Pagina:Werkecarlf02gausrich.djvu/23

$${\begin{aligned}&\Sigma \cos ak\omega =\Sigma \cos a^{\prime }k\omega +\Sigma \cos(n-a^{\prime })k\omega \\&\Sigma \cos bk\omega =\Sigma \cos b^{\prime }k\omega +\Sigma \cos(n-b^{\prime })k\omega \\&\Sigma \sin ak\omega =\Sigma \sin a^{\prime }k\omega +\Sigma \sin(n-a^{\prime })k\omega \\&\Sigma \sin bk\omega =\Sigma \sin b^{\prime }k\omega +\Sigma \sin(n-b^{\prime })k\omega \end{aligned}}$$

Iam quum habeatur ${\textstyle \cos(n-a^{\prime })k\omega =\cos a^{\prime }k\omega }$, ${\textstyle \cos(n-b^{\prime })k\omega =\cos b^{\prime }k\omega }$, ${\textstyle \sin(n-a^{\prime })k\omega =-\sin a^{\prime }k\omega }$, ${\textstyle \sin(n-b^{\prime })k\omega =-\sin b^{\prime }k\omega }$, patet sponte fieri

$${\begin{aligned}&\Sigma \sin ak\omega =\Sigma \sin a^{\prime }k\omega -\Sigma \sin a^{\prime }k\omega =0\\&\Sigma \sin bk\omega =\Sigma \sin b^{\prime }k\omega -\Sigma \sin b^{\prime }k\omega =0\end{aligned}}$$

Summatio cosinuum vero hanc formam assumit

$${\begin{aligned}&\Sigma \cos ak\omega =2\Sigma \cos a^{\prime }k\omega \\&\Sigma \cos bk\omega =2\Sigma \cos b^{\prime }k\omega \end{aligned}}$$

unde fieri debebit

$${\begin{aligned}1+4\Sigma \cos a^{\prime }k\omega &=\pm \surd n\\1+4\Sigma \cos b^{\prime }k\omega &=\mp \surd n\\2\Sigma \cos a^{\prime }k\omega -2\Sigma \cos b^{\prime }k\omega &=\pm \surd n\end{aligned}}$$

II. In casu eo, ubi ${\textstyle n}$ est formae ${\textstyle 4m+3}$, complementum cuiusvis residui ${\textstyle a}$ ad ${\textstyle n}$ erit non-residuum, complementumque cuiusvis ${\textstyle b}$ erit residuum; quocirca omnes ${\textstyle n-a}$ convenient cum omnibus ${\textstyle b}$, omnesque ${\textstyle n-b}$ cum omnibus ${\textstyle a}$. Hinc colligitur

$$\Sigma \cos ak\omega =\Sigma \cos(n-b)k\omega =\Sigma \cos bk\omega$$

quare quum omnes ${\textstyle a}$ et ${\textstyle b}$ iuncti omnes numeros ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots n-1}$ expleant, adeoque fiat ${\textstyle \Sigma \cos ak\omega +\Sigma \cos bk\omega =\cos k\omega +\cos 2k\omega +\cos 3k\omega +{\text{etc.}}+\cos(n-1)k\omega =-1}$, summationes

$${\begin{aligned}&\Sigma \cos ak\omega =-{\frac {1}{2}}\\&\Sigma \cos bk\omega =-{\frac {1}{2}}\end{aligned}}$$

sponte sunt obviae. Perinde erit

$$\Sigma \sin ak\omega =\Sigma \sin(n-b)k\omega =-\Sigma \sin bk\omega$$