De summis serierum Arithmo-Geometricarum infinitarum idque revocando summarum inventiones ad æquationes ordinarias, quod in aliis innumeris procedit (Janvier 1678)

Transcription of De summis serierum Arithmo-Geometricarum infinitarum idque revocando summarum inventiones ad æquationes ordinarias, quod in aliis innumeris procedit (Janvier 1678)

Sint in progressione Geometrica $t.t^2.t^3.t^4.t^5.$ Horum summa, $t+t^2+t^3+t^4+t^5$ &c. $\sqcap \ \frac{t}{1-t} \sqcap S$. Sint in progressione Arithmetica : $0.1.2.3.4.5.6$ &c. ducantur in priores respondentes : fiet : $0t + 1t^2 + 2t^3 + 3t^4 + 4t^5$ &c. quorum etiam quæ ritur summa. Id est momentum figuræ infinitæ seu asympototon habentis $0T\gimel$ ex $0T$, posito $\aleph \beth$ esse $1$ et $0T$ esse $t$ et ita porro.

Hanc ita inveniemus : sit $0t + 1t^2 + 2t^3 + 3t^4$ &c. $\sqcap \ A \sqcap \int y$ et sit momentum figuræ ex $\aleph \beth$ seu $1t + 2t^2 + 3t^3 + 4t^4$ &c. $\sqcap \ B \sqcap \int x$ patet fore $B-A$ æqu. $1+ t + t^2 + t^3 + t^4 + t^5$ &c. $\sqcap \ S$. Rursus $Bt$ æqu. $A$. Ergo $B$ æqu. $\frac{A}{t}$. At eadem $B$ æqu. $S+A$ fiet ergo $\frac{A}{t}$ æqu. $S+A$ seu $St + At$ æqu. $A$. Seu $St$ æqu. $A-At$ seu denique fiet $A$ æqu. $\frac{St}{1-t}$. seu $A$ æqu. $S,\frac{t}{1-t}$ est autem $\frac{t}{1-t}$ æqu. $S$. ergo $A$ æqu. $S^2$. seu $A$ æqu. $\fbox{2}\frac{t}{1-t}$

Inventio serierum per differentias tantum est casus specialis huius methodi

Qui sane ratiocinandi modus est pulcherrimus. Artificium autem medios terminos ac novas incognitas, ut hoc loco $B$ assumendi in eo consistit ut una incognita nova $B$ nobis afferat secum æquationes duas, ut $B$ æqu. $\frac{A}{t}$ et $B \sqcap S+A$ ita enim lucrabimur hac assumtione, et utramque incognitarum $B$ pariter et $A$ inveniemus. Uti hoc loco factum est nam et $A$ inventa statim habetur $B$. Non est autem opus $B$ mediam prius inveniri quam $A$ quæ sitam, sufficit ab illa nobis suppeditari æquationes novas.

Quæritur jam porro summa seriei : $0t + 1t^2 + 4t^3 + 9t^4 + 16t^5$ &c. $\sqcap \ C \sqcap \int y^2$

Assumatur alia ................................. $1t + 4t^2 + 9t^3 + 16t^4 + 25t^5$ &c. $\sqcap \ D \sqcap \int x^2$

[marge] $x^2 - y^2 \sqcap \begin{array}{l} x+y \\ x-y \end{array}$

$\begin{array}{l} \textrm{Erit $D-C$ æqu.} \\ \textrm{addatur $S$ æqu. } \end{array}$$\left.\begin{array}{l} \textrm{$1t + 3t^2 + 5t^3 + 7t^4 + 9t^5$ etc.} \\ \textrm{$1t + 3t^2 + 5t^3 + 7t^4 + 9t^5$ etc.} \end{array}\right\}$ $D-C \sqcap A+\underset{\overbrace{S+A}}{B}$ $\textrm{Ergo $D-C \sqcap S + 2A$}$

fiet $D-C + S$ æqu. $2t + 4t^2 + 6t^3 + 8t^4 + 10t^5$ &c. æqu. $2B$ vel æqu. $2S + 2A$ Ergo $D-C$ æqu. $S + 2A$. rursus $Dt$ æqu. $C$. Ergo $D$ æqu. $\frac{C}{t}$. et rursus $D$ æqu. $S+2A + C$. Ergo $\frac{C}{t}$ æqu. $S+2A + C$ vel erit $C$ æqu. $St + 2At + Ct$ erit denique $C$ æqu. $\frac{St + 2At}{1-t}$. Sive erit $C$ æqu. $S +2A, \frac{t}{1-t}$ (seu $S^2 + 2AS$). est autem $A$ æqu. $S^2$. et et $\frac{t}{1-t}$ æqu. $S$ ergo fiet $C$ æqu. $S^2 + 2S^3$

Quæritur eodem modo summa seriei : $0t + 1t^2 + 8t^3 + 27t^4 + 64t^5$ &c. $\sqcap \ E \sqcap \int y^3$

