Question Number 207314 by universe last updated on 11/May/24
$$\:\mathrm{let}\:\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{then} \\ $$$$\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)\:\forall\:\mathrm{x}\:\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\:\:\mathrm{function} \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{2x}+\mathrm{1}\right)\:\forall\mathrm{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\:\:\mathrm{constant}\:\mathrm{function}\: \\ $$
Answered by Berbere last updated on 11/May/24
![f(x)=f(x^2 )⇒f is continus f(−x)=f(x)⇒ we show njst f∣[0,∞[ is constante if x∈[0,1] let p∈[0,1[ suchat that px∈[0,p[ ⇒f(px)=f(p^2 x^2 )=(p^2^n x^2^n );∀n∈N f(px)=lim_(n→∞) f(p^2^n x^2^n )=f(0) f(1−(1/n))=lim_(n→∞) f(1−(1/n))=0 =lim_(n→∞) f(0)=f(0) if x>1 f((√x))=f(x)⇒∀n∈N^∗ f(x^(1/2^n ) )=f(x)⇒lim_(n→∞) f(x^(1/2^n ) )=f(lim_(n→∞) e^((1/2^n )ln(x)) )=f(1)=f(0) ⇒∀x∈R f(x)=f(0) 2) f(x)=f(2x+1) x→^h 2x+1 hoh....h n times =h^n (x)=a_n x+b_n h^0 (x)=x=f(x)=2x+1 h^(n+1) =2a_n x+2b_n +1=a_(n+1) x+b_(n+1) b_(n+1) =2b_n +1;a_(n+1) =2a_n ⇒a_n =2^n b_n =2^n −1 h^n (x)=2^n x+2^n −1 ⇒y=2^n x+2^n −1; x_n =(y/2^n )−1+(1/2^n )⇒f(x)=f(y) ;∀n y is fixed f(y)=lim_(n→∞) f(((y+1)/(2^n ))−1)=f(−1) f is constante](https://www.tinkutara.com/question/Q207316.png)
$${f}\left({x}\right)={f}\left({x}^{\mathrm{2}} \right)\Rightarrow{f}\:{is}\:{continus} \\ $$$${f}\left(−{x}\right)={f}\left({x}\right)\Rightarrow\:\:{we}\:{show}\:{njst}\:{f}\mid\left[\mathrm{0},\infty\left[\:{is}\:{constante}\right.\right. \\ $$$${if}\:{x}\in\left[\mathrm{0},\mathrm{1}\right]\: \\ $$$${let}\:{p}\in\left[\mathrm{0},\mathrm{1}\left[\:{suchat}\:{that}\:{px}\in\left[\mathrm{0},{p}\left[\right.\right.\right.\right. \\ $$$$\Rightarrow{f}\left({px}\right)={f}\left({p}^{\mathrm{2}} {x}^{\mathrm{2}} \right)=\left({p}^{\mathrm{2}^{{n}} } {x}^{\mathrm{2}^{{n}} } \right);\forall{n}\in\mathbb{N} \\ $$$${f}\left({px}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({p}^{\mathrm{2}^{{n}} } {x}^{\mathrm{2}^{{n}} } \right)={f}\left(\mathrm{0}\right) \\ $$$${f}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)=\mathrm{0} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left(\mathrm{0}\right)={f}\left(\mathrm{0}\right) \\ $$$${if}\:{x}>\mathrm{1} \\ $$$${f}\left(\sqrt{{x}}\right)={f}\left({x}\right)\Rightarrow\forall{n}\in\mathbb{N}^{\ast} \\ $$$${f}\left({x}^{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }} \right)={f}\left({x}\right)\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}^{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }} \right)={f}\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}{e}^{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{ln}\left({x}\right)} \right)={f}\left(\mathrm{1}\right)={f}\left(\mathrm{0}\right) \\ $$$$\Rightarrow\forall{x}\in\mathbb{R}\:{f}\left({x}\right)={f}\left(\mathrm{0}\right)\: \\ $$$$\left.\mathrm{2}\right) \\ $$$${f}\left({x}\right)={f}\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${x}\overset{{h}} {\rightarrow}\mathrm{2}{x}+\mathrm{1} \\ $$$${hoh}….{h}\:{n}\:{times} \\ $$$$={h}^{{n}} \left({x}\right)={a}_{{n}} {x}+{b}_{{n}} \\ $$$${h}^{\mathrm{0}} \left({x}\right)={x}={f}\left({x}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${h}^{{n}+\mathrm{1}} =\mathrm{2}{a}_{{n}} {x}+\mathrm{2}{b}_{{n}} +\mathrm{1}={a}_{{n}+\mathrm{1}} {x}+{b}_{{n}+\mathrm{1}} \\ $$$${b}_{{n}+\mathrm{1}} =\mathrm{2}{b}_{{n}} +\mathrm{1};{a}_{{n}+\mathrm{1}} =\mathrm{2}{a}_{{n}} \Rightarrow{a}_{{n}} =\mathrm{2}^{{n}} \\ $$$${b}_{{n}} =\mathrm{2}^{{n}} −\mathrm{1} \\ $$$${h}^{{n}} \left({x}\right)=\mathrm{2}^{{n}} {x}+\mathrm{2}^{{n}} −\mathrm{1} \\ $$$$\Rightarrow{y}=\mathrm{2}^{{n}} {x}+\mathrm{2}^{{n}} −\mathrm{1};\: \\ $$$${x}_{{n}} =\frac{{y}}{\mathrm{2}^{{n}} }−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\Rightarrow{f}\left({x}\right)={f}\left({y}\right)\:;\forall{n}\:{y}\:{is}\:{fixed} \\ $$$${f}\left({y}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left(\frac{{y}+\mathrm{1}}{\mathrm{2}^{{n}} \:}−\mathrm{1}\right)={f}\left(−\mathrm{1}\right)\: \\ $$$${f}\:{is}\:{constante}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by universe last updated on 11/May/24
$${thank}\:{u}\:{sir} \\ $$
Commented by Berbere last updated on 11/May/24
$${Withe}\:{Pleasur} \\ $$$$ \\ $$