Introductio in analysin infinitorum

24 DE TRANSFORMATIONE LIB. I. $\frac{1-2z+3z^2-4z^3}{1+4z^4} = \frac{\alpha+\zeta z}{1+2z+2z^2} + \frac{\gamma+\delta z}{1-2z+2z^2}$ : addantur actu hæ duæ fractiones, eritque summæ Numerator $+\alpha - 2\alpha z + 2\alpha z^2$ $+\zeta z - 2\zeta z^2 + 2\zeta z^3$ $+\gamma + 2\gamma z + 2\gamma z^2$ $+\delta z + 2\delta z^2 + 2\delta z^3$ Denominator $1+4z^4$ Cum ergo denominator æqualis sit denominatori fractionis propositæ, numeratores quoque æquales reddi debent: quod, ob tot litteras incognitas $\alpha, \zeta, \gamma, \delta$, quot sunt termini æquales efficiendi, utique fieri, idque unico modo poterit: nanciscimur scilicet has quatuor æquationes I. $\alpha + \gamma = 1$ II. $-2\alpha + \zeta + 2\gamma + \delta = -2$ III. $2\alpha - 2\zeta + 2\gamma + 2\delta = 3$ IV. $2\zeta + 2\delta = -4$. Hinc ob $\alpha + \gamma = 1$, & $\zeta + \delta = -2$; æquationes II. & III. dabunt $\alpha - \gamma = 0$ & $\delta - \zeta = \frac{1}{2}$; ex quibus fit $\alpha = \frac{1}{2}$; $\gamma = \frac{1}{2}$; $\zeta = -\frac{5}{4}$; $\delta = -\frac{3}{4}$; ideoque fractio proposita $\frac{1-2z+3z^2-4z^3}{1+4z^4}$, transformatur in has duas $\frac{\frac{1}{2} - \frac{5}{4}z}{1+2z+2z^2} + \frac{\frac{1}{2} - \frac{3}{4}z}{1-2z+2z^2}$. Simili autem modo facile perspicietur resolutionem semper succedere debere: quoniam semper tot litteræ incognitæ introducuntur, quot opus est ad numeratorem propositum eliciendum. Ex doctrina vero fractionum communi intelligitur hanc resolutionem succedere non posse, nisi isti denominatoris Factores fuerint inter se primi. 40. Functio igitur fracta $\frac{M}{N}$ in tot fractiones simplices forma $\frac{A}{p-qz}$ resolvi poterit, quot Factores simplices habet denominator N inter se inæquales. Repræ-