Look at the table below. This is a table that shows the approximate value of the integral
$latex \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
$latex \frac{(b-a)^3}{12} f^{\prime\prime} (c)$
where c is some point not known but in the domain [a,b] of $latex \int_{a}^{b} f(x) dx$. 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
$latex \frac{(b-a)^3}{12n^2} f^{\prime\prime} $
where the $latex f^{\prime\prime}$ 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 $latex f^{\prime\prime}$ 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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