Pagina:Werkecarlf03gausrich.djvu/148

SECTIO SECUNDA.
Fractiones continuae.

12.

Designando

${\frac {F(\alpha ,\beta +1,\gamma +1,x)}{F(\alpha ,\beta ,\gamma ,x)}}\;$ per $\;G(\alpha ,\beta ,\gamma ,x)$

fit

${\frac {F(\alpha +1,\beta ,\gamma +1,x)}{F(\alpha ,\beta ,\gamma ,x)}}={\frac {F(\beta ,\alpha +1,\gamma +1,x)}{F(\beta ,\alpha ,\gamma ,x)}}=G(\beta ,\alpha ,\gamma ,x)$

et proin, dividendo aequationem 19 per ${F}(\alpha ,\beta +1,\gamma +1,x),$

$1-{\frac {1}{G(\alpha ,\beta ,\gamma ,x)}}={\frac {\alpha (\gamma -\beta )}{\gamma (\gamma +1)}}xG(\beta +1,\alpha ,\gamma +1,x)$

sive

[24]	$G(\alpha ,\beta ,\gamma ,x)={\frac {1}{1-{\frac {\alpha (\gamma -\beta )}{\gamma (\gamma +1)}}xG(\beta +1,\alpha ,\gamma +1,x)}}$

et quum perinde fiat

$G(\beta +1,\alpha ,\gamma +1,x)={\frac {1}{1-{\frac {(\beta +1)(\gamma +1-\alpha )}{(\gamma +1)(\gamma +2)}}xG(\alpha +1,\beta +1,\gamma +2,x)}}$

etc., resultabit pro $G(\alpha ,\beta ,\gamma ,x)$ fractio continua

[25]	$\displaystyle {\frac {F(\alpha ,\beta +1,\gamma +1,x)}{F(\alpha ,\beta ,\gamma ,x)}}={\cfrac {1}{1-{\cfrac {ax}{1-{\cfrac {bx}{1-{\cfrac {cx}{1-{\cfrac {dx}{1-{\text{etc.}}}}}}}}}}}}$

ubi

${\begin{alignedat}{2}&a={\frac {\alpha (\gamma -\beta )}{\gamma (\gamma +1)}}&\qquad &b={\frac {(\beta +1)(\gamma +1-\alpha )}{(\gamma +1)(\gamma +2)}}\\&c={\frac {(\alpha +1)(\gamma +1-\beta )}{(\gamma +2)(\gamma +3)}}&\qquad &d={\frac {(\beta +2)(\gamma +2-\alpha )}{(\gamma +3)(\gamma +4)}}\\&e={\frac {(\alpha +2)(\gamma +2-\beta )}{(\gamma +4)(\gamma +5)}}&\qquad &f={\frac {(\beta +3)(\gamma +3-\alpha )}{(\gamma +5)(\gamma +6)}}\end{alignedat}}$

etc., cuius lex progressionis obvia est.

Porro ex aequationibus 17, 18, 21, 22 sequitur