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I have the polynomial $P(x)=x^{2}+2013x+1$ and a number $n\in\mathbb{N}$. Now I have to show that the polynomial $P(P(...P(x)...)$ $(n$ times$)$ has at least one real root. How can I do this?

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2 Answers 2

up vote 6 down vote accepted

Note that the quadratic equation $$P(x)=x$$ has some real root $x_-<0$.

For every $n\ge 1$, denote the $n$-th iteration of $P$ by $P^{\circ n}$. Then $$P^{\circ n}(x_-)=x_-<0\quad\text{and}\quad P^{\circ n}(+\infty)=+\infty,$$ so there exists $x_n\in(x_-,+\infty)$, such that $P^{\circ n}(x_n)=0$.

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I'm sorry, why is $P^{\circ n}(x_{-})= x_-$? –  Eric Auld Jul 27 '13 at 11:47
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@EricAuld: Since $P(x_-)=x_-$, $P(P(x_-))=P(x_-)=x_-$, and so on. –  23rd Jul 27 '13 at 11:49
    
I thought $P(x_-) = 0$. –  Eric Auld Jul 27 '13 at 11:52
    
@Eric The first line of the answer states that $x_{-}$ is the real root of the quadratic equation $P(x)=x$, i.e., $P(x_{-})=x_{-}$ ... –  Amitesh Datta Jul 27 '13 at 11:53
    
Oh, I see. Whoops! –  Eric Auld Jul 27 '13 at 11:54

The polynomial has two real zeros, call them $a$ and $b$, where $a<b$ (this we can tell from taking the discriminant). If you can show that $P([a,b])\supset [a, b]$, then you're home free, because then $a$ has a preimage $c_1\in [a,b]$, then $c_1$ has a preimage $c_2\in [a,b]$, then...

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Presumaby $f(x)=P(x)$, no? –  Nils Matthes Jul 27 '13 at 12:47
    
Whoops! Fixed. -- –  Eric Auld Jul 27 '13 at 12:51

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