}}{3}+\frac{4{x}^{4}}{4}+\frac{{x}^{5}}{5}.x+\frac{5xx}{2}+\frac{10{x}^{3}}{3}+\frac{10{x}^{4}}{4}+\frac{5{x}^{5}}{5}+\frac{{x}^{6}}{6}$. &c As in yethe following table. By whose first terme is represented yethe square of yethe Hyperbola, viz ytthat it is $\begin{array}{}\begin{array}{c}\begin{array}{cc}\begin{array}{l}\begin{array}{lllllllll}\hfill \phantom{\phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{0×}\phantom{-}=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}\\ \phantom{0}\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{x×}a\phantom{.}& a\phantom{.}& a\phantom{.}& a\phantom{.}& \phantom{1.}& \phantom{1.}& \phantom{1.}& \phantom{1.}& \phantom{1.}\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{x×}p\phantom{.}& p\phantom{.}& p\phantom{.}& p\phantom{.}& \phantom{1.}& \phantom{4.}& \phantom{10.}& \phantom{20.}& \phantom{35.}\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{x×}q\phantom{.}& q\phantom{.}& q\phantom{.}& q\phantom{.}& \phantom{0.}& \phantom{1.}& \phantom{5.}& \phantom{15.}& \phantom{35.}\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{x×}d\phantom{.}& e\phantom{.}& f\phantom{.}& g\phantom{.}& \phantom{0.}& \phantom{0.}& \phantom{1.}& \phantom{6.}& \phantom{21.}\\ \hfill \text{=}\phantom{.}& \text{=}\phantom{.}& \text{=}\phantom{.}& \text{=}\phantom{.}& \phantom{0.}& \phantom{0.}& \phantom{1.}& \phantom{6.}& \phantom{21.}\end{array}\\ \begin{array}{lllllllll}\hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}x×1.& 1.& 1.& 1.& 1.& 1.& 1.& 1.& 1.\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\frac{{x}^{2}}{2}×-1.& 0.& 1.& 2.& 3.& 4.& 5.& 6.& 7.\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\frac{{x}^{3}}{3}×1.& 0.& 0.& 1.& 3.& 6.& 10.& 15.& 21.\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\frac{{x}^{4}}{4}×-1.& 0.& 0.& 0.& 1.& 4.& 10.& 20.& 35.\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\frac{{x}^{5}}{5}×1.& 0.& 0.& 0.& 0.& 1.& 5.& 15.& 35.\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\frac{{x}^{6}}{6}×-1.& 0.& 0.& 0.& 0.& 0.& 1.& 6.& 21.\\ \hfill \phantom{-}\phantom{0}\phantom{.}\phantom{0}\frac{{x}^{7}}{7}×1.& 0.& 0.& 0.& 0.& 0.& 0.& 1.& 7.\\ \hfill \phantom{\phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{0×}\phantom{-}=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}& \phantom{=.}\end{array}\\ \begin{array}{l}\phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.\hfill \\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}3333.33333.33333.33333.33333.3\hfill \\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.00000.00000.0\hfill \\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}8571.42857.14285.71428.57142.8\hfill \\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}1111.11111.11111.11111.11111.1\hfill \\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}9090.90909.09090.90909.09090.9\hfill \end{array}\end{array}& \begin{array}{c}\begin{array}{}\phantom{\text{0}}\\ \phantom{\text{0}}\\ 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}\text{. &c.}\end{array}\\ \begin{array}{c}\begin{array}{r}\phantom{0}\\ \begin{array}{r}\begin{array}{c}\hfill \text{As if}\phantom{\rule{0.5em}{0ex}}x=\frac{1}{10}\phantom{\rule{0.5em}{0ex}}\text{. or}\phantom{\rule{0.5em}{0ex}}cq=\frac{11}{10}=1,1\phantom{\rule{0.5em}{0ex}}\text{.}\hfill \\ \hfill {\text{y}}^{\text{n}}\phantom{\rule{0.5em}{0ex}}\text{is}\phantom{\rule{0.5em}{0ex}}dapq=\frac{1}{10}-\frac{1}{200}+\frac{1}{3000}-\hfill \\ \hfill -\frac{1}{40000}+\frac{1}{500000}-\frac{1}{6000000}\phantom{\rule{0.5em}{0ex}}\text{. &c}\hfill \end{array}\end{array}\end{array}\\ \begin{array}{r}{\text{y}}^{\text{t}}\phantom{0}\text{is}\phantom{0}apqd=0,10000.00000.000000\phantom{.}\\ \frac{1}{3}{x}^{3}=+0,00033.33333.333333\phantom{.}\\ \frac{1}{5}{x}^{5}=+0,00000.20000.000000\phantom{.}\\ \frac{1}{7}{x}^{7}=+0,00000.00142.857142\phantom{.}\\ \frac{1}{9}{x}^{9}=+0,00000.00001.111111\phantom{.}\\ \frac{1}{11}{x}^{11}=+0,00000.00000.009090\phantom{.}\\ \frac{1}{13}{x}^{13}=+0,00000.00000.000076\phantom{.}\\ \underset{¯}{\phantom{000}\frac{1}{15}{x}^{15}=+0,00000.00000.000000\phantom{.}}\\ 0,10033.53477.310755.\\ \text{Summa}\phantom{000000000}\end{array}\end{array}\\ \phantom{.}\end{array}\end{array}\\ \begin{array}{l}\begin{array}{l}\phantom{-}\phantom{0}\phantom{.}\phantom{0}9230.76923.07692.30769.23076.9230769230\phantom{.}\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}6666.66666.66666.66666.66666.6666666666.\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0058.82352.94117.64705.88235.2941176470.\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.52631.57947.36842.10526.3157947368\phantom{.}\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00476.19047.61904.76190.4761904761.\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00004.34782.60869.56521.7391304347.\end{array}\end{array}\end{array}\\ \begin{array}{}\begin{array}{l}\phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.04000.00000.00000.0000000000\phantom{.0000000000.00000000}\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.000==\phantom{.}=====\phantom{.}=====.\phantom{0000000000.}\phantom{0000000000.00000000}\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{0000.00000.000}37.03703.70370.3703703703\phantom{.0000000000.00000000}\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.34482.75862.0689655172.4137931034.48275862\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.00===\phantom{.}=====\phantom{.}=\phantom{000000000.0000000000.00000000}\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{0000.00000.00000.}\phantom{00}322.58064.5161290322.5806451612.90322580\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.00003.03030.3030303030.3030303030.30303030\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}\phantom{0000.00000.00000.}\phantom{0000}0\phantom{.}02857.1428571428.5714285714.28571428\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.00000\phantom{.}00027.0270270270.2702702702.70270270\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.00000\phantom{.}00000.2564010256.4010256401.02564010\\ \phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.00000\phantom{.}00000.0024390243.9024390243.90243902\end{array}\\ \begin{array}{c}\phantom{-}=\phantom{.}\phantom{}8063.57265.52007.40736.63159.415063\phantom{\rule{0.5em}{0ex}}\text{&c}=\text{summæ}\hfill \\ \phantom{\phantom{-}\phantom{0}\phantom{.}\phantom{0}0000.00000.00000.00000.00000.0000000000.0000000000.00000000}\end{array}\\ \begin{array}{r}\hfill -0.00500.00000.00000.00000.00000.00000.00000.00000.00000.00000.0\phantom{.}\\ \hfill \phantom{-0.0000}2.50000.00000.00000.00000.00000.00000.00000.00000.00000.0\phantom{.}\\ \hfill \phantom{-0.00000.}\phantom{0}1666.66666.66666.66666.66666.66666.66666.66666.66666.6\phantom{.}\\ \hfill \phantom{-0.00000.}\phantom{000}12.50000.00000.00000.00000.00000.00000.00000.00000.0\phantom{.}\\ \hfill \phantom{-0.00000.00000}10000.00000.00000.00000.00000.00000.00000.00000.0\phantom{.}\\ \hfill \phantom{-0.00000.00000000}83.33333.33333.33333.33333.33333.33333.33333.3\phantom{.}\\ \hfill \phantom{-0.00000.0000000000}71428.57142.85714.28571.42857.14285.71428.5\phantom{.}\\ \hfill \phantom{-0.00000.000000000000}630.55555.55555.55555.55555.55555.55555.5\phantom{.}\\ \hfill \phantom{-0.00000.0000000000000000}5045.45454.54545.45454.54545.45454.5.\\ \hfill \phantom{-0.00000.00000000000000000000}41666.66666.66666.66666.66666.6\phantom{.}\\ \hfill \phantom{-0.00000.0000000000000000000000}384.61538.46153.84615.38461.5\phantom{.}\\ \hfill \phantom{-0.00000.000000000000000000000000}3.57142.85714.28571.42857.1\phantom{.}\\ \hfill \phantom{-0.00000.00000000000000000000000000}3364.58333.33333.33333.3\phantom{.}\\ \hfill \phantom{-0.00000.000000000000000000000000000000}29411.76470.58823.5\phantom{.}\end{array}\\ -0.00502.51679.26750.72059.17744.28779.27385.30147.14044.12586.\phantom{0.}\end{array}\end{array}$ cui addendum $\begin{array}{}\begin{array}{rr}-277.77777.77777.7& \\ \phantom{00}-2.63157.89421.0& \\ \phantom{000.}-02523.80952.4& \\ \phantom{000.00000.