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Let $L_k(t)$ be the Legendre polynomials normalized so that $$\int_{-1}^1 L_k(t)^2\,\frac{1}{2}\,dt = 1.$$ With a few identities (http://en.wikipedia.org/wiki/Legendre_polynomials), one can show that $$ \int_{-1}^1 L_k'(t)\,\frac{1}{2}\,dt \;=\; \left\{ \begin{array}{ll} 0 & \mbox{if $k$ is even,}\\ \sqrt{2k+1} & \mbox{if $k$ is odd,} \end{array} \right. $$ where the prime notation means "derivative." I'm wondering how general this result might be, and if anyone has experience with this? I think the average derivative of the Hermite polynomials are all zero. I wonder if all polynomials orthogonal to a symmetric weight function on a bounded interval might behave like the Legendre polynomials, i.e., 0 if $k$ is even and $\mathcal{O}(\sqrt{k})$ if $k$ is odd.

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  • $\begingroup$ try to check this link mayeb help you to get the answer of your problem :mathworks.com/matlabcentral/fileexchange/… $\endgroup$ – salimmath15 May 24 '15 at 0:07
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    $\begingroup$ The average of the derivative $P'$ of any polynomial $P$ on $(a,b)$ is $(P(b)-P(a))/(b-a)$. So for example if $P$ is the Čebyšev polynomial $T_k$ then we get $0$ or $1$ according as $k$ is even or odd. Not sure what you're claiming about the Hermite polynomials: their interval of orthogonality is $\bf R$, on which the average is a divergent integral for $k>0$. $\endgroup$ – Noam D. Elkies May 24 '15 at 0:48
  • $\begingroup$ Thanks, @NoamD.Elkies. The Chebyshev polynomials are a nice counterexample. For the Hermite polynomials, I mean averaged against a standard Gaussian weight function, like you would define their orthogonality. I think you could use the relationship you mention to prove my about zero average derivatives. $\endgroup$ – Paul Constantine May 25 '15 at 13:04

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