$\textbf{Problem 17}$
Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$.
DO MENTION THE SOURCE
Proof(Not sure this the final version ):
It is easy to see the $f(x)$ is bijectve just fix any $x=x_0$ and the rest is easy.
Now since it is bijective it has a inverse function namely $f^{-1} (x)$ Plugging inverse function onto the original equation $\Rightarrow f(xf^{-1} (x)+y)=x^2 +f^{-1} (y)$ now plugging $x=y=0 \Rightarrow f(0)=f^{-1} (0)$
Let $P(x,y)$ be the assertion in the problem and $f(0)=t$
$P(0,x) \Rightarrow f(f(x))=t^2 +x.......(0)$
$P(0,0) \Rightarrow f(f(0))=f(0)^2$
since $f(0)=f^{-1} (0) \Rightarrow f(f^{-1} (0))=f(0)^2 \Rightarrow f(0)=0$
Plugging the value of $f(0)=t=0$ in $(0) \Rightarrow f(f(x))=x$ and $f(xf(x))=f(x)^2$ now notice that plugging $x=f(x) \Rightarrow f(xf(x))=(f(f(x)))^2=x^2$
$\Rightarrow f(x)^2 =x^2 \Rightarrow f(x)=\pm x, \forall x \in \mathbb{R}$
Re: FE Marathon!
Posted: Mon Mar 01, 2021 3:56 pm
by Asif Hossain
Proble 18:(not serious about this problem just asking
Find all functions $f:$ $\mathbb{R} \rightarrow \mathbb{R}$ such that $f(f(x+1))=f(x)+1$
for all real $x$.
Source: self made or encountered
Update: this has no specific or general solution thanks to Dustan.
$\textbf{Problem 17}$
Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$.
DO MENTION THE SOURCE
Proof(Not sure this the final version ):
It is easy to see the $f(x)$ is bijectve just fix any $x=x_0$ and the rest is easy.
Now since it is bijective it has a inverse function namely $f^{-1} (x)$ Plugging inverse function onto the original equation $\Rightarrow f(xf^{-1} (x)+y)=x^2 +f^{-1} (y)$ now plugging $x=y=0 \Rightarrow f(0)=f^{-1} (0)$
Let $P(x,y)$ be the assertion in the problem and $f(0)=t$
$P(0,x) \Rightarrow f(f(x))=t^2 +x.......(0)$
$P(0,0) \Rightarrow f(f(0))=f(0)^2$
since $f(0)=f^{-1} (0) \Rightarrow f(f^{-1} (0))=f(0)^2 \Rightarrow f(0)=0$
Plugging the value of $f(0)=t=0$ in $(0) \Rightarrow f(f(x))=x$ and $f(xf(x))=f(x)^2$ now notice that plugging $x=f(x) \Rightarrow f(xf(x))=(f(f(x)))^2=x^2$
$\Rightarrow f(x)^2 =x^2 \Rightarrow f(x)=\pm x, \forall x \in \mathbb{R}$
$\textbf{Problem 17}$
Find all functions $f: \mathbb R \to \mathbb R$ such that\[ f( xf(x) + f(y) ) = f(x)^2 + y \]for all $x,y\in \mathbb R$.
DO MENTION THE SOURCE
Proof(Not sure this the final version ):
It is easy to see the $f(x)$ is bijectve just fix any $x=x_0$ and the rest is easy.
Now since it is bijective it has a inverse function namely $f^{-1} (x)$ Plugging inverse function onto the original equation $\Rightarrow f(xf^{-1} (x)+y)=x^2 +f^{-1} (y)$ now plugging $x=y=0 \Rightarrow f(0)=f^{-1} (0)$
Let $P(x,y)$ be the assertion in the problem and $f(0)=t$
$P(0,x) \Rightarrow f(f(x))=t^2 +x.......(0)$
$P(0,0) \Rightarrow f(f(0))=f(0)^2$
since $f(0)=f^{-1} (0) \Rightarrow f(f^{-1} (0))=f(0)^2 \Rightarrow f(0)=0$
Plugging the value of $f(0)=t=0$ in $(0) \Rightarrow f(f(x))=x$ and $f(xf(x))=f(x)^2$ now notice that plugging $x=f(x) \Rightarrow f(xf(x))=(f(f(x)))^2=x^2$
$\Rightarrow f(x)^2 =x^2 \Rightarrow f(x)=\pm x, \forall x \in \mathbb{R}$
Source:BMO 2000
$f(x)^2=x^2$ doesnt imply $f(x)=\pm x, \forall x$
I meant either $f(x)=x$ or $f(x)=-x$ $\forall x \in \mathbb{R}$???
ooooo now i understand thanks to aops (this was subtle )
Here is the rest credit goes le qk's math
Assume there exists $a,b$ such that $f(a)=a ,f(b)=-b$ now we provide a contradiction
$P(a,b) \Rightarrow f(a^2 -b )=a^2 +b$ so, happy contradiction....
Re: FE Marathon!
Posted: Wed Mar 03, 2021 3:33 pm
by Asif Hossain
Nobody posting any problem
Find all functions $f:\mathbb{Q} \rightarrow \mathbb{R}$
$1$. $f(x+y)-yf(x)-xf(y)=f(x)f(y)-x-y+xy; \forall x,y \in \mathbb{Q}$
$2$. $f(x)=2f(x+1)+x+2; \forall x \in \mathbb{Q}$
$3$. $f(1)+1>0$
Source:
BMO 2003, Problem no.3
Re: FE Marathon!
Posted: Thu Mar 04, 2021 9:32 pm
by Asif Hossain
Partial Solution:
Let $P(x,y),Q(x)$ be the assertion for $1,2$
$P(0,0) \Rightarrow f(0)=0$ or $f(0)=1$
If $f(0)=0$ then $Q(0) \Rightarrow f(1)=-1$ which contradicts $3$. so, $f(0)=1$
Now solving the recurrence $a_n =2a_{n+1}+n+2$ with $a_0 =1$ gives $a_n = \frac{1}{2^n} -n$(Recurrence solved by wolfram but you can solve with generating functions or just by a non trivial guessing)
(Note: I didn't checked it please somebody check it)
Doubtful about this arguement: Now since the previous equation is satisfied for $\mathbb{Z}$ which is a sub set of $\mathbb{Q}$ so it follows all $x \in \mathbb{Q}$ also must follow $f(x)= \frac{1}{2^x} -x$ $\forall x \in \mathbb{Q}$ since the the $2$ is only satisfied by that for integers.
Re: FE Marathon!
Posted: Tue Mar 16, 2021 7:02 pm
by Asif Hossain
Since Nobody posting any problem here is one... Problem 19
Determine all functions $\mathbb{Q} \to \mathbb{C}$ ST
1) for any rational $x_1,X_2,...,x_2010,$ $f(x_1+x_2+...+x_2010)=f(x_1)f(x_2)...f(x_2010)$
2) $\forall x \in \mathbb{Q}$ , conjugate of $f(2010)$ $\times$ $f(x)=f(2010)$ $\times$ conjugate of $f(x)$ where $\times$ represent normal multiplication.
2) $\forall x \in \mathbb{Q}$ , conjugate of $f(2010)$ $\times$ $f(x)=f(2010)$ $\times$ conjugate of $f(x)$ where $\times$ represent normal multiplication.
Did you mean $\overline{f(2010)}\cdot f(x)$ or $\overline{f(2010)\cdot f(x)}$? I guess the first one.
.