}-22727.3& \\ \phantom{000.00000.00}-217.4& \text{that is}\\ \phantom{000.00000.0000-}2.1& \end{array}\\ \begin{array}{cc}-280.43459.71098.0& \phantom{\text{that is}}\end{array}\\ \begin{array}{}\end{array}\end{array}$ And so yethe summe will bee $\begin{array}{}\begin{array}{r}+0.10033.53477.31075.58063.57265.52007.40736.63159.41506.3\\ -0.00502.51679.26750.72059.17144.28779.27385.30427.57503.8\end{array}\\ \begin{array}{c}\phantom{-}0.09530.01798.04324.86004.40121.23228.13351.32731.84002.5\end{array}\\ \begin{array}{}\end{array}\end{array}$ wchwhich is yethe quantity of yethe area $adpq$. If $cpab=1$. & $cp=$ $ab=10pq$ & $qde||ap||bc=ap$ . In like manner if I make $x=\frac{1}{100}=pq$ . The opperacotion followeth. $\begin{array}{}\begin{array}{r}\phantom{+}0,01000.03333.33333.33333.33333.33333.33333.33333.33333.3\phantom{0}\\ +\frac{1}{5}{x}^{5}+\frac{1}{7}{x}^{7}=\phantom{\rule{2em}{0ex}}33333.20001.42857.14285.71428.57142.85714.28571.40\\ \phantom{\frac{1}{9}{x}^{9}+\frac{1}{11}{x}^{11}=\phantom{\rule{2em}{0ex}}}==\phantom{.}=====\phantom{.}=====\phantom{.}=====\phantom{.}=====\phantom{.}=\phantom{0202.0\phantom{0}}\\ \frac{1}{9}{x}^{9}+\frac{1}{11}{x}^{11}=\phantom{\rule{2em}{0ex}}11.11202.02020.20202.02020.20202.0\phantom{0}\\ \frac{1}{13}{x}^{13}+\frac{1}{15}{x}^{15}=\phantom{\rule{2em}{0ex}}769.23076.92307.69230.7\phantom{0}\\ \frac{1}{17}{x}^{17}=\phantom{\rule{2em}{0ex}}6666.66666.66666.6\phantom{0}\\ \frac{1}{19}{x}^{19}=\phantom{\rule{2em}{0ex}}58823.52941.1\phantom{0}\\ \frac{1}{21}{x}^{21}=\phantom{\rule{2em}{0ex}}5.26315.7\phantom{0}\\ 47.6\phantom{0}\end{array}\\ \begin{array}{r}+0,01000.03333.53334.76201.58821.07551.40422.38870.97309.\phantom{30}\end{array}\\ \begin{array}{r}-0,00005.00025.00166.66666.66666.66666.66666.66666.66666.6.\\ -\frac{1}{8}{x}^{8}-\frac{1}{10}{x}^{10}-\frac{1}{12}{x}^{12}=\phantom{\rule{2em}{0ex}}-1250.10000.83333.33333.33333.33333.3\phantom{.}\\ -\frac{1}{14}{x}^{14}=\phantom{\rule{7em}{0ex}}-7.14285.71428.57142.8.\\ -\frac{1}{16}{x}^{16}-\frac{1}{18}{x}^{18}=\phantom{\rule{2em}{0ex}}-62.50555.55555.5\phantom{.}\\ =====\phantom{.}=====\phantom{.}=\phantom{.}\\ -\frac{1}{20}{x}^{20}=\phantom{\rule{2em}{0ex}}-5000.4.\end{array}\\ \begin{array}{r}-0,00005.00025.00166.67916.76667.50007.14348.21984.17699.\phantom{0.}\end{array}\\ \begin{array}{}\\ +0,00995.03308.53168.08284.82153.57544.26074.16886.79610\phantom{.0.}\end{array}\\ \begin{array}{}\end{array}\end{array}$ wchwhich is yethe quantity of yethe area $apqd$ if $100p=cp$ . and $abcp=1$ 21 (13 $y=db$. $x=ba$ $aay={x}^{3}$. $a$ $b$ $+$ $y$ $z$ $=y$ $z=bc$ $aab+aa$ $y$ $z$ $={x}^{3}$. $a$ $b$ $-$ $y$ $z$$=y=$ $z=bf$. $aab-aa$ $y$ $z$ $={x}^{3}$. $y$ $z$$-$ $a$ $b$$=y=$ $z=bg$ $aa$ $y$ $z$$-$ $a$ $ab={x}^{3}$. $x=d+\xi$ $\xi =ah$. $aa$$y$$-{d}^{3}-3dd\xi -3d\xi \xi -{\xi }^{3}=0$ ξ $\begin{array}{}aab+aaz\\ aab-aaz\\ aaz-aab\end{array}\right\}=\begin{array}{}\\ {d}^{3}+3dd\xi +3d\xi \xi +{\xi }^{3}\\ \end{array}$. $x=b-\xi$ . $aq=\xi$ $\begin{array}{}aay\\ aab±aaz\\ aaz-aab\end{array}\right\}=\begin{array}{}\\ {d}^{3}-3dd\xi +3d\xi \xi -{\xi }^{3}\\ \end{array}$. $x=\xi -b$ . $ak=x$ $\begin{array}{}aay\\ aab±aaz\\ aaz-aab\end{array}\right\}=\begin{array}{}\\ {\xi }^{3}-3d\xi \xi -3dd\xi -{d}^{3}\\ \end{array}$. $na=\xi$. $nd=z$. Or. ${\xi }^{3}-{a}^{2}z+bb\xi$ $ab:an\colon\colon a:b$. &. $ab:nb\colon\colon a:c$ then ${\xi }^{3}=\frac{{b}^{3}}{a}z-\frac{bbc}{a}\xi$. or if $abn$ is a right angle. & $\xi =d+\zeta$ $\zeta =mn$ ${d}^{3}+3{d}^{2}\zeta$ $+bbc\zeta$ $+3d\zeta \zeta +{\zeta }^{3}-\frac{{b}^{3}}{a}z+\frac{bbcd}{a}$ 22 (15 $db=x$. $ba=y$. $aax={y}^{3}$ . $x=b+z$. $z=bc$. $y=c+\xi$. $\xi =ah$ $\begin{array}{}aab+aaz=\\ \left(x=b-z.z=bf.\right)\\ aab-aaz=\\ \left(x=z-b.z=bg\right)\\ aaz-aab=\end{array}\right\}\begin{array}{l}{\xi }^{3}+3c{\xi }^{2}+3cc\xi +{\xi }^{3}.y=c-\xi .\xi =aq\phantom{.}\\ {c}^{3}-3cc\xi +3c{\xi }^{2}-{\xi }^{2}.y=\xi -c.\xi =ak.\\ {\xi }^{3}-3c{\xi }^{2}+3\xi {c}^{2}-{c}^{3}.\end{array}$ $na=y$. $d$ $n$ $=x$. $ab:an\colon\colon a:b$. $ab:nb\colon\colon a:c$. & ${y}^{3}=\frac{{b}^{3}}{a}x-\frac{bbc}{a}y$. Or ${d}^{2}x=\epsilon \epsilon y+{y}^{3}$. $\begin{array}{c}\left(x=z+o.z=nc\right)\phantom{{\xi }^{3}}\\ {d}^{2}z+{d}^{2}o=\phantom{{\xi }^{3}}\\ \left(x=z-o.z=gn\right)\phantom{{\xi }^{3}}\\ ddz-ddo=\phantom{{\xi }^{3}}\\ \left(x=o-z.z=nf\right)\phantom{{\xi }^{3}}\\ ddo-ddz=\phantom{{\xi }^{3}}\\ ddz\end{array}\right\}\begin{array}{l}\epsilon \epsilon y+{y}^{3}.\left(y=n+\xi .\xi =na\right)\\ {\xi }^{2}n+\epsilon \epsilon \xi +{n}^{3}+3nn\xi +3n{\xi }^{2}+{\xi }^{3}.\\ \left(y=n-\xi .\xi =as\right)\\ \epsilon \epsilon n-\epsilon \epsilon \xi +{n}^{3}-3nn\xi +3n{\xi }^{2}-{\xi }^{3}.\\ \left(y=\xi -n.\xi =av.\right)\\ \epsilon \epsilon \xi -\epsilon \epsilon n+{\xi }^{3}-3n{\xi }^{2}+3\xi {n}^{2}-{n}^{3}.\\ \phantom{ddz}\end{array}$ $al=y$. $dl=x$. $ab:$ $al$ $\colon\colon a:b$ &. $al:bl\colon\colon b:c$. whence ${y}^{3}=\frac{x{b}^{3}+y{b}^{2}c}{a}$ or ${y}^{3}=\frac{y{b}^{2}c}{a}=$ ${y}^{3}-\epsilon \epsilon y=ddx$ &c: as before onely varying yethe signes at $\epsilon \epsilon n$ & $\epsilon \epsilon \xi$. $ao=y$. $do=x$. $a:b\colon\colon bd:do$. $b:c\colon\colon do:ob$. $\frac{{a}^{3}}{b}x={y}^{3}-\frac{3cxyy}{b}+\frac{3ccxxy}{bb}-\frac{{c}^{3}{x}^{3}}{bb}$. 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 ${x}^{3}=♉{a}^{3}♉3aae♉3aee♉{e}^{3}$. or ${x}^{3}=$ $+$ $3aex♉{a}^{3}♉{e}^{3}$. that is making ${a}^{3}+{e}^{3}=q$. ynthen ${x}^{3}=3aex♉{c}^{3}$ . & $3ae=p$. ynthen ${x}^{3}$$×$ $=$ $+px♉q$. Againe suppose $y$ $=♉$ $b$ $a$ $\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}$ $e$ Then ${y}^{3}$ Againe suppose $x=a-e$. ynthen ${x}^{3}={a}^{3}-3aae+3aee-{e}^{3}$. ytthat is making ${a}^{3}-{e}^{3}=♉q$, & $3ae=p$, ynthen ${x}^{3}=-px♉q$. Then in the first of these $p=3ae$. or $\frac{p}{3e}=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}qq-\frac{{p}^{3}}{27}}$. & by yethe same reason ${a}^{3}=\frac{1}{2}q\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{1}{4}qq-\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\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{1}{4}qq-\frac{1}{27}{p}^{3}}}♉\sqrt{c:\frac{1}{2}q♉\sqrt{\frac{1}{4}qq-\frac{1}{27}{p}^{3}}}$. is a rule for resolving yethe equation ${x}^{3}\ast -p\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}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}\ast +px\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}q$. may be resolved by this rule $x=a-e=\sqrt{c:\frac{1}{2}q♉\sqrt{\frac{1}{4}qq+\frac{1}{27}{p}^{3}}}-\sqrt{c:\frac{1}{2}q\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{1}{4}qq+\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}\ast +px♉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×bd=1-16+8x-{x}^{2}=4-{x}^{2}$ therefore $8x=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×bd=4-xx=4-\frac{361}{64}=\frac{256-361}{64}$ or $bd×bd=\frac{-105}{8}$. but it is impossible ytthat a square number should be negative. Thus $x=\sqrt{-b}$ is impossible. square it & tis $xx=-b$. Againe, & tis ${x}^{4}=bb$. Extract yethe roote & tis $xx=b$ or $x=\sqrt{b}$. wchwhich is possible. The reason of this proceeding Event is ytthat ${x}^{4}-bb=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{-aa}$, $-\sqrt{-aa}$ , $\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$ to $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\colon\colon$ area of yethe line $:$ to $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\colon\colon 1:p+2$, then $nmbq$ hung on yethe point $q$ shall counterballance yethe figure in its site &c therefore if $ac:cd\colon\colon p+1:1$, $c$ shall be yethe center of gravity of those figures. Also as the figure is now put extending infinitely towards $\delta$ if $-2p+1:-$$p$ $+1\colon\colon am:ac$. $m$ being yethe center of $qnbd$ ynthen $c$ shall bee yethe center of gravity of yethe whole figure $qndb\delta$ . Demonstration sinccesince yethe lines parallell to $a\delta$ 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 $-2p$. Therefore yethe figure is to yethe inscribed parallelogram as $1$ to $1-p$. & all its moments or whole weight of the Para in this its site to the weight of yethe pgrparallelogram as $1$ to $1-2p$. Therefore if, $am:ap\colon\colon 1-2p:1$, 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 $am:ac\colon\colon 1-2p:1-p$.
