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Question Number 111149 by Aina Samuel Temidayo last updated on 02/Sep/20

Let f_0 (x) = (1/(1−x)) and f_n (x) =f_0 (f_(n−1) (x)), n=1,2,3,... Evaluate f_(2018) (2018)

$$\mathrm{Let}\:\mathrm{f}_{\mathrm{0}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\:\mathrm{and}\:\mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right) \\ $$$$=\mathrm{f}_{\mathrm{0}} \left(\mathrm{f}_{\mathrm{n}−\mathrm{1}} \left(\mathrm{x}\right)\right),\:\mathrm{n}=\mathrm{1},\mathrm{2},\mathrm{3},...\:\mathrm{Evaluate} \\ $$$$\mathrm{f}_{\mathrm{2018}} \left(\mathrm{2018}\right) \\ $$

Commented by kaivan.ahmadi last updated on 02/Sep/20

f_1 (x)=f_0 (f_0 (x))=f_0 ((1/(1−x)))=(1/(1−(1/(1−x))))=(1/((1−x−1)/(1−x)))=((1−x)/(−x))=1−(1/x) f_2 (x)=(1/(1−1+(1/x)))=x f_3 (x)=(1/(1−x))=f_0 (x) ⇒f_n (x)=f_i ;i=0 or 1 or 2 and n≡^3 i 2018≡^3 2⇒f_(2018) (x)=f_2 (x)=x ⇒f_(2018) (2018)=2018

$${f}_{\mathrm{1}} \left({x}\right)={f}_{\mathrm{0}} \left({f}_{\mathrm{0}} \left({x}\right)\right)={f}_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}=\frac{\mathrm{1}}{\frac{\mathrm{1}−{x}−\mathrm{1}}{\mathrm{1}−{x}}}=\frac{\mathrm{1}−{x}}{−{x}}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$${f}_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{1}+\frac{\mathrm{1}}{{x}}}={x} \\ $$$${f}_{\mathrm{3}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}={f}_{\mathrm{0}} \left({x}\right) \\ $$$$\Rightarrow{f}_{{n}} \left({x}\right)={f}_{{i}} ;{i}=\mathrm{0}\:{or}\:\mathrm{1}\:{or}\:\mathrm{2}\:{and}\:{n}\overset{\mathrm{3}} {\equiv}{i} \\ $$$$\mathrm{2018}\overset{\mathrm{3}} {\equiv}\mathrm{2}\Rightarrow{f}_{\mathrm{2018}} \left({x}\right)={f}_{\mathrm{2}} \left({x}\right)={x} \\ $$$$\Rightarrow{f}_{\mathrm{2018}} \left(\mathrm{2018}\right)=\mathrm{2018} \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Please I don′t understand these lines. I will be glad if you can help shed more light on it. ⇒f_n (x)=f_i , i=0 or 1 or 2 and n≡^3 i 2018≡^3 2⇒f_(2018) (x)=f_2 (x)=x

$$\mathrm{Please}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{these}\:\mathrm{lines}.\: \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{be}\:\mathrm{glad}\:\mathrm{if}\:\mathrm{you}\:\mathrm{can}\:\mathrm{help}\:\mathrm{shed}\:\mathrm{more} \\ $$$$\mathrm{light}\:\mathrm{on}\:\mathrm{it}. \\ $$$$\Rightarrow\mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{f}_{\mathrm{i}} ,\:\mathrm{i}=\mathrm{0}\:\mathrm{or}\:\mathrm{1}\:\mathrm{or}\:\mathrm{2}\:\mathrm{and}\:\mathrm{n}\overset{\mathrm{3}} {\equiv}\mathrm{i} \\ $$$$\mathrm{2018}\overset{\mathrm{3}} {\equiv}\mathrm{2}\Rightarrow\mathrm{f}_{\mathrm{2018}} \left(\mathrm{x}\right)=\mathrm{f}_{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{x} \\ $$

Commented by kaivan.ahmadi last updated on 03/Sep/20

f_0 (x)=(1/(1−x)) f_1 (x)=1−(1/x) f_2 (x)=x ⇒f_3 (x)=f_0 (x) f_5 (x)=f_1 (x) f_6 (x)=f_2 (x) so f_n (x) is a one of the f_0 ,f_1 ,f_2 f_n (x)= { ((f_0 (x) ; n=3k)),((f_1 (x) ; n=3k+1 (k∈Z))),((f_2 (x) ; n=3k+2)) :} now since 2018=3(672)+2 we have f_(2018 ) (x)=f_2 (x)=x

$${f}_{\mathrm{0}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${f}_{\mathrm{1}} \left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$${f}_{\mathrm{2}} \left({x}\right)={x} \\ $$$$\Rightarrow{f}_{\mathrm{3}} \left({x}\right)={f}_{\mathrm{0}} \left({x}\right)\:\:\:\:\:{f}_{\mathrm{5}} \left({x}\right)={f}_{\mathrm{1}} \left({x}\right)\:\:\:\:\:{f}_{\mathrm{6}} \left({x}\right)={f}_{\mathrm{2}} \left({x}\right) \\ $$$${so}\:{f}_{{n}} \left({x}\right)\:{is}\:{a}\:{one}\:{of}\:{the}\:{f}_{\mathrm{0}} ,{f}_{\mathrm{1}} ,{f}_{\mathrm{2}} \\ $$$${f}_{{n}} \left({x}\right)=\begin{cases}{{f}_{\mathrm{0}} \left({x}\right)\:\:\:\:\:\:;\:{n}=\mathrm{3}{k}}\\{{f}_{\mathrm{1}} \left({x}\right)\:\:\:\:\:\:\:;\:{n}=\mathrm{3}{k}+\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\left({k}\in\mathbb{Z}\right)}\\{{f}_{\mathrm{2}} \left({x}\right)\:\:\:\:\:\:\:;\:{n}=\mathrm{3}{k}+\mathrm{2}}\end{cases} \\ $$$${now}\:{since}\:\mathrm{2018}=\mathrm{3}\left(\mathrm{672}\right)+\mathrm{2}\:\:{we}\:{have} \\ $$$${f}_{\mathrm{2018}\:} \left({x}\right)={f}_{\mathrm{2}} \left({x}\right)={x} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

I understand now. Thanks.

$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now}.\:\mathrm{Thanks}. \\ $$

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