Question Number 161817 by HongKing last updated on 22/Dec/21
Answered by aleks041103 last updated on 23/Dec/21
$${f}\left({x}\right)=\left({e}^{−\mathrm{1}} \circ{g}\circ{e}\right)\left({x}\right) \\ $$$${e}\left({x}\right)={x}^{\mathrm{2021}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${e}^{−\mathrm{1}} \left({x}\right)={x}^{\mathrm{1}/\mathrm{2021}} \\ $$$$ \\ $$$$\Rightarrow\left({f}^{{n}} \right)\left({x}\right)=\left({e}^{−\mathrm{1}} \circ{g}^{{n}} \circ{e}\right)\left({x}\right) \\ $$$$\Omega_{{n}} ={e}^{−\mathrm{1}} \left(\left({g}^{{n}} \right)\left({e}\left({x}\right)\right)\right) \\ $$$${g}^{−\mathrm{1}} \left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$${g}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}=\frac{{x}−\mathrm{1}}{{x}}=\mathrm{1}−\frac{\mathrm{1}}{{x}}={g}^{−\mathrm{1}} \left({x}\right) \\ $$$$\Rightarrow{g}^{\mathrm{3}} \left({x}\right)={x} \\ $$$$\Rightarrow{g}^{\mathrm{2022}} \left({x}\right)={g}^{\mathrm{3}.\mathrm{674}} \left({x}\right)=\left({g}^{\mathrm{3}} \right)^{\mathrm{674}} \left({x}\right)={id}^{\mathrm{674}} \left({x}\right)={x} \\ $$$$\Rightarrow\Omega_{\mathrm{2022}} =\left({e}^{−\mathrm{1}} \circ{id}\circ{e}\right)\left({x}\right)=\left({e}^{−\mathrm{1}} \circ{e}\right)\left({x}\right)= \\ $$$$={id}\left({x}\right)={x}=\Omega_{\mathrm{2022}} \\ $$$$\Rightarrow{Ans}.\:{x}=\mathrm{2021} \\ $$
Commented by HongKing last updated on 25/Dec/21
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$