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

In summary, using Simpson's Rule with n=10 to approximate the integral of tan(x) from 0 to 1.55 results in an answer of 4.923651704, which differs from the exact answer of -Log[Cos(1.55)] calculated by a calculator. This discrepancy may be due to the singularity of tan(x) at pi/2 and the behavior of the error term in Simpson's Rule near singularities. Further investigation is needed to fully understand the cause of this over-estimation.
  • #1
mathew350z
2
0
[tex]\int tan(x)dx [/tex]

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?
 
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  • #2
mathew350z said:
[tex]\int tan(x)dx [/tex]

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.
 
  • #3
Count Iblis said:
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.
 
  • #4
mathew350z said:
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.
 

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

1. What is Simpson's Rule and how does it work?

Simpson's Rule is a numerical method used to approximate the area under a curve. It works by dividing the area under the curve into smaller, simpler shapes (usually trapezoids) and then summing the areas of these shapes to get an estimate of the total area.

2. Why does Simpson's Rule have a larger margin of error compared to other numerical integration methods?

Simpson's Rule has a larger margin of error because it only uses a few data points to approximate the curve, whereas other methods may use more data points and therefore provide a more accurate estimate of the area under the curve.

3. Can Simpson's Rule be used for any type of curve?

Simpson's Rule can be used for any curve that is smooth and continuous. It may not provide accurate results for curves with sharp corners or discontinuities.

4. How can the margin of error in Simpson's Rule be reduced?

The margin of error in Simpson's Rule can be reduced by using a smaller interval between data points, increasing the number of data points, or using a higher-order version of Simpson's Rule (such as Simpson's 3/8 Rule).

5. Are there any real-world applications for Simpson's Rule?

Yes, Simpson's Rule is commonly used in various scientific fields to estimate areas under curves, such as in calculating the volume of irregularly shaped objects or in analyzing data from experiments. It is also used in computer graphics to render smooth curves and surfaces.

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