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f-x-1-f-x-3f-x-f-x-1-D-f-N-2023-f-1402-1-have-equation-f-x-1-solution-




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

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