Examples:
Estimate the integral of the function
using Simpson’s Method with N=4 and thus h=
/4. This integral is of course equal to 1.
f(0) 0 f( /4)
0.3535533 f( /2)
0.5 f(3 /4)
0.3535533 f( )
0 Thus the error is
Use the same intergral but now use N=8 subintervals and h=
/8.
f(0) 0 f( /8)
0.1913417 f( /4)
0.3535533 f(3 /8)
0.4619397 f( /2)
0.5 f(5 /8)
0.46193975 f(3 /4)
0.3535533 f(7 /8)
0.19134174 f( )
0 We expect that when we increase the number of intervals by a factor of 2 (8/4=2), that our error will be reduced by a factor of 24=16. That means that our answer will be 16 times more accurate, or our error will be 1/16 of the error in the first approximation.
Let's see if that's the case for this example.
Recall, that the error when N=4 was
We can now compute the error when N=8. It is ~0.01%
which is approximately
times better than the approximation with N=4.
This result is comparable to the
improvement we would expect when we increase the number of intervals from 4 to 8.