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Question Number 164299 by mr W last updated on 15/Jan/22

if f(f(f(...f(x)...)))_(n times) =2x+1, find f(1)=?

$${if}\:\underset{{n}\:{times}} {{f}\left({f}\left({f}\left(...{f}\left({x}\right)...\right)\right)\right)}=\mathrm{2}{x}+\mathrm{1},\: \\ $$$${find}\:{f}\left(\mathrm{1}\right)=? \\ $$

Answered by ajfour last updated on 16/Jan/22

y_1 =ax+b y_2 =a(ax+b)+b y_3 =a^2 (ax+b)+ab+b y_n =a^n x+(a^(n−1) +a^(n−2) +...+a+1)b =2x+1 ⇒ a^n =2 (((a^n −1)/(a−1)))b=1 ⇒ b=a−1=2^(1/n) −1 y_1 (1)=f(1)=a+b=2a−1 ⇒ f(1)=2^((n+1)/n) −1 e.g. say n=5 y_5 =a^5 x+b(((a^5 −1)/(a−1)))=2x+1 ⇒ a^5 =2 ; b=2^(1/5) −1 y_1 =(2^(1/5) )x+2^(1/5) −1 y_1 (1)=2^(6/5) −1 (see blue formula).

$${y}_{\mathrm{1}} ={ax}+{b} \\ $$$${y}_{\mathrm{2}} ={a}\left({ax}+{b}\right)+{b} \\ $$$${y}_{\mathrm{3}} ={a}^{\mathrm{2}} \left({ax}+{b}\right)+{ab}+{b} \\ $$$${y}_{{n}} ={a}^{{n}} {x}+\left({a}^{{n}−\mathrm{1}} +{a}^{{n}−\mathrm{2}} +...+{a}+\mathrm{1}\right){b} \\ $$$$\:\:\:\:\:=\mathrm{2}{x}+\mathrm{1} \\ $$$$\Rightarrow\:\:\:{a}^{{n}} =\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\frac{{a}^{{n}} −\mathrm{1}}{{a}−\mathrm{1}}\right){b}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:{b}={a}−\mathrm{1}=\mathrm{2}^{\mathrm{1}/{n}} −\mathrm{1} \\ $$$${y}_{\mathrm{1}} \left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)={a}+{b}=\mathrm{2}{a}−\mathrm{1} \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:{f}\left(\mathrm{1}\right)=\mathrm{2}^{\frac{{n}+\mathrm{1}}{{n}}} −\mathrm{1} \\ $$$${e}.{g}.\:\:\:{say}\:\:\:\:\:{n}=\mathrm{5} \\ $$$$\:\:\:{y}_{\mathrm{5}} ={a}^{\mathrm{5}} {x}+{b}\left(\frac{{a}^{\mathrm{5}} −\mathrm{1}}{{a}−\mathrm{1}}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:{a}^{\mathrm{5}} =\mathrm{2}\:\:\:\:\:;\:\:\:{b}=\mathrm{2}^{\mathrm{1}/\mathrm{5}} −\mathrm{1} \\ $$$$\:\:\:\:{y}_{\mathrm{1}} =\left(\mathrm{2}^{\mathrm{1}/\mathrm{5}} \right){x}+\mathrm{2}^{\mathrm{1}/\mathrm{5}} −\mathrm{1} \\ $$$$\:\:\:{y}_{\mathrm{1}} \left(\mathrm{1}\right)=\mathrm{2}^{\mathrm{6}/\mathrm{5}} −\mathrm{1}\:\:\:\left({see}\:{blue}\:{formula}\right). \\ $$

Commented by mr W last updated on 16/Jan/22

perfect sir!

$${perfect}\:{sir}! \\ $$

Commented by Tawa11 last updated on 16/Jan/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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