# Is there an example where the error of Gauss-Laguerre quadrature does not vanish?

The $n$th Gauss-Laguerre quadrature aims to approximate integral $$\int_{\mathbb{R}_+} f(x) \exp(-x)$$ by the sum $$\sum_{i=1}^n f(x_i) w_i$$ where $x_1,...,x_n$ are the roots of the $n$th Laguerre polynomial $L_n$, and the weights $w_1,...,w_n$ are chosen according to $w_i=\frac{1}{x_i (L_n'(x_i))^2}$.

The intuition is that if $f$ is polynomial of degree at most $2n-1$ then the approximation is exact; In general, the approximation error (as in any Gauss quadrature) is known to be given by $$E_n(f) = \frac{(n!)^2}{(2n)!} f^{(2n)}(\xi)$$ for some $\xi \in (0,\infty)$. My question is simply the following:

Is there function $f$ that is smooth in $(0,\infty)$, such that the the approximation error $E_n(f)$ does not go to zero as $n\to \infty$?

My feeling is that if $f$ derivative grows very fast, or, say, $f$ has infinite derivative at the end point 0, e.g., $f(x)=x \log x$, then maybe $\xi$ is very close to zero and the error will not improve as $n$ grows, but I am no expert in this field. Any comment will be welcomed.

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You need some extra condition to rule out placing bumps at the largest roots of the Laguerre polynomials. – Douglas Zare Mar 24 '14 at 6:37
Douglas, I meant for a fixed function $f$ and let $n$ goes to infinity. So $f$ cannot depend on $n$. – mr.gondolier Mar 24 '14 at 7:24
The Runge function $x\mapsto \frac{1}{1+25x^2}$ is the standard example of hard-to-approximate function for the Gauss-Legendre quadrature in the interval $[-1,1]$ (which is exact for integrating polynomials of degree up to $2n-1$). My first attempt would be trying a change of variable of some form to take that interval into $[0,\infty]$ and Legendre into Laguerre, but I am not an expert in quadrature myself. – Federico Poloni Mar 24 '14 at 7:57
I didn't suggest that $f$ should depend on $n$. I meant that you can easily construct $f$ by placing infinitely many bumps, one at the largest root of the $n$th polynomial, so that the $n$th degree Gauss-Laguerre quadrature estimate is off by $1$, for example. Since the largest roots have no limit point you can put a smooth bump there to adjust the $n$th degree estimate arbitrarily without affecting lower estimates. – Douglas Zare Mar 24 '14 at 16:10
@FedericoPoloni, The Runge function seems to be an example for equidistant case instead of Gauss quadrature. – mr.gondolier Mar 24 '14 at 21:11

that is : $E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\cdot{f^{(2n)}} so the condition is : \frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\ge$$\frac{1}{\epsilon} where, \epsilon=f^{(2n+1)}=C_{2n} is the first coefficient ! so ， \Longrightarrow$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}\ge$$\frac{1}{f(x)-f(a)-\frac{f^{(1)}{}(a)}{1!}(x-a)-\frac{f^{(2)}{(a)}}{2!}(x-a)^{2}-......}$$\cdot$$\frac{(x-a)^{2n+1}}{(2n+1)!}=$$\frac{2}{f(x)-C_{1}-\frac{C_{1}+C_{2}}{1!}x-\frac{C_{1}+C_{2}+C_{3}}{2!}x^{2}-......}$$\cdot$$\frac{x^{2n+1}}{(2n+1)!}$$=C_{2n} \Longrightarrow$$\frac{2}{\sum{C_{i}}-C_{1}-\frac{C_{1}+C_{2}}{1!}-\frac{C_{1}+C_{2}+C_{3}}{2!}-......}$$\cdot$$\frac{1}{(2n+1)!}$$\le$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$
with$,C_{2n}\longrightarrow0$