Problem with tolerances in integral

1 次查看（过去 30 天）
Mario
Mario 2018-8-15
回答： Mario 2018-8-17
While evaluating an integral with expensive computation of the integrand, I tried the adaptive integration provided by the integral function. Unfortunately the function seems not to accept the desired (lower) tolerance argument, if the integral is not improper as in the example from the help file. Instead of taking the desired tolerance, integral seems to work with default tolerance in any case. I give a minimal example for illustration:
if true
% code
f = @(t) log(t);
I_true1 = -1;
I_true2 = 3*log(3)-2*log(2)-1;
for k = 2:2:12
epsi = 10^(-k);
I1 = integral(f,0,1,'RelTol',0,'AbsTol',epsi);
I2 = integral(f,2,3,'RelTol',0,'AbsTol',epsi);
fprintf('%1.1e %1.4e %1.4e\n', epsi, abs(I1-I_true1), abs(I2-I_true2))
end
end
producing the output
1.0e-02 1.7536e-07 1.1102e-16
1.0e-04 1.7536e-07 1.1102e-16
1.0e-06 1.0960e-08 1.1102e-16
1.0e-08 1.7124e-10 1.1102e-16
1.0e-10 6.6880e-13 1.1102e-16
1.0e-12 1.0214e-14 1.1102e-16
Has anyone an idea how to change this behaviour? Thanks in advance for your help!

请先登录，再回答此问题。

采纳的回答

Mario
Mario 2018-8-17
In the meanwhile I found the answer: integral works initially with 150 evaluations of the integrand. In the example I found problematic, machine precision is reached already in this initial configuration.

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Programming 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by