Why is simpsons rule exact for 3rd degree polynomials?

In summary, the conversation discusses the perplexing concept of how Simpson's rule, which uses quadratics to approximate a curve, can be exact when the approximation is being done for a cubic. This is a general property of Newton-Cotes quadrature formulas, where each rule is exact for a certain degree of polynomial. This is due to the fact that the integration of an interpolating polynomial of a certain degree will be equal to the integration of the original polynomial of the same degree. The conversation also mentions that the truncation error for Simpson's rule is proportional to the fourth order derivative, so if the integrand is a cubic, the quadrature will be exact.
  • #1
okkvlt
53
0
this is really perplexing. how can it be exact? simpsons rule uses quadratics to approximate the curve. how can it be exact if I am approximating a cubic with a quadratic?
 
Physics news on Phys.org
  • #2
This is a general property of Newton-Cotes quadrature formulas: trapezoid is exact for degree 1 polynomials, Simpsons' for 3rd degree, the 5-point rule (sometimes called Bode or Boolean quadrature) is exact for 4th degree, etc.

One way to see this is the following: notice that, for simpsons' rule, you may assume that the points are symmetric relative to the origin (and as you have three, an odd number, one of them is in the origin); now let [tex]p_{3}(x)[/tex] be a 3rd degree polynomial and consider the integral:

[tex]\int^{1}_{-1}p_{3}(x)dx = \int^{1}_{-1}(a_{3}x^{3} + p_{2}(x))dx

= \int^{1}_{-1}a_{3}x^{3}dx + \int^{1}_{-1}p_{2}(x)dx = \int^{1}_{-1}p_{2}(x)dx[/tex]

And notice that you are left with the integral of a 2nd order polynomial. But remember that Simpsons' rule is based on the integration of an interpolating 2nd order polynomial, and these are unique; therefore, you are integrating the 2nd order interpolating polynomial of a 2nd order polynomial, and these must be equal.
 
Last edited:
  • #3
The truncation error for the Simpson method is proportional to the f(iv)(x). If the integrand is cubic, then the fourth order derivative is zero. So the quadrature will be exact for this case.
 

Related to Why is simpsons rule exact for 3rd degree polynomials?

1. Why is simpsons rule exact for 3rd degree polynomials?

Simpson's rule is exact for 3rd degree polynomials because it is based on the fundamental theorem of calculus, which states that the definite integral of a function can be calculated by finding the anti-derivative of that function. Since 3rd degree polynomials can be integrated exactly, Simpson's rule is able to accurately calculate the area under the curve.

2. What makes Simpson's rule different from other numerical integration methods?

Simpson's rule differs from other numerical integration methods because it uses a parabolic approximation of the curve, rather than straight line segments or trapezoids. This makes it more accurate for curved functions, particularly 3rd degree polynomials.

3. How does Simpson's rule calculate the area under a curve?

Simpson's rule divides the area under the curve into multiple small parabolic segments and calculates the area of each segment. These areas are then added together to give an approximation of the total area under the curve.

4. Is Simpson's rule always exact for 3rd degree polynomials?

Yes, Simpson's rule is always exact for 3rd degree polynomials. However, it may not be exact for higher degree polynomials or other types of functions.

5. Can Simpson's rule be used for any type of function?

Simpson's rule can be used for any function that is continuous and has a known anti-derivative. However, it may not always give an accurate result for functions with large variations or sharp turns.

Similar threads

Replies
1
Views
1K
Replies
2
Views
1K
Replies
2
Views
933
  • Calculus
Replies
1
Views
4K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
5
Views
2K
Back
Top