Sorry didn’t saw that coming. I made a blunder mistake.....in limit spiritAnindya Biswas wrote: ↑Mon Feb 22, 2021 3:15 pmSorry 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)
FE Marathon!
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Hmm..Hammer...Treat everything as nail
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Re: FE Marathon!
can it be solved using elementary approach?
(I saw the general version of it which is Maxwell's Theorem which is beyond my linear algebra knowledge )
(I saw the general version of it which is Maxwell's Theorem which is beyond my linear algebra knowledge )
Hmm..Hammer...Treat everything as nail
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Re: FE Marathon!
Mehrab4226 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} $
"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
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Re: FE Marathon!
Problem 14 :
Find all functions $f:\mathbb{Z}\mapsto\mathbb{Z}$ that satisfies :
$f(x-f(y))=f(f(x))-f(y)-1$
For all $x,y\in\mathbb{Z}$.
Source :
Find all functions $f:\mathbb{Z}\mapsto\mathbb{Z}$ that satisfies :
$f(x-f(y))=f(f(x))-f(y)-1$
For all $x,y\in\mathbb{Z}$.
Source :
"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
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Re: FE Marathon!
just a silly question: if the domain and range are integers can't we just check all the function like $kx, k+x, k-x,k^x, x^k$Anindya Biswas wrote: ↑Thu Feb 25, 2021 10:18 pmProblem 14 :
Find all functions $f:\mathbb{Z}\mapsto\mathbb{Z}$ that satisfies :
$f(x-f(y))=f(f(x))-f(y)-1$
For all $x,y\in\mathbb{Z}$.
Source :
Hmm..Hammer...Treat everything as nail
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Re: FE Marathon!
Of course you can! But you have to rigorously prove your result. Cause the function can have different forms for different inputs, like piecewise function. There can be multiple solutions. Guessing and checking would help, but won't complete the proof.Asif Hossain wrote: ↑Fri Feb 26, 2021 9:43 amjust a silly question: if the domain and range are integers can't we just check all the function like $kx, k+x, k-x,k^x, x^k$Anindya Biswas wrote: ↑Thu Feb 25, 2021 10:18 pmProblem 14 :
Find all functions $f:\mathbb{Z}\mapsto\mathbb{Z}$ that satisfies :
$f(x-f(y))=f(f(x))-f(y)-1$
For all $x,y\in\mathbb{Z}$.
Source :
"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
Re: FE Marathon!
$\textbf{Solution 14:}$Anindya Biswas wrote: ↑Thu Feb 25, 2021 10:18 pmProblem 14 :
Find all functions $f:\mathbb{Z}\mapsto\mathbb{Z}$ that satisfies :
$f(x-f(y))=f(f(x))-f(y)-1$
For all $x,y\in\mathbb{Z}$.
Source :
Re: FE Marathon!
$\textbf{Problem 15}$
Find all functions $ f: \mathbb{N}\to \mathbb{N}$ satisfying
\[ \left(f(m)^2+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]for any two positive integers $ m$ and $ n$.
Find all functions $ f: \mathbb{N}\to \mathbb{N}$ satisfying
\[ \left(f(m)^2+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]for any two positive integers $ m$ and $ n$.
Re: FE Marathon!
Let $P(m,n)$ be the assertion of
$f(m)^2+f(n)|(m^2+n)^2$
$P(1,1)$:
$f(1)^2+f(1)|4$
$f(1)(f(1)+1)|4$ implies $f(1)|4$
From this $f(1)=1$
$P(1,n)$
$1+f(n)|(1+n)^2$
If,n+1 is a prime. Then
$1+f(p-1)|p^2$
Now, $f(p-1)+1=1 $ or $p $or$ p^2$
1st case : $1+f(p-1)=1$
=> $f(p-1)=0$ contradiction!!
2nd case : $1+f(p-1)=p^2$
Or,$f(p-1)=p^2-1$
$P(p-1,p-1)$
$f(p-1)^2 +f(1)|((p-1)^2 +1)^2$
Or, $(p-1)^2+1|((p-1)^2+1)^2$
This is not true.
So, $f(p-1)=p-1$ for all prime
Now $P(n,p-1)$
Let P be the large prime then
$f(n)^2+p-1|((n^2-p-1)^2=((n^2-f(n)^2)^2$
is not true unless $(n^2-f(n)^2)^2=0$
So, $f(n)=n $ for all natural number.
$f(m)^2+f(n)|(m^2+n)^2$
$P(1,1)$:
$f(1)^2+f(1)|4$
$f(1)(f(1)+1)|4$ implies $f(1)|4$
From this $f(1)=1$
$P(1,n)$
$1+f(n)|(1+n)^2$
If,n+1 is a prime. Then
$1+f(p-1)|p^2$
Now, $f(p-1)+1=1 $ or $p $or$ p^2$
1st case : $1+f(p-1)=1$
=> $f(p-1)=0$ contradiction!!
2nd case : $1+f(p-1)=p^2$
Or,$f(p-1)=p^2-1$
$P(p-1,p-1)$
$f(p-1)^2 +f(1)|((p-1)^2 +1)^2$
Or, $(p-1)^2+1|((p-1)^2+1)^2$
This is not true.
So, $f(p-1)=p-1$ for all prime
Now $P(n,p-1)$
Let P be the large prime then
$f(n)^2+p-1|((n^2-p-1)^2=((n^2-f(n)^2)^2$
is not true unless $(n^2-f(n)^2)^2=0$
So, $f(n)=n $ for all natural number.
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Re: FE Marathon!
DIdin't understand the last arguementDustan wrote: ↑Fri Feb 26, 2021 6:13 pmLet $P(m,n)$ be the assertion of
$f(m)^2+f(n)|(m^2+n)^2$
$P(1,1)$:
$f(1)^2+f(1)|4$
$f(1)(f(1)+1)|4$ implies $f(1)|4$
From this $f(1)=1$
$P(1,n)$
$1+f(n)|(1+n)^2$
If,n+1 is a prime. Then
$1+f(p-1)|p^2$
Now, $f(p-1)+1=1 $ or $p $or$ p^2$
1st case : $1+f(p-1)=1$
=> $f(p-1)=0$ contradiction!!
2nd case : $1+f(p-1)=p^2$
Or,$f(p-1)=p^2-1$
$P(p-1,p-1)$
$f(p-1)^2 +f(1)|((p-1)^2 +1)^2$
Or, $(p-1)^2+1|((p-1)^2+1)^2$
This is not true.
So, $f(p-1)=p-1$ for all prime
Now $P(n,p-1)$
Let P be the large prime then
$f(n)^2+p-1|((n^2-p-1)^2=((n^2-f(n)^2)^2$
is not true unless $(n^2-f(n)^2)^2=0$
So, $f(n)=n $ for all natural number.
You might have used another arguement like $f(n)+1\leq f(n+1)$ $\rightarrow$ $n\leq f(n)$ so $f(n)< f(n+1)$ and the rest is the prime arguement and then use the infinitness of prime to establish $f(n)=n$ for all $n$ in $\mathbb{N}$
Last edited by Asif Hossain on Sat Feb 27, 2021 11:41 am, edited 1 time in total.
Hmm..Hammer...Treat everything as nail