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Question Number 165744 by mnjuly1970 last updated on 07/Feb/22

Answered by TheSupreme last updated on 07/Feb/22

$$\lfloor x\rfloor -\lfloor x^2\rfloor ⩾0$$$$\lfloor x^2\rfloor =\lfloor {(\lfloor x\rfloor +\left\{x\right\})}^2\rfloor =$$$$\lfloor \lfloor x{\rfloor }^2+2\lfloor x\rfloor \left\{x\right\}+{\left\{x\right\}}^2\rfloor =\lfloor x{\rfloor }^2+2\lfloor \lfloor x\rfloor \left\{x\right\}\rfloor$$$$$$$$2\lfloor \lfloor x\rfloor \left\{x\right\}\rfloor ⩽\lfloor x\rfloor -\lfloor x{\rfloor }^2$$$$0⩽2\lfloor \lfloor x\rfloor \left\{x\right\}\rfloor <2\lfloor x\rfloor$$$$0⩽\lfloor x\rfloor -\lfloor x{\rfloor }^2\to 0⩽\lfloor x\rfloor ⩽1$$$$\lfloor x\rfloor =0$$$$f\left(x\right)=0$$$$$$$$\lfloor x\rfloor =1$$$$f\left(x\right)=\sqrt{-\lfloor 2\left\{y\right\}\rfloor }=0\to 0⩽\left\{y\right\}<\frac{1}{2}$$$$f\left(x\right)=0\mathrm{\forall }x\in [0,\frac{3}{2})$$

Answered by mahdipoor last updated on 07/Feb/22

$$getx=k+ak\in \mathrm{Z}0⩽a<1$$$$ifk⩾0,0⩽a<1\{\begin{array}{l}0⩽a^2<1\\ 0⩽2ka<2k\end{array}\Rightarrow$$$$0⩽a^2+2ka<2k+1\Rightarrow 0⩽[2ka+a^2]⩽2k\Rightarrow$$$$k^2⩽[k^2+2ka+a^2]⩽k^2+2k\Rightarrow k^2⩽\left[x^2\right]⩽k^2+2k$$$$so\Rightarrow -k^2-k⩽\left[x\right]-\left[x^2\right]⩽k-k^2$$$$bot\left[x\right]-\left[x^2\right]⩾0\Rightarrow 0⩽\left[x\right]-\left[x^2\right]⩽k-k^2$$$$\Rightarrow 0⩽f\left(x\right)⩽\sqrt{k-k^2}\stackrel{0⩽k\in \mathrm{Z}}{\Rightarrow }f\left(x\right)=0$$$$andifk<0\Rightarrow k-k^2⩽\left[x\right]-\left[x^2\right]⩽-k^2-k$$$$\Rightarrow 0⩽f\left(x\right)⩽\sqrt{-k^2-k}\stackrel{0>k\in \mathrm{Z}}{\Rightarrow }f\left(x\right)=0$$$$R_f=0$$

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