consider-f-function-Df-0-1-f-0-f-1-c-0-1-2-show-that-f-c-f-c-1-2-

Question Number 164039 by ArielVyny last updated on 13/Jan/22

$${consider}\:{f}\:{function}\:{Df}=\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\:{c}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right]\:{show}\:{that}\:{f}\left({c}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Answered by Ar Brandon last updated on 13/Jan/22

$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({c}+\mathrm{0}\right)={f}\left({c}+\mathrm{1}\right)\:,\:\mathrm{since}\:{f}\:<\:\mathrm{1}-\mathrm{periodic} \\ $$$${f}\left({c}\right)={f}\left(\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{Hence}\:{f}\left({c}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Commented by Ar Brandon last updated on 13/Jan/22

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