Pagina:Werkecarlf03gausrich.djvu/189

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

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

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

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

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

9.

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

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

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

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

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

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

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

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