Question Number 201681 by jabarsing last updated on 10/Dec/23
$${f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=\mathrm{3}{f}\left({x}\right)×{f}\left({x}+\mathrm{1}\right) \\ $$$${D}_{{f}} ={N} \\ $$$$\mathrm{2023}×{f}\left(\mathrm{1402}\right)=\mathrm{1} \\ $$$${have}\:{equation}\:{f}\left({x}\right)=\mathrm{1}\:{solution}? \\ $$
Answered by Frix last updated on 10/Dec/23
$${f}\left({x}+\mathrm{1}\right)=\frac{{f}\left({x}\right)}{\mathrm{3}{f}\left({x}\right)+\mathrm{1}} \\ $$$${f}\left(\mathrm{0}\right)={c} \\ $$$$\Rightarrow \\ $$$${f}\left(\mathrm{1}\right)=\frac{{c}}{\mathrm{3}{c}+\mathrm{1}} \\ $$$${f}\left(\mathrm{2}\right)=\frac{{c}}{\mathrm{6}{c}+\mathrm{1}} \\ $$$$… \\ $$$${f}\left({n}\right)=\frac{{c}}{\mathrm{3}{nc}+\mathrm{1}} \\ $$$${f}\left(\mathrm{1402}\right)=\frac{\mathrm{1}}{\mathrm{2023}}=\frac{{c}}{\mathrm{3}×\mathrm{1402}{c}+\mathrm{1}}\:\Rightarrow\:{c}=−\frac{\mathrm{1}}{\mathrm{2103}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}{x}−\mathrm{2183}} \\ $$$${f}\left(\mathrm{728}\right)=\mathrm{1} \\ $$