Obliczyć dla n całkowitych.... Question from @KFKuba

KFKuba @KFKuba

September 2018 1 19 Report

Obliczyć $\lfloor\frac{n+1}{\sqrt{2}}\rfloor+ \lfloor\frac{n+1}{2 + \sqrt{2}}\rfloor$ dla n całkowitych

Roma

Def. Funkcji podłoga

$\lfloor x \rfloor = max \{k \in \mathbb{Z} : \ k \leq x \}$

Własności:

$\lfloor x+y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor$

$Dla \ k \in \mathbb{Z}, \ x \in \mathbb{R}: \ \lfloor k+x \rfloor = k + \lfloor x \rfloor$

$Dla \ x \in \mathbb{Z}: \ \lfloor x \rfloor = x$

$\lfloor\frac{n+1}{\sqrt{2}}\rfloor+\lfloor\frac{n+1}{2+\sqrt{2}}\rfloor=\lfloor\frac{(n+1) \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}\rfloor+\lfloor\frac{(n+1)(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}\rfloor =$

$=\lfloor\frac{\sqrt{2}n+\sqrt{2}}{2}\rfloor+\lfloor\frac{2n-\sqrt{2}n+2-\sqrt{2}}{4-2}\rfloor = \lfloor\frac{\sqrt{2}n+\sqrt{2}}{2}\rfloor+\lfloor\frac{2n-\sqrt{2}n+2-\sqrt{2}}{2}\rfloor$

$\lfloor\frac{\sqrt{2}n+\sqrt{2}}{2}\rfloor+\lfloor\frac{2n-\sqrt{2}n+2-\sqrt{2}}{2}\rfloor \leq \lfloor\frac{\sqrt{2}n+\sqrt{2}}{2} + \frac{2n-\sqrt{2}n+2-\sqrt{2}}{2}\rfloor$

$\lfloor\frac{\sqrt{2}n+\sqrt{2}}{2} + \frac{2n-\sqrt{2}n+2-\sqrt{2}}{2}\rfloor = \lfloor\frac{\sqrt{2}n+\sqrt{2} + 2n-\sqrt{2}n+2-\sqrt{2}}{2} \rfloor =$