Question Number 200004 by Mingma last updated on 12/Nov/23
Answered by aleks041103 last updated on 12/Nov/23
$$−\mathrm{1}<{x}<\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{{x}+\mathrm{2}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }=\frac{{x}}{\mathrm{2}^{{k}+\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }<{f}\left({x}\right)<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} } \\ $$$$\Rightarrow\lfloor\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\rfloor\leqslant\lfloor{f}\left({x}\right)\rfloor<\lfloor\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\rfloor \\ $$$${k}=\mathrm{0}:\:\mathrm{0}\leqslant\lfloor{f}\left({x}\right)\rfloor<\mathrm{1},\:\Rightarrow\lfloor{f}\left({x}\right)\rfloor=\mathrm{0} \\ $$$${k}\geqslant\mathrm{1}:\:\lfloor{f}\left({x}\right)\rfloor=\mathrm{0} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\lfloor\frac{{x}+\mathrm{2}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }\rfloor=\begin{cases}{\mathrm{1},\:{x}=\mathrm{1}}\\{\mathrm{0},\:{x}\in\left[−\mathrm{1},\mathrm{1}\right)}\end{cases} \\ $$$$\Rightarrow\int_{−\mathrm{1}} ^{\:\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\lfloor\frac{{x}+\mathrm{2}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }\rfloor{dx}=\mathrm{0}\neq−\mathrm{1} \\ $$
Commented by Mingma last updated on 12/Nov/23
Nice solution