Introductio in analysin infinitorum

274 DE PARTITIONE LIB. I. geometricæ duplæ $1, 2, 4, 8, 16, 32, \&c.$, diversis per additionem formari posse, hocque unico modo. Nota est hæc proprietas in praxi ponderandi, si enim habeantur pondera $1, 2, 4, 8, 16, 32, \&c.$, librarum; his solis ponderibus omnia onera ponderari poterunt, nisi partes libræ requirant. Sic his decem ponderibus, nempe $1$ lb, $2$ lb, $4$ lb, $8$ lb, $16$ lb, $32$ lb, $64$ lb, $128$ lb, $256$ lb, $512$ lb, omnia pondera usque ad $1024$ lb, librari possunt, & si unum pondus $1024$ lb, addatur omnibus oneribus usque ad $2048$ lb, ponderandis sufficient. 330. Ostendi autem insuper solet in praxi ponderandi paucioribus ponderibus, quæ scilicet in ratione geometrica tripla progrediantur, nempe $1, 3, 9, 27, 81, \&c,$ librarum pariter omnia onera ponderari posse, nisi opus sit fractionibus. In hac autem praxi pondera non solum uni lanci, sed ambabus, uti necessitas exigit, imponi debent. Nititur ergo ista praxis hoc fundamento, quod ex terminis progressionis geometricæ triplæ $1, 3, 9, 27, 81, \&c.$, diversis semper sumendis per additionem ac subtractionem omnes omnino numeri produci queant; erit scilicet. $1 = 1$ $5 = 9 - 3 - 1$ $9 = 9$ $2 = 3 - 1$ $6 = 9 - 3$ $10 = 9 + 1$ $3 = 3$ $7 = 9 - 3 + 1$ $11 = 9 + 3 - 1$ $4 = 3 + 1$ $8 = 9 - 1$ $12 = 9 + 3$ &c. 331. Ad hanc veritatem ostendendam considero hoc productum infinitum $(x^{-1}+1+x^{1})(x^{-3}+1+x^{3})(x^{-9}+1+x^{9})(x^{-27}+1+x^{27}) \&c.$ $= P,$ quod evolutum alias non dabit Potestates ipsius $x$, nisi quarum Exponentes formari possint ex numeris $1, 3, 9, 27, 81, \&c.$, sive