Clear Filters
Clear Filters

Reimann sum and numerical integration

20 views (last 30 days)
I am trying to solve the functions below numerically using MATLAB
This is my code where I have used "integral" command to solve for the second integral, while I used Reimann sum to solve for the first. Take for example c=1/128.
x= -1/128:0.0001:1/128;
for j=1:length(x)
den= @(y) pi^2.* sqrt(1- (x(j)+y).^2) .* sqrt(1- y.^2);
fun= @(y) 1./den(y);
ymin= max(-0.999,-0.999-x(j));
ymax= min(0.999,0.999-x(j));
for j=1:length(omegat)
Do you think it is correct? Are there any other ways to make it more accurate?

Accepted Answer

Star Strider
Star Strider on 25 Jan 2015
I would likely use either trapz or cumtrapz to calculate ‘F’, but otherwise I see no problems.
DM on 25 Jan 2015
Also did you notice I have 0.999 instead of 1, so as to avoid the singularity point, is that OK ?
Star Strider
Star Strider on 25 Jan 2015
Personal preference, really. The trapz function is designed for numerical integration:
F = trapz(x, pdfX)
Avoiding the singularity is likely a good move. I doubt that substituting -0.999 would produce significant error. You might want to experiment with (1-1E-8) instead if you are curious.

Sign in to comment.

More Answers (0)

Community Treasure Hunt

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

Start Hunting!