Introductio in analysin infinitorum

122 DE INVESTIGATIONE LIB. I. $(1 + \frac{xx}{(2\pi+g)^2})(1 + \frac{xx}{(4\pi-g)^2})(1 + \frac{xx}{(4\pi+g)^2})$ $(1 + \frac{xx}{(6\pi-g)^2})(1 + \frac{xx}{(6\pi+g)^2}) \&c.$ Atque, si loco $x$ ponatur $z\sqrt{-1}$, erit $\frac{\cos.z-\cos.g}{1-\cos.g} = (1-\frac{z}{g})(1+\frac{z}{g})$ $(1 - \frac{z}{2\pi-g})(1 + \frac{z}{2\pi-g})(1 - \frac{z}{2\pi+g})(1 + \frac{z}{2\pi+g})$ $(1 - \frac{z}{4\pi-g})(1 + \frac{z}{4\pi-g}) \&c., = 1 - \frac{zz}{1.2(1-\cos.g)} +$ $\frac{z^4}{1.2.3.4(1-\cos.g)} - \frac{z^6}{1.2....6(1-\cos.g)} + \&c.$ Hujus adeo Seriei in infinitum continuatae Factores omnes cognoscuntur. 16 [TRUNCATED]