Steps / Historical AI

Leonhard Euler · 1748 · Internet Archive

Introductio in analysin infinitorum

All 320 leaves · leaf 288

Transcription

260 DE PARTITIONE LIB. I. $P = \frac{x}{1-x}$, dat Seriem geometricam $x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + \&c.,$ ex qua quidem manifestum est quemvis numerum semel in Serie numerorum integrorum contineri. 309. Expressio secunda $\frac{x^3}{(1-x)(1-x^2)}$, dat hanc Seriem $x^3 + x^4 + 2x^5 + 2x^6 + 3x^7 + 3x^8 + 4x^9 + 4x^{10} + \&c.,$ in qua cujusvis termini coëfficiens indicat quot modis Exponens ipsius $x$ in duas partes inæquales dispertiri possit. Sic terminus $4x^9$ indicat, numerum $9$ quatuor modis in duas partes inæquales secari posse. Quod si hanc Seriem per $x^3$ dividamus, prodibit Series, quam præbet ista fractio $\frac{1}{(1-x)(1-x^2)}$, quæ erit $1 + x + 2x^2 + 2x^3 + 3x^4 + 3x^5 + 4x^6 + 4x^7 + \&c.,$ cujus terminus generalis fit $= Nx^n$; atque ex genesi hujus Seriei intelligitur coëfficientem $N$ indicare, quot variis modis Exponens $n$ ex numeris $1$ & $2$ per additionem nasci queat. Cum igitur prioris Seriei terminus generalis fit $= Nx^{n+3}$, deducitur hinc istud theorema. Quot variis modis numerus $n$ per additionem ex numeris $1$ & $2$ produci potest, totidem variis modis numerus $n+3$ in duas partes inæquales secari poterit. 310. Expressio tertia $\frac{x^6}{(1-x)(1-x^2)(1-x^3)}$ in Seriem evoluta dabit $x^6 + x^7 + 2x^8 + 3x^9 + 4x^{10} + 5x^{11} + 7x^{12} + 8x^{13} + \&c.,$ in qua cujusvis termini coëfficiens indicat quot variis modis Exponens Potestatis $x$ adjunctæ in tres partes inæquales dispertiri
confidence 0.98 23 expressions source scan at the Internet Archive

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