This has come up with a math modeling project I am doing. To try and boil it down, suppose I am using a numerical quadrature such that

$ \int_0^1 f(x) dx = \sum_{i=1}^n f(x_i) + O(h^n)$

where $O(h^n)$ is the error associated with the quadrature. Additionally, suppose that the function being integrated is rather nasty to evaluate itself, hence we replace it with a numerical estimate $g(x)$ that has it's own error, e.g.

$f(x) = g(x) + err(x)$

Then I would like to consider the error of using the quadrature on $g(x)$. Naively,

\begin{align*} \int_0^1 f(x) dx =& \int_0^1 g(x) + err(x) dx \\ =&\int_0^1 g(x) dx + \int_0^1 err(x) dx\\ =& \sum_{i=1}^n g(x_i) + O(h^n) + \int_0^1 err(x) dx \\ \end{align*}

so then it seems like my end estimate is

$\sum_{i=1}^n g(x_i) = \int_0^1 f(x) dx - \left(+ O(h^n) + \int_0^1 err(x) dx\right)$

and the error is $\left(+ O(h^n) + \int_0^1 err(x) dx\right)$. Does something like this seem reasonable? Can anybody point me in the direction of references that consider such things? Unfortunately, my actual problem involves an infinite volume integral to boot.


Yes, this is correct. Now the usual technique is to say the error is bounded by the sum of the absolute values of your terms by the triangle inequality and compute the size of each term. Usually one will dominate over the other. You might well be able to bound $err(x)$ by some constant. Then you can say the integral from $0$ to $1$ is also bounded by that constant. If it is small compared to $O(h^n)$ you can ignore it. To really do this you need the constant in front of $h^n$ for the comparison.

  • $\begingroup$ Would you have any references you could suggest? Thank you $\endgroup$ – Fractal20 Jul 27 '18 at 16:47
  • $\begingroup$ Also, I am a little unsure because the integral of the error is whatever it is which doesn't necessarily depend on $h$ so it seems like it would always dominate as $h$ goes to zero. But I suppose that makes sense. Controlling the quadrature error won't influence the error in the approximation of the original problem. So I guess the point is for efficiency you'd want to choose $h$ to make them be on the same order. $\endgroup$ – Fractal20 Jul 27 '18 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.