Pagina:Werkecarlf03gausrich.djvu/150

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Operae pretium erit, quosdam casus speciales huc adscripsisse. Ex formula I art. 5 sequitur, statuendo $t=1,$ $\beta =1$

[31]	$(1+u)^{n}={\cfrac {1}{1-{\cfrac {nu}{1+{\cfrac {{\frac {n+1}{2}}u}{1-{\cfrac {{\frac {n-1}{2.3}}u}{1+{\cfrac {{\frac {2(n+2)}{3.4}}u}{1-{\cfrac {{\frac {2(n-2)}{4.5}}u}{1+{\text{etc.}}}}}}}}}}}}}}$

E formulis VI, VII art. 5 sequitur

[32]	$\log(1+t)={\cfrac {t}{1+{\cfrac {{\frac {1}{2}}_{}t}{1+{\cfrac {{\frac {1}{6}}_{}t}{1+{\cfrac {{\frac {2}{6}}_{}t}{1+{\cfrac {{\frac {2}{10}}_{}t}{1+{\cfrac {{\frac {3}{10}}_{}t}{1+{\cfrac {{\frac {3}{14}}_{}t}{1^{}+{\text{etc.}}}}}}}}}}}}}}}}$
[32]	$\log {\frac {1+t}{1-t}}={\cfrac {2t}{1-{\cfrac {{\frac {1}{3}}tt}{1-{\cfrac {{\frac {2.2}{3.5}}_{}tt}{1-{\cfrac {{\frac {3.3}{5.7}}_{}tt}{1-{\cfrac {{\frac {4.4}{7.9}}_{}tt}{1^{}-{\text{etc.}}}}}}}}}}}}$

Mutando hic signa $-$ in $+$ prodit fractio continua pro $\operatorname {arc.tang} t.$

Porro habemus

[34]	$e^{t}={\cfrac {1}{1-{\cfrac {t}{1+{\cfrac {{\frac {1}{3}}t}{1-{\cfrac {{\frac {1}{6}}t}{1+{\cfrac {{\frac {1}{6}}t}{1-{\cfrac {{\frac {1}{10}}t}{1+{\cfrac {{\frac {1}{10}}t}{1-{\text{etc.}}}}}}}}}}}}}}}}$