Pagina:Werkecarlf03gausrich.djvu/189

ita ut ${\textstyle T^{\prime }}$ sit functio integra ipsius ${\textstyle t}$ in hoc producto contenta, ${\textstyle T^{\prime \prime }}$ vero pars altera, scilicet series descendens in infinitumque excurrens. Quo facto valor integralis ${\textstyle \int {\frac {T\operatorname {d} t}{t-a}}}$ a ${\textstyle t=0}$ usque ad ${\textstyle t=1}$ aequalis erit valori functionis ${\textstyle T^{\prime }}$ pro ${\textstyle t=a.}$ Quodsi itaque valores determinatos functionis

$${\frac {T^{\prime }}{({\frac {\operatorname {d} T}{\operatorname {d} t}})}}$$

pro ${\textstyle t=a,}$ ${\textstyle t=a^{\prime },}$ ${\textstyle t=a^{\prime \prime },}$ ${\textstyle t=a^{\prime \prime \prime }}$ etc. usque ad ${\textstyle t=a^{(n)}}$ resp. per ${\textstyle R,}$ ${\textstyle R^{\prime },}$ ${\textstyle R^{\prime \prime },}$ ${\textstyle R^{\prime \prime \prime }\ldots R^{(n)}}$ denotamus, integrale ${\textstyle \int Y\operatorname {d} }$ a ${\textstyle t=0}$ usque ad ${\textstyle t=1}$ fiet

$$=RA+R^{\prime }A^{\prime }+R^{\prime \prime }A^{\prime \prime }+{\text{etc.}}+R^{(n)}A^{(n)}$$

quod per ${\textstyle \Delta }$ multiplicatum exhibebit valorem vel verum vel approximatum integralis ${\textstyle \int y\operatorname {d} x}$ ab ${\textstyle x=g}$ usque ad ${\textstyle x=g+\Delta .}$

9.

Hae operationes aliquanto facilius perficiuntur, si loco indeterminatae ${\textstyle t}$ introducitur alia ${\textstyle u=2t-1.}$ Scribimus quoque brevitatis caussa ${\textstyle b=2a-1,}$ ${\textstyle b^{\prime }=2a^{\prime }-1,}$ ${\textstyle b^{\prime \prime }=2a^{\prime \prime }-1}$ etc. Transeat ${\textstyle T,}$ substituto pro ${\textstyle t}$ valore ${\textstyle {\frac {1}{2}}u+{\frac {1}{2}},}$ in ${\textstyle {\frac {U}{2^{n+1}}},}$ sive sit

$$U=(u-b)(u-b^{\prime })(u-b^{\prime \prime })\ldots (u-b^{n})$$

Erit itaque ${\textstyle {\frac {\operatorname {d} T}{\operatorname {d} t}}={\frac {1}{2^{n}}}.{\frac {\operatorname {d} U}{\operatorname {d} u}},}$ adeoque ${\textstyle M,}$ ${\textstyle M^{\prime },}$ ${\textstyle M^{\prime \prime }}$ etc. valores determinati ipsius ${\textstyle {\frac {1}{2^{n}}}.{\frac {\operatorname {d} U}{\operatorname {d} u}},}$ si deinceps statuitur ${\textstyle u=b,}$ ${\textstyle u=b^{\prime },}$ ${\textstyle u=b^{\prime \prime }}$ etc.

Quum series ${\textstyle t^{-1}+{\frac {1}{2}}t^{-2}+{\frac {1}{3}}t^{-3}+{\frac {1}{4}}t^{-4}+}$ etc. nihil aliud sit quam ${\textstyle \log {\frac {1}{1-t^{-1}}}=\log {\frac {1+u^{-1}}{1-u^{-1}}}{:}}$ per substitutionem ${\textstyle t={\frac {1}{2}}u+{\frac {1}{2}}}$ necessario transibit in ${\textstyle 2u^{-1}+{\frac {2}{3}}u^{-3}+{\frac {2}{5}}u^{-5}+{\frac {2}{7}}u^{-7}+}$ etc. Quodsi itaque statuimus

$$U(u^{-1}+{\frac {1}{3}}u^{-3}+{\frac {1}{3}}u^{-5}+{\frac {1}{7}}u^{-7}+{\text{etc.}})=U^{\prime }+U^{\prime \prime }$$

ita ut ${\textstyle U^{\prime }}$ sit functio integra ipsius ${\textstyle u}$ in hoc producto contenta, ${\textstyle U^{\prime \prime }}$ vero pars altera, puta series descendens infinita, patet esse

$$T^{\prime }+T^{\prime \prime }={\frac {1}{2^{n}}}(U^{\prime }+U^{\prime \prime })$$

Sed manifesto ${\textstyle T^{\prime },}$ tamquam functio integra ipsius ${\textstyle t,}$ per substitutionem ${\textstyle t={\frac {1}{2}}u+{\frac {1}{2}}}$ necessario functionem integram ipsius ${\textstyle u}$ producet: contra ${\textstyle T^{\prime \prime },}$ quae non continet nisi potestates negativas ipsius ${\textstyle t,}$ per eandem substitutionem tantummodo potesta-