Pagina:Werkecarlf03gausrich.djvu/31

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Lemma. Si quantitas $r$ angulusque $\varphi$ ita sunt determinati, ut habeantur aequationes $${\begin{array}{rlr}r^{m}\cos m\varphi +Ar^{m-1}\cos(m-1)\varphi +Br^{m-2}\cos(m-2)\varphi +{\text{etc.}}&&\\+Krr\cos 2\varphi +Lr\cos \varphi +M&=0&[1]\end{array}}$$ $${\begin{array}{rlr}r^{m}\sin m\varphi +Ar^{m-1}\sin(m-1)\varphi +Br^{m-2}\sin(m-2)\varphi +{\text{etc.}}&&\\+Krr\sin 2\varphi +Lr\sin \varphi &=0&[2]\end{array}}$$ functio $x^{m}+Ax^{m-1}+Bx^{m-2}+{\text{etc.}}+Kxx+Lx+M=X$ divisibilis erit per factorem duplicem $xx-2\cos \varphi .rx+rr,$ si modo $r\sin \varphi$ non $=0;$ si vero $r\sin \varphi =0,$ eadem functio divisibilis erit per factorem simplicem $x-r\cos \varphi .$

Demonstr. I. Ex art. praec. omnes sequentes quantitates divisibiles erunt per $xx-2\cos \varphi .rx+rr:$ $${\begin{array}{lll}\sin \varphi .rx^{m}&-\sin m\varphi .r^{m}x&+\sin(m-1)\varphi .r^{m+1}\\A\sin \varphi .rx^{m-1}&-A\sin(m-1)\varphi .r^{m-1}x&+\sin(m-2)\varphi .r^{m}\\B\sin \varphi .rx^{m-2}&-B\sin(m-2)\varphi .r^{m-2}x&+\sin(m-3)\varphi .r^{m-1}\\&{\text{etc.}}&{\text{etc.}}\\K\sin \varphi .rx^{2}&-K\sin 2\varphi .rrx&+\sin(m-1)\varphi .r^{3}\\L\sin \varphi .rx&-L\sin \varphi .rx&\\M\sin \varphi .r&&+M\sin(-\varphi ).r\end{array}}$$

Quamobrem etiam summa harum quantitatum per $xx-2\cos \varphi .rx+rr$ divisibilis erit. At singularum partes primae constituunt summam $\sin \varphi .rX;$ secundae additae dant $0,$ propter [2]; tertiarum vero aggregatum quoque evanescere, facile perspicitur, si [1] multiplicatur per $\sin \varphi ,$ [2] per $\cos \varphi ,$ productumque illud ab hoc subducitur. Unde sequitur, functionem $\sin \varphi .rX$ divisibilem esse per $xx-2\cos \varphi .rx+rr,$ adeoque, nisi fuerit $r\sin \varphi =0,$ etiam functionem $X.$ Q.E.P.

II. Si vero $r\sin \varphi =0,$ erit aut $r=0$ aut $\sin \varphi =0.$ In casu priori erit $M=0,$ propter [1], adeoque $X$ per $x$ sive per $x-r\cos \varphi$ divisibilis; in posteriori erit $\cos \varphi =\pm 1,$ $\cos 2\varphi =+1,$ $\cos 3\varphi =\pm 1$ et generaliter $\cos n\varphi =\cos \varphi ^{n}.$ Quare propter [1] fiet $X=0,$ statuendo $x=r\cos \varphi ,$ et proin functio $X$ per $x-r\cos \varphi$ erit divisibilis. Q.E.S.