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Look at the table below. This is a table that shows the approximate value of the integral

\int_{8}^{30} 2000 ln \frac{140000}{140000-2100t}-9.8t dt 

as a function of the number of segments used in the Trapezoidal rule and the corresponding true error.

n

Value

Et

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

\frac{(b-a)^3}{12} f^{\prime\prime} (c)

where c is some point not known but in the domain [a,b] of \int_{a}^{b} f(x) dx. 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

\frac{(b-a)^3}{12n^2} f^{\prime\prime}

where the f^{\prime\prime} 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 f^{\prime\prime} 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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This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu

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