Pagina:Werkecarlf02gausrich.djvu/27

sive generaliter

$$(m,\mu )=(m,m-\mu )$$

II. Porro facile confirmatur, haberi generaliter

$$(m,\mu +1)=(m-1,\mu +1)+x^{m-\mu -1}(m-1,\mu )$$

quamobrem, quum perinde sit

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

quae series continuari poterit usque ad

$${\begin{aligned}(\mu +2,\mu +1)&=(\mu +1,\mu +1)+x(\mu +1,\mu )\\&=(\mu ,\mu )+x(\mu +1,\mu )\end{aligned}}$$

siquidem ${\textstyle m}$ est integer positivus maior quam ${\textstyle \mu +1}$, erit

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

Hinc patet, si pro aliquo valore determinato ipsius ${\textstyle \mu }$ quaevis functio ${\textstyle (m,\mu )}$ integra sit, existente ${\textstyle m}$ integro positivo, etiam quamvis functionem ${\textstyle (m,\mu +1)}$ integram evadere debere. Quare quum suppositio illa pro ${\textstyle \mu =1}$ locum habeat, eadem etiam pro ${\textstyle \mu =2}$ valebit, atque hinc etiam pro ${\textstyle \mu =3}$ etc., i.e. generaliter pro valore quocunque integro positivo ipsius ${\textstyle m}$ erit ${\textstyle (m,\mu )}$ functio integra, sive productum

$$(1-x^{m})(1-x^{m-1})(1-x^{m-2})\ldots (1-x^{m-\mu +1})$$

divisibile per

$$(1-x)(1-x^{2})(1-x^{3})\ldots (1-x^{\mu })$$

6.

Duas iam progressiones considerabimus, quae ambae ad scopum nostrum ducere possunt. Progressio prima haec est