Is it just a coincidence – true error in multiple segment Trapezoidal rule gets approximately quartered as the number of segments is doubled?

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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