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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
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 Originally Posted by idontknow How to solve the equation ? $\displaystyle f(x)=\sqrt{f(x^2 )}$ .
Hello,

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
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 Originally Posted by Integrator 1) If $\displaystyle f(x)\in \mathbb R$, then we get the equation $\displaystyle f(x)=|f(x)|$
Why? That doesn't seem to be implied. July 18th, 2019, 08:26 AM   #4
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 Originally Posted by skipjack Why? That doesn't seem to be implied.
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
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Quote:
 Originally Posted by idontknow How to solve the equation ? $\displaystyle f(x)=\sqrt{f(x^2 )}$ .
Hello,

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 $\displaystyle x=\sqrt6-3$.
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. Thanks from topsquark and idontknow July 18th, 2019, 08:52 PM   #8
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 Originally Posted by skipjack 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.
Hello,

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. Tags equation, functional Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Elementary Math 4 July 30th, 2018 09:25 AM Dacu Real Analysis 4 June 7th, 2015 05:41 AM nukem4111 Real Analysis 1 September 16th, 2013 03:44 AM georgij Algebra 3 October 9th, 2011 02:55 AM llambi Math Events 4 April 22nd, 2011 05:57 AM

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