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I have got an interesting exercise. Proof that for all positive integer $a$ and $p(x) = x^2+2013x + 1$, $\underbrace{p(p(\dots p}_{a \ \ \text{times}}(x)\dots )) = 0$ has got at least 1 real solution $x_0$.

Hope you can help me. I haven't got any ideas to proof that. Thanks.

beep-boop
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Daifus
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3 Answers3

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Consider the solutions of $p(x)=x$ this gives, $$x^2+2012x+1=0 \implies x = \frac{-2012-\sqrt{2012^2-4}}{2} , \frac{-2012+\sqrt{2012^2-4}}{2} $$ Let $$ m= \frac{-2012-\sqrt{2012^2-4}}{2} , n= \frac{-2012+\sqrt{2012^2-4}}{2}$$ Note that, $m<n<0$.

Now let $f(x)= {}\underbrace{p(p(\ldots(p}_{a\ {\rm times}}(x))\ldots))$.

Then, note that, $f(n)=n<0$

And, also the biggest term(term with largest degree) will be $x^{2^a}$ with positive coeffecient. So when $x \rightarrow \infty, f(x) >0$

Since $f(x)$ is a polynomial , so it must be continuous . And so there will be a solution of $f(x)=0$ in range $(m,\infty)$ $\Box$

Shivang jindal
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  • thank you very much for your quick answer, but I have got a question. why can you verify p(x) = x and f(n) = n? Do the polynomials have those value? – Daifus Aug 26 '14 at 12:06
  • Can you explain your question please? What i did is that i checked are roots of $p(x)=x$ , and noticed that root is negative. – Shivang jindal Aug 26 '14 at 12:10
  • but if you check if the roots of p(x) negative, why didn't you check it with p(x) = 0? Because has p(x) the value x? And you let f(x) = $\underbrace{p(p(...(p}_{a-times}(x))...))$ and noticed that f(n) = n but why is it so? you set n = $\frac{-2012 + \sqrt{2012^2 - 4}}{2}$ or? – Daifus Aug 26 '14 at 12:20
  • Checking roots of $p(x)=0$ will of no use. I would have reached nowhere by that. But since $f(x)$ is composition of $p(x)$ , and if i have some value of such that $p(n)=n$ then obviously $f(n)=n$ , like if you take a =3 then you have $f(x)=p(p(p(x))) \implies f(n) = p(p(p(n)))= p(p(n))=p(n)=n$ . – Shivang jindal Aug 26 '14 at 12:24
  • @Daifus If $p(x_1)=x_1$ and $x_1<0$, then $\underbrace{p(p(\dots p}{a-times}(x_1)\dots )) < 0$, and he also showed that there exists an $x_2$ such that $\underbrace{p(p(\dots p}{a-times}(x_2)\dots )) > 0$, hence by the Intermediate Value Theorem we have that there exists an $x_3$ such that $\underbrace{p(p(\dots p}_{a-times}(x_3)\dots )) = 0$. Checking the case $p(y)=y$ is useful, since we then have $f(y)=y$ and so it helps with the observations of what values $f(x)$ can possibly gain. The solution would've been even easier if we'd had both negative and positive possible values of $y$. – user26486 Aug 26 '14 at 14:06
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Possibly you could show there is $x_1 < 0$ such that $p(x_1)=x_1$. And there is $x_2 > x_1$ such that $p(x_2)>0$. Then for each natural $a$:
$$p^{a}(x_1) = x_1 <0$$
and
$$p^{a}(x_2)>0$$

CiaPan
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$p(x)$ attains its minimum at $(x_{\min}, y_{\min})$, where $x_{\min} = -\frac{2013}{2}$ and $y_{\min} = p(x_{\min}) = x_{\min}^2 + 2013x_{\min} + 1$. The exact values of $x_{\min}$ and $y_{\min}$ are not important; what is important is that $y_{\min} < x_{\min} < 0$, which you can check for yourself.

Now let $q(x)$ be the polynomial $p(x)$ restricted to the domain $[x_{\min},\infty)$. So $q(x)$ is defined on the interval $[x_{\min},\infty)$, and takes the value $p(x)$ on that interval. Because $q(x)$ is monotonic increasing, it has a well-defined inverse on its range, which is $[y_{\min},\infty)$; and because $x_{\min} > y_{\min}$, it has a fortiori a well-defined inverse $q^{-1}(x)$ on $[x_{\min},\infty)$.

So to solve $\underbrace{p(p(\dots p}_{a-times}(x)\dots )) = 0$, just iterate this inverse $a$ times starting from $0$:

$$x = \underbrace{q^{-1}(q^{-1}(\dots q^{-1}}_{a-times}(0)\dots ))$$

TonyK
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  • How can $x_{min}=-\frac{2013}{2}$ be positive? I can't see where this fact is used so it seems to be an irrelevant mistake. But I can't edit. – Taladris Aug 26 '14 at 13:48
  • @Taladris: I have fixed it now. Thank you for pointing it out! – TonyK Aug 26 '14 at 14:09