Pagina:Werkecarlf02gausrich.djvu/32

$${\begin{aligned}&\Sigma \cos ak\omega =\cos({\frac {1}{2}}(n+1))^{2}k\omega +\cos({\frac {1}{2}}(n+3))^{2}k\omega +{\text{etc.}}+\cos(n-1)^{2}k\omega \\&\Sigma \sin ak\omega =\sin({\frac {1}{2}}(n+1))^{2}k\omega +\sin({\frac {1}{2}}(n+3))^{2}k\omega +{\text{etc.}}+\sin(n-1)^{2}k\omega \end{aligned}}$$

Statuendo itaque

$${\begin{alignedat}{2}&T=&1+\cos k\omega +\cos 4k\omega +\cos 9k\omega +{\text{etc.}}+\cos(n-1)^{2}k\omega \\&U=&\sin k\omega +\sin 4k\omega +\sin 9k\omega +{\text{etc.}}+\sin(n-1)^{2}k\omega \end{alignedat}}$$

erit

$${\begin{aligned}1+2\Sigma \cos ak\omega &=T\\2\Sigma \sin ak\omega &=U\end{aligned}}$$

Hinc patet, summationes, quales in art. 1. propositae sunt, pendere a summatione serierum ${\textstyle T}$ et ${\textstyle U}$, quocirca, missis illis, disquisitionem nostram his adaptabimus, eaque generalitate absolvemus, ut non modo valores primos ipsius ${\textstyle n}$, sed quoscunque compositos complectatur. Numerum ${\textstyle k}$ autem supponemus ad ${\textstyle n}$ primum esse: nullo enim negotio casus is, ubi ${\textstyle k}$ et ${\textstyle n}$ divisorem communem haberent, ad hunc reduci poterit.

11.

Designemus quantitatem imaginariam ${\textstyle \surd {-1}}$ per ${\textstyle i}$, statuamusque

$$\cos k\omega +i\sin k\omega =r$$

unde erit ${\textstyle r^{n}=1}$, sive ${\textstyle r}$ radix aequationis ${\textstyle r^{n}-1=0}$. Facile perspicietur, omnes numeros ${\textstyle k}$, ${\textstyle 2k}$, ${\textstyle 3k\ldots (n-1)k}$ per ${\textstyle n}$ non divisibiles atque inter se secundum modulum ${\textstyle n}$ incongruos esse: hinc potestates ipsius ${\textstyle r}$

$$1,r,rr,r^{3}\ldots r^{n-1}$$

omnes erunt inaequales, singulae vero quoque aequationi ${\textstyle x^{n}-1=0}$ satisfacient. Hanc ob caussam hae potestates omnes radices aequationis ${\textstyle x^{n}-1=0}$ repraesentabunt.

Hae conclusiones non valerent, si ${\textstyle k}$ divisorem communem haberet cum ${\textstyle n}$. Si enim ${\textstyle \nu }$ esset talis divisor communis, foret ${\textstyle k.{\frac {n}{\nu }}}$ per ${\textstyle n}$ divisibilis, adeoque potestas inferior quam ${\textstyle r^{n}}$, puta ${\textstyle r^{\frac {n}{\nu }}}$, unitati aequalis. In hoc itaque casu potestates ipsius ${\textstyle r}$ ad summum ${\textstyle {\frac {n}{\nu }}}$ radices aequationis ${\textstyle x^{n}-1=0}$ exhibebunt, et quidem revera tot radices diversas sistent, si ${\textstyle \nu }$ est divisor communis maximus nume-