Pagina:Werkecarlf03gausrich.djvu/152

${\begin{aligned}&RF({\tfrac {1}{2}},{\tfrac {5}{2}},{\tfrac {9}{2}},x)=F({\tfrac {1}{2}},{\tfrac {3}{2}},{\tfrac {7}{2}},x)\\&QF({\tfrac {5}{2}},{\tfrac {1}{2}},{\tfrac {9}{2}},x)=F({\tfrac {5}{2}},-{\tfrac {1}{2}},{\tfrac {7}{2}},x)\end{aligned}}$

sive permutando elementum primum cum secundo

$QF({\tfrac {1}{2}},{\tfrac {5}{2}},{\tfrac {9}{2}},x)=F(-{\tfrac {1}{2}},{\tfrac {5}{2}},{\tfrac {7}{2}},x)$

Sed per aequationem 21 habemus

$F(-{\tfrac {1}{2}},{\tfrac {5}{2}},{\tfrac {7}{2}},x)-F({\tfrac {1}{2}},{\tfrac {3}{2}},{\tfrac {7}{2}},x)=-{\tfrac {4}{7}}xF({\tfrac {1}{2}},{\tfrac {5}{2}},{\tfrac {9}{2}},x)$

unde fit $Q={R}-{\tfrac {4}{7}}x,$ quo valore in formula supra data substituto prodit

$\xi ={\frac {{\frac {2}{35}}xx}{R-{\frac {18}{35}}x}}$

quae est formula posterior.

Statuendo in formula 30, $\alpha ={\tfrac {m}{n}},$ $x=-\gamma nt,$ fit pro valore infinite magno ipsius $\gamma$

[36]	$F({\tfrac {m}{n}},1,\gamma ,-\gamma nt)=1-mt+m(m+n)tt-m(m+n)(m+2n)t^{3}+{\text{etc. }}$
	$\quad ={\cfrac {1}{1+{\cfrac {mt}{1+{\cfrac {nt}{1+{\cfrac {(m+n)t}{1+{\cfrac {2nt}{1+{\cfrac {(m+2n)t}{1+3nt{\text{ etc.}}}}}}}}}}}}}}$

---

SECTIO TERTIA.
De summa seriei nostrae statuendo elementum quartum $=1,$ ubi simul quaedam aliae functiones transscendentes discutiuntur.

15.

Quoties elementa $\alpha ,$ $\beta ,$ $\gamma$ omnia sunt quantitates positivae, omnes coëfficientes potestatum elementi quarti $x$ positivi evadunt: quoties vero ex illis elementis unum alterumve negativum est, saltem inde ab aliqua potestate $x^{m}$ omnes coëfficientes eodem signo affecti erunt, si modo $m$ accipitur maior quam valor absolutus elementi negativi maximi. Porro hinc sponte patet, seriei șummam pro