How can Simpson's Rule have a large margin of error?

[mathew350z]

\int tan(x)dx

with the limits of 0 to 1.55, with n=10. Using Simpson's Rule, my answer was 4.923651704. But I don't understand why Simpson's Rule varies from my calculator's answer (TI-84 Plus) which is 3.873050987. I thought the higher N with Simpson's Rule would make your answer even more accurate?

 

[Count Iblis]

\int tan(x)dx

with the limits of 0 to 1.55, with n=10. Using Simpson's Rule, my answer was 4.923651704. But I don't understand why Simpson's Rule varies from my calculator's answer (TI-84 Plus) which is 3.873050987. I thought the higher N with Simpson's Rule would make your answer even more accurate?

The calculator seems to give you the exact answer, which is -Log[Cos(1.55)]. When you do numerical integration, you have to be careful when you are close to singularities. In this case 1.55 is close to pi/2 where tan diverges.

 

[mathew350z]

The calculator seems to give you the exact answer, which is -Log[Cos(1.55)]. When you do numerical integration, you have to be careful when you are close to singularities. In this case 1.55 is close to pi/2 where tan diverges.

Yes, I understand that tan(pi/2) diverges but I still am having a little trouble as to why that Simpson's Rule over-estimates the actual answer.

 

[Count Iblis]

Yes, I understand that tan(pi/2) diverges but I still am having a little trouble as to why that Simpson's Rule over-estimates the actual answer.

I guess you have to do a detailed investigation of the error term in Simpson's rule, study how it behaves near an 1/x -like singularity.