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Question Number 40090 by maxmathsup by imad last updated on 15/Jul/18
![let f(x)= 1−[x]−[1−x] 1) prove that f is periodic with period 1 2) give a expression of f(x) when x∈[0,1[](Q40090.png)
$${let}\:{f}\left({x}\right)=\:\mathrm{1}−\left[{x}\right]−\left[\mathrm{1}−{x}\right] \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}\:{is}\:{periodic}\:{with}\:{period}\:\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{give}\:{a}\:{expression}\:{of}\:{f}\left({x}\right)\:{when}\:\:{x}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$
Commented by math khazana by abdo last updated on 29/Jul/18
![1) first we have f(x)=1−[x]−1−[−x] =−[x]−[−x] ⇒ f(x+1) =−[x+1]−[−(x+1)] =−[x]−1−[−x]+1=f(x) so f is periodic with T=1 2) if 0≤x<1 [x]=0 and −1<−x≤0 ⇒[−x]=−1 ⇒ f(x)=−[−x]=−(−1)=1 .](Q40926.png)
$$\left.\mathrm{1}\right)\:{first}\:{we}\:{have}\:{f}\left({x}\right)=\mathrm{1}−\left[{x}\right]−\mathrm{1}−\left[−{x}\right] \\ $$$$=−\left[{x}\right]−\left[−{x}\right]\:\Rightarrow \\ $$$${f}\left({x}+\mathrm{1}\right)\:=−\left[{x}+\mathrm{1}\right]−\left[−\left({x}+\mathrm{1}\right)\right] \\ $$$$=−\left[{x}\right]−\mathrm{1}−\left[−{x}\right]+\mathrm{1}={f}\left({x}\right)\:{so}\:{f}\:{is}\:{periodic}\:{with} \\ $$$${T}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{if}\:\mathrm{0}\leqslant{x}<\mathrm{1}\:\left[{x}\right]=\mathrm{0}\:{and}\:−\mathrm{1}<−{x}\leqslant\mathrm{0}\:\Rightarrow\left[−{x}\right]=−\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)=−\left[−{x}\right]=−\left(−\mathrm{1}\right)=\mathrm{1}\:. \\ $$$$ \\ $$