# Error in Gauss-Laguerre numerical quadrature scheme

The $$n$$-th Gauss-Laguerre quadrature scheme aims to approximate integral of exponentially decreassing function over $$[0 ; \infty[$$ by a finite sum, according to: $$\int _0 ^{+ \infty} e^{-x} f(x) \ dx \approx \sum _{i = 1} ^n \omega_i f(x_i) \ ,$$ where $$x_1$$, $$\cdots$$, $$x_n$$ are the roots of the $$n$$-th Laguerre polynomial $$L_n$$ and the weights $$\omega_1$$, $$\cdots$$, $$\omega_n$$ are chosen according to $$\omega_i = \dfrac{1}{x_i \big ( L_n'(x_i) \big )^2}$$, $$1 \leq i \leq n$$.

$$\bullet$$ Let us denotes by $$E_n(f)$$ the error of the quadrature: $$E_n(f) = \displaystyle { \int _0 ^{+ \infty} e^{-x} f(x) \ dx } - \displaystyle { \sum _{i = 1} ^n \omega_i f(x_i) }$$. The general estimation of the error of gaussian quadratures, specialized in Gauss-Laguerre scheme, is the following:

For all $$n \in \mathbb{N}$$, there exists $$\xi \in ]0 ; \infty[$$ such that $$E_n(f) = \dfrac{n!^2}{(2n)!} f^{(2n)}(\xi)$$.

Unfortunately, this is unusable in many case, since we know nothing on this $$\xi$$. Therefore, we shall consider functions $$f$$ with derivatives satisfyng $$||f^{2n}||_{\infty, \mathbb{R}^+} = \mathcal{o}\left ( \dfrac{1}{n!^2} \right )$$... This is a bit restrictive...

$$\bullet$$ We also know the Uspensky theorem:

$$E_n(f) \underset{n \longrightarrow + \infty}{\longrightarrow} 0$$ for functions $$f$$ satisfying $$|f(x)| \leq c \dfrac{e^x}{x^{1 + \rho}}$$ for large $$x >> 1$$, where 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)

But, this do not gives explicit result on the convergence speed to $$0$$, nor a upper bound of $$E_n(f)$$ going to $$0$$.

$$\bullet$$ So, what is the most precise result about the error term? In particular, how can we know which degree $$n$$ should we use to find a numerical approximation of $$\displaystyle { \int _0 ^{+ \infty} e^{-x + \sqrt{x}} \ dx}$$.

• Maybe this helps ? Mar 15, 2021 at 17:49
• Unfortunately, not really... I was aware of the result of Mastroianni and Monegato, but I'm not able to use it... Especially to compute the infinite norm $E_n(f, \omega)$... Mar 16, 2021 at 14:51

The function $$f(x)=e^{-x+\sqrt{x}}$$ belongs to the space $$C_{0}^{3}[0,\infty)$$ defined, for $$q\geq p\geq0$$, by $$C _ { p } ^ { q } [ 0,\infty ) : = \{ f \in C ^ { p } [ 0,\infty ) \cap C ^ { q } ( 0,\infty ),~x ^ { i } f ^ { ( p + i ) } ( x ) \in C [ 0,\infty ),~i=1,\ldots,q-p\}.$$ According to the result given in , the error rate behaves like $$\begin{equation}\label{rate-quad} \mathcal{O}(n^{-1})E_{n-1}(\Phi^{(3)}(x),e^{-x/2})=\mathcal{O}(n^{-1})E_{n-1}(\Phi^{(3)}(2x),e^{-x}), \end{equation}$$ where $$\Phi ( x ) : = x ^ { 3} f ( x )=x^{3}e^{-x+\sqrt{x}}$$, and $$E _ { n } ( f ; w )$$ is the rate of weighted polynomial approximation, $$E _ { n } ( f ; w ) : = \inf_ { p _ { n } } \| w (f - p _ { n })\| _ { \infty ,[ 0,\infty ) }.$$ Here, $$\Phi^{(3)}(x)=P_{6}(\sqrt{x})e^{-x+\sqrt{x}}$$, where $$P_{6}$$ is a polynomial of degree $$6$$.

