3
$\begingroup$

First edition was: Let $f(x)$ be a polynomial such that $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that $f(f(x))-x$ also has only one real root?

Second edition: Let $f(x)=a_3 x^3 + a_2 x^2 + a_1 x + a_0$. All $a_i \neq 0$. $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that $f(f(x))-x$ also has only one real root?

$\endgroup$
5
  • 6
    $\begingroup$ It isn't true. Let $f(x)=-x$. $\endgroup$ Jan 13, 2012 at 6:40
  • $\begingroup$ I suspect that the statement is true if we revise the question to require $f(f(x)) - x$ not be the zero function, but I am unable to prove it. $\endgroup$ Jan 13, 2012 at 7:06
  • 3
    $\begingroup$ One thing that is true is that if $f(x)$ is nondecreasing, then all real roots of $f(f(x)) - x$ are roots of $f(x)-x$. $\endgroup$ Jan 13, 2012 at 7:39
  • $\begingroup$ Yes! it was my first idea, but i cant proof this. Can you write more detailed proof of this: "if f(x) is nondecreasing, then all real roots of f(f(x))−x are roots of f(x)−x" ? $\endgroup$
    – Mike
    Jan 13, 2012 at 8:00
  • $\begingroup$ If $f(f(r)) = r$ but $f(r) \ne r$, let $f(r) = s$, so $f(s) = r$. Either $r < s$ and $f(r) > f(s)$ or $s < r$ and $f(s) > f(r)$. $\endgroup$ Jan 13, 2012 at 15:54

3 Answers 3

6
$\begingroup$

More generally, $f(x)=-x^{2n-1}$ gives a counterexample for each positive integer $n$.

A counterexample for the second edition: $f(x)=-x^3-x^2-x-1$.

$\endgroup$
1
  • $\begingroup$ Thanks its really help me. $\endgroup$
    – Mike
    Jan 13, 2012 at 8:01
2
$\begingroup$

As Jonas Meyer has pointed out, the problem is actually not true.

As suggested, let $f(x)= -x$. Then $f(x) -x = -x -x = -2x$. Considering $f(f(x)) -x= f(-x) -x = -(-x) -x = 0$. That is to say, for $f(x)=-x$, we have shown $f(x) -x$ has only one real root, namely $x=0$, while $f(f(x))-x=0$ has infinitely many real roots.

Response to comments: Let $f(x)=x^3 - 2 x^2 + 5 x - 1$. Then, $$f(x) -x = x^3 -2x^2 +5x -1 =x^3 -2x^2 +4x -1$$ Solving for $x$ (by Wolfram Alpha) we have that $x \approx 0.284775$ and that there is only one real root. Considering $f(f(x))-x$ we have the following, $$f(f(x))-x = (x^3 -2x^2 +4x -1)^3 - 2(x^3 -2x^2 +4x -1)^2 +4(x^3 -2x^2 +4x -1) -1$$ which can be "simplifed" to, $$f(f(x))-x = x^9-6 x^8+24 x^7-61 x^6+116 x^5-156 x^4+155 x^3-102 x^2+44 x-8$$ Which has only one real root $x \approx 0.379555$. So, you have shown that there does exist a polynomial, namely $f(x) = x^3 -2x^2 +5x -1$ for which it is true. But this does not show in general that your conjecture is true, since there exists at least one counterexample.

@Jonas: I can't think of another counterexample off the top of my head, I don't want to try to hard at thinking about this.

$\endgroup$
8
  • $\begingroup$ I have no idea what I was thinking, after thinking about it for more than 2 seconds it is clear that the intermediate value theorem would not help at all to show that there is only one real root. I was absently thinking that you were trying to show there is at least one real root. $\endgroup$ Jan 13, 2012 at 6:53
  • $\begingroup$ Ok. I see. But can you help for example with f(x)=x^3 - 2 x^2 + 5 x - 1 $\endgroup$
    – Mike
    Jan 13, 2012 at 6:56
  • 1
    $\begingroup$ @user22867: If your question is actually about a particular polynomial, will you please edit that into your question? $\endgroup$ Jan 13, 2012 at 6:59
  • 1
    $\begingroup$ @user22867: If you already know that the degree is odd (or if it is an assumption included in the problem), that information would also be helpful in the problem. $\endgroup$ Jan 13, 2012 at 7:06
  • 1
    $\begingroup$ @user22867: If you have any additional information about the problem I suggest you revise your question as I can edit my answer to accommodate. Please always post a full question with all of the information you know when you ask a question. $\endgroup$ Jan 13, 2012 at 7:09
2
$\begingroup$

Here's a more general way of making counterexamples. Choose any real numbers $a < b$. Let $g(x)$ be any polynomial of odd degree whose leading coefficient is positive. Take $f(x) = a + b - x - \varepsilon (x-a)(x-b) g(x)$, choosing $\varepsilon \ge 0$ small enough that $f$ is decreasing, and thus $f(x)$ and $f(x) - x$ each have only one zero. But $f(a) = b$ and $f(b)=a$, so $a$ and $b$ are zeros of $f(f(x))-x$.

$\endgroup$
1
  • $\begingroup$ Can you please explain how to prove: "if f(x) is nondecreasing, then all real roots of f(f(x))−x are roots of f(x)−x"? $\endgroup$
    – Mike
    Jan 13, 2012 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.