Pagina:Werkecarlf02gausrich.djvu/40

$$W^{2}=2r^{{\frac {1}{4}}n}n\dots \dots \dots [9]$$

Iam facile perspicietur, ${\textstyle r^{{\frac {1}{4}}n}}$ esse vel ${\textstyle =+i}$ vel ${\textstyle =-i}$, prout scilicet ${\textstyle k}$ vel formae ${\textstyle 4\mu +1}$ sit, vel formae ${\textstyle 4\mu +3}$. Et quum sit

$$2i=(1+i)^{2},\quad -2i=(1-i)^{2}$$

erit in casu eo, ubi ${\textstyle k}$ est formae ${\textstyle 4\mu +1}$,

$$W=\pm (1+i)\surd n,{\text{ adeoque }}\quad T=U=\pm \surd n$$

in casu altero autem, ubi ${\textstyle k}$ est formae ${\textstyle 4\mu +3}$,

$$W=\pm (1-i)\surd n,{\text{ adeoque }}T=-U=\pm \surd n$$

18.

Methodus art. praec. valores absolutos functionum ${\textstyle T}$, ${\textstyle U}$ suppeditavit, conditionesque assignavit, sub quibus signa aequalia vel opposita illis tribuenda sint: sed signa ipsa hinc nondum determinantur. Hoc pro eo casu, ubi statuitur ${\textstyle k=1}$, sequenti modo supplebimus.

Statuamus ${\textstyle \rho =\cos {\frac {1}{2}}\omega +i\sin {\frac {1}{2}}\omega }$, ita ut fiat ${\textstyle r=\rho \rho }$, patetque, propter ${\textstyle \rho ^{n}=-1}$ aequationem [8] ita exhiberi posse

$$W=2(1+\rho ^{n-2})(1+\rho ^{-4})(1+\rho ^{n-6})(1+\rho ^{-8})\ldots (1+\rho ^{-n+4})(1+\rho ^{2})$$

sive factoribus alio ordine dispositis

$$W=2(1+\rho ^{2})(1+\rho ^{-4})(1+\rho ^{6})(1+\rho ^{-8})\ldots (1+\rho ^{-n+4})(1+\rho ^{n-2})$$

Iam fit

$${\begin{aligned}&1+\rho ^{2}=2\rho \cos {\frac {1}{2}}\omega \\&1+\rho ^{-4}=2\rho ^{-2}\cos \omega \\&1+\rho ^{+6}=2\rho ^{3}\cos {\frac {3}{2}}\omega \\&1+\rho ^{-8}=2\rho ^{-4}\cos 2\omega {\text{ etc.}}\end{aligned}}$$

usque ad

$${\begin{aligned}&1+\rho ^{-n+4}=2\rho ^{-{\frac {1}{2}}n+2}\cos({\frac {1}{4}}n-1)\omega \\&1+\rho ^{n-2}=2\rho ^{{\frac {1}{2}}n-1}\cos({\frac {1}{4}}n-{\frac {1}{2}})\omega \end{aligned}}$$

Quamobrem habetur