Page:Joseph Louis de Lagrange - Œuvres, Tome 4.djvu/321

D’où, en supposant $m$ infiniment petit ou infiniment grand, on déduira les suivantes

$\log(1+x)=x-{\frac {\cfrac {x^{2}}{2}}{1+{\cfrac {2x}{3}}-{\cfrac {{\cfrac {2}{9}}{\cfrac {x^{2}}{4}}}{1+{\cfrac {2.4.x}{3.5}}-{\cfrac {{\cfrac {2.3}{25}}{\cfrac {x^{2}}{4}}}{1+{\cfrac {2.9.x}{5.7}}-{\cfrac {{\cfrac {3.4}{49}}{\cfrac {x^{2}}{4}}}{1+\ldots }}}}}}}},$

$e^{x}=1+x+{\frac {\cfrac {x^{2}}{2}}{1-{\cfrac {x}{3}}+{\cfrac {{\cfrac {1}{9}}{\cfrac {x^{2}}{4}}}{1-{\cfrac {x}{3.5}}+{\cfrac {{\cfrac {1}{25}}{\cfrac {x^{2}}{4}}}{1-{\cfrac {x}{5.7}}-{\cfrac {{\cfrac {1}{49}}{\cfrac {x^{2}}{4}}}{1-\ldots }}}}}}}},$

$\log(1+x)={\frac {x}{1+{\cfrac {x}{2}}-{\cfrac {{\cfrac {1}{3}}{\cfrac {x^{2}}{4}}}{1+{\cfrac {x}{2}}-{\cfrac {{\cfrac {4}{3.5}}{\cfrac {x^{2}}{4}}}{1+{\cfrac {x}{2}}-{\cfrac {{\cfrac {9}{5.7}}{\cfrac {x^{2}}{4}}}{1+\ldots }}}}}}}},$

$e^{x}=1+{\frac {x}{1-{\cfrac {x}{2}}+{\cfrac {{\cfrac {1}{3}}{\cfrac {x^{2}}{4}}}{1+{\cfrac {{\cfrac {1}{3.5}}{\cfrac {x^{2}}{4}}}{1+{\cfrac {{\cfrac {1}{5.7}}{\cfrac {x^{2}}{4}}}{1+{\cfrac {{\cfrac {1}{7.9}}{\cfrac {x^{2}}{4}}}{1+\ldots }}}}}}}}}}.$