Pagina:Werkecarlf02gausrich.djvu/155

${\textstyle =\varphi (a-b,a+b)}$, tertius ${\textstyle =-\varphi (a,b)}$; secundus vero fit

$$={\frac {1}{4}}bb+\sum \left[{\frac {aa-2ax}{2b}}\right]$$

Sed fit, scribendo terminos inverso ordine,

$$\Sigma \left[{\frac {aa-2ax}{2b}}\right]=\left[{\frac {a}{2b}}\right]+\left[{\frac {3a}{2b}}\right]+\left[{\frac {5a}{2b}}\right]+\ldots +\left[{\frac {(b-1)a}{2b}}\right]=\varphi (2b,a)-\varphi (b,a)$$

Formula itaque nostra sequentem induit formam:

$$g=\varphi (a-b,a+b)+\varphi (2b.a)-\varphi (a,b)-\varphi (b,a)+{\frac {1}{4}}bb$$

Consideremus primo terminum ${\textstyle \varphi (a-b,a+b)}$, qui protinus transmutatur in ${\textstyle \varphi (a-b,2b)+1+2+3+{\text{etc.}}+{\frac {1}{2}}(a-b-1)}$ sive in

$$\varphi (a-b,2b)+{\frac {1}{8}}((a-b)^{2}-1)$$

Dein quum per theorema generale fiat ${\textstyle \varphi (t,u)+\varphi (u,t)=\left[{\frac {1}{2}}t\right]\cdot \left[{\frac {1}{2}}u\right]}$, dum ${\textstyle t,u}$ sunt integri positivi inter se primi, habemus

$$\varphi (a-b,2b)={\frac {1}{2}}b(a-b-1)-\varphi (2b,a-b)$$

adeoque

$$\varphi (a-b,a+b)={\frac {1}{8}}(aa+2ab-3bb-4b-1)-\varphi (2b,a-b)$$

Disponamus partes ipsius ${\textstyle \varphi (2b,a-b)}$ sequenti modo

$${\begin{aligned}&{\left[{\frac {a-b}{2b}}\right]+\left[{\frac {3(a-b)}{2b}}\right]+\left[{\frac {5(a-b)}{2b}}\right]+{\text{etc.}}+\left[{\frac {(b-1)(a-b)}{2b}}\right]}\\+&{\left[{\frac {a-b}{b}}\right]+\left[{\frac {2(a-b)}{b}}\right]+\left[{\frac {3(a-b)}{b}}\right]+{\text{etc.}}+\left[{\frac {{\frac {1}{2}}b(a-b)}{b}}\right]}\end{aligned}}$$

Series secunda manifesto fit

$$=\varphi (b,a-b)=\varphi (b,a)-1-2-3-{\text{etc.}}-{\frac {1}{2}}b=\varphi (b,a)-{\frac {1}{8}}(bb+2b)d$$

seriem primam ordine terminorum inverso ita exhibemus:

$$\left[{\frac {1}{2}}(a+1-b)-{\frac {a}{2b}}\right]+\left[{\frac {1}{2}}(a+3-b)-{\frac {3a}{2b}}\right]+\left[{\frac {1}{2}}(a+5-b)-{\frac {5a}{2b}}\right]+{\text{etc.}}+\left[{\frac {1}{2}}(a-1)-{\frac {(b-1)a}{2b}}\right]$$

quae expressio, quum denotante ${\textstyle t}$ numerum integrum, ${\textstyle u}$ fractum, generaliter sit ${\textstyle [t-u]=t-1-[u]}$, mutatur in sequentem

$${\begin{aligned}&{\frac {1}{8}}b(2a-4-b)-\left[{\frac {a}{2b}}\right]-\left[{\frac {3a}{2b}}\right]-\left[{\frac {5a}{2b}}\right]-{\text{etc.}}-\left[{\frac {(b-1)a}{2b}}\right]\\=&{\frac {1}{8}}b(2a-4-b)-\varphi (2b,a)+\varphi (b,a)\end{aligned}}$$