Page:Philosophical Transactions - Volume 003.djvu/131

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$1-a+a^{2}-a^{3}$	$+a^{4}\;\mathrm {\&c.}$
$1-2a+4a^{2}-8a^{3}$	$+16a^{4}\;\mathrm {\&c.}$
$1-3a+9^{2}-27a^{3}$	$+81a^{4}\;\mathrm {\&c.}$
& sic deinceps usque ad
$1-A+A^{2}-A^{3}$	$+A^{4}\,\mathrm {\&c.}$

Cum itaque sint

$1+{}$	$1$	${}+1\mathrm {\&c.}$ (usque ad ultimum)	${}=A$
$a+{}$	$2a$	${}+3a\mathrm {\&c.}$ (usque ad $A$ )	${}={\frac {1}{2}}A^{2}$
$a^{2}+{}$	$4a^{2}$	${}+9a^{2}\mathrm {\&c.}$ (usque ad $A^{2}$ )	${}={\frac {1}{3}}A^{3}$
$a^{3}+{}$	$8a^{3}$	${}+2a^{3}\mathrm {,\;\&c.}$ (usque ad $A^{3}$ )	${}={\frac {1}{4}}A^{4}$
& sic deinceps:

(quod ostendit ille prop. 16, estque à me alibi demonstratum:) Recte colligit, prop. 17. Expositum spatium Hyperbolicum $BIru=A-{\frac {1}{2}}A^{2}+{\frac {1}{3}}A^{3}-{\frac {1}{4}}A^{4}+{\frac {1}{5}}A^{5}$ , &c. Adeoque si (assignato, ipsi $A=Ir$ , valore suo in numeris, ut res postulaverit,) distribuantur ih duas classes $A$ , ${\frac {1}{3}}A^{3}$ , ${\frac {1}{5}}A^{5}$ , &c. (potestates affirmatæ,) & ${\frac {1}{2}}A^{2}$ , ${\frac {1}{4}}A^{4}$ , &c. (potestates negatæ;) harumque Aggregatum, ex Aggregato illarum, subducatur; Residuum erit ipsum $BIru$ spatium Hyperbolicum.

Nequis autem operam lusum iri existimet,, propter Addendorum seriem in utraque classe infinitam; adeoque non absolvendam: Hinc incommodo medelam (tacitus) adhibet: ponendo $A=0.\;1$ , vel $A=0$ , 21, aliive fractioni decimali æqualem, adeoque minorem quam $1$ : (Hoc est, sumpta $Ir$ minore quam $AI=1$ .) Quo fit, ut postriores ipsius $A$ potesiates tot gradibus infra Integrorum sedem descendant, ut merito negligi possint.

Exempli gratia; positis $AI=1$ , & $IR=0$ 2i. erit

	$A$	${}=0,21$
${\frac {1}{3}}$	$A^{3}$	${}=0,003087$	${\frac {1}{2}}$	$A^{2}$	${}=0,02205$
${\frac {1}{5}}$	$A^{5}$	${}=0,0000081682$	${\frac {1}{4}}$	$A^{4}$	${}=0,000486202$
${\frac {1}{7}}$	$A^{7}$	${}=0,000002572$	${\frac {1}{6}}$	$A^{6}$	${}=0,000014294$
${\frac {1}{9}}$	$A^{9}$	${}=0,000000088$	${\frac {1}{8}}$	$A^{8}$	${}=0,000000472$
${\frac {1}{11}}$	$A^{11}$	${}=0,000000003$	${\frac {1}{10}}$	$A^{10}$	${}=0,000000016$
	${}+{}$	$0,213171345$ ———			$0,022550984$	${}=0,190620361=BIru$

Quæ est brevis Synopsis Quadraturæ suæ satis elegans.

Dissimulandum interim non est; sequis totius $BIHF$ spatii (cujus latus $IH$ longius intelligatur quam $AI$ ) quadraturam postulet; rem non ita feliciter successuram: propter medelam, quam modo diximus, malo minus sufficientem. Cum enim jam ponenda $A>1$ ; manifestum est, posteriores ipsius potestates, altius in Integrorum sedes penetraturas, adeoque minime negligendas.