Assumatur alia ..................................... $1t + 8t^2 + 27t^3 + 64t^4 + 125t^5$ &c. $\sqcap \ F \sqcap \int x^3$

erit $F-E$ æqu. .................................... $1t + 7t^2 + 19t^3 + 37t^4 + 61t^5$ &c. $\sqcap \ \int \overline{x^3 - y^3} \sqcap \int \overline{x^2 + xy + y^2}$

[marge] : $x^3 - y^3 \sqcap \begin{matrix} x^2 + xy + y^2 \\ x-y \end{matrix}$

termini seriei $E$ sint $y^3$, termini seriei sint $x^3$ patet fore $x-y$ æqu. $1$. Ergo $x^3 - y^3$ æqu. $x^2 + xy + y^2$ et $\int \overline{x^2} \sqcap D$. et $\int \overline{y^2} \sqcap C$. quia $x \sqcap y + 1$ erit $yx$ æqu. $y^2 + y$. Ergo $\int \overline{yx}$ æqu. $\int \overline{y^2} + \int \overline{y}$. at $\int \overline{y^2}$ æqu. $C$ et $\int \overline{y}$ æqu. $A$. Ergo $F-E \sqcap D+2C + A$. rursus $Ft$ æqu. $E$.

[marge] $\begin{array}{r} F - E \ \sqcap \int \overline{x^2 + xy + y^2} \\ D+C+\underset{C+A}{\int\overline{xy}} \\ D+2C+A \end{array}$

Ergo $F$ æqu. $\frac{E}{t}$. Et $F$ æqu. $D + 2C + A +E$. Ergo $E$ æqu. $Dt + 2Ct+At+Et$. sive $E$ æqu. $\frac{t}{1-t}, \cap \fbox{$\underset{\overbrace{S+2}A + C}{D+2C} + A$} \cap S + \underset{\overbrace{3S^2}}{3A} + \underbrace{2C}_{2S^2 + 4S^3}$ Ergo $E$ æqu. $\frac{t}{1-t}\cap 1S + 5S^2 + 4S^3$. Seu $E$ æqu. $1S^2 + 5S^3 + 4S^4$

 

Quæritur eodem modo summa seriei : $0t + 1t^2 + 16t^3 + 81t^4 + 256t^5$ &c. $\sqcap \ G \sqcap \int y^4$

Assumatur alia ..................................... $1t + 16t^2 + 81t^3 + 256t^4 + 625t^5$ &c. $\sqcap \ H \sqcap \int x^4$

[marge] $\begin{array}{ccccccc} y^2x & \sqcap & y^3 & + & y^2 & & \\ yx^2 & \sqcap & y^3 & + & 2y^2 & + & y \\ \hline & & 2y^3 & + & 3y^2 & + & y \end{array}$

$H-G \sqcap \int \overline{x^4 - y^4} \sqcap \int \overline{x^3 + x^2y + xy^2 +y^3} \sqcap \underbrace{F+E + \underbrace{\int \overline{x^2y + xy^2}}_{\underbrace{\int \overline{2y^3 + 3y^2 +y}}_{2E+3C+A}}}_{\sqcap \underbrace{F + 3E + 3C + A}_{\odot}}$

jam $H-G \sqcap \odot$. et $Ht\sqcap G$ vel $H\sqcap \frac{G}{t}$ vel $H \sqcap \odot + G$. Ergo $\odot + G \sqcap \frac{G}{t}$. Ergo $G \sqcap \odot \cap \frac{t}{1-t}$ vel $G \sqcap \odot S$

$\odot \sqcap \underset{\overbrace{\frac{|+E +2C + A +}{}\underset{\overbrace{\overset{A}{A} + \overset{\underbrace{+B}}{S+A}}}{\frac{D|}{}}}}{F}+3E+3C+A$ Ergo $\odot \sqcap 4E + 5C + 4A + S$ Eodem modo supra : $F-E~\sqcap ☽ \sqcap \underset{\overbrace{\underset{\overbrace{A+S}}{+B} + A}}{D}+2C + A $ Ergo $ ☽ \sqcap~2C + 3A + S$.

Unde patet si quis prosequi velit, nova semper compendia.

Et methodus inveniendi compendia in talibus huc redit, primum invenienda in ipsis $x^3 + x^2y +xy^2 +y^3$, et similibus inde transitu facto quæ rendum compendium seu progressio in ipsis $\odot$ vel ☽. Id est $H-G$ vel $ F-E$ vel $D-C$ vel $B-A$ simpliciter expressus retentis $B$. $D$. $F$. $H$ &c. inde quæ renda progressio eliminatis his $B$. $D$. $F$. $H$ &c. re ad solus $S$. $A$. $C$. $E$. $G$. reducta. atque ita denique apparebit etiam progressis eliminatis $A$. $C$. $E$. $G$. seu re ad solas ipsius $S$. potentias reducta ita enim per gradus apparebit quomodo progressio una ex alia derivetur. Cum alioqui non statim liceat ad progressiones et compendia perveniat per saltum.

