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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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