Pagina:Werkecarlf02gausrich.djvu/28

$$1-{\frac {1-x^{m}}{1-x}}+{\frac {(1-x^{m})(1-x^{m-1})}{(1-x)(1-xx)}}-{\frac {(1-x^{m})(1-x^{m-1})(1-x^{m-2})}{(1-x)(1-xx)(1-x^{3})}}+{\text{etc.}}$$

sive

$$1-(m,1)+(m,2)-(m,3)+(m,4)-{\text{etc.}}$$

quam brevitatis caussa per ${\textstyle f(x,m)}$ denotabimus. Primo statim obvium est, quoties ${\textstyle m}$ sit numerus integer positivus, hanc seriem post terminum suum ${\textstyle {m+1}^{\text{tum }}}$ (qui fit ${\textstyle =\pm 1}$) abrumpi, adeoque in hoc casu summam fieri debere functionem finitam integram ipsius ${\textstyle x}$. Porro per art. 5. II. patet, generaliter pro valore quocunque ipsius ${\textstyle m}$ haberi

$${\begin{aligned}1&=1\\-(m,1)&=-(m-1,1)-x^{m-1}\\+(m,2)&=+(m-1,2)+x^{m-2}(m-1,1)\\-(m,3)&=-(m-1,3)-x^{m-3}(m-1,2){\text{ etc.}}\end{aligned}}$$

adeoque

$${\begin{aligned}f(x,m)=1-x^{m-1}&-(1-x^{m-2})(m-1,1)+(1-x^{m-3})(m-1,2)\\&-(1-x^{m-4})(m-1.3)+{\text{etc.}}\end{aligned}}$$

Sed manifesto fit

$${\begin{aligned}&(1-x^{m-2})(m-1,1)=(1-x^{m-1})(m-2,1)\\&(1-x^{m-3})(m-1,2)=(1-x^{m-1})(m-2,2)\\&(1-x^{m-4})(m-1,3)=(1-x^{m-1})(m-2,3){\text{ etc.}}\end{aligned}}$$

unde deducimus aequationem

$${\begin{aligned}f(x,m)=(1-x^{m-1})f(x,m-2)&\dots \dots \dots [1]\end{aligned}}$$

7.

Quum pro ${\textstyle m=0}$ fiat ${\textstyle f(x,m)=1}$, per formulam modo inventam erit

$${\begin{aligned}&f(x,2)=1-x\\&f(x,4)=(1-x)(1-x^{3})\\&f(x,6)=(1-x)(1-x^{3})(1-x^{5})\\&f(x,8)=(1-x)(1-x^{3})(1-x^{5})(1-x^{7}){\text{ etc.}}\end{aligned}}$$

sive generaliter pro valore quocunque pari ipsius ${\textstyle m}$

$$f(x,m)=(1-x)(1-x^{3})(1-x^{5})\ldots (1-x^{m-1})\dots \dots \dots [2]$$