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Question Number 207314 by universe last updated on 11/May/24

$$\mathrm{let}\mathrm{f}:\mathbb{R}\to \mathbb{R}\mathrm{be}\mathrm{a}\mathrm{continuous}\mathrm{function}\mathrm{then}$$$$\mathrm{show}\mathrm{that}$$$$\left(1\right)\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left({\mathrm{x}}^2\right)\mathrm{\forall }\mathrm{x}\in \mathbb{R}\mathrm{then}\mathrm{f}\mathrm{is}\mathrm{a}\mathrm{constant}$$$$\mathrm{function}$$$$\left(2\right)\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(2\mathrm{x}+1)\mathrm{\forall }\mathrm{x}\in \mathbb{R}\mathrm{then}\mathrm{f}\mathrm{is}\mathrm{a}$$$$\mathrm{constant}\mathrm{function}$$

Answered by Berbere last updated on 11/May/24

$$f\left(x\right)=f\left(x^2\right)\Rightarrow fiscontinus$$$$f(-x)=f\left(x\right)\Rightarrow weshownjstf\mid [0,\mathrm{\infty }[isconstante$$$$ifx\in [0,1]$$$$letp\in [0,1[suchatthatpx\in [0,p[$$$$\Rightarrow f\left(px\right)=f\left(p^2{x}^2\right)=\left(p^{2^n}{x}^{2^n}\right);\mathrm{\forall }n\in \mathbb{N}$$$$f\left(px\right)=\underset{n\to \mathrm{\infty }}{\lim }f\left(p^{2^n}{x}^{2^n}\right)=f\left(0\right)$$$$f(1-\frac{1}{n})=\underset{n\to \mathrm{\infty }}{\lim }f(1-\frac{1}{n})=0$$$$=\underset{n\to \mathrm{\infty }}{\lim }f\left(0\right)=f\left(0\right)$$$$ifx>1$$$$f\left(\sqrt{x}\right)=f\left(x\right)\Rightarrow \mathrm{\forall }n\in {\mathbb{N}}^{\ast }$$$$f\left(x^{\frac{1}{2^n}}\right)=f\left(x\right)\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }f\left(x^{\frac{1}{2^n}}\right)=f\left(\underset{n\to \mathrm{\infty }}{\lim }{e}^{\frac{1}{2^n}ln\left(x\right)}\right)=f\left(1\right)=f\left(0\right)$$$$\Rightarrow \mathrm{\forall }x\in \mathbb{R}f\left(x\right)=f\left(0\right)$$$$2)$$$$f\left(x\right)=f(2x+1)$$$$x\stackrel{h}{\to }2x+1$$$$hoh....hntimes$$$$=h^n\left(x\right)=a_{n}x+b_n$$$$h^0\left(x\right)=x=f\left(x\right)=2x+1$$$$h^{n+1}=2a_{n}x+2b_n+1=a_{n+1}x+b_{n+1}$$$$b_{n+1}=2b_n+1;a_{n+1}=2a_n\Rightarrow a_n=2^n$$$$b_n=2^n-1$$$$h^n\left(x\right)=2^{n}x+2^n-1$$$$\Rightarrow y=2^{n}x+2^n-1;$$$$x_n=\frac{y}{2^n}-1+\frac{1}{2^n}\Rightarrow f\left(x\right)=f\left(y\right);\mathrm{\forall }nyisfixed$$$$f\left(y\right)=\underset{n\to \mathrm{\infty }}{\lim }f(\frac{y+1}{2^n}-1)=f(-1)$$$$fisconstante$$$$$$$$$$$$$$$$$$

Commented by universe last updated on 11/May/24

$$thankusir$$

Commented by Berbere last updated on 11/May/24

$$WithePleasur$$$$$$

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