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Four-point Gaussian quadrature rule

  1. Jul 21, 2011 #1
    I need to use the four-point Gaussian quadrature rule to do some intense numerical calculations. Could anyone link to this page where it's written out explicitly over an [a,b] interval. I haven't been able to find it, I'm trying to derive it now but it's crucial that I'm 100% correct. I haven't used a Gaussian quadrature before so seeing clearly what it should look like would make me feel a lot better.

    This isn't really homework but a part of a much larger project I'm working on for school so you will not be doing my homework for my or anything like that.

    thanks
     
  2. jcsd
  3. Jul 21, 2011 #2
    This is for single-integral calculations? Also, are you looking for the theoretical derivation or the pseudo-code?
     
  4. Jul 21, 2011 #3
    This book gives a decent derivation:

    Numerical Analysis, 8th Edition, Burden and Faires, Brooks/Cole 2005, 978-0534382162

    This book has a list of the coefficients and roots tabulated for general n-point quadratures:

    Gaussian Quadrature Formulas, Stroud and Secrest, Prentice Hall 1966,
     
  5. Jul 21, 2011 #4
    I need to use this on a integral of two dimensions x and x'. But I figured that I would just use it first on x and then look at that result as a function of x' and use it again. I don't really care for a theoretical derivation I just want the formula. I'm trying to calculate the kernel of a integro-differential equation. This means heavy numerical calculations so my professor recommended this rule because it uses the fewest points with best accuracy. But I' very new to it.
     
  6. Jul 21, 2011 #5
    Suppose that [itex] x_1,\ldots, x_n [/itex] are the roots of the nth Legendre polynomial, and that for each [itex] i=1,2,\ldots, n [/itex] we define [itex] c_i [/itex] by
    [tex] c_i = \int_{-1}^1 \prod_{j=1, j\neq i}^n \frac{ x- x_j}{x_i-x_j} dx [/tex]
    If P(x) is any polynomial of degree less than 2n then
    [tex] \int_{-1}^1 P(x) dx = \sum_{i=1}^n c_i P(x_i) [/tex].

    This gives a (2n)th order approximation to the integral of any function. You can look up the roots of the nth Legendre polynomial (or calculate them computationally if you like). The c_i can then easily be calculated, but again, there are extensive tables with this information already computed. Hence

    [tex] \int_{-1}^1 f(x) dx \approx \sum_{i=1}^n c_i f(x_i) [/tex]

    To make the algorithm work between [a,b] rather than just [-1,1], apply the transformation
    [tex] t = \frac{2x-a-b}{b-a}, \qquad x = \frac12[(b-a)t + a + b] [/tex]
    and calculate
    [tex] \int_a^b f(x) dx = \int_{-1}^1 f\left( \frac{(b-a)t + (b+a) }{2} \right) \frac{b-a}2 dt [/tex]

    In the particular case when you want a 4 point approximation, the following is the table of the values (you can get better values by computing them using the formula I've given above)

    [tex] \begin{array}{|r|r|} \hline
    \text{Roots}, x_i & \text{Coefficients}, c_i \\ \hline
    0.8611363116 & 0.3478548451 \\
    0.339981436 & 0.6521451549 \\
    -0.339981436 & 0.6521451549 \\
    -0.8611363116 & 0.3478548451 \\ \hline
    \end{array}[/tex]
     
  7. Jul 21, 2011 #6

    Thank you so much!

    This is a cool method of doing numerical integrations, I wonder why they didn't teach this in my numerical analysis class.
     
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