- Why is numerical integration necessary in the first place?
- Where do these formulas come from (especially Simpson’s Rule)?
- How can I do all of these formulas quickly?
- Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?
- Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?
- Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?
where is the number of subintervals and is the width of each subinterval, so that . Also, is the midpoint of the th subinterval.
For this special case, the true area under the curve on the subinterval will be
In the above, the shorthand can be formally defined, but here we’ll just take it to mean “terms that have a factor of or higher that we’re too lazy to write out.” Since is supposed to be a small number, these terms will small in magnitude and thus can be safely ignored. I wrote the above formula to include terms up to and including because I’ll need this later in this series of posts. For now, looking only at the Midpoint Rule, it will suffice to write this integral as
.Using the midpoint of the subinterval, the left-endpoint approximation of is . Using the Binomial Theorem, this expands as
Once again, this is a little bit overkill for the present purposes, but we’ll need this formula later in this series of posts. Truncating somewhat earlier, we find that the Midpoint Rule for this subinterval gives
Subtracting from the actual integral, the error in this approximation will be equal to
In other words, unlike the left-endpoint and right-endpoint approximations, both of the first two terms and cancel perfectly, leaving us with a local error on the order of .
.So that this formula doesn’t appear completely mystical, this actually matches the numerical observations that we made earlier. The figure below shows the left-endpoint approximations to for different numbers of subintervals. If we take and , then the error should be approximately equal to
,which, as expected, is close to the actual error of . Let , so that the error becomes
,where is the average of the . Clearly, this average is somewhere between the smallest and the largest of the . Since is a continuous function, that means that there must be some value of between and — and therefore between and — so that by the Intermediate Value Theorem. We conclude that the error can be written as
,Finally, since is the length of one subinterval, we see that is the total length of the interval . Therefore,
,where the constant is determined by , , and . In other words, for the special case , we have established that the error from the Midpoint Rule is approximately quadratic in — without resorting to the generalized mean-value theorem and confirming the numerical observations we made earlier.
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