Pagina:Werkecarlf03gausrich.djvu/205

20.

Quum in functione ${\textstyle U}$ potestates ${\textstyle u^{n},}$ ${\textstyle u^{n-2},}$ ${\textstyle u^{n-4}}$ etc. absint, e radicibus aequationis ${\textstyle U=0}$ binae semper erunt magnitudine aequales signis oppositae, quibus pro valore pari ipsius ${\textstyle n}$ adhuc associare oportet radicem singularem ${\textstyle 0.}$ Inventis radicibus, valores coëfficientium ${\textstyle R,}$ ${\textstyle R^{\prime },}$ ${\textstyle R^{\prime \prime }}$ etc. secundum methodum art. 11 habebuntur per functionem integram ipsius ${\textstyle u,}$ quae pro valore impari ipsius ${\textstyle n}$ erit formae

$$\gamma u^{n-1}+\gamma ^{\prime }u^{n-3}+\gamma ^{\prime \prime }u^{n-5}+{\text{etc. }}$$

pro valore pari autem, si excluditur coëfficiens radici ${\textstyle u=0}$ respondens, formae

$$\gamma u^{n-2}+\gamma ^{\prime }u^{n-4}+\gamma ^{\prime \prime }u^{n-6}+{\text{etc. }}$$

Exemplum art. 12 ipsam hanc reductionem exhibet pro ${\textstyle n=6.}$ Manifesto igitur valoribus oppositis ipsius ${\textstyle u}$ semper respondent coëfficientes aequales. Ceterum in casu eo, ubi ${\textstyle n}$ est par, coëfficiens radici ${\textstyle u=0}$ respondens facile generaliter a priori assignari potest. Habebitur hic coëfficiens, si in ${\textstyle {\frac {U^{\prime }}{({\frac {dU}{du}})}}}$ substituitur ${\textstyle u=0.}$ Valorem numeratoris ${\textstyle U^{\prime }}$ pro ${\textstyle u=0}$ iam in art. 18 tradidimus, valor denominatoris autem ibinde erit

$$=\pm {\frac {3.5.7\ldots ..(n+1)}{(n+3)(n+5)\ldots .(2n+1)}}=\pm {\frac {3.3.5.5.7.7\ldots .(n+1)(n+1)}{3.5.7.9.11\ldots .(2n+1)}}$$

adeoque coëfficiens quaesitus

$$=({\frac {2.4.6.8\ldots \ldots n}{3.5.7.9\ldots \ldots (n+1)}})^{2}$$

21.

Functio integra ipsius ${\textstyle u}$ coëfficientes ${\textstyle {R},}$ ${\textstyle {R}^{\prime },}$ ${\textstyle {R}^{\prime \prime }}$ etc. repraesentans in eo quem hic tractamus casu etiam independenter a methodo generali art. 11 erui potest sequenti modo. Differentiando aequationem

$$\varphi -{\frac {U^{\prime }}{U}}={\frac {U^{\prime \prime }}{U}}$$

substituendo dein ${\textstyle {\frac {\operatorname {d} \varphi }{\operatorname {d} u}}={\frac {1}{1-uu}},}$ ac multiplicando per ${\textstyle UU(uu-1),}$ obtinemus

$$(uu-1)U^{\prime }{\frac {\operatorname {d} U}{\operatorname {d} u}}-U({\frac {\operatorname {d} U^{\prime }}{\operatorname {d} u}}.(uu-1)+U)=(uu-1)UU{\frac {\operatorname {d} ({\frac {U^{\prime \prime }}{U}})}{\operatorname {d} u}}$$