Question Number 179336 by mathlove last updated on 28/Oct/22
$${f}\left({x}\right)=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$$\left(\underset{\mathrm{2004}} {\underbrace{{fofofo}……..{of}}}\right)_{{x}} =? \\ $$
Commented by a.lgnaoui last updated on 28/Oct/22
$${fof}\left({x}\right)=\frac{\mathrm{1}−\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}=\frac{\mathrm{2}{x}}{\mathrm{2}}={x}\:\:\left({ordre}\:\mathrm{2}\right) \\ $$$${fof}\left({f}\left({x}\right)\right)={f}\left({x}\right)=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:\:\:\:\:\:\:\:\:\left({ordre}\:\mathrm{3}\right) \\ $$$$……… \\ $$$${fofof}…..{f}\left({x}\right)\:\:={x}\:\:\:\:\:\:{for}\:{n}\:{pair}\:{n}=\mathrm{2}{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\left({for}\:{n}\:{impair}\:{n}=\mathrm{2}{k}+\mathrm{1}\:{n}\in\mathbb{N}\right) \\ $$$${donc}\:\: \\ $$$${fofofof}…..{f}\left({x}\right)_{\mathrm{2004}\:} \:={x} \\ $$
Commented by mathlove last updated on 28/Oct/22
$${thanks} \\ $$