My Math Forum Functional Equation

 Real Analysis Real Analysis Math Forum

 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:
 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
Global Moderator

Joined: Dec 2006

Posts: 20,921
Thanks: 2203

Quote:
 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
Member

Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Quote:
 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
Member

Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

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
Member

Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Quote:
 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top