Polynomial interpolation whose roots are real and simple - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:56:26Z http://mathoverflow.net/feeds/question/111492 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111492/polynomial-interpolation-whose-roots-are-real-and-simple Polynomial interpolation whose roots are real and simple driss-alamilouati 2012-11-04T22:42:04Z 2012-11-05T00:50:42Z <p>Let $\lambda_1,\ldots,\lambda_m$ real numbers pairwise distinct and $\mu_1,\ldots,\mu_m$ real numbers all nonzero. We know from the Lagrange polynomial interpolation that there exists an unique polynomial $R$ of degree less than $(m+1)$ such that $R(\lambda_i)=\mu_i,1\leq i\leq m$. We can prove by the means of an adequate use of the intermediate value Theorem, the existence of a real $\alpha$ such that the polynomial $P_{\alpha}(x)=R(x)+(x-\alpha)(x-\lambda_1)\ldots(x-\lambda_m)$ admits $(m+1)$ real roots pairwise distinct.</p> <p>I wonder if it is possible to generalize this property. Precisely, let $1\leq r\leq m$, we denote by $Q$ the polynomial of degree less than $(m+r+1)$ such that $Q(\lambda_i)=\mu_i,1\leq i\leq m$ and $Q'(\lambda_i)=0,1\leq i\leq r$. Is it possible to find $(\alpha,\beta)\in\mathbb{R}^2$ such that the polynomial $$S_{\alpha,\beta}(x)=Q(x)+(x-\alpha)(x-\beta)\displaystyle\prod_{i=1}^{r}(x-\lambda_i)^2\displaystyle\prod_{i=r+1}^{m}(x-\lambda_i)$$ admits $(m+r+2)$ real roots pairwise distinct?</p> http://mathoverflow.net/questions/111492/polynomial-interpolation-whose-roots-are-real-and-simple/111499#111499 Answer by Alexandre Eremenko for Polynomial interpolation whose roots are real and simple Alexandre Eremenko 2012-11-05T00:46:35Z 2012-11-05T00:46:35Z <p>The answer is no. (Assuming that $Q$ is indeed the unique interpolation polynomial of degree $m-1$). Indeed, take all your $\mu_j=1$. then $Q$ will be identically equal to $1$. Then choose your other data $\lambda_j$ in such a way that the polynomial $$p(x)=\prod_{j=1}^r(x-\lambda_j)^2\prod_{j=r+1}^m(z-\lambda_j)$$ has the following shape: when you look at the graph from left to right, it has a series (say, more than 3) of multiple zeros, then a single simple zero, then a series of multiple zeros again, and a simple zero, and again a series of multiple zeros. All zeros are real. Than you see that on each series of multiple zeros, $p^{\prime\prime}(\lambda_j)$ has the same sign, and this sign alternates: one sign in the first series, opposite sign in the second series, and so on.</p> <p>Multiplying by a second degree factor $(x-\alpha)(x-\beta)$, will alter this signs pattern somewhat, but not too much. In any case some double root with POSITIVE second derivative will remain. This root will be a local minimum of $(x-\alpha)(x-\beta)p(x)$.</p> <p>Now it is evident that when you have an equation of the form $1+q(x)=0$, where $q$ is a polynomial with all roots real, and one of them $x_0$ is of multiplicity $2$, and the $p^{\prime\prime}(x_0)>0$, then this equation must have some non-real roots. Just graph on the real line to see it.</p>