FE Marathon!
Problem 9: $f:R\rightarrow R $ such that
$f(xf(x-y))+yf(x)=x+y+f(x^2)$ for all real x,y.
$f(xf(x-y))+yf(x)=x+y+f(x^2)$ for all real x,y.
Re: FE Marathon!
$\textbf{Problem 10 :}$
Determine all the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that:
\[f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))\ ,\ \forall x,y\in \mathbb{R}\]
Determine all the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that:
\[f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))\ ,\ \forall x,y\in \mathbb{R}\]
Re: FE Marathon!
Problem 11: $f:R\rightarrow R$ and
$f(t^2+u)=tf(t)+f(u)$ for all real valued t,u.
$f(t^2+u)=tf(t)+f(u)$ for all real valued t,u.
- Mehrab4226
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Re: FE Marathon!
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
- Mehrab4226
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Re: FE Marathon!
Problem 12:
Find all functions $f$: $\mathbb{R} \to \mathbb{R}$ defined by,
$$f(\sqrt{x^2+y^2})=f(x)f(y)$$ $ \forall x,y \in \mathbb{R} $
Find all functions $f$: $\mathbb{R} \to \mathbb{R}$ defined by,
$$f(\sqrt{x^2+y^2})=f(x)f(y)$$ $ \forall x,y \in \mathbb{R} $
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
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Re: FE Marathon!
$Claim$: $ \forall x \in \mathbb{R} $ $f(x) $ = 0,1Mehrab4226 wrote: ↑Tue Jan 26, 2021 9:58 pmProblem 12:
Find all functions $f$: $\mathbb{R} \to \mathbb{R}$ defined by,
$$f(\sqrt{x^2+y^2})=f(x)f(y)$$ $ \forall x,y \in \mathbb{R} $
Plugging $x$ := 0, $y$ :=0 implies $f(0) $ = 0 or $f(0) $ = 1
$f(0) $ = 0 implies $ \forall x \in \mathbb{R} $ $f(x) $ = 0 which is obviously true.
Now let's assume f(x) is not equal to 0
Plugging $x=y$ implies $ \forall x \in \mathbb{R} $ $f(x) $ $>0 $
It can be proved by induction very easily that $f^n (x) = f(\sqrt{n}x) $
By repeating the process we can derive $f^n (x) =f(\sqrt{n} x) =(f((\sqrt{n})^2 x)^{1/n}= ...=(f((\sqrt{n})^n x)^{1/n^{n-1}}=... $
Since $f(x) >0 $ we can say as $n \to \infty $, $1/n^{n-1} \to 0$ so $f(x)=1 $
Which proves the claim
(sorry for my first time my latexcode tanmoy bhaiya ektu confirm den problem to hoiche kina)
Last edited by Asif Hossain on Sat Feb 20, 2021 11:34 pm, edited 5 times in total.
Hmm..Hammer...Treat everything as nail
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Re: FE Marathon!
New proble find all $f$: $\mathbb{R} \to \mathbb{R} $ ST
$f(f(x+1))=f(x)+1 $
Self Discovered problem
$f(f(x+1))=f(x)+1 $
Self Discovered problem
Hmm..Hammer...Treat everything as nail
- Anindya Biswas
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Re: FE Marathon!
Sorry to disappoint, but $f(x)=e^{kx^2}$ is also a valid solution.Asif Hossain wrote: ↑Sat Feb 20, 2021 10:12 pm$Claim$: $ \forall x \in \mathbb{R} $ $f(x) $ = 0,1Mehrab4226 wrote: ↑Tue Jan 26, 2021 9:58 pmProblem 12:
Find all functions $f$: $\mathbb{R} \to \mathbb{R}$ defined by,
$$f(\sqrt{x^2+y^2})=f(x)f(y)$$ $ \forall x,y \in \mathbb{R} $
Plugging $x$ := 0, $y$ :=0 implies $f(0) $ = 0 or $f(0) $ = 1
$f(0) $ = 0 implies $ \forall x \in \mathbb{R} $ $f(x) $ = 0 which is obviously true.
Now let's assume f(x) is not equal to 0
Plugging $x=y$ implies $ \forall x \in \mathbb{R} $ $f(x) $ $>0 $
It can be proved by induction very easily that $f^n (x) = f(\sqrt{n}x) $
By repeating the process we can derive $f^n (x) =f(\sqrt{n} x) =(f((\sqrt{n})^2 x)^{1/n}= ...=(f((\sqrt{n})^n x)^{1/n^{n-1}}=... $
Since $f(x) >0 $ we can say as $n \to \infty $, $1/n^{n-1} \to 0$ so $f(x)=1 $
Which proves the claim
(sorry for my first time my latexcode tanmoy bhaiya ektu confirm den problem to hoiche kina)
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann