real roots for polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:35:34Z http://mathoverflow.net/feeds/question/70609 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70609/real-roots-for-polynomials real roots for polynomials mosen 2011-07-18T10:59:40Z 2011-07-18T15:45:41Z <p>Is there any necessary and sufficinet condition for a complex polynomial to have a real root?</p> <p>A complex polynomial has a real root if and only if...? </p> http://mathoverflow.net/questions/70609/real-roots-for-polynomials/70613#70613 Answer by mathphysicist for real roots for polynomials mathphysicist 2011-07-18T11:51:11Z 2011-07-18T11:51:11Z <p>This is just a bit too long for a comment :)</p> <p>Let $P$ be your polynomial and $x$ its real root. Obviously, $P(x)=0$ if and only if $$Q(x)\equiv (\mathrm{Re} P(x))^2+(\mathrm{Im} P(x))^2=0.$$ Now, $Q(x)$ is a polynomial with <em>real</em> coefficients, which reduces your question to finding criteria for a <em>real-coefficients</em> polynomial to have a real root, and these are discussed <a href="http://mathoverflow.net/questions/20946/criteria-to-determine-whether-a-real-coefficient-polynomial-has-real-root" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/70609/real-roots-for-polynomials/70614#70614 Answer by Denis Serre for real roots for polynomials Denis Serre 2011-07-18T12:26:50Z 2011-07-18T12:26:50Z <p>Yes. If your polynomial is not yet real, replace $P$ by $P\bar P$ ($\bar P$ has complex conjugated coefficients). Therefore we may suppose that $P\in{\mathbb R}[X]$.Using the Euclid algorithm, you may find the g.c.d of $P$ and $P'$. Dividing $P$ by this g.c.d, your are left with the case where $P$ is real and has simple roots.</p> <p>Now, you use te Euclid algorithm : $P_0=P$, $P_1=P'$ and $P_{k-1}=Q_kP_k-P_{k+1}$. The sequence $(P_k)_k$ ends with a constant polynomial. Take $a>0$ large enough that $P$ may not have a root in $[a,+\infty)$. Let $V(a)$ be the number of sign changes in the sequence $(P_k(a))_k$. Likewise, take $b&lt;0$ such that $P$ has no root in $(-\infty,b)$ and compute $V(b)$. </p> <p>Theorem : the number of real roots of $P$ equals $V(b)-V(a)$.</p> http://mathoverflow.net/questions/70609/real-roots-for-polynomials/70634#70634 Answer by mosen for real roots for polynomials mosen 2011-07-18T15:45:41Z 2011-07-18T15:45:41Z <p>since, $P$ has a real root if and only if $p\bar{p}$ (Denis introduced this above) has. $p\bar{p}$ is a real polynomial and one can use Tarski's theorem.</p>