26 Of Refractions. 1 If yethe ray $ac$ bee refracted at the center of yethe circle $acdg$ towards $d$ & $ab⟂be⟂gc\parallel ed$. Then suppose $ab:ed\colon\colon d:e$. See Cartes Dioptricks 2 If there be an hyperbola whose the distance of whose foci $bd$ are to its transverse axis $hf$ as $d$ to $e$ . Then yethe ray $ac\parallel bd$ is refracted to yethe exterior focus $\left(d\right)$. See C: Dioptr 3 Having yethe proportion of $d$ to $e$, or. $bd:hf$. The Hyperbola may bee thus described. 1 Upon yethe centers $a$, $b$ let yethe instrument $adbtec$ bee moved in wchwhich instrumntinstrument observe ytthat $ad$ $⟂de⟂$ $c$ $et$ & ytthat the beame $cet$ is not in yethe same plane wthwith $adbe$ but intersects it at yethe angle $tev$ soe ytthat if $tv⟂ev$, then $d:e\colon\colon et:tv$. Or $d:e\colon\colon Rad:sine of \angle tev$. Also make $de=\frac{q}{2}$, i.e half yethe transverse diameter. Then place the fiduciall side of plate $chm$ in the same plaine wthwith $ab$ . & moving yethe instrument $adbect$ to & fro its edge $cet$ shall cut or weare its into yethe shape of yethe desired Parabola. Or the plate $chm$ may bee filed away untill yethe edgedge $cet$ exactly touch it everywhere. 2 By the same proceeding Des=Cartes concave Hyperbolicall wheele may bee described by beeing turned wthwith a chissell $d$ $tec$ whose edge is a streight line inclined to the edge axis of the mandrill by yethe $tev$ wchwhich angle is found by making $d:e\colon\colon et:tv\colon\colon Rad:sine of etv$. 3 By the same reason a wheele may be turned Hyperbolically concave yethe Hyperbola being convex. Or a Plate may bee turned Hyperbolically concave Also Des=Cartes his Convex wheele $B$ may be turned or ground trew a concave wheele $A$ being made use of instead of a patterne 5 In turning the concave wheele $A$ it will perhaps bee best to weare it wthwith a stone $p$ & let the streight edged chissell $d$ serve for a patterne. And it may bee convenient to grind yethe stone (or iron &c ) $p$ into yethe fashion of a cone $S$ That it may fit yethe hollow of the wheele $A$. The angle of wchwhich concecone being a right one or something greater it will almost grind the wheele to a tre figur Same done by helpe of a Cone. 6 Draw $ab=q$; $ac$ & $cb$ of any length or intersecting one another at any angles. to make up the triangle $abc$ . [Suppose ytthat $ac$ bee called $b$, & that $e=q=$ then is $d$ the distance of yethe foci]. produce $ac$ to $d$ soe that $ad=\frac{dd-ee}{b}$. Then draw $bk$ through yethe point $d$, & draw $eh$ parallell to $bc$ , lastly wthwith the sides $he$ , $ek$ & angle $hek$ describe the cone $heklm$ . Then produce $ba$ to $g$ &c indefinitely & $ag$ being yethe axis of a section $mal$ shall be yethe sought Par Hyperbola 7 Since the proportion of $cb$ to $ab$ & ∠ $cba$ is not determined it will be most convenient to make 27 $cb=ab=e=q$ . & $cb⟂ab⟂ah$. And then there will bee little danger of error at yethe vertex of the Hyperbola. And yethe calculation is readier for drawing $bp⟂ca$, Then is $cp=bp=ap=\frac{e}{\sqrt{2}}=\sqrt{\frac{ee}{2}}$ & $pd=\frac{dd}{e\sqrt{2}}$. Soe that $ee:dd\colon\colon cp=bp:pd\colon\colon$$Rad:tangen$ $\colon\colon Radius:Tangent of pbd$ soe ytthat yethe $\angle cbk=hek$ is easily found. 9 Halving such a cone smoothly pollished wthwithin & wthwithout, by the helpe of a square set yethe plate perpendicular to one side $hae$ the fiduciall edge being distant from yethe vertex the length of $ae=\frac{edd-{e}^{3}}{dd+ee}$ & if yethe edge of yethe plaine every where touch the cone, tis trew. 10 The exact distance $\left(ae\right)$ of yethe plate from the vertex of yethe cone neede not bee much regarded for that changeth onely the shape bigness not yethe shape of yethe figure. [By yethe broken lookinglasse I find in glasse refraction, ytthat $d:e\colon\colon 43:28\colon\colon 1000:651+\colon\colon 1536$$:1000$. These are insensib almost insensibly different from truth $d:e\colon\colon 20:13\colon\colon 100:65\colon\colon 153-:1000$. Or $d:e\colon\colon$$23:$ $23:15$ $\colon\colon 100:652+$ $d:e\colon\colon 66:43\colon\colon 100:651,5151+$. Or $d:e\colon\colon 100:651$ For yethe Ellipsids $\frac{dd-ee}{d}x+\frac{ee-dd}{dd}xx=yy$ The former $\left\{\begin{array}{}\text{descriptions}\\ \text{propositions}\end{array}\right\}$ demonstrated. Lemma. If in yethe Opposite Hyperbolas one of $abc$ $edf$ (one of wchwhich are to bee described) supposing $bd=d$. $hf=e$. $gh=x$ $gc=y$. $gc⟂ghd$ & point & $gc$ terminated by yethe hyperbola Then is $\frac{dd-ee}{ee}xx+\frac{dd-ee}{e}x=yy$. $b$ $d$ $h$ $=\frac{d-e}{2}$ . $dh=\frac{d+e}{2}$ . $d$ $b$ $g=\frac{2x-d+e}{2}$. $gd=\frac{2x+d+e}{2}$. ${dc}^{2}=\frac{4xx+4dx+4ex+dd+2ed+ee+4yy}{4}=gd×gd+gc×gc$. ${bc}^{2}=\frac{4xx-4dx+4ex+dd-2ed+ee+4yy}{4}={gb}^{2}+{gc}^{2}$. And since $af$ $dc=bc+hf$. Or ${dc}^{2}={bc}^{2}+2bc×hf+{hf}^{2}$ Therefore $2dx+ed-ee=e\sqrt{4xx-4dx+4ex+dd-2ed+ee+4yy}$. Both parts of wchwhich □edsquared & ordered yethe result is $4ddxx-4eexx+4eddx-4{e}^{3}x-4eeyy=0$. That is $\frac{dd-ee}{e}x+\frac{dd-ee}{ee}xx=yy$. DesciptionDescription yethe 1st demonstrated Synthetically. See ytthat Scheame Naming yethe quantitys $ed=\frac{e}{2}=dh$ $d:e\colon\colon$ $:$ e $tv$ Nameing yethe quantitys $ed=dh=\frac{e}{2}$. $gh=x$. $gc=be=y$. $dg=x+\frac{e}{2}=nc$. $cg⟂dhg$. ${cd}^{2}={x}^{2}+ex+\frac{ee}{4}+y$ ${ce}^{2}=xx+ex+yy$. ${eg}^{2}=xx+$$ex$. $d:e:$ Also $d:e\colon\colon et:tv\colon\colon ce:eg$, therefore $ddxx+ddex=ee{x}^{2}+{e}^{3}x+{e}^{2}y$ That is $\frac{dd-ee}{e}x+\frac{dd-ee}{ee}xx=yy$. As in yethe lem̄a The Same demonstrated synthe Analytically. Nameing yethe quantitys, $de=dh=$$a$ . $gh=x$. $gc=y$. $dg=a+x$ ${dc}^{2}=aa+2ax+xx+yy$. ${ce}^{2}=2ax+{x}^{2}+yy$. ${eg}^{2}=xx+ex$. Supose ytthat $b:c\colon\colon et:tv\colon\colon ce:eg$. 28 Then is $bbxx+bbex=2ccax+ccxx+ccyy$. That is $\frac{bb-cc}{cc}xx+\frac{bbe-2cca}{cc}x=yy$. Therefore yethe line $chm$ is a Conick Section & since $\left(bb\right)$ is greater ynthan $\left(cc\right)$ tis an Hyperbola, wchwhich ytthat it may bee yethe same wthwith ytthat in yethe lemma, Their correspondntspondent termes are to bee compared together & soe I find ytthat $\frac{bb-cc}{cc}xx=\frac{dd-ee}{ee}xx$. & $\frac{bbe-2cca}{cc}x=\frac{dd-ee}{e}x$ by yethe 1st =tionequation $bb=\frac{ccdd}{ee}$. Or $b=\frac{cd}{e}$. ytthat is $b:c\colon\colon d:e$. by yethe 2nd $ccee-ccdd+bbee=2ccea$. And by substituting $\frac{ccdd}{ee}$ into the place of $bb$ And ordering it tis 2 $ccee=2ccea$. Or $\frac{e}{2}=a$. Therefore if I take $\frac{e}{2}=a=de$. & $d:e\colon\colon b:c\colon\colon ct:tv$. then shall $chm$ bee yethe ParaHyperbola 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$. ☞ Instead of yethe 6th & 7th Descriptions wchwhich are false use these 6 Draw 2 concentrick circles ($na$ & $cd$) wthwith yethe Radij $e$ & $d$. Then from yethe commmon center $b$ draw 2 lines $bc$ & $\left\{\begin{array}{}ba\\ bd\end{array}\right\}$ at the given angle $\left\{\begin{array}{}bae=abc\\ aed=cbd\end{array}\right\}$ of $\left\{\begin{array}{}\text{section}\\ {\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{Cone}\end{array}\right\}$ then draw a line $cad$ thr from $c$ throug by yethe end of yethe Rad $\left\{\begin{array}{}ba\\ bd\end{array}\right\}$ & to yethe intersection of ytthat line wthwith yethe circle $\left\{\begin{array}{}cd\\ na\end{array}\right\}$ draw $\left\{\begin{array}{}bd\\ ba\end{array}\right\}$ & so the angle of $\left\{\begin{array}{c}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{Cone,}\phantom{\rule{0.5em}{0ex}}hek=cbd\\ \text{section,}\phantom{\rule{0.5em}{0ex}}eab=abc\end{array}\right\}$ is found. Or wchwhich is the same make $ab=e$. $bd=d$ & then if ytthat cone is sought make $cd=2r$ the angle $cba$ being given, make $ac=a$. Then is $cd=\frac{dd-ee+aa}{a}$. & soe yethe $\angle cbc=aed$ is knowne & also $ae=ed=\frac{{d}^{3}-dee}{dd-ee+aa}$, & $ad=\frac{dd-ee}{a}$. Buf But if yethe $\angle bae=abc$ of yethe section is sought yethe cone being given ynthan make $cd=2b$. And it will bee $ac=b+\sqrt{ee-dd+bb}$. & soe $\angle abc=bae$ is given also $ad=b-\sqrt{ee-dd+bb}$. & $ae=\frac{db-d\sqrt{ee-dd+bb}}{2b}$ In generall observe ytthat in any cone cut any ways $bd=be+ea=d$. & $b$$a=e$. 7. DesCartes his wheele thus described cut by any plaine produceth one of yethe Conick=Sections. Description yethe 6th Demonstrated. Synthetically. Call, $bd=d$. $ba=e$. $cp=pd=a$. $bp=\sqrt{dd-aa}$. $ag=x$ $ap=\sqrt{ee-dd+aa}$. $ac=a+\sqrt{ee-dd+aa}$. $ad=a-\sqrt{ee-dd+aa}$. $ba:ac\colon\colon ag:gh=\frac{ax+x\sqrt{ee-dd+aa}}{e}$. $ba:ad\colon\colon bg:gk=\frac{ea+ax}{e}\phantom{\rule{1em}{0ex}}\frac{-e-x}{e}\sqrt{ee-dd+aa}$. $gk×gh={gm}^{2}={y}^{2}$. Therefore $\frac{dd-ee}{ee}xx+\frac{dd-ee}{e}x=yy$. by ordering yethe result of $gk×gh$. wchwhich is like ytthat in the lemma. The 7th Proposition may be easyly demonstrated after the same manner. If the two equall cones $bad$ $bcd$ intersect the one the other soe ytthat $ab=bc$ their intersection $\left(bf\right)$ shall bee one of yethe Conick sections as they had each beene intersected by the plane $bf$. 