Look at the table below. This is a table that shows the approximate value of the integral
as a function of the number of segments used in the Trapezoidal rule and the corresponding true error.
n |
Value |
E_{t} |
1 |
11868 |
-807 |
2 |
11266 |
-205 |
3 |
11153 |
-91.4 |
4 |
11113 |
-51.5 |
5 |
11094 |
-33.0 |
6 |
11084 |
-22.9 |
7 |
11078 |
-16.8 |
8 |
11074 |
-12.9 |
The true error for n=1 is -807 and for n=2 is -205. As you can see the quarter of -807 is approximately -201.75 and close to the true error for n=2. Is this a coincidence?
Look at the true error for n=2 which is -205 and for n=4 is -51.5. As you can see the quarter of -205 is approximately -51.75 and close to the value of the true error for n=4. Is this a coincidence?
No. This is because the true error in a single segment trapezoidal rule is
where c is some point not known but in the domain [a,b] of . It can be then shown (see page 14 of this pdf file for full proof) that for the multiple segment trapezoidal rule, the true error is
where the is an average value of the second derivative of the function f(x) calculated at some point within each of the n segments. Since a and b are constant, and becomes almost a constant as n increases, the true error is approximately inversely proportional to the square of the number of segments.
Note to the reader: Develop a similar table as given above for an integral of your choice and see it for yourself if the true error gets approximately quartered as the number of segments is doubled.
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ihsan said:
Please solve the whole question on the site.
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