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Using Hermite-Gauss Quadrates to approximate the integral $I = \int_0^\infty e^{ -x^2} f(x) \, dx $, the error is given as $$ E = \frac{m!\sqrt{\pi}}{2^m (2m)!} f^{(2m)}(\theta) $$ with $0 < \theta < \infty $. However, this estimation is not practical since in some cases we cant know $f^{(2m)}(\theta)$. So are there any ways to determine that error or bounds of that one?

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  • $\begingroup$ I changed {f^{{\left(2m\right)}}} to f^{(2m)} and large numbers of things like {{{x}}{{{y}}}} to xy, etc. $\endgroup$ – Michael Hardy Feb 8 '13 at 19:02
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Many times we know that $|f^{(2 m)}(\theta)|$ is bounded. For example, $f(x) = \sin{(a x)}$ implies that $|f^{(2 m)}(\theta)| < a^{2 m}$. In this case, the error is bounded and we may say that the error in using Hermite-Gauss quadrature to this order is less than some number, which is valuable.

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