Pagina:Werkecarlf03gausrich.djvu/180

$${\begin{aligned}&A.{\frac {(nt-1)(nt-2)(nt-3)\ldots (nt-n)}{(-1)(-2).(-3)\ldots .(-n)}}\\+&A^{\prime }.{\frac {nt.(nt-2).(nt-3)\cdots \cdots .(nt-n)}{1.(-1)(-2)\cdots \cdots (1-n)}}\\+&A^{\prime \prime }.{\frac {nt.(nt-1).(nt-3)\ldots \ldots (nt-n)}{2.1.(-1)\ldots \ldots (2-n)}}\\+&A^{\prime \prime \prime }.{\frac {nt.(nt-1).(nt-2)\ldots \ldots (nt-n)}{3.2.1\ldots \ldots (3-n)}}\\+&{\text{ etc. }}\\+&A^{(n)}.{\frac {nt(nt-1).(nt-2)\ldots \ldots (nt-n+1)}{n.(n-1)(n-2)\ldots \ldots .1}}\end{aligned}}$$ sive ${\textstyle \Sigma {\frac {A^{(\mu )}T^{(\mu )}}{M^{(\mu )}}},}$ ubi repraesentante ${\textstyle \mu }$ singulos integros ${\textstyle 0,1,2,3\ldots n,}$

${\textstyle \displaystyle T^{(\mu )}={\frac {nt(nt-1)(nt-2)(nt-3)\ldots (nt-n)}{nt-\mu }}}$

${\textstyle M^{(\mu )}}$ valor ipsius ${\textstyle T}$ pro ${\textstyle nt=\mu .}$

Manifestum erit, ${\textstyle {Y}}$ exhibere functionem algebraicam integram ipsius ${\textstyle t}$ ordinis ${\textstyle n,}$ atque eius valores pro singulis ${\textstyle n+1}$ valoribus ipsius ${\textstyle t,}$ puta ${\textstyle 0,}$ ${\textstyle {\frac {1}{n}},}$ ${\textstyle {\frac {2}{n}},}$ ${\textstyle {\frac {3}{n}}\ldots 1}$ aequales esse valoribus ipsius ${\textstyle y.}$ Porro patet, si ${\textstyle Y^{\prime }}$ sit functio alia integra pro iisdem valoribus cum ${\textstyle y}$ conspirans, ${\textstyle {Y}^{\prime }-{Y}}$ pro iisdem evanescere, adeoque per factores ${\textstyle t,}$ ${\textstyle t-{\frac {1}{n}},}$ ${\textstyle t-{\frac {2}{n}},}$ ${\textstyle t-{\frac {3}{n}}\ldots t-1}$ et proin etiam per eorum productum (quod est ordinis ${\textstyle n+1}$) divisibilem esse, unde patet, ${\textstyle Y^{\prime },}$ nisi prorsus identica sit cum ${\textstyle {Y},}$ certo ad altiorem ordinem ascendere debere, sive ${\textstyle {Y}}$ ex omnibus functionibus integris ordinem ${\textstyle n}$ haud egredientibus unicam esse, quae pro illis ${\textstyle n+1}$ valoribus cum ${\textstyle y}$ conspiret. Quodsi itaque ${\textstyle y,}$ in seriem secundum potestates ipsius ${\textstyle t}$ progredientem evoluta, ante terminum qui implicat ${\textstyle t^{n+1}}$ omnino abrumpitur, cum ${\textstyle {Y}}$ identica erit: si vero saltem tam cito convergit, ut terminos sequentes spernere liceat, functio ${\textstyle Y}$ inter limites ${\textstyle t=0,}$ ${\textstyle t=1}$ sive ${\textstyle x=g,}$ ${\textstyle x=h}$ ipsius ${\textstyle y}$ vice fungi poterit.

3.

Iam integrale nostrum ${\textstyle \int y\operatorname {d} x}$ transit in ${\textstyle \Delta \int y\operatorname {d} t}$ a ${\textstyle t=0}$ usque ad ${\textstyle t=1}$ sumendum, cuius loco per ea, quae modo monuimus, adoptabimus ${\textstyle \Delta \int Y\operatorname {d} t.}$ Evolvendo itaque ${\textstyle T^{(\mu )}}$ in $$\alpha t^{n}+\beta t^{n-1}+\gamma t^{n-2}+\delta t^{n-3}+{\text{etc. }}$$ erit ${\textstyle \int T^{(\mu )}{d}t,}$ a ${\textstyle t=0}$ usque ad ${\textstyle t=1,}$