I wonder if there is anything known (formula, asymptotics, etc) of computing the remainder

$R_{k,m} \equiv H_{k} ~ \mod H_m$

for $k > m$, where $H_m$ denotes the $m$th Hermite polynomial (orthogonal under the weight $w(x) = e^{-x^2}$) and $\deg R_{k,m} \leq m-1$. I haven't been able to find anything online, neither could compute it through the recurrence relation of Hermite polynomials...


The motivation for my question is as follows. The $m$-point Gauss-quadrature is obtained by placing the nodes at the roots of $H_m$ and choosing the weights accordingly such that integrating any polynomial (with respect to weight $w$) of order $\leq 2m-1$ is exact. Now I want to know the error formula for polynomials of degree $k \geq m$, especially $H_k$. By computing $H_k$ modulo $H_m$, the integration error is given by the integration of the remainder $R_{k,m}$.

  • $\begingroup$ Why do you want to know this? And why ask this question about the Hermite polynomials? The thing is that Hermite polynomials have a lot of special properties, of which some they share with other polynomials / functions, but always only some. So it might be interesting which special properties you are interested in the proof using. $\endgroup$
    – Helge
    Aug 16, 2010 at 11:36
  • $\begingroup$ OK. I have added the motivation in the original post. $\endgroup$
    – gondolier
    Aug 16, 2010 at 11:42
  • $\begingroup$ math.ucla.edu/~cbm/aands//page_890.htm , particularly the formula in the upper right of that page. $\endgroup$ Aug 16, 2010 at 11:59
  • $\begingroup$ i am aware of this estimate. it does not help in this situation for $k > 2m$, because the derivative of $H_k$ is unbounded. Yes it is tight when $k=2m$ but in that case the integral can be easily calculated... $\endgroup$
    – gondolier
    Aug 16, 2010 at 12:25
  • $\begingroup$ Wait... so what you want to do is to calculate the integral of $\exp\left(-x^2\right)H_K(x)$ with an m-point Gauss-Hermite formula? I don't understand why you wish to do this; it's not that hard to generate the nodes and weights of the Gauss-Hermite quadrature formula these days (e.g. via Golub-Welsch). $\endgroup$ Aug 16, 2010 at 13:50

1 Answer 1


Here is how to obtain a formula for $R_{k,m}$. Denote by $x_1, \dots, x_m$ the zeros of $H_m$. We have that $$ H_k(x) = P(x) H_{m}(x) + R_{k,m}(x) $$ for all $x$, so in particular for $x_j$ that $$ H_k(x_j) = R_{k,m}(x_j) $$ since $H_m(x_j) = 0$. So we know $R_{k,m}(x_j)$ at $m$ points, and it is a polynomial of degree $m-1$, so it is uniquely determined. In particular, one could write down a formula for it using some interpolation method.

I guess to understand this gadget, one would need to understand $H_k(x_j)$, which my first guess would be to check the already linked Abramowitz--Stegun book...


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