Introductio in analysin infinitorum

FUNCTIONUM. 35 atque ex Factore $1+z$ oritur haec fractio partialis $-\frac{1}{4(1+z)}$. Jam sumatur Factor quadratus $(1-z)^2$ qui dat $\frac{P}{q}=1$. $M=1$, & $Z=z^3+z^4$; positis ergo fractionibus partialibus hinc ortis $\frac{A}{(1-z)^2}+\frac{B}{1-z}$, erit $A=\frac{1}{z^3+z^4}$, posito $z=1$; ergo $A=\frac{1}{2}$; fiat $P=\frac{M-\frac{1}{2}Z}{1-z}=\frac{1-\frac{1}{2}z^3-\frac{1}{2}z^4}{1-z}=1+z+zz+\frac{1}{2}z^3$; eritque $B=\frac{P}{Z}=\frac{1+z+zz+\frac{1}{2}z^3}{z^3+z^4}$, posito $z=1$; ergo $B=\frac{7}{4}$ & fractiones partiales quaesitae $\frac{1}{2(1-z)^2}+\frac{7}{4(1-z)}$. Denique tertius Factor cubicus $z^3$ dat $\frac{P}{q}=0$; $M=1$; [TRUNCATED]