Introductio in analysin infinitorum

120 DE INVESTIGATIONE LIB. I. autem cof. $\frac{2k+1}{i}\pi = 1 - \frac{(2k+1)^2\pi\pi}{2ii}$, unde forma Facto- ris erit $= \frac{4xx}{ii} + \frac{(2k+1)^2\pi^2}{ii}$, evanescente termino cujus de- nominator est $i^4$. Quoniam ergo omnis Factor expressionis $1+\frac{xx}{1.2}+\frac{x^4}{1.2.3.4}+\&c.$ hujusmodi formam habere debet $1+axx$, quo Factor inventus ad hanc formam reducatur, dividi debet per $\frac{(2k+1)^2\pi^2}{ii}$: hinc Factor formæ propositæ erit $1+\frac{4xx}{(2k+1)^2\pi\pi}$, ex eoque omnes Factores infiniti inve- nientur, si loco $2k+1$ successive omnes numeri impares substituantur. Hanc ob rem erit $\frac{e^x+e^{-x}}{2} = 1+\frac{xx}{1.2}+\frac{x^4}{1.2.3.4}+\frac{x^6}{1.2.3.4.5.6}+\&c. =$ $(1+\frac{4xx}{\pi\pi})(1+\frac{4xx}{9\pi\pi})(1+\frac{4xx}{25\pi\pi})(1+\frac{4xx}{49\pi\pi}) \&c.$ 158. Si $x$ fiat quantitas imaginaria, formulæ hæ exponen- tiales in Sinum & Cosinum cujuspiam Arcus realis abeunt. Sit enim $x = z\sqrt{-1}$; erit $\frac{e^{z\sqrt{-1}}-e^{-z\sqrt{-1}}}{2\sqrt{-1}} =$ $sin. z = z - \frac{z^3}{1.2.3} + \frac{z^5}{1.2.3.4.5} - \frac{z^7}{1.2.3....7} + \&c.$, quæ adeo expressio hos habet Factores numero infinitos $z(1-\frac{zz}{\pi\pi})(1-\frac{zz}{4\pi\pi})(1-\frac{zz}{9\pi\pi})(1-\frac{zz}{16\pi\pi})(1-\frac{zz}{25\pi\pi})$ $\&c.$, seu erit $sin. z = z(1-\frac{z}{\pi})(1+\frac{z}{\pi})(1-\frac{z}{2\pi})$ $(1+\frac{z}{2\pi})(1-\frac{z}{3\pi})(1+\frac{z}{3\pi}) \&c.$ Quoties ergo Ar- cus $z$ ita est comparatus, ut quispiam Factor evanescat, quod fit si $z=0, z=\pm\pi; z=\pm 2\pi$, & generaliter si $z=$ $\pm k\pi$, denotante $k$ numerum quemcunque integrum, simul Si- nus