2
$\begingroup$

I've recently been looking into optimisation and have found it quite fascinating. During my readings, I came across a proof for the fundamental theorem of algebra.

I follow the proof for the most part, but I get lost at one step. The proof takes a polynomial of degree $n$, denoted as $p(z)$ where $z$ is complex and shows that $\left|p(z)\right|$ is coercive and thus at least one minimiser $\hat{x}$ exists.

What I do not follow is the proof then saying that we can assume $\hat{x}=0$ without loss of generality and thus show that $p(\hat{x})=0$. Could someone please explain the justification behind this step?

$\endgroup$
  • $\begingroup$ If $\hat{x} \neq 0$, consider the polynomial $p(z- \hat{x})$. $0$ is a minimizer for $|p(z- \hat{x})|$, and so if you prove that $0$ is a zero of $p(z- \hat{x})$, then $\hat{x}$ is a zero of $p$. $\endgroup$ – Crostul Aug 13 '16 at 13:12
1
$\begingroup$

If $\hat{x}$ is a non-zero number $a$, then the polynomial $q(z):=p(z+a)$ reaches the minimum at $0$ (thus $\hat{x}=0$ for $q$).

Now, if you prove that $q(0)=0$, then you have $p(a)=0$ and replacing $p$ by $q$ doesn't change the nature of the problem.

$\endgroup$
  • $\begingroup$ Thank you very much for that! It sorted out my issue completely. $\endgroup$ – vs9734 Aug 13 '16 at 14:25

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.