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

Question

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.

Answer 1

This is to answer the question that the OP asked in the comment, about convergence of the Gauss-Laguerre quadrature when $f(x)=x\log x$.

Uspensky has shown that the Gauss-Laguerre quadrature rule for computing integrals of the form $$\int_{0}^{\infty}x^{\alpha}e^{-x}f(x)dx$$ converges, as $n\to\infty$, if $f$ satisfies the following growth at infinity, $$|f(x)|\leq c x^{-\alpha-1-\rho}e^{x},\qquad|x|\text{ large},$$ for some $\rho>0$, see

J. V. Uspenksy, On the convergence of quadrature formulas related to an infinite interval, Trans. Amer. Math. Soc. 30 (1928), 542-559.

Thus, the G-L quadrature indeed converges in the case of $f(x)=x\log x$. Actually, even the rate of convergence can be estimated.

The case where $f(x)$ is not smooth at the origin, as for $x\log x$, was studied in

Mastroianni G., Monegato G., Convergence of product integration rules over $(0,\infty)$ for functions with weak singularities at the origin. Math. Comp. 64, (1995), 237-249.

The following is proved. For $q\geq p\geq0$, let $$C _ { p } ^ { q } [ 0,\infty ) : = \{ f \in C ^ { p } [ 0,\infty ) \cap C ^ { q } ( 0,\infty ),~x ^ { i } f ^ { ( p + i ) } ( x ) \in C [ 0,\infty ),~i=1,\ldots,q-p\}.$$ Then, the remainder term $R_{n}(f)$ of the Gauss-Laguerre quadrature satisfies $$ |R_{n}(f)|=\begin{cases} \mathcal O(n^{-q/2}) E_{n-p-1}(\Phi^{(q)},e^{-x/2}) & \text{if }q\leq 2p+1, \\ \mathcal O(n^{-p-1}\log n) E_{n-p-1}(\Phi^{(q)},e^{-x/2}) & \text{if }q= 2p+2, \\ \mathcal O(n^{-p-1}) E_{n-p-1}(\Phi^{(q)},e^{-x/2}) & \text{if }q\geq 2p+3, \end{cases} $$ where $ \Phi ( x ) : = x ^ { q - p } f ( x ) \in C ^ { q } [ 0,\infty ) $, and $ E _ { n } ( f ; w ) : = \operatorname{inf} _ { p _ { n } } \| w \left[ f - p _ { n } \right] \| _ { \infty ,[ 0,\infty ) } $.

Hence, for the function $f(x)=x\log x\in C_{0}^{3}[0,\infty)$, $$\Phi(x)=x^{4}\log x,\qquad\Phi^{(3)}(x)=24x\log x+26x,$$ and thus $$|R_{n}(f)|=\mathcal O(n^{-1})E_{n-1}(x\log x, e^{-x/2}).$$ Since, it is also known that $E_{n-1}(x\log x, e^{-x/2})=\mathcal O(n^{-1})$ for $n$ large, one finally gets for the function $f(x)=x\log x$, $$|R_{n}(f)|=\mathcal O(n^{-2}), \qquad n\text{ large}.$$

Answer 2

I think you misunderstand the error term in gauss-laguerre quadrature !

that is $$E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}{f^{(2n)}},$$

see : http://mathworld.wolfram.com/Legendre-GaussQuadrature.html.

Hence 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， \begin{align*} \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}-......}\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}-......}\frac{x^{2n+1}}{(2n+1)!}=C_{2n}, \end{align*} and thus $$\frac{2}{\sum{C_{i}}-C_{1}-\frac{C_{1}+C_{2}}{1!}-\frac{C_{1}+C_{2}+C_{3}}{2!}-......}\frac{1}{(2n+1)!}\le\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3},$$

with $C_{2n}\to0$.