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.


Comments

Post a Comment

If Any Doubt Ask Me

Popular posts from this blog

Accuracy Of Numbers

-
Image
              Accuracy Of Numbers 1.  Approximate Numbers :            There are two types of numbers exact and approximate . Exact numbers are 2,4,9,13,7,2,6.45....etc . But there are numbers such as 4/3 =1.3333....  ,          √(2)= 1.41413 and finite number of digits . These may be approximated by numbers 1.3333, 1.4141 , 3.1416 respectively . Such numbers which represent the given numbers to certain degree of accuracy are called approximate numbers .  2. Significant Figures :                     The digits used to express a number is called significant digits (figures) . Thus each of the numbers 7845, 3.589, 0.4758 contains four significant figures while the numbers 0.00389 , 0.000587 and 0.0000296 contain only three significant figures since zeros only help to fix the position of the decimal point . Similarly the numbers 45000 and 7300.00 have two significant figures only . 3. Rounding Off :                   There are numbers with
Read more

PFAFFIAN Differential Equations And It's Solutions

-
Image
PFAFFIAN Differential Equations And It's Solutions :-      The relation              n             Σ  Fᵢ(x₁,x₂,......xₙ) dxᵢ = 0            i=1   where Fᵢ's (i=1,2,....n) are functions of all or some of the n independent variables x₁,x₂,.....xₙ is called a Pfaffian differential equation . Here we shall study Pfaffian differential equation in three independent variables which is of the form              P dx + Q dy + R dz = 0    ...........(1)    where P, Q , R are functions of x , y and z . Setting -->                        -->               X = (P,Q,R)  and dr = (dx,dy,dz), equation (1) can be written in the vector notation as                   -->  -->                  X . dr    = 0                  ..............(2)  If there exists a function μ(x,y,z) such that μ(P dx + Q dy + R dz) is an exact differential dΦ , the equation is said to be integrable and to possess an integrating factor μ(x,y,z) . In such case(1) becomes
Read more

Derivation Of Composite Trapezoidal Rule

-
Image
Derivation Of Composite Trapezoidal Rule :                                                   b       Let us find the value of ∫ f(x) dx                                                  a numerically by composite trapezoidal rule. Solution :              Let [a,b] be divided into n equal sub intervals of length h each . Let                            a= x₀,x₁,x₂,.......xₙ=b  be the points of sub division .  i.e x₁ = x₀+ih  for i= 0,1,2,......n By Newton's Forward Difference Interpolation  f(x) = f(x₀)+uΔf(x₀) +u(u-1)/2!  Δ²f(x₀)+........        where x = x₀+uh b                 ∫ f(x) dx = a       xₙ          ∫ [f(x₀)+uΔf(x₀)+u(u-1)/2!  Δ²f(x₀)+...]dx           x₀              n        = h∫ [f(x₀)+uΔf(x₀)+u(u-1)/2  Δ²f(x₀)+...]dx             0     =h[uf(x₀) +u²/2 Δf(x₀) +                                                                                                       n                                  (u³/6 -u²/4)Δ²f
Read more