FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
tanmoy
Posts: 305
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Location: Rangpur,Bangladesh

Re: FE Marathon!

Unread post by tanmoy » Wed Dec 23, 2020 9:45 pm

Mehrab4226 wrote:
Wed Dec 23, 2020 7:15 pm
Putting y= 0 we get,
$f(x^4)=x^3f(x) $
How? putting $y = 0$ gives, $\ f(x^4)=x^3f(x) + f(f(0))$ and you didn't prove that $f(f(0)) = 0$.
Mehrab4226 wrote:
Wed Dec 23, 2020 7:15 pm
Or, $f(x^4)=f(t)=x^3f(x)\cdots (2)$[Let $x^4=t$]

Now from the question we get,

$f(x^4+y)=x^3f(x)+f(f(y))$

Or, $f(t+y)=f(t)+f(y)$[Using (1) and (2)]

Which is the cauchy function,

$\therefore f(x)=cx$

Using it in the question's equation and solving it we get c=1

$\therefore f(x)=x$ which is indeed the answer.
First of all here $t$ is a positive number and as a consequence, doesn't cover all real values. I mean $f(t) + f(y) \neq f(t + y )$ for all real values of $t$ and $y$.

Finally and most importantly, a Cauchy Equation has solution $f(x) = cx$ if the domain is Rational Number and if the function's domain is Real Number and it fulfils at least one of the following conditions:

(1) The function is monotonous on some interval
(2) The function is continuous
(3) The function is bounded on some interval
(4) $f(x) \geq 0$ for $x \geq 0$.
"Questions we can't answer are far better than answers we can't question"

Dustan
Posts: 64
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Thu Dec 24, 2020 6:15 pm

Mehrab4226 wrote:
Wed Dec 23, 2020 8:24 pm
If my solution is correct then the next problem in the marathon, if my solution is not correct :cry: then ignore this message and plz tell me where I am wrong.


Problem 6:
Find all functions $f : \Bbb R \to \Bbb R$ that satisfies,

\[f((x-y)^2)=f(x)^2-2xf(y)+y^2\]
Solution 6: Let $y=x=0$
$f(0)=0 $ or $f(0)=1$
First case, $f(0)=0)$
Let $y=x$
$f(0)$=0=$f(x)^2-2xf(x)+x^2$
Solving this equation we get $f(x)=x$
Which is indeed a solution.
Second case $f(0)=1$
$0=f(x)^2-2xf(x)+x^2-1$
Solving this we get $f(x)=x+1 $ and $f(x)=x-1$
But they don't satisfy the equation. Hence $f(x)=x$

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Mehrab4226
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Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Thu Dec 24, 2020 11:42 pm

Mehrab4226 wrote:
Wed Dec 23, 2020 7:15 pm
This one kinda looks like an :| okay solution.[This looks big because of the spacings]
$f(x^4+y)=x^3f(x)+f(f(y))$

Putting x=0 we get,

$f(f(y))=f(y)\cdots (1)$

Putting y= 0 we get,

$f(x^4)=x^3f(x) $

Or, $f(x^4)=f(t)=x^3f(x)\cdots (2)$[Let $x^4=t$]

Now from the question we get,

$f(x^4+y)=x^3f(x)+f(f(y))$

Or, $f(t+y)=f(t)+f(y)$[Using (1) and (2)]

Which is the cauchy function,

$\therefore f(x)=cx$

Using it in the question's equation and solving it we get c=1

$\therefore f(x)=x $ which is indeed the answer.
Bug 1:$f(0)=?$
If we put $x=1$ and $y=0$ in the question we get,
$f(1)=f(1)+f(f(0))$
Or,$f(f(0))=0$
Or,$f(0)=0$[Using 1]
Bug 2: (t can be negative as well)
If we take $y= -x^4$ in the question we get,
$f(x^4 -x^4)=x^3f(x)+f(f(-x^4))$
Or,$f(0)=f(x^4)+f(-x^4)$
Or,$f(-x^4)=-f(x^4)$
Or,$f(-t)=-f(t)$
Thus we now have the negative values for t.
Bug 3: Cauchy function application criteria
But how do I do that?
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

Dustan
Posts: 64
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Fri Dec 25, 2020 11:54 am

