Question Number 173613 by mr W last updated on 14/Jul/22
$${if}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}, \\ $$$${solve}\:\underset{{n}\:{times}} {{f}\left({f}\left({f}\left({f}\left(...{f}\left({x}\right)\right)\right)\right)\right)}=\mathrm{0} \\ $$
Answered by mahdipoor last updated on 14/Jul/22
$${f}\left({x}\right)=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\Rightarrow \\ $$$${f}\left({f}\left({x}\right)\right)=\left(\left(\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\right)+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}=\left({x}+\mathrm{2}\right)^{\mathrm{4}} −\mathrm{2} \\ $$$${f}\left({f}\left({f}\left({x}\right)\right)\right)=\left(\left(\left({x}+\mathrm{2}\right)^{\mathrm{4}} −\mathrm{2}\right)+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}=\left({x}+\mathrm{2}\right)^{\mathrm{8}} −\mathrm{2} \\ $$$$... \\ $$$$\left.{f}\left({f}...{f}\left({x}\right)\right)\right)=\left({x}+\mathrm{2}\right)^{\mathrm{2}^{{n}} } −\mathrm{2}\:\:\:\left({n}\:{times}\right) \\ $$$$, \\ $$$$\left.{f}\left({f}...{f}\left({x}\right)\right)\right)=\left({x}+\mathrm{2}\right)^{\mathrm{2}^{{n}} } −\mathrm{2}=\mathrm{0}\:\Rightarrow \\ $$$$\pm\mathrm{2}^{\mathrm{2}^{−{n}} } −\mathrm{2}=\pm^{\:\:\mathrm{2}^{{n}} } \sqrt{\mathrm{2}}−\mathrm{2}={x} \\ $$
Commented by mr W last updated on 14/Jul/22
$${perfect}\:{sir}!\:{thanks}! \\ $$
Commented by Tawa11 last updated on 14/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$