Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why does the following approximation work very well for small theta's but not for large theta's?


$$ I = \int_0^\infty x (1 + \theta x)^{-2 \left(\frac{1}{\theta} + 1 \right)} \, \mathrm{d}x $$

Exact evaluation using integration by part

$$ I = \frac{1}{2(\theta + 2)} $$

Approximation using Gauss-Laguerre quadrature $$ I \approx \sum_{k=1}^N w_k \exp(x_k) \, x_k (1 + \theta x_k)^{-2 \left(\frac{1}{\theta} + 1 \right)} $$ where the $x_k$'s and the $w_k$'s can be found, e.g., here.

Comparison (N = 100)

  • theta = 0.5806:

    true = 0.1937534 and approx = 0.1937522

  • theta = 15:

    true = 0.02941176 and approx = 0.01662866


my R-code

#------ package ------
library(gaussquad)
#---------------------

#------ f(x) ------
f <- function(x, theta)
{
  x * (1 + theta * x)^(-2 * ((1 / theta) + 1))
}
#------------------

#------ xk and wk ------
laguerre <- glaguerre.quadrature.rules(n=100, alpha=0, normalized=FALSE)[[100]]
x <- laguerre[, 1]
w <- laguerre[, 2]
#-----------------------

#------ approx ------
approx <- function(theta, x, w)
{
  sum(w * exp(x) * f(x=x, theta=theta))
}
#--------------------

> approx(theta=0.5806, x=x, w=w)
[1] 0.1937522
> approx(theta=15, x=x, w=w)
[1] 0.01662866
share|cite|improve this question
    
Sorry, your integral is not showing up in the post. – Ron Gordon Apr 22 '13 at 10:35
    
@RonGordon: Should be fixed now. Thanks! – Marco Apr 22 '13 at 10:38
    
Sorry again, I still cannot see it. – Ron Gordon Apr 22 '13 at 10:40
    
@RonGordon: Sorry for the trouble! Hope it is ok now – Marco Apr 22 '13 at 11:14
    
You should recall that Gauss-Laguerre quadrature works best if the "infinite extent" portion of your integral behaves very much like $\exp(-x)$. This for instance is the reason why Gauss-Laguerre is not very useful for computing integrals like $\int_0^\infty \frac{\mathrm dx}{1+x^2}$. – J. M. Apr 22 '13 at 12:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.