Steps / Historical AI

Leonhard Euler · 1748 · Internet Archive

Introductio in analysin infinitorum

All 320 leaves · leaf 187

Transcription

SINUUM EXPLICATIONIBUS INFINITIS. 159 tangentium, & Secantium, Cosecantiumque, quamobrem pro CAP. XI. his peculiaribus formulis non erit opus. Ceterum notandum est numerorum $m$, $n$, $n-m$, $n+m$, &c. Logarithmos hyperbolicos accipi oportere, cum Logarithmi hyperbolici Sinuum Cosinuumque quæruntur, vulgares autem, cum tales ope posteriorum formularum sunt indagandi. Præterea $m:n$ denotat rationem, quam Angulus propositus habet ad Angulum rectum; sicque, cum Sinus Angulorum semirecto majorum æquentur Cosinibus Angulorum semirecto minorum ac vicissim, fractio $\frac{m}{n}$ nunquam major accipienda erit quam $\frac{1}{2}$, hancque ob rem termini illi multo magis convergent, ut semissis instituto sufficere possit. 197. Antequam hoc argumentum relinquamus, aptiorem aperiamus modum Tangentes & Secantes quorumvis Angulorum inveniendi, quam Caput præcedens suppeditat. Quanquam enim Tangentes & Secantes per Sinus & Cosinus determinantur; tamen hoc fit per divisionem, quæ operatio in tantis numeris nimis est operosa. Ac Tangentes quidem & Cotangentes jam supra (§. 136.) exhibuimus, verum illo loco rationem formularum reddere non licuit, quam huic Capiti reservavimus. 198. Ex §. 181. ergo primum expressionem pro Tangente Anguli $\frac{m}{2n}\pi$ elicimus. Cum enim fit $\frac{1}{nn-mm} + \frac{1}{9nn-mm} + \frac{1}{25nn-mm} + \&c. = \frac{\pi}{4mn} \text{tang.} \frac{m}{2n}\pi$ erit $\text{tang.} \frac{m}{2n}\pi = \frac{4mn}{\pi}(\frac{1}{nn-mm} + \frac{1}{9nn-mm} + \frac{1}{25nn-mm} + \&c.)$. Cum deinde fit $\frac{1}{nn-mm} + \frac{1}{4nn-mm} + \frac{1}{9nn-mm} + \&c. = \frac{1}{2mm} - \frac{\pi}{2mn} \text{cot.} \frac{m}{n}\pi$, si pro $n$ scribamus $2n$ erit $\text{cot.} \frac{m}{2n}\pi = \frac{2n}{m\pi} - \frac{4mn}{\pi}(\frac{1}{4nn-mm} + \frac{1}{16nn-mm} + \frac{1}{36nn-mm} + \&c.)$
confidence 0.60 14 expressions source scan at the Internet Archive

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