Remainder term for Gauss-Laguerre quadrature

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I need to calculate a quadrature rule with maximum degree of accuracy that looks like this:

$$ \int_0^\infty e^{-x}f(x)dx = \sum_{i=0}^n A_if(x_i) + R_n(f) $$

where $n=2$.

For $R_n(f)$ I have this formula:

$$ R_n(f) = \frac{f^{(2n)}(\xi)}{(2n)!} \, (\pi_n,\pi_n) $$

I've already calculated $A_n$ and $x_n$ and I already know that $\pi_n$ is a Laguerre polynomial.

As joriki pointed out, Laguerre polynomials are orthogonal so I'm left with this:

$$ R_2(f) = \frac{f^{(2n)}(\xi)}{24} $$

My question is: how do I choose $\xi$? Or do I just leave it like that?

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So, to answer my own question, you're not supposed to calculate $\xi$. The remainder term is just a way to gauge how big the approximation error is.

For example, if $f(x)=cos(x)$, the remainder is:

$$ R = \frac{cos(\xi)}{24} $$

which is not so bad, since $cos(x) \in [-1, 1]$.