Page:Philosophical Transactions - Volume 003.djvu/20

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My Method for Squaring the Hyperbola is this:

Let AB be one Asymptote of the Hyperbola EdC; and let AE and BC be parallel to th'other: Let also AE be to BC as 2 to 1; and let the Parallelogram ABDE equal 1. See Fig. 1. And note, that the Letter x every where stands for Multiplication.

Supposing the Reader knows, that EA. αζ. KH. βη. dθ. γχ. δλ. εμ. CB. &c. are in an Harmonic series, or a series reciproca primanorum sue arithmetice proportionalium (otherwise he is referr'd for satisfaction to the 87, 88, 89, 90, 91, 92, 93, 94, 95, prop. Arithm. Infinitor. Wallisij:)

I say	$ABCdEA={\frac {1}{1\times 2}}+{\frac {1}{3\times 4}}+{\frac {1}{5\times 6}}+{\frac {1}{7\times 8}}+{\frac {1}{9\times 10}}$ &c.	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \ \end{matrix}}\right\}\,}$ infinitum.
	$EdCDE={\frac {1}{2\times 3}}+{\frac {1}{4\times 5}}+{\frac {1}{6\times 7}}+{\frac {1}{8\times 9}}+{\frac {1}{10\times 11}}$ &c.
	$EdCyE={\frac {1}{2\times 3\times 4}}+{\frac {1}{4\times 5\times 6}}+{\frac {1}{6\times 7\times 8}}+{\frac {1}{8\times 9\times 10}}$ &c.

For (in Fig. 2, & 3) the Parallelog.		And (in Fig. 4.) the Triangl.	Note.
	$CA={\frac {1}{1\times 2}}$	$EdC={\frac {1}{2\times 3\times 4}}={\frac {\Box dD-\Box dF}{2}}$	${\tfrac {1}{2}}CA=dD+dF$
$dD={\frac {1}{2\times 3}}$	$dF={\frac {1}{3\times 4}}$	$Ebd={\frac {1}{4\times 5\times 6}}={\frac {\Box br-\Box bn}{2}}$	${\tfrac {1}{2}}dD=br+bn$
$br={\frac {1}{4\times 5}}$	$bn={\frac {1}{5\times 6}}$	$dfC={\frac {1}{6\times 7\times 8}}={\frac {\Box fG-\Box fk}{2}}$	${\tfrac {1}{2}}dF=fG+fk$
$fG={\frac {1}{6\times 7}}$	$fk={\frac {1}{7\times 8}}$	$Eab={\frac {1}{8\times 9\times 10}}={\frac {\Box aq-\Box ap}{2}}$	${\tfrac {1}{2}}br=aq+ap$
$aq={\frac {1}{8\times 9}}$	$ap={\frac {1}{9\times 10}}$	$bcd={\frac {1}{10\times 11\times 12}}={\frac {\Box cs-\Box cm}{2}}$	${\tfrac {1}{2}}bn=cs+cm$
$cs={\frac {1}{10\times 11}}$	$cm={\frac {1}{11\times 12}}$	$def={\frac {1}{12\times 13\times 14}}={\frac {\Box et-\Box el}{2}}$	${\tfrac {1}{2}}fG=et+el$
$et={\frac {1}{12\times 13}}$	$el={\frac {1}{13\times 14}}$	$fgC={\frac {1}{14\times 15\times 16}}={\frac {\Box gu-\Box gh}{2}}$	${\tfrac {1}{2}}fk=gu+gh$
$gu={\frac {1}{14\times 15}}$	$gh={\frac {1}{15\times 16}}$	&c.	&c.
&c.	&c.

And