FE Marathon!

[Dustan]

Problem 9: $f:R\rightarrow R $ such that
$f(xf(x-y))+yf(x)=x+y+f(x^2)$ for all real x,y.

[~Aurn0b~]

Problem 9: $f:R\rightarrow R $ such that
$f(xf(x-y))+yf(x)=x+y+f(x^2)$ for all real x,y.

$\textbf{Solution :}$

Let $P(x,y)$ be the assertion of the functional equation.
$P(0,y) \Rightarrow f(0)=1 $.
$P(x,x) \Rightarrow f(x)+xf(x)=2x+f(x^2) \dots \dots (1)$
Plugging in $x=1$ in $(1)$, we have $f(1)=2$.
And Plugging in $(1)$ in the main equation we have,$P(x,y) \Rightarrow f(xf(x-y))= f(x)(x-y+1) + (y-x) $

$P(1,1-y) \Rightarrow f(f(y))=f(1)(y+1) - y= y+2 \Rightarrow f(f(y-2))=y$,
We also have, $f(f(y))=y+2 \Rightarrow f(f(f(y)))=f(y+2) \Rightarrow f(y)+2=f(y+2)\dots \dots (2)$
$P(x,x-y) \Rightarrow f(xf(y))= f(x)(y+1) -y \dots \dots (3)$
Plugging $y=f(y-2)$ in $(3)$ and using the fact that $f(f(y-2))=y$ ,
$f(xy)=f(x)(f(y-2)+1) -f(y-2)$
Again, from $(2)$, we know that, $f(y-2)=f(y)-2$, using this we get,
$f(xy)+f(x)+f(y)=f(x)f(y)+2$,
Let $Q(x,y)$ be the assertion of $f(xy)+f(x)+f(y)=f(x)f(y)+2$,
$Q(x,x) \Rightarrow f(x^2)+2f(x)=f(x)^2+2 \dots \dots (4)$
Now, $(2)-(4) \Rightarrow (f(x)-x-1)(f(x)-2)=0$
$f$ clearly cannot be a constant function, so $f(x)=x+1.\blacksquare$

[~Aurn0b~]

$\textbf{Problem 10 :}$
Determine all the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that:
\[f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))\ ,\ \forall x,y\in \mathbb{R}\]

[Dustan]

Solution 10:
Let, $g(x)=f(x)-f(0)$ $\Rightarrow f(x)=g(x)+f(0)$
By using this we get the similar equation. So,WLOG $f(0)=0$
$P(x,0)$:
$g(x^3)=x^2g(x)$
From this
$x^2g(x)-y^2g(y)=(x^2+xy+y^2)(g(x)-g(y))$
Or, $(x+y)yg(x)=x(x+y)g(y)$
Or,$\frac{g(x)}{x}=\frac{g(y)}{y}$
Seting, $y=1$
$g(x)=ax$
So,$f(x)=ax+b$

[Dustan]

Problem 11: $f:R\rightarrow R$ and
$f(t^2+u)=tf(t)+f(u)$ for all real valued t,u.

[Mehrab4226]

Let, $p(t^2,u)=f(t^2+u)=tf(t)+f(u)$
$p(1,0)$ yields f(0)=0
Now,
$p(t^2,0)=f(t^2)=tf(t) \cdots (1)$
Now let us put $t= \pm x$ in (1)

$f(t^2)=t(f(t)=-tf(-t)$
$\therefore f(-t)=-f(t) \cdots(2)$

Now,
$p(x,u)$ gives us
$f(x+u)=f(x)+f(u)$ for $x \geq 0 \cdots (3)$ [Skipped a line, but it is obvious]

Finally, For $t \geq 0$
$(t+1)(f(t)+f(1)=(t+1)f(t+1)$
$=f((t+1)^2)[Using(1)]$
$=f(t^2+2t+1)=f((t^2)+(2t+1))=f(t^2)+f(2t+1)=f(t^2)+f(2t)+f(1)$
$f(2t)=2f(t)$[Got by putting x,u=t,t in (3)]
Now,

$(t+1)(f(t)+f(1)) = f(t^s)+2f(t)+f(1)$
Now doing some algebraic operations we will get,
$f(t)=f(1)t$, let, $f(1)=m$
$\therefore f(x)=mx$ for $x \geq 0$[$x$ looks better than t]
putting x = -x we get,
$f(-x)=-f(x)$[from (2)]$ = -mx = m(-x) for x \geq 0$
This should solve the problem of being only positive,
$\therefore f(x)=mx$ For $x \in \mathbb{R}$ and $m=f(1)$

[Mehrab4226]

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} $

[Asif Hossain]

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} $

$Claim$: $ \forall x \in \mathbb{R} $ $f(x) $ = 0,1
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)

[Asif Hossain]

New proble find all $f$: $\mathbb{R} \to \mathbb{R} $ ST
$f(f(x+1))=f(x)+1 $
Self Discovered problem

[Anindya Biswas]

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} $

$Claim$: $ \forall x \in \mathbb{R} $ $f(x) $ = 0,1
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)

Sorry to disappoint, but $f(x)=e^{kx^2}$ is also a valid solution.