Pagina:Werkecarlf02gausrich.djvu/35

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

Multiplicando hanc aequationem per [5] in forma primitiva, prodit

$$W^{2}=(-1)^{{\frac {1}{2}}(n-1)}(r-r^{-1})(r^{2}-r^{-2})(r^{3}-r^{-3})\ldots (r^{n-1}-r^{-n+1})$$

ubi ${\textstyle (-1)^{{\frac {1}{2}}(n-1)}}$ est vel ${\textstyle =+1}$ vel ${\textstyle =-1}$, prout ${\textstyle n}$ est formae ${\textstyle 4\mu +1}$, vel formae ${\textstyle 4\mu +3}$. Hinc

$$W^{2}=\pm r^{{\frac {1}{2}}n(n-1)}(1-r^{-2})(1-r^{-4})(1-r^{-6})\ldots (1-r^{-2(n-1)})$$

Sed nullo negotio perspicitur, ${\textstyle r^{-2}}$, ${\textstyle r^{-4}}$, ${\textstyle r^{-6}\ldots r^{-2n+2}}$ exhibere omnes radices aequationis ${\textstyle x^{n}-1=0}$, radice ${\textstyle x=1}$ excepta, unde locum habere debebit aequatio identica indefinita

$$(x-r^{-2})(x-r^{-4})(x-r^{-6})\ldots (x-r^{-2n+2})=x^{n-1}+x^{n-2}+x^{n-3}+{\text{etc.}}+x+1$$

Quamobrem statuendo ${\textstyle x=1}$, fiet

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

et quum manifesto sit ${\textstyle r^{{\frac {1}{2}}n(n-1)}=1}$, aequatio nostra transit in hanc

$$W^{2}=\pm n\dots \dots \dots [6]$$

In casu itaque eo, ubi ${\textstyle n}$ est formae ${\textstyle 4\mu +1}$, fiet

$$W=\pm \surd n,{\text{ et proin }}T=\pm \surd n,\quad U=0$$

Contra in casu altero, ubi ${\textstyle n}$ est formae ${\textstyle 4\mu +3}$, fiet

$$W=\pm i\surd n,{\text{ adeoque }}T=0,\quad U=\pm \surd n$$

14.

Methodus art. praec. valorem tantummodo absolutum aggregatorum ${\textstyle T}$, ${\textstyle U}$ assignat, ambiguumque linquit, utrum statuere oporteat ${\textstyle T}$ in casu priori atque ${\textstyle U}$ in casu posteriori ${\textstyle =+\surd n}$, an ${\textstyle =-\surd n}$. Hoc autem, saltem pro casu eo ubi ${\textstyle k=1}$, ex aequatione [5] sequenti modo decidere licebit. Quum sit, pro ${\textstyle k=1}$,