Let $n$ be a positive integer. Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such
that, for all reals $y$ and all nonzero reals $x$
\[x^nf(y)-y^nf(x)=f(\frac{y}{x})\]
IMO Mock-1 Problem 1
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
বড় ভালবাসি তোমায়,মা
-
- Posts:461
- Joined:Wed Dec 15, 2010 10:05 am
- Location:Dhaka
- Contact:
Re: IMO Mock-1 Problem 1
My solution steps:
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: IMO Mock-1 Problem 1
Let $P(x,y)$ be the assertion.$P(x,0)\Rightarrow f(0)=0$Now take $x\neq 0,y\neq 1,-1$.
$P\left (\displaystyle \frac {y}{x},y\right )\Longrightarrow f(x)=\left (\displaystyle \frac {y}{x}\right )^n\displaystyle \frac {1-x^{2n}}{1-y^{2n}}f(y)$
$\Longrightarrow f(x)=a\left (x^n-\displaystyle \frac {1}{x^n}\right )\; or\; -a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$ where $a$ is a fixed real.Now assume $\exists c,d$ such that $f(c)=a\left (c^n-\displaystyle \frac {1}{c^n}\right ), f(d)=-a\left (d^n-\displaystyle \frac {1}{d^n}\right )$.Now take $P(c,d)$ to get a contradiction.So all the functions are:
$(i)f(0)=0, \; f(x)=a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$
$(ii)f(0)=0, \; f(x)=-a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$
$P\left (\displaystyle \frac {y}{x},y\right )\Longrightarrow f(x)=\left (\displaystyle \frac {y}{x}\right )^n\displaystyle \frac {1-x^{2n}}{1-y^{2n}}f(y)$
$\Longrightarrow f(x)=a\left (x^n-\displaystyle \frac {1}{x^n}\right )\; or\; -a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$ where $a$ is a fixed real.Now assume $\exists c,d$ such that $f(c)=a\left (c^n-\displaystyle \frac {1}{c^n}\right ), f(d)=-a\left (d^n-\displaystyle \frac {1}{d^n}\right )$.Now take $P(c,d)$ to get a contradiction.So all the functions are:
$(i)f(0)=0, \; f(x)=a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$
$(ii)f(0)=0, \; f(x)=-a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$