Three _ Point Gauss_Legendre Rule

Three_Point Gauss_Legendre Rule:

       We have from  Gauss Quadrature Rule that 

 Rₙ₊₁(f) = w₀f(x₀) + w₁d(x₁) + ......+wₙf(xₙ) ...(1)

   ⇒R₃(f) = w₀f(x₀) + w₁f(x₁) + w₂f(x₂) .....(2)

                        1

Again , I(f) = ∫ f(x)dx .........(3)

                       -1

Now , I(f) is given by 

    I(f) = R₃(f) + E₃(f) ...........(4)

      where E₃(f) is its error . 

In order to determine the six parameters w₀, w₁,w₂,x₀,x₁,x₂  we make R₃(f) exact for all monomials if degree ≤ 5 , i.e 1,x,x²,x³,x⁴,x⁵ with  E₅(xʲ) = 0 , j= 0,1,2,3,4,5 

                                                        ...............(5)

Equation (4) can be written as 

     1

     ∫ f(x)dx = w₀f(x₀)+w₁f(x₁)+w₂f(x₂)+E₃(xʲ)

     -1

                                              ..................(6)

For f(x) = 1:

      We get from equation(6) 

     1

     ∫ 1dx= w₀.1 +w₁.1+w₂.1+E₃(1)

    -1

    ⇒2 = w₁+w₂+w₃+0

   ⇒ 2 = w₁+w₂+w₃

    For f(x) = x :

       We get from equation(4) 

   1

   ∫ xdx= w₀x₀+w₁x₁+w₂x₂+E₃(x)

  -1

   ⇒0 = w₀x₀+w₁x₁+w₂x₂+0 

   ⇒0 = w₀x₀+w₁x₁+w₂x₂ 

For f(x) = x² :

       We get from equation(6) 

    1

    ∫ f(x)dx  = w₀x₀²+w₂x₁²+w₂x₂²+E₃(x²)

   -1

  ⇒2/3 = w₀x₀²+w₁x₂²+w₂x₂² 

For f(x) = x³ :

     We get from equation(4)
1
∫ f(x)dx = w₀x₀³+w₁x₁³+w₂x₂³+E₃(x³)
-1

⇒0 = w₀x₀³+w₁x₁³+w₂x₂³+0

For f(x) = x⁴ :

     We get from equation(4) 

   1

   ∫ f(x)dx = w₀x₀⁴+w₁x₁⁴+w₂x₂⁴+E(x⁴)

  -1

 ⇒2/5 = w₀x₀⁴+w₁x₁⁴+w₂x₂⁴+0

 ⇒2/5 = w₀x₀⁴+w₁x₁⁴+w₂x₂⁴

For f(x) = x⁵ :

      We get from equation(4)

   1

   ∫ f(x)dx = w₀x₀⁵+w₁x₁⁵+w₂x₂⁵+E(x⁵)

  -1

 ⇒0 = w₀x₀⁵+w₁x₁⁵+w₂x₂⁵+0

 ⇒0 = w₀x₀⁵+w₁x₁⁵+w₂x₂⁵

Now we have six equations 

  [w₀+w₁+w₂=2

w₀x₀+w₁x₁+w₂x₂=0

w₀x₀²+w₁x₁²+w₂x₂²=2/3

w₀x₀³+w₁x₁³+w₂x₂³=0

w₀x₀⁴+w₁x₁⁴+w₂x₂⁴=2/5

w₀x₀⁵+w₁x₁⁵+w₂x₂⁵=0 ]       ..........(6)

Equations (6) are satisfied by 

  w₀=8/9 , w₁=w₂=5/9

  x₀=0 ,x₁= -x₂= √(3/5)

Hence the three _ point Gauss Legendre rule is given by 

R₃(f)= 1/9[8f(0) + 5{f(-√(3/5)+f(√(3/5)}]

Which is required rule.