29 To describe yethe Parabola (& other figures after yethe same manner) pretty exactly. Thake a squire $cbe$ , soe ytthat $cb=\frac{r}{2}$ (for then the circle described by $\left(bc\right)$ will bee as crooked as yethe Parabola at the vertex $d$ ). Divide yethe other leg $\left(be\right)$ of yethe Squire into any number of ptspoints, Then get a plate of Brasse &c: $lkfd$ streight & eaven. And taking one point $d$ for yethe vertex of it & another point $c$ for yethe Squire to moven soe ytthat $cd=cb=\frac{r}{2}$ , & weareing away yethe edge of the plate untill (yethe Squire being erected) $ab=qd$. the squire touching yethe plate at $a$ . thus shall yethe edge $adf$ become Parabolicall. wthwith yethe $Rad: ab$ describe a circle $adg$ & by that meanes it may bee knowne when $ab=a$. Instead of yethe leg $be$ a Demonstracocircle may be used Supose $aq=y$. $cd=cb=\frac{r}{2}$ then . Then is $\sqrt{\frac{rr}{4}}+$ $zz$ $cq$ $ad=\sqrt{xx+yy}=ab$. & $ac=\sqrt{\frac{rr}{4}+xx+yy}$. And $cq=\sqrt{\frac{rr}{4}+xx}$ . & $dq=x$. Demonstracon. $qd=x$. $cd=\frac{r}{2}=cb$. $cq=\frac{r}{2}-x$. $aq=\sqrt{rx}=y$. ${ac}^{2}=\frac{rr}{4}+xx$ ${ab}^{2}=\frac{rr}{4}+xx-\frac{rr}{4}$ & $ab=x$. Q.E.D. Another description of yethe Parabola ywethwith yethe compasses. Make $ab=bc=\frac{r}{4}$. Make $ce=cd$ & $ce⟂bd$. Make $af=ae$, & $bf=bd$ then shall $f$ be a point in yethe Parabola. Another. Make $ab=\frac{r+x}{2}=ac$. $eb=x⟂ce$ & yethe point $c$ shall bee in yethe parabola. This like yethe first by calculation may bee made use of in other lines. 29 The manner whereby any kind of little lines may be described very accurately. And that the same Instrument serve for all lines (though never so small) differing in quantity but not in quality. Make yethe plate $d$ of yethe figure required (by some of yethe former meanes) the larger the better. Then hold the streight steele staffe $b$ against the center $a$ & rouleroute it to & fro it shall grind $c$ into yethe same figure but soe much lesse as $ac$ is lesse ynthan $ad$. Soe if yethe glass $c$ bee fastened upon yethe mandrill $f$, it may be ground acording to yethe sollid f figure $d$ by yethe helpe of a stick of steele (as a cilind cone) whose cuspis is in yethe hole $a$ upon wchwhich it is moved as on a center. when yethe stick cone $b$ leanes uppon yethe vertices of $d$ & $c$ it must be perpendicular to the mandrill $f$. Perhaps it may be convenient to cause yethe cone $b$ to turne about its axis. Or it may bee better instead of yethe nutt at $a$ wthwith a hole in it to make a sharpe pointed nutt, & instead of yethe cilinder cone $b$ to make use of a broad plate to cover $a$, $c$ & $d$ & move every way upon them 30 Another way to describe lines on plates Suppose yethe plate bee $abc$, whose edgedge $boc$ is to be made into yethe fashion of a given crooked line suppoessuppose $\left(o\right)$ is its vertex & ytthat a circle described wthwith yethe Radius $eo$ would bee as crooked as yethe given line at its vertex. Againe suppose two streight rulers $mn$ & $pq$ to bee very trew & steddyly fastened together wchwhich must a very little incline yethe one to yethe other, soe as that being produced they would meete at $ar$. Then are yethe lines $pn=a$ , & $pr=b$ given. Suppose ynthen yethe point $d$ in yethe crooked line is to bee found ynthen is $dc$ given by supposition, & consequently (supposing $dk$ to bee a tangent) $dg=y$. $gc=x$. $fg=v$. $fd=s$ $ec=c$. $fk=v+\frac{yy}{v}$ . $ef=v+x-c$. $ek=c-x+\frac{yy}{v}$. & (if $eh⟂dk⟂df$) then is $eh=\frac{cv-xv+yy}{\sqrt{vv+yy}}=d$. $\left(eh\right)$ being thus found, supposing ytthat $pn=a=ec$, then I take $re=\frac{bd}{a}$. that is $pe=b-\frac{bd}{a}$. haveing thus found yethe point $e$ lay yethe plate twixt two rule the two rulers so ytthat yethe point of it, fall upon yethe point $e$ ynthen should yethe line $mn$ touch yethe plate in $d$. But note ytthat $pn⟂mn$. In both telescopes & microscopes tis most convenient to have a convex glasse next yethe eye for by that meanes yethe angle of vision will bee much greater ynthan it will bee wthwith a concave one (though both doe magnifie alike). If yethe convex glasse be Hyperbolicall (&c) make it soe bigg ytthat yethe penecilli may crosse in yethe pupill; ytthat is, yethe exterior focus will be as far distant from yethe vertex as yethe eye is. let yethe glass bee as thinn ytthat yethe as may bee ytthat yethe eye bee not too farrfar from yethe vertex yetytthat it should bee about as thick as yethe distance of yethe interior focus from yethe vertex. And by this meanes also, (yethe focus of yethe objectglasse being within yethe telescope twixt yethe glasses) there may bee placed at that focus yethe edge of a steele ruler accurately divided into equall parts (to measure yethe diameters or distances of starrs &c) wchwhich should bee soe made ytthat by a pinne or handle it may be placed in any posture & in any parte of yethe focus, wthwithout otherwise altering yethe Telescope in observations. Note that were not yethe 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 parte of yethe pupill must bee more exactly refracted to one point of yethe Tunica Retina ynthan in ordinary visioon in wchwhich each of yethe penicilli spreads over all the pupill. ☞ Note also that ytthat yethe glasse $a$ may be ground Parabolic Hyperbolicall by yethe line $cb$, if it turne on yethe mandrill $e$ whilst $c$$b$ turnes on yethe axis $rd$ being inclined to it as was shewed before. If the edge $\left(cb\right)$ bee not durable enough, inough instead thereof use a long small cilinder: wchwhich I conceive to bee the best way, of all. For a Cilinder of all sollids is most easily made exact (being turned, as in the figure, by a gage untill its thicknesse bee every where equall). 2 the Cilinder may bee made to slip up & downe & turne round whereby it will not onely grinde yethe glase crosse wise to take of all hubbes, but also yethe glasse & cilinder will grinde yethe one yethe other truer & truer. All yethe difficulty is in placing yethe axis $rd$ perpendicular to the Mandrill $ae$ & vertex to vertex, wchwhich yet may bee done exactly severall ways. & untill ynthen the glasse & Cilinder will not fit. & should yethe axis not intersect yethe 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 hang over the glasse. 31 To Grinde Sphæricall optick Glasses If yethe glasse $\left(bc\right)$ is to bee ground sphærically hollow: naile a steele plate to yethe beame $\left(fg\right)$, on yethe upper side: In wchwhich make a center hole for yethe steele point $\left(f\right)$ of yethe shaft $\left(def\right)$: to wchwhich shaft fasten a plugg $\left(a\right)$ of stone or leade or leather &c: (wthwith wchwhich you intend to grinde yethe glasse $\left(bc\right)$): wchwhich shaft & plugg being swung to & fro upon yethe center $f$ will grind yethe glasse $bc$ sphærically hollow. The manner whereby glasses may bee ground sfphærically convex may appeare by yethe annexed figure (being yethe former way inverted). Also yethe plugg $\left(a\right)$, in yethe $\begin{array}{}\text{first}\\ \text{second}\end{array}$ figure, is ground sphærically $\begin{array}{}\text{convex.}\\ \text{concave.}\end{array}$ But if this way bee not exact enough yet hereby may bee growndground plates of mettall well nigh sphæricall, And by those plates may bee ground glasses after yethe usual manner; If a circular hoope of steele $\left(abc\right)$ bee put about yethe edge of yethe glasse $\left(d\right)$ to keepe it from grinding away at yethe edges faster ynthan in yethe middle. But the best way of all will bee to tiurne yethe glass circularly upon a mandrill whilest yethe plate is steadily rubbed upon it or else 31 to turne yethe plate upon a mandrill whilest yethe glasse is rubbed upon it or let sometimes yethe one, sometimes yethe 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 wchwhich weares more difficultly then glasse to defend yethe glasse from wearing more at its edges then in yethe middle. Perhaps it may doe well first to weare yethe plate sphæricall by yethe hoope alone wthwithout the glasse. The same meanes may bee used for grinding 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 yethe other, & turne yethe convexest side towards yethe object. 