Question Number 164039 by ArielVyny last updated on 13/Jan/22
![consider f function Df=[0,1] f(0)=f(1) c∈[0,(1/2)] show that f(c)=f(c+(1/2))](Q164039.png)
$${consider}\:{f}\:{function}\:{Df}=\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\:{c}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right]\:{show}\:{that}\:{f}\left({c}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
Answered by Ar Brandon last updated on 13/Jan/22
$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({c}+\mathrm{0}\right)={f}\left({c}+\mathrm{1}\right)\:,\:\mathrm{since}\:{f}\:<\:\mathrm{1}-\mathrm{periodic} \\ $$$${f}\left({c}\right)={f}\left(\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{Hence}\:{f}\left({c}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
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