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$S \sqcap t+t^2 + t^3 + t^4 + t^5$ &c. $\sqcap \ \int \overline{t^{y+1}} \sqcap \frac{t}{1-t}$ Nimirum in serie $S$, quilibet terminus est $t^{y+1}$ id est potestas a $t$ cuius exponens sit $y+1$ positus $y$ esse $0$ vel $1$ vel $2$ vel $3$ &c. Ergo seriei summa est $\int \overline{t^{y+1}}$ quam aliunde constat esse $\frac{t}{1-t}$

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$A \sqcap 0t+ 1t^2 + 2t^3 + 3t^4 + 4t^5$ &c. $\sqcap \ \int \overline{y,t^{y+1}}$ $B \sqcap 1t + 2t^2 + 3t^3$ &c. seu $\sqcap \ \int \overline{y+1,t^{\underline{y+1}}}$ $C \sqcap 0t+ 1t^2 + 4t^3 + 9t^4 + 16t^5$ &c. $\sqcap \ \int \overline{y^2,t^{y+1}}$ et $D \sqcap 1t+ 4t^2 + 9t^3$ &c. seu $\sqcap \int \overline{\fbox{$2$}\overline{y+1},t^{\underset{^{.......}}{y+1}}}$

Et ita porro

Sit ergo $S \sqcap \int \overline{t^{y+1}} \sqcap \frac{t}{1-t}$ Appellamus $y+1$ æqu. $x$

$A \sqcap \int \overline{y,t^{y+1}}$. $B \sqcap \int \overline{y+1,t^{y+1}}$ Ergo $B-A \sqcap \int \overline{t^{y+1}} \sqcap S$. seu $B\sqcap S+A$ vel $B \sqcap \int \overline{y,t^y - 0t^0}$ (quia primum $y,t^y$ seu $0,t^0$ non potest etiam esse $y+1,t^{y+1}$) Ergo $B \sqcap \int \overline{y,t^y}$ quia, $0t^0$ æqu. $0$. Ergo $B \sqcap \frac{A}{t}$

$\frac{A}{t}$ æqu. $\int \overline{y,t^y}$ Ergo $A \sqcap \frac{t}{1-t}, S$ seu $A \sqcap S^2$

$C$ æqu. $\int \overline{y^2,t^{y+1}}$. $D$ æqu. $\int \overline{\overline{\underbrace{y^2+2y+1}_{x^2}},t^{y+1}}$. $D-C$ æqu. $\int \overline{\overline{2y+1},t^{y+1}}$ æqu. $2A+S$ Ergo $2S^2+S$ seu $D = C+2A+S$

$\frac{C}{t}$ æqu. $\int \overline{y^2,t^y}$ $D$ æqu. $\int \overline{y^2,t^y - 0^2t^0}$. Ergo $D\sqcap \frac{C}{t}$. Ergo $C \sqcap 2A + S, \cap\frac{t}{1-t}$ seu $\overline{2A+S},S$ seu $2S^3+ S^2$

$E$ æqu. $\int \overline{y^3,t^{y+1}}$. $F$ æqu. $\int \overline{x^3,t^{y+1}}$. $F$ æqu. $\frac{E}{t}$ et $E$ æqu. $\overline{3C +3A+S},S$

$G$ æqu. $\int \overline{y^4,t^{y+1}}$. $H$ æqu. $\int \overline{x^4,t^{y+1}}$. $H$ æqu. $\frac{G}{t}$ et $G$ æqu. $\overline{4E + 6C + 4A + S},S$

Et ita porro Sed ut inde fiat Tabula : Sit :

Progressio ubi inventa fuerit quod continuato labore tabulæ facile erit tunc utilissimum apparebit hoc inventum ad summas serierum, quales sunt $\frac{t^1}{1} - \frac{t^3}{3} + \frac{t^5}{5} - \frac{t^7}{7}$ &c. ac similes vide schedam adjectam, signo $\odot$.

series non nisi duobus modis scilicet additionibus et substractionibus in constantium et multiplicationibus ac divisionibus constantium tractari possunt item divulsionibus in alias series.

Hac tabula paulo nitidius scripta et si opus continuata, facile progressio apparebit

Nota unam quæ que linea æquivalet sed summa continet explicationem perfectam, hac explicationum gradatione series optime inveniuntur.

Nota : patet eadem methodo etiam inveniri posse summas serierum $y^2,t^{y^2}$ v.g $01^0+1t^1 + 4t^4 + 9t^9$ &c. vel $0t^1+1t^4 + 4t^9 + 9t^{16}$. Imo video id difficilius, quæ ne quidem habemus? Opus ergo erit altera illa arte per quam a momentis seu summis et summis summarum &c. transitur ad potentias seu quadratæ cubos, &c.