Timeline for Showing that a family of polynomials has positive and real roots.
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2012 at 16:24 | history | edited | john mangual | CC BY-SA 3.0 |
formula was incorrect
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| Jul 17, 2012 at 6:51 | comment | added | Enzo | Sorry the integral should be $$\int_{\rho}\frac{z+1}{(z^2+1+2z(x-1))(z-1)^k}dz$$ | |
| Jul 17, 2012 at 6:48 | comment | added | Enzo | 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. | |
| Jul 16, 2012 at 23:43 | history | answered | john mangual | CC BY-SA 3.0 |