IMO-1992-2
- Phlembac Adib Hasan
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Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that \[f(x^2+f(y))=y+[f(x)]^2\]
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Re: IMO-1992-2
Here are the steps of my solution.
1.Put $x=0$ to deduce that $f$ is surjective.
2.Let,$f(a)=f(b)$.Now add $a^2$ in both sides and use the given equation to conclude that f is injective.
3.Let,$f(0)=c$,puting $x=0$ we have $f(f(y))=y + c^2$
4.as $f$ is surjective,there is a $k$ such that f(k)=0.Put $x=k$ and y=$f(y)$.Then use equation in the 3rd step and injectivity of $f$ to deduce that $c=0$
So the equation in the 3rd step reduces to $f(f(y))=y$
5.Now put $x=y$ and u will get an equation.In that equation put $x=f(x)$ to conclude that $f(x)^2=x^2$
6.let there are $c,d$ such that $f(c)=c$ and $f(d)=-d$,now put $x=c$ and $y=d$ and use the fact that $f(0)=0$ to deduce that for all real $x$ either $f(x)=x$ or $f(x)=-x$.
Now,checking kills the 2nd case
1.Put $x=0$ to deduce that $f$ is surjective.
2.Let,$f(a)=f(b)$.Now add $a^2$ in both sides and use the given equation to conclude that f is injective.
3.Let,$f(0)=c$,puting $x=0$ we have $f(f(y))=y + c^2$
4.as $f$ is surjective,there is a $k$ such that f(k)=0.Put $x=k$ and y=$f(y)$.Then use equation in the 3rd step and injectivity of $f$ to deduce that $c=0$
So the equation in the 3rd step reduces to $f(f(y))=y$
5.Now put $x=y$ and u will get an equation.In that equation put $x=f(x)$ to conclude that $f(x)^2=x^2$
6.let there are $c,d$ such that $f(c)=c$ and $f(d)=-d$,now put $x=c$ and $y=d$ and use the fact that $f(0)=0$ to deduce that for all real $x$ either $f(x)=x$ or $f(x)=-x$.
Now,checking kills the 2nd case
Last edited by *Mahi* on Fri May 18, 2012 9:23 pm, edited 1 time in total.
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Re: IMO-1992-2
দুঃখিত,আমার চতুর্থ স্টেপের পর ভুল আছে।
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Re: IMO-1992-2
After 4th step,
5.put $y=0$ to get $f(x^2)=f(x)^2$
6.Now put $y=f(x)$ and use the equation in the 5th step to get a Cauchy's equation.
Hence,$f(x)=cx$ for some constant $c$.Checking ensures that $c=1$
5.put $y=0$ to get $f(x^2)=f(x)^2$
6.Now put $y=f(x)$ and use the equation in the 5th step to get a Cauchy's equation.
Hence,$f(x)=cx$ for some constant $c$.Checking ensures that $c=1$
Re: IMO-1992-2
$f$ is $\mathbb R \mapsto \mathbb R$. So you can not use cauchy equation without proving some additional properties of $f$shehab ahmed wrote: 6.Now put $y=f(x)$ and use the equation in the 5th step to get a Cauchy's equation.
Hence,$f(x)=cx$ for some constant $c$.Checking ensures that $c=1$
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Re: IMO-1992-2
হায়রে,মোবাইল থেকে কিচ্ছু বুঝতে পারছি না।আন্দাজের উপর ডোমেন এবং কো ডোমেন ধরে নিয়ে অঙ্ক করা শুরু করেছি।একটু সোজা বাংলায় ডোমেন এবং কোডোমেন কী তা বলে দিলে ভাল হত
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Re: IMO-1992-2
ও আচ্ছা।ধন্যবাদ মাহি।এটা আমার মনে ছিল না।
additional property of f:
$f(x)^2=f(x^2)$ now if $x$ is non negative, then putting $x=x^{\frac {1}{2}}$ we have $f(x)$ is non negative.So we can use cauchy.
additional property of f:
$f(x)^2=f(x^2)$ now if $x$ is non negative, then putting $x=x^{\frac {1}{2}}$ we have $f(x)$ is non negative.So we can use cauchy.
Re: IMO-1992-2
That should be enough. But please never forget it in contests (which may cost you severely ).shehab ahmed wrote:additional property of $f$:
$f(x)^2=f(x^2)$ now if $x$ is non negative then putting $x=x^{1/2}$ we have $f(x)$ is non negative.So we can use cauchy
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- Phlembac Adib Hasan
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