32 If the Glasses of a Telescope bee not truely ground, how to find where the fault is, & consequently to rectify it. Take two plates $abfh$, $dgkh$ of wood or brasse iat yethe midst of yethe sides of wchwhich boare two very small holes $c$ & $e$ (viz whose diameters are about yethe 20th or 30th pteparte of an inch, that they may transmit but soe much light as may serve to see yethe edge of yethe sunne or a starre of yethe first magnitude) Also make two other plates $rs$, & $tv$ like yethe former but yethe holes in ymthem must be as small as can bee (viz about yethe 100th pteparte of an inch in diameter or lesse). For yethe small end of yethe tube Also make another plate $wzxy$ wthwith a hole in the midst of it about yethe 5th or sixt pteparte of an inch in diameter (viz equall to yethe diameter of yethe pupill of yethe eye or eyeglasse). Make yethe like plate $A$ (but wthwith a very small hole) for yethe eyeglaesseyeglasse) First cover yethe object glasse wthwith yethe plates $af$, & $gh$ distant about yethe yethe holes in ymthem being distant about yethe sixt parte of an inch; & placed neare yethe center of yethe object glasse. Also cover yethe eye glasse wthwith yethe plate $A$ soe ytthat its hole exactly respect yethe center of yethe eye glasse then turne yethe tube to a Starre wchwhich will appeare like two starrs if yethe tube bee twotoo long or short, wchwhich bee shortned or lengthned untill there appeare but one, And then is yethe Tube of a good distance length for yethe vertices of yethe Glasses. Secondly remove those plates, and instead thereof cover yethe object glasse wthwith yethe plate $wz$, its hole exactly respecting yethe center of yethe glasse. (Fig 2) If yethe Glasses of a Telescope bee not truely ground Theire errors may bee may bee thus found. Because an error is much more easily discernable in yethe object glasse ynthan in yethe eyeeye glasse let us first suppose yethe eye glasse to bee ground true towards its center, (tis exact enough if it be sphericall, & not Hyperbolicall), & so wee may find & rectifie yethe errors of yethe object glasse. First make a thin plate $\left(A\right)$ of brasse & in the center of it a Small hole (whose diameter perhaps may bee about yethe 50th or 100dth parte of an inch. With wchwhich plate cover yethe eye glass yethe center of it respecting yethe center of yethe glasse. Secondly make two other plates the one $B$ wthwith two holes as neare to its edge as may bee theire distance being about yethe 5tth pteparte of an inch or lesse, & yethe other $C$ wthwith one hole close to yethe midst of its edge. Let yethe diameters of these 3 holes bee about a 20th pteparte of an inch or lesse. And theire edges must bee true that they may slide one upon another, & yet not let yethe suns rays passe through, to wchwhich purpose make ymthem oblique. Wthwith these two plates cover yethe object glaseglasse (first stopping yethe hole of $C$ yethe holes of yethe other plate respecting yethe center of yethe glasse & looke at a stare (or yethe edge of yethe sunne &c) & if yethe object appeare double (like two starrs &c) make yethe Tube longer or shorter untill it appeare single. Then open yethe hole of $C$ , & yethe plate $B$ being fixed, slide yethe plate $C$ up & downe still looking at yethe starre, When then appeares 33but one starre ytthat part of yethe glasse under yethe hole of $C$ is truely ground in respect of yethe 2 parts of yethe glasse under yethe two holes of $B$. But no when yethe starre appeares double. And yethe position of yethe starre caused by yethe hole of $C$ in repectrespect of yethe starre caused by yethe holes of $B$, shews yethe error of yethe incli wchwhich way yethe glasse under yethe hole of $C$ is erroneously inclined; the distance of yethe two starres giving yethe quantity of ytthat inclinati error. Thus yethe errors of yethe of object glasse being found in every place of it they may bee all rectified, & found againe, & againe rectified, untill they almost or altogether vanish. Then may yethe eye=glasse bee rectified much after yethe same manner, in every parte of it, & if it bee necessary yethe object glasse may bee aganieagaine rectified & againe yethe eye=glasse untill yethe Telescope bee as perfect as yethe workeman can make: Whome perhaps experience & other may teach by this & yethe former rules to make telescopes as perfect as men can hope to make them. These glasses may also bee rectified whilst on yethe Mandrill by observing yethe images mabde by reflection from yethe vertex & all other parts of yethe glasse wtwhat proportion they have one to another & how much they are longer ynthan broader in one place then another. &c. $\begin{array}{ccccc}\begin{array}{}\phantom{\text{BLANK}}\\ \text{The sines measuring refractions are in}\end{array}& \begin{array}{}\text{Aere}\\ 42\end{array}& \begin{array}{}\text{water}\\ 56\end{array}& \begin{array}{}\text{Glasse}\\ 65\end{array}& \begin{array}{}\text{christall}\\ 70\end{array}\\ \begin{array}{}\text{The proportions of the motions of the}\\ \text{extreamely heterogeneous rays are in}\end{array}\right\}& 39,4.40,4.& 70\frac{3}{8}.71\frac{3}{8}.& \phantom{0}95\frac{1}{10}.\phantom{0}96\frac{1}{10}& 110\frac{1}{3}.111\frac{1}{3}\hfill \\ \begin{array}{}\begin{array}{c}\text{The proportions of}\phantom{\rule{0.5em}{0ex}}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{sines of}\\ \text{refraction of the extreamely hetero-}\phantom{\frac{0}{0}}\\ \text{geneous rays into aire out of}\end{array}\\ \text{Their common sine of incidence}\phantom{\frac{0}{0}}\\ \text{Which substracted the difference is}\phantom{\frac{0}{0}}\end{array}& \begin{array}{}\begin{array}{}\phantom{\text{000Ty000}}\\ \text{c}\phantom{\frac{0}{0}}\\ \phantom{\text{000gf000}}\end{array}\\ \phantom{\frac{0}{0}}\\ \phantom{\frac{0}{0}}\end{array}& \begin{array}{}\begin{array}{}\phantom{\text{000Ty000}}\\ 90\frac{2}{3}.91\frac{2}{3}\phantom{.}\\ \phantom{\text{000Ty000}}\end{array}\\ 68\frac{1}{3}\\ 22\frac{1}{3}.23\frac{1}{3}\phantom{.}\end{array}& \begin{array}{}\begin{array}{}\phantom{\text{000Ty000}}\\ \phantom{0}68.\phantom{\frac{0}{0}}\phantom{0}69\phantom{\frac{0}{0}}\\ \phantom{\text{000Ty000}}\end{array}\\ 44\frac{1}{4}\\ \phantom{0}23\frac{3}{4}.\phantom{0}24\frac{3}{4}\end{array}& \begin{array}{}\begin{array}{}\phantom{\text{000Ty000}}\\ \phantom{0}61\frac{4}{5}.\phantom{0}62\frac{4}{5}\phantom{.}\hfill \\ \phantom{\text{000Ty000}}\end{array}\\ 36\frac{4}{5}\\ \phantom{0}24.\phantom{\frac{0}{0}}\phantom{0}25\phantom{.\frac{0}{0}}\end{array}\\ \begin{array}{}\text{The like proportions for refrac-}\\ \text{tions made into water out of}\end{array}& \phantom{}& \phantom{}& \begin{array}{}275\frac{4}{5}.276\frac{4}{5}\\ 238\frac{2}{5}\\ \phantom{0}37\frac{2}{5}.\phantom{0}38\frac{2}{5}\end{array}& \begin{array}{}196\frac{1}{3}.197\frac{1}{3}\\ 157\frac{4}{9}\\ \phantom{0}39\frac{1}{9}.\phantom{0}40\frac{1}{9}\end{array}\hfill \\ \phantom{}& \phantom{}& \phantom{}& \phantom{}& \phantom{}\end{array}$ of the method of infinite series I Newton 35 Theoremata varia. Circa angulorum æqualitates. si ang $DAB$ $=$ & $DAE$ bisecentur a rectis $FH$ et $IG$ et ducatur quævis $KLMN$ . Erit 1. $AK.AM\colon\colon KL.LM\colon\colon$ $KN.MN$ Euclid 6 3 2. $AK×AM=ALq+KL×LM=$ $A$ $+$$KL×LM-ALq$. Schooteen de concis æquis 3. $AM+AK.PK\colon\colon AQ.AL$ posito $AP=AM$. Si in angulo quovis $PAQ$ inseribantur æquales $AB$, $BC$, $CD$, $DE$, $EF$, $FG$, $GH$ &c anguli $BA$ $\mathrm{P}$ erit angulus $CBQ$ dulplus $DCP$ tripl, $ED$ $\mathrm{P}$$\mathrm{Q}$ quadr $FEQ$ quint, $GFQ$ sext, $HGP$ sept. $IHQ$ oct &c. Horuum vero angulorum positio radio $AB$ sinus erunt $B\beta$ , $C\chi$ &c cosinus $AB$ , $B\chi$ , $C\delta$ &c. Ergo si $AB=r$, & $AB=x$ erit $AC=2x$ $A\chi =\frac{2xx}{r}$. $B\chi =\frac{2xx-4x}{r}$. $AD=\left(2A\chi -AB\right)=\frac{4xx-rr}{r}$. $A\delta =$ $\frac{4{x}^{3}-rrx}{rr}$ &c To find the sume of yethe squares cubes &c. of yethe rootes of an equation If $a$, $bg$, $c$, $d$, $e$, $f$ &c be the rootes of yethe equation ${x}^{6}+p{x}^{5}+q{x}^{4}+r{x}^{3}+sxx+tx+v=0$. ynthen is $a+b+c+d+e+f=p\left(=g\right)$ ${a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}+{e}^{2}+{f}^{2}=pp-2q.\left(=pg-2q$ $=h\right)$ ${a}^{3}+{b}^{3}+{c}^{3}+{d}^{3}+{e}^{3}+{f}^{3}={p}^{3}-3pq+3r.\left(=ph-qg+3r=k\right)$ ${a}^{4}+{b}^{4}$ &c $={p}^{4}-4ppq+4pr+2qq-4s.\left(=pk-qh+rg-4s=l\right)$ ${a}^{5}$ &c $={p}^{5}-5{p}^{3}q+5pqq+5ppr-5ps-5qr+5t.\left(=pl-qk+rh-sg+5t$ $=$ $m\right)$ ${a}^{6}$ &c: $={p}^{6}-6{p}^{4}q+9ppqq+6{p}^{3}r-12pqr-6pps+6pt-2{q}^{3}+3rr+6qs-6v$. 