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July 18th, 2019, 02:26 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  Functional Equation
How to solve the equation ? $\displaystyle f(x)=\sqrt{f(x^2 )}$ . 
July 18th, 2019, 07:57 AM  #2  
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 6  Quote:
1) If $\displaystyle f(x)\in \mathbb R$ , then we get the equation $\displaystyle f(x)=f(x)$ and so we get an identity $\displaystyle f(x)=f(x)$ for $\displaystyle f(x)\geq 0$ and $\displaystyle f(x)=0$ for $\displaystyle f(x)<0$. 2) If $\displaystyle f(x)\in \mathbb C$ with $\displaystyle Re(f(x))\neq 0)$ and $\displaystyle Im(f(x))\neq 0$ , then we get the equation $\displaystyle f(x)=f(x)\cdot (1)^{\bigg[\frac{1}{2}arctg \bigg(\frac{Im(f(x)}{\pi\cdot Re(f(x))}\bigg)\bigg]}$ because $\displaystyle 1=e^{i\pi}$ where $\displaystyle i^2=1$ and $\displaystyle \bigg[\frac{1}{2}arctg \bigg(\frac{Im(f(x)}{\pi\cdot Re(f(x))}\bigg)\bigg]$ is the integer part of $\displaystyle \frac{1}{2}arctg \bigg(\frac{Im(f(x)}{\pi\cdot Re(f(x))}\bigg)$. All the best, Integrator P.S. Thousands of apologies!The reasoning above is for $\displaystyle f(x)=\sqrt{(f(x))^2}$ and no for $\displaystyle f(x)=\sqrt{f(x^2 )}$.Thousands of apologies! Last edited by Integrator; July 18th, 2019 at 08:36 AM.  
July 18th, 2019, 08:12 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203  
July 18th, 2019, 08:26 AM  #4 
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 6  Hello, Thousands of apologies! The reasoning above is for $\displaystyle f(x)=\sqrt{(f(x))^2}$ and no for $\displaystyle f(x)=\sqrt{f(x^2)}$. Thousands of apologies! Thank you very much for the correction! All the best, Integrator Last edited by Integrator; July 18th, 2019 at 08:37 AM. 
July 18th, 2019, 09:03 AM  #5  
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 6  Quote:
Did not You want to write $\displaystyle f(x)=\sqrt{(f(x))^2}$?If not, then it depends on the expression of the function $\displaystyle f(x)$.For exemple , if $\displaystyle f(x)=2x+3$ , then the equation $\displaystyle 2x+3=\sqrt{2x^2+3}$ has the solution $\displaystyle x=\sqrt63$. All the best, Integrator Last edited by Integrator; July 18th, 2019 at 09:19 AM.  
July 18th, 2019, 10:04 AM  #6 
Senior Member Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 
I think the problem is to find the set of functions $\displaystyle f(x)$ such that the equation is satisfied either $\displaystyle \forall x \in \mathbb{R}$ or $\displaystyle \forall x \in \mathbb{C}$. Correct?

July 18th, 2019, 12:05 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
Let's assume that the function is from $\mathbb{R}$ to $\mathbb{R}$. As $f(0)$ is 0 or 1, one can combine either of those possibilities with a definition of $f(x)$ for other values of $x$. For example, $f(x) = 0$ or $f(x) = x^r$, where $r$ is a real number. 
July 18th, 2019, 08:52 PM  #8  
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 6  Quote:
Correct, but if You want to solve the proposed equation as a functional equation, then I think that $\displaystyle f(x)=x^c$ is valid $\displaystyle \forall x,c \in \mathbb C$ with except $\displaystyle x=0$ and $\displaystyle c=0$ at the same time or $\displaystyle x\in \bigg[ \frac{1}{5},0 \bigg)\cup \bigg(0,+\frac{1}{5} \bigg]$. Correct? Thank you very much! All the best, Integrator Last edited by skipjack; July 19th, 2019 at 12:33 AM.  

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