I have been trying to characterize the number of roots on $\theta$ for the following polynomial

$$ \sum_{i=1}^n \frac{\theta- x_i}{1+ \left(x_i - \theta \right)^2} = 0$$

If we were to put everything under a common denominator then we would see that the polynomial is of degree $2n -1 $ with coefficients depending on the $x_i$ for $i=1,2,...,n$ . Taking limits as $\theta \to \infty$ and $\theta \to -\infty$ we find that the polynomial has at least one real root as it tends to zero through positive and negative values respectively.

What I do not understand is how one reaches the conclusion that the number of real roots is odd in this case. I know that the complex roots are $2 n -1$ but I'm talking about the real roots. Is there a theorem that guarantees that or is that common sense hiding in plain sight, at least for me?

Thank you.

  • $\begingroup$ What are $x_i$'s? They are real numbers? $\endgroup$ Feb 27 '16 at 16:38
  • $\begingroup$ @hamidkamali Yes, they are reals. $\endgroup$
    – JohnK
    Feb 27 '16 at 16:38
  • $\begingroup$ Well, first, I get a polynomial of degree $2n+1$, not $2n-1$. Moreover, if all the $x_i$s are real then the polynomial has of course an even number of complex roots. So there's only an odd number left for the real roots. $\endgroup$ Feb 27 '16 at 16:43
  • $\begingroup$ Try with n=2 and you will see 2n-1 is the good. The problem is actually very easy if you discard the expression (which is not a polynomial!) and you consider just the resulting polynomial equation written in general form.The complex roots are necessarily even in their set. $\endgroup$
    – Piquito
    Feb 27 '16 at 17:01

If you write your equation in the form of $p(\theta)=\theta^m+a_{m-1}\theta^{m-1}+...+a_1\theta+a_0=0$, you get that $m=2n+1$ is an odd number and $a_i\in \mathbb R$. So, $p(\theta)$ is a polynomial with real coefficient by odd degree. Thus, $p(\theta)$ has at least one real root. But if $z\in \mathbb C$ is a root of a polynomial with real, then $\overline z$ is to. So the number of complex root of any polynomial with real coefficient, is even. From here, the number of real root of a polynomial with real coefficient, is odd.

  • $\begingroup$ Thanks, that makes sense. If I could you another question, under what conditions does a third degree polynomial have two real solutions? Is that even possible? $\endgroup$
    – JohnK
    Feb 27 '16 at 16:57
  • $\begingroup$ Your welcom. If one of this roots is by repeated order 2. It is mean if $p(x)=a(x-\lambda_1)^2(x-\lambda_2)$ for some $a,\lambda_1,\lambda_2\in \mathbb R$. $\endgroup$ Feb 27 '16 at 17:05
  • 1
    $\begingroup$ Thanks again for your help. $\endgroup$
    – JohnK
    Feb 27 '16 at 17:08
  • $\begingroup$ " under what conditions does a third degree polynomial have two real solutions?" if z is a solution, so it $\overline{z}$ so every polynomial will have an even number of of non-real solutions. So if a third degree polynomial has two real solutions, it can have at most one non-real solution which means it must have zero non-real solutions so it must have three real solutions. So to have two real roots it must have one double root (such as $(x-2)^2(3x + 1)$). One can quibble (I would) that a double root is really two roots so this doesn't count. But it's the only possibility. So there.... $\endgroup$
    – fleablood
    Feb 28 '16 at 17:56

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