Functional Equation

Dec 2015
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166
Earth
How to solve the equation ?
\(\displaystyle f(x)=\sqrt{f(x^2 )}\) .
 
Aug 2018
137
7
România
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:
Aug 2018
137
7
România
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
 
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Aug 2018
137
7
România
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:
Jun 2019
493
262
USA
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?
 

skipjack

Forum Staff
Dec 2006
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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.
 
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Aug 2018
137
7
România
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
 
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