roots of sum of two polynomials - MathOverflow most recent 30 from http://mathoverflow.net2013-06-20T06:11:43Zhttp://mathoverflow.net/feeds/question/30072http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/30072/roots-of-sum-of-two-polynomialsroots of sum of two polynomialsvilvarin2010-06-30T16:32:32Z2010-07-01T17:35:58Z
<p>I believe that there is no common theory for finding roots of polynomial sum. In my case I have
$$P_{n}(x)+AQ_{n}(x)$$.
I am wondering how roots of this sum depend on $A$?</p>
http://mathoverflow.net/questions/30072/roots-of-sum-of-two-polynomials/30075#30075Answer by jc for roots of sum of two polynomialsjc 2010-06-30T17:01:56Z2010-07-01T17:35:58Z<p>Though in general you won't have a closed-form expression for the roots of your polynomials, it's possible to write down perturbation series for roots of a polynomial in a single variable in terms of the coefficients. These are basically the Puiseux series mentioned in <a href="http://mathoverflow.net/questions/15528/asymptotic-series-for-roots-of-polynomials" rel="nofollow">this question</a>.</p>
<p><a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.44.6531&rep=rep1&type=pdf" rel="nofollow">This paper by Bernd Sturmfels</a> (<a href="http://www.ams.org/mathscinet-getitem?mr=1731613" rel="nofollow">MR</a>) sketches out the "global picture" of such series, though it's fairly complicated and I personally am not clear on whether there's a simple algorithm to decide which is the proper choice of series that will converge. See also the article "Algebraic equations and hypergeometric series" by M Passare, A Tsikh in the book: <em>The Legacy of Niels Henrik Abel</em> (<a href="http://www.ams.org/mathscinet-getitem?mr=2077589" rel="nofollow">MR</a>).</p>
<p>What I've just written is probably a little unclear so I'll describe the simplest example. Suppose you'd like to write down a series for the roots of $a_2x^2+a_1x+a_0=0$. There are a pair of series which converges when $\left|\frac{a_1^2}{4a_0a_2}\right|<1$ and a pair which converges when $\left|\frac{a_1^2}{4a_0a_2}\right|>1$, and you can derive the first pair of series by treating $a_1x$ as a perturbation to the equation $a_2x^2+a_0=0$ and you can derive one of the second pair of series by treating $a_2x^2$ as a perturbation to $a_1x+a_0=0$ and the other by treating $a_0$ as a perturbation to the equation $a_2x^2+a_1x=0$.</p>
<p>By plugging in the coefficients of $P_n(x)+AQ_n(x)$ into the appropriate series I just described and looking at the leading order terms as functions of $A$, you will be able to derive the scaling of the corrections to the roots of $P_n(x)$.</p>
<p>Apologies for the rather unexplicit answer, but this is just at the limit of what I understand. </p>
http://mathoverflow.net/questions/30072/roots-of-sum-of-two-polynomials/30086#30086Answer by SandeepJ for roots of sum of two polynomialsSandeepJ2010-06-30T18:13:32Z2010-06-30T18:13:32Z<p>If they are complex polynomials or can be treated as such, then you could apply <a href="http://en.wikipedia.org/wiki/Rouch%25C3%25A9%2527s_theorem" rel="nofollow">Rouche's theorem</a>, where the location of the zeros is determined by the dominant polynomial within the sum. (<em>"Walk the dog on the leash"</em>)</p>
<p>Possibly related: you could use the <a href="http://en.wikipedia.org/wiki/Wronskian" rel="nofollow">Wronskian</a> to determine the values of A that make $P_n(x)$ and $Q_n(x)$ linearly independent.</p>
<p>Your question is related to <a href="http://mathworld.wolfram.com/MasonsTheorem.html" rel="nofollow">Mason's theorem</a>. There are a few papers which explore this specifically</p>
<ol>
<li><em>MR1923392 (2003j:30012) Kim, Seon-Hong . Factorization of sums of
polynomials. Acta Appl. Math. 73
(2002), no. 3, 275--284.</em></li>
<li><em>MR2103113 (2005h:30011) Kim,
Seon-Hong . On zeros of certain sums
of polynomials. Bull. Korean Math.
Soc. 41 (2004), no. 4, 641--646</em></li>
<li><em>MR1911767 (2003d:11036) Pintér, Á.
Zeros of the sum of polynomials. J.
Math. Anal. Appl. 270 (2002), no.
1, 303--305.</em></li>
</ol>