Introductio in analysin infinitorum

56 DE EXPLICATIONE FUNCTIONUM LIB. I. $(P+Q)^{-\frac{1}{3}} = P^{-\frac{1}{3}} - \frac{1}{3} P^{-\frac{4}{3}} Q + \frac{1 \cdot 4}{3 \cdot 6} P^{-\frac{7}{3}} Q^2 - \frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9} P^{-\frac{10}{3}} Q^3 + \&c.$ $(P+Q)^{\frac{2}{3}} = P^{\frac{2}{3}} + \frac{2}{3} P^{-\frac{1}{3}} Q - \frac{2 \cdot 1}{3 \cdot 6} P^{-\frac{4}{3}} Q^2 + \frac{2 \cdot 1 \cdot 4}{3 \cdot 6 \cdot 9} P^{-\frac{7}{3}} Q^3 - \&c.$ &c. 72. Hujusmodi ergo Serierum termini ita progrediuntur ut quilibet ex antecedente formari possit: sit enim Seriei, quae ex $(P+Q)^{\frac{m}{n}}$ nascitur, terminus quilibet $= M P^{\frac{m-kn}{n}} Q^k$ erit sequens $= \frac{m-kn}{(k+1)n} M P^{\frac{m-(k+1)n}{n}} Q^{k+1}$. Notandum autem est in quovis termino sequente exponentem ipsius P unitate decrescere, contra vero exponentem ipsius Q unitate crescere. Quo autem haec facilius ad quemvis casum accommodentur, forma generalis $(P+Q)^{\frac{m}{n}}$ ita exponi potest $P^{\frac{m}{n}} (1+\frac{Q}{P})^{\frac{m}{n}}$: evoluta enim formula $(1+\frac{Q}{P})^{\frac{m}{n}}$ Serieque resultante per $P^{\frac{m}{n}}$ multiplicata, prodibit ipsa Series ante data. Tum vero si $m$ non solum numeros integros denotet, sed etiam fractos, pro $n$ tuto unitas collocari poterit. Quibus factis, si pro $\frac{Q}{P}$, quae est Functio ipsius $z$, ponatur $Z$, habebitur $(1+Z)^m = 1 + \frac{m}{1} Z + \frac{m(m-1)}{1 \cdot 2} Z^2 + \frac{m(m-1)(m-2)}{1 \cdot 2 \cdot 3} Z^3 + \&c.$ Ad sequentes progressionum leges autem observandas conveniet hanc formulae generalis in Seriem conversionem notasse