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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 $\displaystyle { \int _0 ^{+ \infty} e^{-x + \sqrt{x}} \ dx}$.

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  • $\begingroup$ Maybe this helps ? $\endgroup$ – user111 Mar 15 at 17:49
  • $\begingroup$ 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 $E_n(f, \omega)$... $\endgroup$ – MathTolliob Mar 16 at 14:51
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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 [1], 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 [2] 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 $f(x)=e^{-x+\sqrt{x}}$.

[1] 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.

[2] 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.

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  • $\begingroup$ 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 $\endgroup$ – MathTolliob 2 days ago
  • $\begingroup$ Is the underlying constant of the $\mathcal{O}(n^{-1})$ obtain by applying the result of [1] can be explicitely computed or estimated ? Is the constant C used in Proposition 4.1 of [2] can be explicitely computed or estimated? $\endgroup$ – MathTolliob 2 days ago
  • $\begingroup$ 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 [2], p. 109) $\endgroup$ – MathTolliob yesterday

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