Introductio in analysin infinitorum

40 DE TRANSFORMATIONE FUNCTIONUM LIB. I. neque vicissim $z$ per $y$ exhiberi potest. Quo igitur huic incommodo remedium afferatur; ponatur $y = x^m z^n$, eritque $ax^{am} z^{an} + bz^\zeta + cx^{\gamma m} z^{\gamma n + \delta} = 0$. Determinetur nunc exponens $n$ ita ut ex hac aequatione valor ipsius $z$ definiri queat: quod tribus modis praestari potest. I. Sit $an = \zeta$; ideoque $n = \frac{\zeta}{a}$; erit, aequatione per $z^\zeta$ divisa, $ax^{am} + b + cx^{\gamma m} z^{\gamma n - \zeta + \delta} = 0$; unde oritur $z = (\frac{-ax^{am}-b}{cx^{\gamma m}})^{\frac{1}{\gamma n - \zeta + \delta}}$, sive $z = (\frac{-ax^{am}-b}{cx^{\gamma m}})^{\frac{\alpha}{\alpha\gamma n - \alpha\zeta + \alpha\delta}}, \&$ $y = x^m (\frac{-ax^{am}-b}{cx^{\gamma m}})^{\frac{an}{\alpha\gamma n - \alpha\zeta + \alpha\delta}}$ II. Sit $\zeta = \gamma n + \delta$ seu $n = \frac [TRUNCATED]