User enzo - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:07:47Z http://mathoverflow.net/feeds/user/25157 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102384/showing-that-a-family-of-polynomials-has-positive-and-real-roots Showing that a family of polynomials has positive and real roots. Enzo 2012-07-16T19:56:50Z 2012-08-01T03:43:10Z <p>Hi everybody, for my research I am dealing with the following function: </p> <p>$$\alpha_n(x):=\left.\frac{\partial^{2n+1}}{\partial z^{2n+1}}\frac{\sinh(z)}{\cosh(z)-1+x}\right|_{z=0},\quad n\in \mathbb{N},$$</p> <p>It is possible to show that $$\alpha_n(x)=\frac{P_n(x)}{x^{n+1}},$$ where $P_n(\cdot)$ is a polynomial of order $n$ in $x$, having integer coeffients.</p> <p>To make few concrete examples $$\alpha_0(x)=\frac{1}{x}$$ $$\alpha_1(x)=\frac{-3+x}{x^2}$$ $$\alpha_2(x)=\frac{30-15 x+x^2}{x^3}$$ $$\alpha_3(x)=\frac{-630+420 x-63 x^2+x^3}{x^4}$$ $$\alpha_4(x)=\frac{22680-18900 x+4410 x^2-255 x^3+x^4}{x^5}$$ and so on.</p> <p>What I would need to show (and it is veryfied for all the special cases I was able to compute, like those above) is that all the roots of $P_n(x)$ (and therefore those of $\alpha_n(x)$) are real and strictly greater than 2. </p> <p>An explicit albeit complicated expression for $\alpha_n(x)$ can be obtained, namely:</p> <p><code>\begin{equation*} \begin{array}{ll} \alpha_n(x)=&amp;x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j+1}^{n} (2k)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k}\sum_{i=j}^{k}\left[ {i+1\choose j+1}\binom{2k}{2 i+1}-{i\choose j+1}\binom{2k}{2 i}\right](-2)^{j+1-2k}+\\ &amp;x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j}^{n}{(2k+1)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k+1}}\sum_{i=j}^{k} \left[ {i+1\choose j+1}\binom{2k+1}{2 i+1}-{i\choose j+1}\binom{2k+1}{2 i}\right](-2)^{j-2k}, \end{array} \end{equation*}</code></p> <p>where the number between the curly brakets are the Stirling number of the second kind; Moreover, </p> <p>$$\alpha_n(2)=-\sum_{k=1}^{2n+1}{k!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{k}}(-2)^{-k}\neq0.$$</p> <p>If someone is interested, I can post more on how I got these expressions. Thanks in advance to everybody that will try to help me.</p> <p>Best Regards </p> <p>Enzo</p> http://mathoverflow.net/questions/102384/showing-that-a-family-of-polynomials-has-positive-and-real-roots/102403#102403 Comment by Enzo Enzo 2012-07-17T06:51:43Z 2012-07-17T06:51:43Z Sorry the integral should be $$\int_{\rho}\frac{z+1}{(z^2+1+2z(x-1))(z-1)^k}dz$$ http://mathoverflow.net/questions/102384/showing-that-a-family-of-polynomials-has-positive-and-real-roots/102403#102403 Comment by Enzo Enzo 2012-07-17T06:48:32Z 2012-07-17T06:48:32Z Hi John, that is how I got the explicit formula. Since $$f(z,x)=\left.\frac{\sinh(z)}{\cosh(z)-1+x}=\frac{(y+1)(y-1)}{y^2+1+2y(x-1)}\right|_{y=e^z}$$ one can use the Faa di Bruno's formula $$\alpha_n(x)=\sum_{k=1}^{2n+1}k!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{k}\frac{1}{2\pi i}\int_\rho{z+1}{(z^2+1+2z(x-1))(z-1)^{k}}$$ where $\rho$ is a closed curve encircling $z=1$. Then the integral can be obtained by Cauchy Residue formula. http://mathoverflow.net/questions/102384/showing-that-a-family-of-polynomials-has-positive-and-real-roots/102404#102404 Comment by Enzo Enzo 2012-07-17T06:42:28Z 2012-07-17T06:42:28Z Hi again David. Nice suggestion! I will check the details, but it seems to be right. Regards, Enzo http://mathoverflow.net/questions/102384/showing-that-a-family-of-polynomials-has-positive-and-real-roots Comment by Enzo Enzo 2012-07-16T21:43:40Z 2012-07-16T21:43:40Z Hi David, thanks for the comment. I have noticed that as well, and it reminded me the proof that the Hermite polynomial have real roots. Unfortunately, the same procedure does not seems to work here, since we have that the derivative is taken in $z$ and the roots are in $x$.