Tell me more ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I am having the following issue.

I am evaluating a numerical integral, and I need high precision to get good values of the function being integrated.

NIntegrate[
  Integrand1P12[Q1, Q2, a, b, c, massrho^2] /. solsR[[1]], {Q1, 0, ∞}, {Q2, 0, ∞},
  MaxRecursion -> 2000, 
  WorkingPrecision -> 30,
  MinRecursion -> 40,
  AccuracyGoal -> 8]

Where previously I have defined the function Integrand1P12 and

solsR = {a -> ..., b -> ..., c -> ...}

are replacement rules where a, b, and c are to be replaced with exact quantities such as 234/23.

Up to this point there is no problem; my code works perfectly. However, next, I would like to perform some Monte Carlo where the replacement rules are obtained from a random number generation. To do this I need two steps:

  1. Generate random parameteres through:

    para = RandomReal[NormalDistribution[a[[1]], a[[2]]], 10, WorkingPrecision -> 34];
    

    where a[[1]] and a[[2]] are the relevant parameters of the distribution.

  2. Finally, I want to plug these random numbers into a formula containing simple sums, multiplications, and divisions to obtain the replacement rules tp use in the integral.

However, when Mathematica evaluates the integral, it complains that the precision of the argument is smaller than the requested one (30)

Can anybody tell me where the precision is lost and how to fix it?

share|improve this question
2  
Complete but minimal code that exhibits the problem would assist those who wish to help. – Mr.Wizard Mar 3 at 11:42
I am sory, the problem is code is rather long to obtain everuthing is rather long, I will try to reduce and produce something meaningful What I discovered is that increasing more the working precision in RandomReal, to 40, is solving the issue I will find the simplest example and post – pablo Mar 3 at 13:56
perhaps Rationalise (with dx=0) would be useful? You would need to study if that does something unacceptable to your statistics. – george2079 Mar 3 at 14:22

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

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.