Mehrab4226 wrote:
Thu Dec 24, 2020 11:42 pm
Mehrab4226 wrote:
Wed Dec 23, 2020 7:15 pm
This one kinda looks like an :| okay solution.[This looks big because of the spacings]
$f(x^4+y)=x^3f(x)+f(f(y))$

Putting x=0 we get,

$f(f(y))=f(y)\cdots (1)$

Putting y= 0 we get,

$f(x^4)=x^3f(x) $

Or, $f(x^4)=f(t)=x^3f(x)\cdots (2)$[Let $x^4=t$]

Now from the question we get,

$f(x^4+y)=x^3f(x)+f(f(y))$

Or, $f(t+y)=f(t)+f(y)$[Using (1) and (2)]

Which is the cauchy function,

$\therefore f(x)=cx$

Using it in the question's equation and solving it we get c=1

$\therefore f(x)=x $ which is indeed the answer.
Bug 1:$f(0)=?$
If we put $x=1$ and $y=0$ in the question we get,
$f(1)=f(1)+f(f(0))$
Or,$f(f(0))=0$
Or,$f(0)=0$[Using 1]
Bug 2: (t can be negative as well)
If we take $y= -x^4$ in the question we get,
$f(x^4 -x^4)=x^3f(x)+f(f(-x^4))$
Or,$f(0)=f(x^4)+f(-x^4)$
Or,$f(-x^4)=-f(x^4)$
Or,$f(-t)=-f(t)$
Thus we now have the negative values for t.
Bug 3: Cauchy function application criteria
But how do I do that?

I think if we will show that $f(x+y)=f(x)+f(y)$ for all x greater than 0
And $f(x+y)=f(x)+f(y)$ for x less then 0 then it will be ok maybe.

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Mehrab4226
Posts: 208
Joined: Sat Jan 11, 2020 1:38 pm
Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Sat Dec 26, 2020 8:01 pm

Dustan wrote:
Fri Dec 25, 2020 11:54 am
Mehrab4226 wrote:
Thu Dec 24, 2020 11:42 pm
Mehrab4226 wrote:
Wed Dec 23, 2020 7:15 pm
This one kinda looks like an :| okay solution.[This looks big because of the spacings]
$f(x^4+y)=x^3f(x)+f(f(y))$

Putting x=0 we get,

$f(f(y))=f(y)\cdots (1)$

Putting y= 0 we get,

$f(x^4)=x^3f(x) $

Or, $f(x^4)=f(t)=x^3f(x)\cdots (2)$[Let $x^4=t$]

Now from the question we get,

$f(x^4+y)=x^3f(x)+f(f(y))$

Or, $f(t+y)=f(t)+f(y)$[Using (1) and (2)]

Which is the cauchy function,

$\therefore f(x)=cx$

Using it in the question's equation and solving it we get c=1

$\therefore f(x)=x $ which is indeed the answer.
Bug 1:$f(0)=?$
If we put $x=1$ and $y=0$ in the question we get,
$f(1)=f(1)+f(f(0))$
Or,$f(f(0))=0$
Or,$f(0)=0$[Using 1]
Bug 2: (t can be negative as well)
If we take $y= -x^4$ in the question we get,
$f(x^4 -x^4)=x^3f(x)+f(f(-x^4))$
Or,$f(0)=f(x^4)+f(-x^4)$
Or,$f(-x^4)=-f(x^4)$
Or,$f(-t)=-f(t)$
Thus we now have the negative values for t.
Bug 3: Cauchy function application criteria
But how do I do that?

I think if we will show that $f(x+y)=f(x)+f(y)$ for all x greater than 0
And $f(x+y)=f(x)+f(y)$ for x less then 0 then it will be ok maybe.
But we showed that $f(x+y)=f(x)+f(y) $ for all real $x,y$. That would include your 2 statements together, wouldn't it?
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

Dustan
Posts: 64
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Fri Jan 08, 2021 5:57 pm

Problem 7: Find all function such that
$f:R\rightarrow R $ and
$f(x^2+yf(x))=xf(x+y)$ for all real valued x,y.

~Aurn0b~
Posts: 43
Joined: Thu Dec 03, 2020 8:30 pm

Re: FE Marathon!

