Asymptotics of Hermite and hypergeometric function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:59:17Z http://mathoverflow.net/feeds/question/35648 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35648/asymptotics-of-hermite-and-hypergeometric-function Asymptotics of Hermite and hypergeometric function mr.gondolier 2010-08-15T12:32:23Z 2010-09-12T14:22:18Z <p>I am looking for the asymptotics of the following integral</p> <blockquote> <p>$\int_{\mathbb{R}} H_m^2(x) {\rm e}^{-2 \alpha^2 x^2} {\rm d} x = 2^{m-1/2} \alpha^{-2m -1} (1-2\alpha^2)^m \ \Gamma(m+1/2) ~ _2F_1\left(-m,m,1/2-m,\frac{\alpha^2}{2\alpha^2-1}\right)$</p> </blockquote> <p>where $H_m$ is the $m^{\rm th}$ Hermite polynomial (orthogonal under the weight ${\rm e}^{-x^2}$), and $_2F_1$ is the hypergeometric function.</p> <p>I found this formula from p. 803 of "Table of Integrals, Series, and Products" by Gradshteyn-Ryzhik. However, I have idea about the asymptotics of the $_2F_1$ term. Can anyone enlighten me on the asymptotics of</p> <blockquote> <p>$_2F_1\left(-m,m,1/2-m,\beta\right)$</p> </blockquote> <p>when $m$ is large? In fact I tried mathematica and it seems $_2F_1\left(-m,m,1/2-m,\beta\right) \sim |4 \beta|^m$. Any reference on this issue?</p> <p>Now given the above asymptotics is true, observe that the norm of $H_m$ under the weight ${\rm e}^{-2 \alpha^2 x^2}$ has the same exponent for all $alpha$, including the original weight ($\alpha^2 = 1/2$). Is this a common phenomenon for orthogonal polynomials?</p> http://mathoverflow.net/questions/35648/asymptotics-of-hermite-and-hypergeometric-function/35656#35656 Answer by J. M. for Asymptotics of Hermite and hypergeometric function J. M. 2010-08-15T13:52:42Z 2010-08-15T13:52:42Z <p>I suppose I should be more precise than what I wrote in my comment: as already mentioned, whenever one of the "numerator parameters" of a hypergeometric function is a negative integer -m, the series terminates (the Pochhammer symbols in the terms of degree higher than m vanish), i.e. your hypergeometric function becomes a polynomial.</p> <p>Using this, the degree m term of the polynomial is</p> <p>$\frac{(-m)_m (m)_m}{\left(\frac1{2}-m\right)_m}\frac{\beta^m}{m!}$</p> <p>or, using the properties of the Pochhammer symbols and the factorial:</p> <p>$2^{2m-1}\beta^m$</p> <p>Replacing $\beta$ with $(2-\alpha^{-2})^{-1}$ and multiplying by the factors in front of the hypergeometric expression nets you</p> <p>$\frac{(-1)^m 2^{3m-\frac{3}{2}}\Gamma\left(\frac1{2}+m\right)}{\alpha}$</p> <p>What complicates things, however, are the factors in front of the hypergeometric function, which when expanded is a polynomial in odd powers of the variable $\alpha^{-1}$.</p> <p>It's a bit late here, so I suppose I'll let someone else finish the analysis...</p>