Vietnam 2012- Surjective function

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*Mahi*
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Vietnam 2012- Surjective function

Unread post by *Mahi* » Sat Jan 14, 2012 9:05 pm

Find all function $ f:\mathbb{R}\rightarrow\mathbb{R} $
1)$f$ is surjective
2)For $x>y$, $ f(x)>f(y) $
3)$ f(f(x))=f(x)+12x $
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zadid xcalibured
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Re: Vietnam 2012- Surjective function

Unread post by zadid xcalibured » Thu Feb 28, 2013 3:16 am

$f(f(0))=f(0)$ and strictly increasing property yields $f(0)=0$.
let $a_n=f^{n}(x)$ and let $a_0=x$
then $a_n=ap^n+bq^n$ where $a$ and $b$ are the solutions of $x^2-x-12=0$
$a_2=f(x)=ap+bq=-3p+4q=-3x+7q$
but $f(0)=0$ implies $q=0$
so $f(x)=-3x$ which is surjective and strictly increasing.

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Re: Vietnam 2012- Surjective function

Unread post by *Mahi* » Thu Feb 28, 2013 10:09 am

zadid xcalibured wrote: so $f(x)=-3x$ which is surjective and strictly increasing.
$f(x) = -3x$ is strictly decreasing.
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SANZEED
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Re: Vietnam 2012- Surjective function

Unread post by SANZEED » Thu Feb 28, 2013 11:11 am

The mistake in Zadid vai's solution was perhaps to write $a_n=ap^n+bq^n$. As far as I know the right form should be $a_n=pa^n+qb^n$ and this yields a valid solution $f(x)=4x$ :|
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zadid xcalibured
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Re: Vietnam 2012- Surjective function

Unread post by zadid xcalibured » Thu Feb 28, 2013 1:42 pm

There are two cases.I missed it.One case gives $a=0$ and $b=x$ another cases gives $b=0$ and $a=x$.One case gives no solution while the other case gives $f(x)=4x$ :mrgreen:

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