80 $ag=a$. $ab=x$. $bh=\frac{dx}{c}$. $bc=y$. $bg=\sqrt{xx-aa}$ $gh=\sqrt{\frac{ddxx-eexx+eeaa}{ee}}$. $dg=b$. $ce=\frac{y}{dx}\sqrt{ddxx-eexx+eeaa}=fg$. $cf=\frac{dx+ey}{dx}\sqrt{xx-aa}$. $df=b-\frac{y}{dx}\sqrt{ddxx-eexx+eeaa}$. ${z}^{2}={dc}^{2}=\left\{\begin{array}{c}xx-aa+\frac{2eyx}{d}-\frac{2eyaa}{dx}+bb+yy\\ \frac{-2by}{dx}\sqrt{ddxx-eexx+eeaa}\hfill \end{array}$ $\frac{2by}{dx}\sqrt{ddxx-eexx+eeaa}=xx+yy-zz+bb-aa+\frac{2eyxx-2eyaa}{dx}$. $4bbddxx$ $\begin{array}{ccccccccccccc}dd{x}^{6}& +& 4dey{x}^{5}& +& 2ddyy{x}^{4}& -& 4deaay{x}^{3}& -& 4bbddyyxx& -& 4de{y}^{3}aax& +& 4{e}^{2}{a}^{4}{y}^{2}\hfill \\ & & & -& 2ddzz{x}^{4}\hfill & +& 4de{y}^{3}\hfill & +& 4bbeeyy\hfill & +& 4dey{z}^{2}{a}^{2}\hfill \\ & & & +& 2ddbb\hfill & -& 4deyzz\hfill & -& 8aaeeyy\hfill & -& 4deybb{a}^{2}\hfill \\ & & & -& 2ddaa\hfill & +& 4deybb\hfill & +& 2dd{y}^{4}\hfill & +& 4dey{a}^{4}\hfill & -& 4bb{e}^{2}{a}^{2}{y}^{2}\hfill \\ & & & +& 4eeyy\hfill & -& 4deyaa\hfill \end{array}$ Ad constructionem Canonis angularis. ${90}^{gr}$ ${\frac{90}{5}}^{gr}={18}^{gr}$. ${\frac{18}{5}}^{gr}={3}^{gr}+{36}^{\prime }$. Et ${\frac{60}{3}}^{gr}={20}^{gr}$. ${\frac{20}{3}}^{gr}={6}^{gr}+{40}^{\prime }$ $\frac{{6}^{gr}+{40}^{\prime }}{2}={3}^{gr}+{20}^{\prime }$. ${3}^{gr}+{36}^{\prime }-{3}^{gr}-{20}^{\prime }={16}^{\prime }$. $\frac{{16}^{\prime }}{2}={8}^{\prime }$. $\frac{{8}^{\prime }}{2}={4}^{\prime }$. $\frac{{4}^{\prime }}{2}={2}^{\prime }$ $\frac{{2}^{\prime }}{2}={1}^{\prime }$. If $r=$ radius. Then $\begin{array}{}\begin{array}{c}{\text{y}}^{\text{e}}\phantom{\rule{0.5em}{0ex}}\text{sine of}\\ \phantom{\text{0}}\\ \phantom{\text{0}}\\ \phantom{\text{0}}\end{array}\begin{array}{c}{78}^{degr}\phantom{\rule{0.5em}{0ex}}\text{is,}\phantom{\rule{1em}{0ex}}\frac{r\sqrt{5}-r+r\sqrt{30+6\sqrt{5}}}{8}.\hfill \\ {66}^{degr}\phantom{\rule{0.5em}{0ex}}\text{is,}\phantom{\rule{1em}{0ex}}\frac{r\sqrt{5}-r+r\sqrt{30-6\sqrt{5}}}{8}.\hfill \\ {42}^{degr}\phantom{\rule{0.5em}{0ex}}\text{is,}\phantom{\rule{1em}{0ex}}\frac{-\sqrt{5rr}+r+\sqrt{30rr+6rr\sqrt{5}}}{8}.\hfill \\ {\phantom{0}6}^{degr}\phantom{\rule{0.5em}{0ex}}\text{is,}\phantom{\rule{1em}{0ex}}\frac{\sqrt{30rr-6rr\sqrt{5}}-\sqrt{5rr}-r}{8}.\hfill \end{array}\end{array}$ Suppose $gh=x$. $nh=r$. Then $\begin{array}{ccc}gh& =& x\phantom{\rule{0.5em}{0ex}}\text{. unisectio}\hfill \\ ab×r& =& 2rr-xx\phantom{\rule{0.5em}{0ex}}\text{. bisectio}\hfill \\ \mathrm{h}{}^{2}\mathrm{b}×{r}^{2}& =& 3rrx-{x}^{3}\phantom{\rule{0.5em}{0ex}}\text{. trisectio}\hfill \\ \mathrm{a}{}^{3}\mathrm{b}×{r}^{3}& =& 2{r}^{4}-4rrxx+{x}^{4}\phantom{\rule{0.5em}{0ex}}\text{.}\phantom{\rule{0.5em}{0ex}}{\text{quadrisec}}^{\text{o}}\text{.}\hfill \\ \mathrm{h}{}^{4}\mathrm{b}×{r}^{4}& =& 5{r}^{4}x-5rr{x}^{3}+{x}^{5}\phantom{\rule{0.5em}{0ex}}\text{.}\phantom{\rule{0.5em}{0ex}}{\text{quintusect}}^{\text{o}}\text{.}\hfill \\ \mathrm{a}{}^{5}\mathrm{b}×{r}^{5}& =& 2{r}^{6}-9{r}^{4}{x}^{2}+6rr{x}^{4}-{x}^{6}\phantom{\rule{0.5em}{0ex}}\text{.}\hfill \\ hb×{r}^{6}& =& 7{r}^{6}x-14{r}^{4}{x}^{3}+7rr{x}^{5}-{x}^{7}\phantom{\rule{0.5em}{0ex}}\text{.}\hfill \\ ab×{r}^{7}& =& 2{r}^{8}-16{r}^{6}{x}^{2}+20{r}^{4}{x}^{4}-8rr{x}^{6}+{x}^{8}\phantom{\rule{0.5em}{0ex}}\text{.}\hfill \\ hb×{r}^{8}& =& 9{r}^{8}x-30{r}^{6}{x}^{3}+27{r}^{4}{x}^{5}-9rr{x}^{7}+{x}^{9}\phantom{\rule{0.5em}{0ex}}\text{.}\hfill \\ ab×{r}^{9}& =& 2{r}^{10}-25{r}^{8}{x}^{2}+50{r}^{6}{x}^{4}-35{r}^{4}{x}^{6}+10rr{x}^{8}-{x}^{10}\phantom{\rule{0.5em}{0ex}}\text{.}\hfill \\ hb×{r}^{10}& =& 11{r}^{10}x-55{r}^{8}{x}^{3}+77{r}^{6}{x}^{5}-44{r}^{4}{x}^{7}+11rr{x}^{9}-{x}^{11}\phantom{\rule{0.5em}{0ex}}\text{.}\hfill \end{array}$ As on yethe other leafe excepting some signes here changed. If $gh=x$. $bh=y$.Then $\begin{array}{c}y=bh\phantom{\rule{0.5em}{0ex}}\text{. duplicatio anguli}\phantom{\rule{0.5em}{0ex}}hag\\ yy-xx=x×\mathrm{h}{}^{2}\mathrm{b}\phantom{\rule{0.5em}{0ex}}\text{. triplicatio anguli}\phantom{\rule{0.5em}{0ex}}hag\phantom{\rule{0.5em}{0ex}}\text{.}\\ {y}^{3}-2xxy=xx×\mathrm{h}{}^{3}\mathrm{b}\phantom{\rule{0.5em}{0ex}}\text{. quadruplicatio.}\\ {y}^{4}-3xxyy+{x}^{4}={x}^{3}×{}^{4}\mathrm{b}\mathrm{h}\phantom{\rule{0.5em}{0ex}}{\text{. quint}}^{\text{o}}\\ {y}^{5}-4xx{y}^{3}+3{x}^{4}y={x}^{4}×\mathrm{h}{}^{5}\mathrm{b}\phantom{\rule{0.5em}{0ex}}{\text{. sext}}^{\text{o}}\\ {y}^{6}-5xx{y}^{4}+6{x}^{4}yy-{x}^{6}={x}^{5}×hb\phantom{\rule{0.5em}{0ex}}{\text{. sept}}^{\text{o}}\\ {y}^{7}-6xx{y}^{5}+10{x}^{4}{y}^{3}-4{x}^{6}y={x}^{6}×hb\phantom{\rule{0.5em}{0ex}}{\text{. oct}}^{\text{o}}\\ {y}^{8}-7xx{y}^{6}+15{x}^{4}{y}^{4}-10{x}^{6}yy-{x}^{8}={x}^{7}×hb\phantom{\rule{0.5em}{0ex}}\text{. nonc}\\ {y}^{9}-8xx{y}^{7}+21{x}^{4}{y}^{5}-20{x}^{6}{y}^{3}+5{x}^{8}y={x}^{8}×hb\phantom{\rule{0.5em}{0ex}}\text{. dec}\\ {y}^{10}-9xx{y}^{8}+28{x}^{4}{y}^{6}-35{x}^{6}{y}^{4}+15{x}^{8}{y}^{2}-{x}^{10}={x}^{9}×hb\phantom{\rule{0.5em}{0ex}}\text{. und,}\\ {y}^{11}-10xx{y}^{9}+36{x}^{4}{y}^{7}-56{x}^{6}{y}^{5}+35{x}^{8}{y}^{3}-6{x}^{10}y={x}^{10}×hb\phantom{\rule{0.5em}{0ex}}\text{. duod}\end{array}$ 81 Of Angular sections Suppose $ab=q$. $\frac{ah}{2}=r$. & $ag=x$. & ytthat yethe arches $hg$, $gb$, $bb$ are equall. By yethe following Equations an angle $bah$ may bee divided into any number of partes. $\begin{array}{c}x=q\phantom{\rule{0.5em}{0ex}}\text{. unisectio.}\\ {x}^{2}-2rr=rq\phantom{\rule{0.5em}{0ex}}\text{. bisectio.}\\ {x}^{3}-3rrx=rrq\phantom{\rule{0.5em}{0ex}}\text{. trisectio.}\\ {x}^{4}-4rrxx+2{r}^{4}={r}^{3}q\phantom{\rule{0.5em}{0ex}}\text{. quadrisectio.}\\ {x}^{5}-5rr{x}^{3}+5{r}^{4}x={r}^{4}q\phantom{\rule{0.5em}{0ex}}\text{. quintusectio.}\\ {x}^{6}-6rr{x}^{4}+9{r}^{4}xx-2{r}^{6}={r}^{5}q\phantom{\rule{0.5em}{0ex}}\text{. sextusectio.}\\ {x}^{7}-7rr{x}^{5}+14{r}^{4}{x}^{3}-7{r}^{6}x={r}^{6}q\phantom{\rule{0.5em}{0ex}}\text{. septusectio.}\\ {x}^{8}-8rr{x}^{6}+20{r}^{4}{x}^{4}-16{r}^{6}{x}^{2}+2{r}^{8}={r}^{7}q\phantom{\rule{0.5em}{0ex}}\text{.}\\ {x}^{9}-9rr{x}^{7}+27{r}^{4}{x}^{5}-30{r}^{6}{x}^{3}+9{r}^{8}x={r}^{8}q\phantom{\rule{0.5em}{0ex}}\text{.}\\ {x}^{10}-10rr{x}^{8}+35{r}^{4}{x}^{6}-50{r}^{6}{x}^{4}+25{r}^{8}{x}^{2}-2{r}^{10}={r}^{9}q\phantom{\rule{0.5em}{0ex}}\text{.}\\ {x}^{11}-11rr{x}^{9}+44{r}^{4}{x}^{7}-77{r}^{6}{x}^{5}+55{r}^{8}{x}^{3}-11{r}^{10}x={r}^{10}q\phantom{\rule{0.5em}{0ex}}\text{.}\\ {x}^{12}-12rr{x}^{10}+54{r}^{4}{x}^{8}-112{r}^{6}{x}^{6}+105{r}^{8}{x}^{4}-36{r}^{10}{x}^{2}+2{r}^{12}={r}^{11}q\phantom{\rule{0.5em}{0ex}}\text{.}\\ {x}^{13}-13rr{x}^{11}+65{r}^{4}{x}^{9}-156{r}^{6}{x}^{7}+182{r}^{8}{x}^{5}-91{r}^{10}{x}^{3}+13{r}^{12}x={r}^{12}q\phantom{\rule{0.5em}{0ex}}\text{.}\\ {x}^{14}-14rr{x}^{12}+77{r}^{4}{x}^{10}-210{r}^{6}{x}^{8}+294{r}^{8}{x}^{6}-196{r}^{10}{x}^{4}+49{r}^{12}{x}^{2}-\phantom{\rule{0.5em}{0ex}}\text{&c}\\ {x}^{15}-15rr{x}^{13}+90{r}^{4}{x}^{11}-275{r}^{6}{x}^{9}+450{r}^{8}{x}^{7}-318{r}^{10}{x}^{5}+140{r}^{12}{x}^{3}-\phantom{\rule{0.5em}{0ex}}\text{&c}\\ {x}^{16}-16rr{x}^{14}+104{r}^{4}{x}^{12}-352{r}^{6}{x}^{10}+660{r}^{8}{x}^{8}-672{r}^{10}{x}^{6}+336{r}^{12}{x}^{4}-\phantom{\rule{0.5em}{0ex}}\text{&c}\\ {x}^{17}-17rr{x}^{15}+119{r}^{4}{x}^{13}-442{r}^{6}{x}^{11}+935{r}^{8}{x}^{9}-1122{r}^{10}{x}^{7}+714{r}^{12}{x}^{5}-\phantom{\rule{0.5em}{0ex}}\text{&c}\\ {x}^{18}-18rr{x}^{16}+135{r}^{4}{x}^{14}-546{r}^{6}{x}^{12}+1287{r}^{8}{x}^{10}-1782{r}^{10}{x}^{8}+1386{r}^{12}{x}^{6}-\phantom{\rule{0.5em}{0ex}}\text{&c}\\ {x}^{19}-19rr{x}^{17}+152{r}^{4}{x}^{15}-665{r}^{6}{x}^{13}+1729{r}^{8}{x}^{11}-2717{r}^{10}{x}^{9}+2508{r}^{12}{x}^{7}-\phantom{\rule{0.5em}{0ex}}\text{&c}\\ {x}^{20}-20rr{x}^{18}+170{r}^{4}{x}^{16}-800{r}^{6}{x}^{14}+2275{r}^{8}{x}^{12}-4604{r}^{10}{x}^{10}+4290{r}^{12}{x}^{8}-\phantom{\rule{0.5em}{0ex}}\text{&c}\end{array}$ This scheame is yethe former inversed. versâ paginâ. Suppose yethe perifery $bgh$ to bee $a$ & yethe whole perifery to bee $p$. The line $bh$ subtends these arches. $a$. $p-a$. $p+a$. $2p-a$. $2p+a$. $3p-a$. $3p+a$. $4p-a$. $4p+a$. $5p-a$. $5p+a$. $6p-a$. $6p+a$. &c: All wchwhich are bisected, trisected, quadrisected, quintusected &c after same manner. As for example The rootes of yethe equation $\mathrm{h}{}^{2}\mathrm{b}×rr=3rrx-{x}^{3}$. are 3. The first whereof subtends yethe arches $\frac{a}{3}$. $\frac{3p-a}{3}$. $\frac{3p+a}{3}$. $\frac{6p-a}{3}$. $\frac{6p+a}{3}$. $\frac{9p-a}{3}$. $\frac{9p+a}{3}$ &c. The second subtends yethe arches $\frac{p-a}{3}$. $\frac{2p+a}{3}$. $\frac{4p-a}{3}$. $\frac{5p+a}{3}$. $\frac{7p-a}{3}$. &c. The 3d $\frac{p+a}{3}$. $\frac{2p-a}{3}$. $\frac{4p+a}{3}$. $\frac{5p-a}{3}$. $\frac{7p+a}{3}$ &c. Soe yethe rootes of yethe equation ~ $hb×{r}^{4}=5{r}^{4}x-5rr{x}^{3}+{x}^{5}$, doe yethe first subtend yethe arches $\frac{a}{5}$. $\frac{5p-a}{5}$. $\frac{5p+a}{5}$ &c: yethe 2d $\frac{p-a}{5}$. $\frac{4p+a}{5}$. $\frac{6p-a}{5}$. yethe 3d $\frac{p+a}{5}$. $\frac{4p-a}{5}$. $\frac{6p+a}{5}$ &c. yethe 4th $\frac{2p-a}{5}$. $\frac{3p+a}{5}$. &c yethe 5t $\frac{2p+a}{5}$. $\frac{3p-a}{5}$. $\frac{7p+a}{5}$. &c. Hence may appeare yethe reason of yethe number of rootes in these equations & ytthat yethe points of yethe circuferencecircumference to wchwhich they are extended æquidistant. & by yethe lower scheme may bee known wchwhich rootes are affirmative & wchwhich negative. The numerall cöefficients of yethe afforesaid equations may bee deduced from this progression (if $\angle :\angle \colon\colon 1:n$.) $1×\frac{-0+n\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}-1+n}{1\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}1-n}×\frac{n-2\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}n-3}{2\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}2-n}×\frac{n-4\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}n-5}{3\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}3-n}×\frac{n-6\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}n-7}{4\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}4-n}×\frac{n-8\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}n-9}{5\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}5-n}×\frac{n-10\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}n-11}{6\phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}6-n}$ &c. As if $x$ $n=10$. yethe ꝑreprogression $1×-10×\frac{-7}{2}×\frac{-10}{7}×\frac{-1}{2}×\frac{2}{25}×0$. And yethe coefficients $1.-10.+35.-50.+25.-2$. 82 1663 /4 January. All yethe parallell lines wchwhich can be understoode to bee drawne uppon any superficies are equivalent to it, as all yethe lines drawne from $\left(ao\right)$ to $\left(co\right)$ may be used instead of yethe superficies $\left(aco\text{.}\right)$ If all yethe parallell lines drawne uppon any superficies be multiplied by another line they produce a Sollid like ytthat wchwhich relultsresults from yethe superficies drawne into yethe lamesame line as if either all yethe lines in yethe superficies $\left(oac\right)$ or if yethe superficies $oac$ be drawne into yethe line $\left(b\right)$ they both produce yethe same lollidsollid $\left(d\right)$ whence All yethe parallell superficies wchwhich can bee understoode to bee in any sollid are equivalent to ytthat Sollid. And If all yethe lines in any triangle, wchwhich are parallell to one of yethe sides, be squared there results a Pyramid. if those in a square, there results a cube. If those in a crookelined figure there resutsresults a sollid wthwith 4 sides terminated & bended after according to yethe fasshionfashion of yethe crookelined figure. If each line in one superficies bee drawne into each correspondent line in another superficies as in $aebk$, & $omnc$ if $ae×dh$. $bk×cn$. $qv×wx$. &c. they produce a sollid whoswhose opposite sides are fashioned by one of yethe superfic as figureSollid $fpsrg$. where all yethe lines drawne from $fr$ 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 yethe correspondent lines drawne from $qz$ to $vz$. Theorema. 1 If in the Circle $abcdeP$ there be inscribed any Poligon $abcde$ wthwith an odd number of sides, & from any point in yethe circumference $P$ there bee drawne lines $Pe$, $Pa$, $Pb$, $Pc$, $Pd$ to every corner of yethe Polygon: yethe summ of every other line is equall to yethe summ of yethe rest, $Pa+Pb+Pc$ $=Pd+Pe$. & soe are their cubes ${Pa}^{3}+{Pb}^{3}+{Pc}^{3}={Pd}^{3}+{Pe}^{3}$. unless yethe figure be a Trigon TheorTheorema 2 If from yethe points of yethe Polygon then bee drawne perpendicular $ap$, $br$, $ct$, $ds$, $\mathrm{e}$$\mathrm{q}$ to any Diameter $pt$: yethe summe of yethe Perpendiculars on one side yethe Diameter is equallequal to their summe on yethe other $ap+br+ct=eq+ds$. & soe is yethe summe of their cubes (unlesse wnwhen yethe figure is a Trigon), ${ap}^{3}+{br}^{3}+{ct}^{3}=$ ${eq}^{3}+{ds}^{3}$ . & of theire square cubes (except wnwhen yethe figure is a Trigon or Pentagon. &c. TheorTheorema 3 If yethe 2 circles (fig 1 & 2) be equall wthwith like Poligons inscribed, & $Pa$ in fig 1 be assumed double to $pa$ in fig 2. then are all yethe yethe other corresponding lines in fig 1 double to those in fig 2 viz $Pb=2rb$, $Pc=2tc$, $Pd=2sd$, $Pe=2qe$. 83 To square yethe Parabola In yethe Parabola $cae$ suppose yethe Parameter $ab=r$. $ad=y$. $dc=x$. & $ry=xx$ or $\frac{xx}{r}=y$. Now suppose every yethe lines called $x$ doe increase in arithmeticall proportion all yethe $x$'s taken together yethe supp make yethe superficies $dch$ wchwhich is halfe a square let every line drawne from $cd$ to $hd$ be square & they produce a Pyramid equall to every $xx=\frac{{x}^{3}}{3}$. wchwhich if divided by y $r$ there remaines every $\frac{{x}^{3}}{3r}=$ $=\frac{yx}{3}$ equall to every $\frac{xx}{r}$ equall to every $\left(y\right)$ or all yethe lines drawne from $ag$ to $accc$ equall to yethe superficies $ag$ $\mathrm{c}$ equall to a 3d pteparte of yethe superficies $adcg$ & yethe superficie $acd=\frac{2yx}{3}$. Otherwise. suppose $ce=b$. $co=x$. $to=y$. & $ry=bx-xx$ yethe lines $x$ increasing in arithmeticall proportion every $x$ is equall to 4 times yethe superficies $cdh=\frac{bb}{2}$ wchwhich drawne into $b$ produceth yethe sollid $\frac{{b}^{3}}{2}$ but if every $x$ be squareredsquared they poduceproduce a pyramid equall to $\frac{{b}^{3}}{3}$. wherefore every $bx-xx=\frac{{b}^{3}}{6}$ equall to every $ry$ equall to yethe superfiescies $adce$ drawne into $r$ & $\frac{{b}^{3}}{6r}=$ to $cade$ as before. * $p\delta =pd=\frac{q+y}{5}$ & $\frac{qy+yy}{5}=xx$ 84 To Square yethe Hyperbola In yethe Hyperbola $eqaw$. suppose $ef=a$. $fa=b$. $ap=rq=y$ $a\lambda =d$ $pq=ar=x$. $ad=q=$$5$$oa=$$5$$ac=$$5$ $d=$ $r=$. & $da+ar:ar\colon\colon ar:rq$. $xx=dy+$$yx$. In wchwhich equation Every $x$ taken together is equall to yethe triangle $a\beta b$ equall to $\frac{aa}{2}$ & every $xx$ taken together is a pyramid $=\frac{{a}^{3}}{3}$ . Every $y$ taken together is equall to yethe superficies $eba=mkt$ If ynthen $gh=lm=ns=a\lambda =d$. every $dy$ is equall to yethe solid $nglmhs$. If yethe angle $mhk$ is a right one & if $mh=gl=$$ba=ef=a$ $f=kh$, then all yethe lin that is if yethe triangle $mhk=ab\beta$. every $y$ $x$ will be equall to yethe sollid $mhstk$ Joyne these two sollids together as in $lmtng=\frac{{a}^{3}}{3}$ . * ⊛ Againe Suppose every $x$ taken together to be equall to yethe superficies $aef$ , yethe line $q\gamma$ squared is $4xx$ every $4xx$ composeth a Sollid like $\left(♊♉♌\right)$ an quarter eighth pteparte whereof (wchwhich is equall to every $\frac{xx}{2}$) being like $♊♉♓o=$ $xyzv$; $xv$ will be equall to $♊$ $♓$ $=ef$ $=a=st=xz=o♓=hm$. & $vz=♊♓$ $=a\beta$$km$ . whence yethe covexeconvexe superficies $xyv$ of yethe figure $xyvz$ will fitly joyne wthwith yethe concave superficies $mst$ of yethe figure $shmkt$ . If every $x$ is equall to yethe superficies $aef$ , every $y$ shall be equall to yethe triangle $af\pi =\frac{bb}{2}$. every $yy=\frac{bbb}{3}$ every $qy=\frac{qbb}{2}$ & therefor yethe Sollid $yxzv=\frac{{b}^{3}}{30}+\frac{qbb}{20}=\frac{2{b}^{3}+3qbb}{60}$. Joyne yethe Sollid $shmkt$ to $yxvz$ & there resulteth $shmztk=$ $=\frac{aab}{2}$ from wchwhich againe substract $xvzy=\frac{2{b}^{3}+3qbb}{60}$ & there remaines yethe sollid $mhstk=\frac{30aab-2{b}^{3}-3qbb}{60}$ wchwhich substract from yethe sollid $ntmlg=\frac{{a}^{3}}{3}$ & there remaines $nglmhs=\frac{20{a}^{3}+2{b}^{3}+3qbb-30aab}{60}$ b wchwhich being divided by $\theta =\sqrt{\frac{5rr}{4}}$. there remaines $\frac{40{a}^{3}+4{b}^{3}+6qbb-60aab}{r\sqrt{5}}=$ to yethe superficies $abe$ 85 The squareing of severall crokedcrooked lines of yethe Seacond kind. In a any two crooked lines I call yethe Parameter or right side of yethe greater. $\left(r\right)$. but of yethe lesse $\left(s\right)$ Transverse side $\left(q\right)$. yethe right axis as $cf$ $\left(x\right)$ or $\mathcal{ef}=v$ . $y$ Transverse axis as $fe$ $y$, or $fd$ $z$. Suppose in yethe Parab: $\mathrm{d}\mathcal{d}\mathrm{c}$: $ac=r$. & in $\mathrm{e}\mathcal{e}\mathrm{c}$: $bc=s$ $rx=zz={df}^{2}$. $sx=yy={fe}^{2}$. $\sqrt{rx}-\sqrt{sx}=de=$ $=p$. $rx=sx+pp+2p\sqrt{sx}$. $rx-sx-pp=2p\sqrt{sx}$ $rrxx-2rsxx+ssxx-2pprx-2ppsx+{p}^{4}=4ppsx$. Or ${p}^{4}-2rxpp-6sxpp+rrxx-2rsxx+ssxx=0$. if $p=y$. $xx=\frac{\begin{array}{c}+2ryy\\ +6syy\end{array}x-{y}^{4}}{rr-2rs+ss}$. make $cf=a$. $fd=b$. $fe=c$. $\mathrm{c}\mathcal{e}ef\mathcal{f}=\frac{2ac}{3}$. & $\mathrm{c}\mathcal{d}def\mathcal{f}=\frac{2ab}{3}$ therefore $\frac{2ab-2ac}{3}=\mathrm{c}\mathcal{d}de\mathcal{e}$ yethe square of yethe crooked line $\mathrm{c}\mathcal{d}\mathrm{d}$ (when yethe line $\mathrm{c}\mathcal{e}\mathrm{e}$ is psupposed totoo close wthwith yethe line $cf$ ) whose nature is exprest by yethe foregoing Equation. 