Next, the rate of approximation $$E_{n}(f,w)$$, $$w(x)=e^{-x}$$, is known from  p.112, available here, namely $$\begin{equation}\label{rate-En} E_{n}(f,e^{-x})\leq C\frac{\sqrt{a_{n}}}{n}\left\|\sqrt{x}f'(x)e^{-x}\right\|_{\infty}, \end{equation}$$ for $$f\in W_{1}^{\infty}(e^{-x})$$, where $$a_{n}\sim n$$ is the so-called Mhaskar-Rakhmanov-Saff number, and, for $$r\geq1$$, $$W_{r}^{\infty}(e^{-x})$$ is the Sobolev-type space, $$W_{r}^{\infty}(e^{-x})=\left\{f \in L_{e^{-x}}^{\infty}, f^{(r-1)}\text{ abs. continuous on }(0,\infty)\text { and }\left\|f^{(r)}(x) x^{r/2} e^{-x}\right\|_{\infty}<\infty\right\},$$ $$L_{e^{-x}}^{\infty}=\{f\in C((0,\infty)),~\lim_{x\to0,~x\to\infty} f(x)e^{-x}=0\}.$$ Note that the function $$f(x)=e^{-2x+\sqrt{2x}}$$ belongs to $$W_{1}^{\infty}(e^{-x})$$, but not to any of the more regular spaces $$W_{r}^{\infty}(e^{-x})$$, $$r\geq2$$ (for which the rate of approximation is improved to $$(\sqrt{a_{n}}/n)^{r}$$).

Finally, taking into account the two previous estimates, we get $$\mathcal{O}(n^{-3/2})$$ for the error rate of the Gauss-Laguerre quadrature for $$f(x)=e^{-x+\sqrt{x}}$$.

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

 G. Mastroianni, J. Szabados, Polynomial approximation on the real semiaxis with generalized Laguerre weights. Stud. Univ. Babes-Bolyai Math. 52 (2007), nr.4, 105--128.

• Thanks you for your developped answer. Just for completness, the n-th Mhaskar-Rakhmanov-Saff number $a_n$ associated to the weight $w(x)=e^{-Q(x)}$ is the positive root of the equation $$n=\dfrac{2}{\pi} \displaystyle{\int_0^1\dfrac{a_n tQ'(a_n t)}{\sqrt{1-t^2}}dt}$$ If $Q(x) = |x|^\alpha$, $a_n = \left ( 2^{\alpha - 2} \dfrac{\Gamma(\frac{\alpha}{2})^2}{\Gamma(\alpha)} \right )^{\frac{1}{\alpha}} n^{\frac{1}{\alpha}}$. Here, $\alpha = 1$ and $a_n = \dfrac{\pi}{2}n$. See DS Lubinsky, A Survey of Weighted Approximation for Exponential Weights (arxiv.org/abs/math/0701099), p.11 May 11, 2021 at 13:43
• Is the underlying constant of the $\mathcal{O}(n^{-1})$ obtain by applying the result of  can be explicitely computed or estimated ? Is the constant C used in Proposition 4.1 of  can be explicitely computed or estimated? May 11, 2021 at 13:53
• Is the space $L^{\infty}_{e^{-x}}$ well-defined? Shoudn't it be $$L^{\infty}_{e^{-x}} = \{ f \in \mathcal{C}((0,∞)), \underset{x \longrightarrow + \infty}{\text{lim}} f(x) \cdot e^{-x} = 0\} \ ?$$ Otherwise, the fonction $x \longmapsto \Phi^{(3)}(2x)$ do not belong to $W_1^{\infty}(e^{-x})$, according to $\Phi^{(3)}(0) = 6$.  I suspect that your definition is actually a valid one for generalized Laguerre weight (see , p. 109) May 12, 2021 at 7:48