Solving Guassian Quadrature: Why Must \mu_i Be Roots of P_N?

  • Thread starter Pengwuino
  • Start date
In summary, the roots of the Legendre polynomials must be the quadrature points in order for the quadrature to be accurate and satisfy N even-moment conditions.
  • #1
Pengwuino
Gold Member
5,124
20

Homework Statement



Why must the [itex]\mu_i[/itex] be the roots of [itex]P_N(\mu_i) = 0[/itex] to satisfy N even-moment conditions? Consider [itex]2\pi \int d\mu \mu^{N+n}[/itex] and write [itex]\mu^{N+n}[/itex] as [itex]P_N(\mu)q_N(\mu) + R_k(\mu)[/itex] which involves [itex]P_N[/itex] and the polynomials [itex]q_N[/itex] (quotient) and [itex]R_k[/itex] (remainder). If n<N, these are both of degree less than N.

Homework Equations



The [itex]P_N(\mu)[/itex] are the Legendre polynomials of order N.

The Attempt at a Solution



So my understanding is that the quadrature means that you can write some polynomial as [itex]\int_{-1}^1 H(\mu) = \sum_{i=0}^{N} w_i P_i(\mu_i)[/itex]. I really have no idea how to go about this. Is my understanding of quadrature even correct in the first place?
 
Physics news on Phys.org
  • #2


Your understanding of quadrature is correct. Quadrature is a method for approximating the integral of a function by using a weighted sum of function evaluations at certain points. In this case, the function is H(\mu) and the points are the roots of the Legendre polynomials, \mu_i.

In order for the quadrature to be accurate, the weights w_i must satisfy N even-moment conditions, which means that they must be able to exactly integrate polynomials of degree up to 2N-1. This is where the roots of the Legendre polynomials come in.

To see why the \mu_i must be the roots of P_N(\mu_i) = 0, let's consider the integral 2\pi \int d\mu \mu^{N+n}. We can write \mu^{N+n} as P_N(\mu)q_N(\mu) + R_k(\mu), where q_N and R_k are polynomials of degree less than N.

Since the roots of P_N(\mu_i) = 0 are the quadrature points, we can write the integral as 2\pi \int d\mu \mu^{N+n} = \sum_{i=0}^{N} w_i P_N(\mu_i)q_N(\mu_i) + R_k(\mu_i). Since P_N(\mu_i) = 0, this simplifies to 2\pi \int d\mu \mu^{N+n} = R_k(\mu_i).

Since n<N, R_k is a polynomial of degree less than N. This means that it can be exactly integrated using the weights w_i, which are determined by the roots of P_N(\mu_i) = 0. Therefore, in order for the quadrature to be accurate, the roots of P_N(\mu_i) = 0 must satisfy N even-moment conditions.
 

1. What is Gaussian quadrature?

Gaussian quadrature is a numerical method for approximating definite integrals. It involves choosing a set of points, known as quadrature points, and corresponding weights to approximate the integral of a function. These points and weights are chosen in a way that results in a more accurate approximation compared to other numerical integration methods.

2. What is \mu_i in Gaussian quadrature?

\mu_i refers to the quadrature points in Gaussian quadrature. These points are chosen to be the roots of a specific polynomial, known as the orthogonal polynomial, in order to achieve the most accurate approximation of the integral. The subscript i represents the index of the quadrature point.

3. Why must \mu_i be roots of P_N in Gaussian quadrature?

In Gaussian quadrature, the quadrature points are chosen to be the roots of the orthogonal polynomial, P_N, of degree N. This is because these roots are known to have certain desirable properties that result in a more accurate approximation of the integral. These properties include evenly distributing the quadrature points and minimizing the error in the approximation.

4. How are the weights determined in Gaussian quadrature?

The weights in Gaussian quadrature are determined by solving a system of equations known as the moment equations. These equations involve the function being integrated and the orthogonal polynomial, P_N. The resulting weights are used in the calculation of the integral approximation.

5. Can Gaussian quadrature be used for any type of integral?

Gaussian quadrature can be used for any definite integral, as long as the limits of integration are finite. It is especially useful for integrals involving functions that are difficult to integrate analytically or that require a high degree of accuracy in the approximation. However, the appropriate orthogonal polynomial and number of quadrature points must be chosen based on the specific integral being approximated.

Similar threads

Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Differential Equations
Replies
11
Views
2K
Replies
6
Views
885
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
Back
Top