2 $lk=b$. $li=x$. $qi=y$. $in=z$. $+bx$$-xx$ $=ry$ $-$$bx$ $-xx$ $=sz$. $\frac{-xx+bx}{r}=y$. $\frac{-xx+bx}{s}=z$ $qn=\frac{-sxxs+bsx+rxxr-brx}{rs}=v=$$y$ or, $r$$rx-$$sxx+b$$b$$s$$x-b$$b$$r$$x-rsy=0$. Or $xx-bx-\frac{rsy}{r-s}=0$ 3 $tg=x$. $dg=z$. $gp=y$. $rx-rz={dp}^{2}$. $rx-rz+zz={y}^{2}$. $zz=rz-rx+yy$ $z=-\frac{r}{2}⊻\sqrt{\frac{1}{4}rr-rx+yy}$. $r:a\colon\colon \sqrt{xr-rz}:z$. $\frac{1}{2}rr⊻r\sqrt{\frac{1}{4}rr-rx+yy}=a\sqrt{rx-rz}$. $\frac{1}{2}{r}^{4}+rryy-{r}^{3}x$ $-aarx+rzaa=⊻\frac{r}{2}\sqrt{\phantom{0}}$ $\frac{1}{2}r=⊻r\sqrt{\frac{1}{4}rr-rx+yy}$ 86 $\begin{array}{ccccccccccc}\frac{1}{4}{r}^{6}& +& {r}^{4}yy& +& rr{y}^{4}& -& {r}^{5}x& =& 2{r}^{3}yyx& +& {r}^{4}xx\\ & & & & & -& {r}^{2}aax& & 2ryyaax& +& 2rraaxx\\ & & & & & & & & & +& {a}^{4}xx\end{array}$ $r:a\colon\colon rx-rz:zz$. $zz=arx-arz$. $zz=-az+ax$. $z$ $=-\frac{1}{2}a+\sqrt{\frac{1}{4}aa+ax}$. $ax-az=rz-rx+yy$ Or $-yy+rx+ax=-\frac{1}{2}aa-\frac{1}{2}ar+\frac{a}{r}\sqrt{\frac{1}{4}aa+ax}$. $\begin{array}{}186\\ 31\end{array}$ $\begin{array}{}\begin{array}{}354\\ 59\end{array}\\ \begin{array}{}944\end{array}\end{array}$ $\begin{array}{}\begin{array}{cccccccccc}{y}^{4}& -& 2rx& yy& +& rrxx\hfill & +& \frac{1}{4}{a}^{4}\hfill & +& 2aarx\\ & -& 2ax& & +& 2arxx& +& \frac{1}{2}{a}^{3}\hfill & +& arrx\hfill \\ & -& aa\hfill & & +& aaxx\hfill & +& \frac{1}{4}aarr& +& {a}^{3}x\hfill \\ & -& ar\hfill \end{array}=\frac{1}{4}{a}^{4}+{a}^{3}x+\frac{1}{4}rraa+rrax\\ \phantom{0}\end{array}$ $\begin{array}{cccccccc}{y}^{4}& -& 2rx& yy& +& rrxx\hfill & =& 0\\ & -& 2ax& & +& 2arxx& & \\ & -& aa\hfill & & +& aaxx\hfill & & \\ & -& ar\hfill & & +& \frac{1}{2}{a}^{3}r\hfill & & \\ & & & & +& 2aarx\hfill & & \\ \phantom{\text{.}}\end{array}$. $\begin{array}{ccc}\begin{array}{c}xx\end{array}& \begin{array}{}\begin{array}{ccccc}+& 2aar\hfill & x& +& {y}^{4}\hfill \\ -& 2ryy& & -& aayy& & \\ -& 2ayy& & -& aryy& & \\ & & & +& \frac{1}{2}{a}^{3}r\hfill & & \end{array}\\ rr+2ar+aa\end{array}& \begin{array}{}=0\end{array}\end{array}$. $17548875.\left(5849625$ $\begin{array}{c}51\right)358\left(7,019608.\\ \phantom{00\right)}\underset{¯}{357}\phantom{\left(0,000}-58.\\ \phantom{00\right)00}\underset{¯}{100}\phantom{\left(0,0-00.}\\ \phantom{00\right)000}490\phantom{\left(0,-00.}\\ \phantom{00\right)000}\underset{¯}{459}\phantom{\left(0,-00.}\\ \phantom{00\right)0000}310\phantom{\left(,-00.}\\ \phantom{00\right)0000}\underset{¯}{306}\phantom{\left(,-00.}\\ \phantom{00\right)000000}400\phantom{\left(,-.}\\ \phantom{0}\\ \phantom{0}\\ \phantom{0}\\ \phantom{0}\\ \phantom{0}\\ \phantom{0}\\ \phantom{0}\end{array}$ $\begin{array}{c}51\right)197\left(3,862745\\ \phantom{00\right)}\underset{¯}{153}\phantom{\left(0,00}+392\\ \phantom{000\right)}440\phantom{\left(0,0+000}\\ \phantom{000\right)}\underset{¯}{408}\phantom{\left(0,0+000}\\ \phantom{0000\right)}320\phantom{\left(0,0+00}\\ \phantom{0000\right)}\underset{¯}{306}\phantom{\left(0,0+00}\\ \phantom{00000\right)}140\phantom{\left(0,0+0}\\ \phantom{00000\right)}\underset{¯}{102}\phantom{\left(0,0+0}\\ \phantom{000000\right)}380\phantom{\left(0,0+}\\ \phantom{000000\right)}\underset{¯}{357}\phantom{\left(0,0+}\\ \phantom{0000000\right)}230\phantom{\left(0,+}\\ \phantom{0000000\right)}\underset{¯}{204}\phantom{\left(0,+}\\ \phantom{00000000\right)}26\phantom{\left(0,+}\end{array}$ $\begin{array}{c}\phantom{00\right)}\underset{¯}{93}\phantom{\left(0000000.}\\ 53\right)372\left(7018868\phantom{.}\\ \phantom{00\right)}\underset{¯}{371}\phantom{\left(000}+682.\\ \phantom{0000\right)}\underset{¯}{100}\phantom{\left(000+0.}\\ \phantom{00000\right)}470\phantom{\left(00+0.}\\ \phantom{00000\right)}\underset{¯}{424}\phantom{\left(00+0.}\\ \phantom{000000\right)}460\phantom{\left(00+.}\\ \phantom{000000\right)}\underset{¯}{424}\phantom{\left(00+.}\\ \phantom{0000000\right)}36\phantom{0}\phantom{\left(0+.}\\ \phantom{0000000\right)}\underset{¯}{318}\phantom{\left(0+.}\\ \phantom{0000000\right)}42\phantom{0}\phantom{\left(+.}\\ \phantom{0}\\ \phantom{0}\\ \phantom{0}\end{array}$ $\begin{array}{c}212\right)\underset{¯}{819}\left(3863207\\ \phantom{00\right)}636\phantom{\left(00000}70\\ \phantom{00\right)}1830\phantom{\left(000000}\\ \phantom{00\right)}\underset{¯}{1696}\phantom{\left(000000}\\ \phantom{000\right)}1340\phantom{\left(00000}\\ \phantom{000\right)}\underset{¯}{1272}\phantom{\left(00000}\\ \phantom{000000\right)}680\phantom{\left(0000}\\ \phantom{000000\right)}636\phantom{\left(0000}\\ \phantom{0000000\right)}44\phantom{\left(0000}\\ \phantom{0000000\right)}\underset{¯}{424}\phantom{\left(000}\\ \phantom{00000000\right)}160\phantom{\left(00}\end{array}$ $\begin{array}{}\stackrel{●}{0}·0·9·7·5·4·5·\stackrel{●}{5}·\stackrel{●}{5}·\stackrel{●}{9}·\stackrel{●}{7}·\stackrel{●}{7}·\stackrel{●}{7}·\phantom{0}·\phantom{0}·3·\phantom{0}·\phantom{0}·\\ \phantom{\stackrel{●}{0}}·\phantom{8}·\phantom{9}·\stackrel{●}{0}·\stackrel{●}{0}·\stackrel{●}{0}·\stackrel{●}{2}·\stackrel{●}{0}·\stackrel{●}{0}·\stackrel{●}{2}·\stackrel{●}{0}·\stackrel{●}{0}·\end{array}$ 87 4 In yethe Hyperbola Parabola $cb=a$. $be=x$. $2aa-2ax$ $-aa+2ax-xx={ed}^{2}$ $aa-xx={ed}^{2}=yy$. $\frac{aa-xx}{r}=y$. $cp=cb$ $\sqrt{}$ $eb×df=fg$ $×c$ $=$ $zc$. $\frac{aax-{x}^{3}}{r}+xx=zc$ ${x}^{3}-rxx-aax+rcz=0$. Since all $eb×df=\frac{1}{8}$ all ${co}^{2}=\frac{1}{4}ab×ab×r$. $ab=b$. all ${eb}^{2}=\frac{{a}^{3}}{3}$ therefore $bgpf$ $\frac{1}{4}bbr+\frac{{a}^{3}}{3}$. 88 $ab=$ $\mathrm{b}$ $\mathrm{e}=y$. $bd=x$. $bq=dg=b$. $nb=c$. $\frac{ybx}{yc}=\frac{bx}{c}=ef$ Then shall $\mathrm{b}\mathrm{q}&c$: be the axis of gravity in $\mathrm{f}\mathrm{e}\mathrm{b}&c$ & $bqgd$. 89 In yethe 1st figure. $gc:cd\colon\colon cfed:ckhg=\frac{cd×cfed}{gc}$. $ac=gc$. $x:z\colon\colon za:xy$. $\frac{zza}{x}=ckhg$. or $\frac{xxy}{z}=cdef$. Suppose $cd:ca\colon\colon ac:bc\colon\colon$ yethe swiftnesse of $de$ $:$ to yethe swiftnesse of $gh$ . $de×\text{its swiftnes}:gh×\text{its swiftness}\colon\colon gc:cd$. $de×cd:gh×ca\colon\colon de×ac:gh×bc\colon\colon gc×cd$. Fig 2d2d. 3d. $\mathrm{c}$ $d$ $\theta$ $:ca\colon\colon ac:bc\colon\colon nm:am\colon\colon \text{swiftness}\phantom{\rule{0.5em}{0ex}}\theta e:\text{swiftnesse}\phantom{\rule{0.5em}{0ex}}gh$. $de×$its swiftnesse $:gh×$ itits swifnesswiftness $\colon\colon$ $\mathrm{c}$ $\mathrm{d}$ $\mathrm{k}$$×de:gh$ $\mathrm{c}$ $×$ $ac\colon\colon de×ac:gh×bc$ $de×$ $\mathrm{c}$ $\mathrm{k}$ $:gh×ac\colon\colon de×ac:gh×bc$ $\colon\colon$ $de×nm:gh×am\colon\colon gc:cd$. $de×ck$ $de×ck×cd=gh×ac×gc$. $de×cd=gh×bc$. $de×nm=gh×gc$. Fig 4 $ck:ca\colon\colon$motion of yethe point $a$ from $c$ $:$ motion of yethe point $a$ from $m$ $\colon\colon$ $ck:ca\colon\colon$increasing of $ac=gc$$:$increasing of $cd$ $\colon\colon$motion of $gh$ $:$ motion of $de$. &c as before. These are to find such figures $cghk$, $cfed$, as doe equiponderate in respect of yethe axis $acfk$. 90 Reasonings concerning chance. If by one of yethe equall chances $a$, I gaine $p$, by yethe chance the equall chances $a$ , $b$ , $c$ , &c are such ytthat one of ymthem must necessarily happen, & ytthat if one of yethe chances $a$ happen I gaine $p$ thereby, or $q$ by one of yethe chances $b$ , or $r$ by one of yethe chances $c$ . My chance or expectation is worth $\frac{pa+qb+rc}{a+b+c}$ 1 If $p$ is yethe number of chances by one of wchwhich I may gaine $a$, & $q$ those by one of wchwhich I may gaine $b$, & $r$ those by one of which I may gaine $c$; soe ytthat those chances are all equall & one of them must necessarily happen: My hopes or chance is worth $\frac{pa+qb+rc}{p+q+r}=A$. The same is true if $p$, $q$, $r$ signify anyye proportion of chances for $a$, $b$, $c$. 2. If I bargaine for more ynthan one chacnce (viz: ytthat after I have taken yethe gaines by my first chance, from the stake $a+b+c$; I will venter another chance at yethe remaining stake &c) my second lott is worth $\frac{pa+qb+rc}{a+b+c×p+q+r}$ $A$ $A\frac{-AA}{a+b+c}=A-\frac{AA}{a+b+c}=B$. My third lot is worth $A\frac{-AA-AB}{a+b+c}=C$. My Fourth lot is worth $A\frac{-AA-AB-AC}{a+b+c}=D$. My Fift lot is worth $A\frac{-AA-AB-AC-AD}{a+b+c}=E$. My sixt lot is worth $A-A×\frac{A+B+C+D+E}{a+b+c}$. &c As if 6 men $\left(1.2.3.4$