Pagina:Werkecarlf03gausrich.djvu/198

tibus ${\textstyle u^{-1},}$ ${\textstyle u^{-2},}$ ${\textstyle u^{-3},}$ ${\textstyle u^{-4}\ldots u^{-(n+1)}.}$ Modus posterior aliquanto simplicior erit: quum enim facile perspiciatur, coëfficientes ipsius ${\textstyle U,}$ ut conditioni praescriptae satisfiat, alternatim evanescere debere, sive statui ${\textstyle \beta =0,}$ ${\textstyle \beta ^{\prime \prime }=0,}$ ${\textstyle \beta ^{\prime \prime \prime \prime }=0}$ etc., laboris dimidia fere pars iam absoluta censenda erit. Evolvamus casus quosdam simpliciores.

I. Pro ${\textstyle n=0,}$ coëfficiens unicus ipsius ${\textstyle t^{-1}}$ in producto

$$(t+\alpha )(t^{-1}+{\frac {1}{2}}t^{-2}+{\frac {1}{3}}t^{-3}+{\text{etc.}})$$

evanescere debet. Qui quum fiat ${\textstyle ={\frac {1}{2}}+\alpha ,}$ habemus ${\textstyle \alpha =-{\frac {1}{2}},}$ sive ${\textstyle T=t-{\frac {1}{2}}.}$ Perinde ${\textstyle U=u.}$
II. Pro ${\textstyle n=1,}$ determinatio ipsius ${\textstyle T}$ pendet a duabus aequationibus

$${\begin{aligned}&0={\frac {1}{3}}+{\frac {1}{2}}\alpha +\alpha ^{\prime }\\&0={\frac {1}{4}}+{\frac {1}{3}}\alpha +{\frac {1}{2}}\alpha ^{\prime }\end{aligned}}$$

unde deducimus ${\textstyle \alpha =-1,}$ ${\textstyle \alpha ^{\prime }=+{\frac {1}{6}},}$ sive ${\textstyle T=tt-t+{\frac {1}{6}}.}$ Determinatio functionis ${\textstyle U}$ unicam aequationem affert

$$0={\frac {1}{3}}+\beta ^{\prime }$$

unde ${\textstyle \beta ^{\prime }=-{\frac {1}{3}},}$ sive ${\textstyle U=uu-{\frac {1}{3}}.}$

III. Pro ${\textstyle n=2,}$ functio ${\textstyle T}$ determinatur adiumento trium aequationum

$${\begin{aligned}&0={\frac {1}{4}}+{\frac {1}{3}}\alpha +{\frac {1}{2}}\alpha ^{\prime }+\alpha ^{\prime \prime }\\&0={\frac {1}{5}}+{\frac {1}{4}}\alpha +{\frac {1}{3}}\alpha ^{\prime }+{\frac {1}{2}}\alpha ^{\prime \prime }\\&0={\frac {1}{6}}+{\frac {1}{5}}\alpha +{\frac {1}{4}}\alpha ^{\prime }+{\frac {1}{3}}\alpha ^{\prime \prime }\end{aligned}}$$

unde nanciscimur ${\textstyle \alpha =-{\frac {3}{2}},\alpha ^{\prime }={\frac {3}{5}},\alpha ^{\prime \prime }=-{\frac {1}{20}},}$ adeoque ${\textstyle T=t^{3}-{\frac {3}{2}}tt+{\frac {3}{5}}t-{\frac {1}{20}}.}$ Ad determinandam ${\textstyle U}$ unica aequatio sufficit

$$0={\frac {1}{5}}+{\frac {1}{3}}\beta ^{\prime }$$

unde ${\textstyle \beta ^{\prime }=-{\frac {3}{5}}}$ sive ${\textstyle U=u^{3}-{\frac {3}{5}}u.}$

Attamen hunc modum, qui calculos continuo molestiores adducit, hic ulterius non persequemur, sed ad fontem genuinum solutionis generalis progrediemur.