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What a quadrature rule actually promises, how the promise is verified monomial by monomial, and the counting argument that caps every rule with $n$ points at degree $2n - 1$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to determine a rule's degree of exactness, explain why testing monomials suffices, connect the degree to the error term and to polynomial approximation, and prove the ceiling that no rule with a given number of points can pass.
You can apply the trapezoid, midpoint and Simpson rules and their composite versions, and you have met their error formulas. This lesson asks the question behind those formulas: what exactly does a rule promise, and what is the most any rule of its size could promise?
A quadrature rule is a sum $\sum_i w_i f(x_i)$ standing in for an integral; the $x_i$ are its nodes and the $w_i$ its weights. Its degree of exactness is the largest $d$ such that it integrates every polynomial of degree $d$ with no error. A rule is interpolatory when its weights are the integrals of the Lagrange basis functions for its nodes.
The degree of exactness of a rule is the largest $d$ for which it integrates every polynomial of degree $d$ exactly. Because both the integral and the rule are linear in $f$, testing the monomials $1, x, x^{2}, \ldots$ in turn settles it, and the degree is one below the first failure. That first failure also names the derivative in the error term: a rule exact through degree $d$ has an error proportional to $f^{(d+1)}$. The reason the number matters is a bridge to approximation theory: if $p$ is any polynomial of degree $d$, the rule integrates it exactly, so the whole error comes from $f - p$ — and it is bounded by the length of the interval times the sum of the weight sizes times $\|f - p\|_\infty$. A rule is as good as polynomials are at approximating the integrand, and no better. That also bounds the whole subject. An $n$-point rule has $n$ nodes and $n$ weights, $2n$ numbers, so it can satisfy at most $2n$ exactness conditions: degree $2n - 1$. The bound is sharp in the other direction too — $\prod_i (x - x_i)^{2}$ has degree $2n$, is positive, and every rule returns zero for it — so no $n$-point rule, however constructed, exceeds $2n - 1$.
Another way: steps
Another way: example
The trapezoid rule on $[-1, 1]$: weights $1, 1$ at $\pm 1$. For $f = 1$ it gives $2$, correct. For $f = x$ it gives $0$, correct. For $f = x^{2}$ it gives $2$ against the true $\tfrac23$ — wrong, so the degree of exactness is $1$, and the error term involves $f''$.
Degree of exactness is read as a measure of accuracy. It is a statement about polynomials only, and it reaches a general integrand through one further step: how well polynomials of that degree approximate it. For a function with a corner, a pole near the interval, or an endpoint singularity, that step supplies nothing, and a rule of degree nineteen can be worse than the trapezoid rule. The second half of the mistake is assuming a higher degree is always better. High-order Newton-Cotes rules have negative weights, which turn rounding errors into a growing sum, and are not used at all.
Simpson on $[-1, 1]$: $\tfrac13(f(-1) + 4f(0) + f(1))$. For $f = x^{2}$: $\tfrac13(1 + 0 + 1) = \tfrac23$, exact.
The parabola it integrates.
For $f = x^{3}$: $\tfrac13(-1 + 0 + 1) = 0$, and the true value is $0$ too.
Free, by symmetry.
For $f = x^{4}$: $\tfrac23$ against $\tfrac25$. Degree of exactness $3$.
The first failure.
Any two-point rule: try $p(x) = (x - x_1)^{2}(x - x_2)^{2}$, degree four.
Positive except at two points.
Its integral is positive; the rule returns $0$.
So degree four is impossible.
Degree $3$ is therefore the ceiling for two points, and the two-point Gauss rule attains it.
Sharp.
The rule is $2f(0)$: one node, weight $2$.
For $f = 1$ it gives $2$, exact; for $f = x$ it gives $0$, exact.
For $f = x^{2}$ it gives $0$ against $\tfrac23$ — so the degree is $1$, which is the one-point ceiling $2n - 1$ attained.
Simpson's rule on $[-3, 3]$ is $\dfrac{3}{3}\left(f(-3) + 4f(0) + f(3)\right)$. Apply it to $f(x) = x^{4}$ and give the error, as a fraction if it is not whole.
Answer:
Give the degree of exactness of the Gauss-Legendre rule with $3$, $4$ and $5$ points.
| Points | Degree of exactness | |
|---|---|---|
| $3$ points | 3 | |
| $4$ points | 4 | |
| $5$ points | 5 |
Match each quadrature rule to its degree of exactness.
| Degree $1$, from two fixed endpoints | Degree $1$, from a single fixed point | Degree $3$, from three fixed points | Degree $3$, from two points the rule chose itself | |
|---|---|---|---|---|
| The trapezoid rule | ||||
| The midpoint rule | ||||
| Simpson's rule | ||||
| The two-point Gauss rule |
Put the five steps of establishing a rule's degree of exactness into order.
Number the steps in order (write the number in the box):
A quadrature rule $\sum_{i=1}^{4} w_i f(x_i)$ may choose its nodes as well as its weights. How many numbers is it free to choose?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Select every statement that follows from a rule having degree of exactness $d$.
This task has no paper form; do it on a device.
You can find and interpret a degree of exactness and state the ceiling for a given number of points. Say in your own words why a high degree of exactness does not promise an accurate answer.
8. Your turn: the degree of exactness of the midpoint rule on $[-1, 1]$, step 3