Unread post by ~Aurn0b~ » Wed Jan 13, 2021 11:14 pm

Dustan wrote:
Fri Jan 08, 2021 5:57 pm
Problem 7: Find all function such that
$f:R\rightarrow R $ and
$f(x^2+yf(x))=xf(x+y)$ for all real valued x,y.
$\textbf{Solution :}$
Let $P(x,y)$ be the assertion of the functional equation
$P(x,0) \Rightarrow f(x^2)=xf(x) \dots \dots (1)$
So $P(0,0)\Rightarrow f(0)=0$.

$\textbf{Case 1:}$ Assume that there exists $a_0 \in \mathbb{R}$, and $a_0 \neq 0$, such that, $f(a_0)=0$,
Then, $P(a_0,y)\Rightarrow f(a_0^2)=a_0 f(a_0 + y)$
Hence, $a_0 f(a_0 + y) = f(a_0^2)= a_0 f(a_0)=0$
which implies,$f(a_0 + y) =0$ for all $y\in \mathbb{R}$.(we can divide both sides by $a_0$ beacuse it is not $0$.)
So, we can say that for all real numbers, $f(x)=0.$, which indeed is a solution.

$\textbf{Case 2:}$ $x=0$ is the only input for which $f(x)=0$ ,
$P(x,-x) \Rightarrow f(x^2 -xf(x))=0$
$\Rightarrow x^2 =xf(x)$
$\Rightarrow f(x)=x$
Which satifies the main function.

So all the solution for the equation is, $f(x)=x,0$.$\blacksquare$

Dustan
Posts: 64
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Thu Jan 14, 2021 10:30 pm

~Aurn0b~ wrote:
Wed Jan 13, 2021 11:14 pm
Dustan wrote:
Fri Jan 08, 2021 5:57 pm
Problem 7: Find all function such that
$f:R\rightarrow R $ and
$f(x^2+yf(x))=xf(x+y)$ for all real valued x,y.
$\textbf{Solution :}$
Let $P(x,y)$ be the assertion of the functional equation
$P(x,0) \Rightarrow f(x^2)=xf(x) \dots \dots (1)$
So $P(0,0)\Rightarrow f(0)=0$.

$\textbf{Case 1:}$ Assume that there exists $a_0 \in \mathbb{R}$, and $a_0 \neq 0$, such that, $f(a_0)=0$,
Then, $P(a_0,y)\Rightarrow f(a_0^2)=a_0 f(a_0 + y)$
Hence, $a_0 f(a_0 + y) = f(a_0^2)= a_0 f(a_0)=0$
which implies,$f(a_0 + y) =0$ for all $y\in \mathbb{R}$.(we can divide both sides by $a_0$ beacuse it is not $0$.)
So, we can say that for all real numbers, $f(x)=0.$, which indeed is a solution.

$\textbf{Case 2:}$ $x=0$ is the only input for which $f(x)=0$ ,
$P(x,-x) \Rightarrow f(x^2 -xf(x))=0$
$\Rightarrow x^2 =xf(x)$
$\Rightarrow f(x)=x$
Which satifies the main function.

So all the solution for the equation is, $f(x)=x,0$.$\blacksquare$
Post Problem 8

~Aurn0b~
Posts: 43
Joined: Thu Dec 03, 2020 8:30 pm

Re: FE Marathon!

Unread post by ~Aurn0b~ » Thu Jan 14, 2021 11:48 pm

$\textbf{Problem 8 :}$
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ satisfy:
$$f(f(x)+2f(y))=f(x)+y+f(y)$$

Dustan
Posts: 64
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Fri Jan 15, 2021 1:58 am

~Aurn0b~ wrote:
Thu Jan 14, 2021 11:48 pm
$\textbf{Problem 8 :}$
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ satisfy:
$$f(f(x)+2f(y))=f(x)+y+f(y)$$
Solution 8: $P(x,y)$ be the assertion.
From $P(a,b)$ and $P(b,a)$
$f$ is injective.
$P(0,-f(0))$:
$f(f(0)+2f(-f(0)))=f(-f(0))$
$f(0)+2f(-f(0))=-f(0)$
$f(-f(0))=-f(0)$...(1)
Again, $P(-f(0),0)$ gives
$f(f(-f(0))+2f(0))=f(-f(0))+f(0)=0$[from (1)]
$f(f(0))=0$
$P(f(0),0)$ again gives us
$f(2f(0))=f(0)$
From this $f(0)=0$

Now, $P(x,0)$ gives
$f(f(x))=f(x)$
From injectivity
$f(x)=x$ which indeed fits the given equation.

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