Polynomials having a common root with their derivatives - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:54:27Z http://mathoverflow.net/feeds/question/27851 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27851/polynomials-having-a-common-root-with-their-derivatives Polynomials having a common root with their derivatives algori 2010-06-11T18:52:37Z 2011-05-28T22:50:01Z <p>Here is a question someone asked me a couple of years ago. I remember having spent a day or two thinking about it but did not manage to solve it. This may be an open problem, in which case I'd be interested to know the status of it.</p> <p>Let \$f\$ be a one variable complex polynomial. Supposing \$f\$ has a common root with every \$f^{(i)},i=1,\ldots,\deg f-1\$, does it follow that \$f\$ is a power of a degree 1 polynomial?</p> <p>upd: as pointed out by Pedro, this is indeed a conjecture (which makes me feel less badly about not being able to do it). But still the question about its status remains.</p> http://mathoverflow.net/questions/27851/polynomials-having-a-common-root-with-their-derivatives/27859#27859 Answer by Pedro Teixeira for Polynomials having a common root with their derivatives Pedro Teixeira 2010-06-11T19:43:49Z 2010-06-11T19:43:49Z <p>That is known as the Casas-Alvero conjecture. Check this out, for instance:</p> <p><a href="http://front.math.ucdavis.edu/0605.5090" rel="nofollow">http://front.math.ucdavis.edu/0605.5090</a></p> <p>Not sure of its current status, though.</p> http://mathoverflow.net/questions/27851/polynomials-having-a-common-root-with-their-derivatives/27865#27865 Answer by Andrey Rekalo for Polynomials having a common root with their derivatives Andrey Rekalo 2010-06-11T20:27:25Z 2010-06-12T12:22:02Z <p>The strongest result in this direction that I've heard of is Sudbery's theorem (which was originally conjectured by Popoviciu and Erdös).</p> <blockquote> <p><strong>Theorem.</strong> Let \$P(z)\$ be a polynomial of degree \$n\geq 2\$ and let \$\Pi(z)=\prod\limits_{k=0}^{n-1}P^{(k)}(z)\$ where \$P^{(k)}\$ is the \$k\$th derivative of \$P\$. Then either \$\Pi(z)\$ has exactly one distinct root or \$\Pi(z)\$ has at least \$n+1\$ distinct roots.</p> </blockquote> <p>See the <a href="http://blms.oxfordjournals.org/cgi/reprint/5/1/13" rel="nofollow">original paper by Sudbery</a>.</p>