Lagrangian Interpolation Formula
Lagrangian Interpolation Formula :
Let y = f(x) be a real valued function which is defined in an interval [a,b] . Let
x₀ , x₁ ,x₂ ,............xₙ be n+1 distinct points in that interval at which the respective values y₀ ,y₁,y₂ ............yₙ are tabulated.
Now our aim is to construct a polynomial Φ(x) of degree ≤ n , which interpolates f(x) such that
Φ(xᵢ) = y(xᵢ) , i = 1,2,3,.......,n .........(1)
Let us suppose that the polynomial Φ(x)
n
be given by Φ(x) = Σ lᵢ(x) y(xᵢ) .........(2)
i= 0
where each lᵢ(x) is a polynomial of degree ≤n in xᵢ , called Lagrangian function.
The function given in equation (1) if each lᵢ (x) satisfied lᵢ(xⱼ) = 0 where i ≠ j ,lᵢ(xⱼ)=1 when i=j .........(3)
Now ,
the polynomial lᵢ(x) vanishes at the (n+1) points xₒ , x₁ , ........., xₙ therefore , we can write this polynomial in the following form
lᵢ(x) = cᵢ (x-x₀)(x-x₁)........(x-xᵢ₋₁)(x-xᵢ₊₁)....(x-xₙ)
........(4)
where cᵢ's are constant co_efficients
putting x = xᵢ in equation (4) , we get
lᵢ(xᵢ) =cᵢ(xᵢ-x₀)(xᵢ-x₁).....(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)...(xᵢ-xₙ)
or, 1 = cᵢ(xᵢ-x₀)(xᵢ-x₁).....(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)
......(xᵢ-xₙ) as lᵢ(xᵢ) =1
or,cᵢ = 1/(xᵢ-x₀)(xᵢ-x₁)...(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)...(xᵢ-xₙ)
..........(5)
putting the value of cᵢ from equation (5) in equation (4) , we get
lᵢ(x) = (x-x₀)(x-x₁)...(x-xᵢ₋₁)(x-xᵢ₊₁) /
(xᵢ-x₀)(xᵢ-x₁)....(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁).....(xᵢ-xₙ)
.......(6)
putting the values of lᵢ(x) from equation (6) in equation (2) we get ,
Φ(x) =
n
Σ (x-x₀)(x-x₁)...(x-xᵢ₋₁)(x-xᵢ₊₁)....(x-xₙ) y(xᵢ) /
i=1 (xᵢ-x₀)(xᵢ-x₁)...(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)...(xᵢ-xₙ)
.......(8)
Equation (8) is known as Lagrangian Interpolation Formula.
Let us suppose that the polynomial Φ(x)
n
be given by Φ(x) = Σ lᵢ(x) y(xᵢ) .........(2)
i= 0
where each lᵢ(x) is a polynomial of degree ≤n in xᵢ , called Lagrangian function.
The function given in equation (1) if each lᵢ (x) satisfied lᵢ(xⱼ) = 0 where i ≠ j ,lᵢ(xⱼ)=1 when i=j .........(3)
Now ,
the polynomial lᵢ(x) vanishes at the (n+1) points xₒ , x₁ , ........., xₙ therefore , we can write this polynomial in the following form
lᵢ(x) = cᵢ (x-x₀)(x-x₁)........(x-xᵢ₋₁)(x-xᵢ₊₁)....(x-xₙ)
........(4)
where cᵢ's are constant co_efficients
putting x = xᵢ in equation (4) , we get
lᵢ(xᵢ) =cᵢ(xᵢ-x₀)(xᵢ-x₁).....(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)...(xᵢ-xₙ)
or, 1 = cᵢ(xᵢ-x₀)(xᵢ-x₁).....(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)
......(xᵢ-xₙ) as lᵢ(xᵢ) =1
or,cᵢ = 1/(xᵢ-x₀)(xᵢ-x₁)...(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)...(xᵢ-xₙ)
..........(5)
putting the value of cᵢ from equation (5) in equation (4) , we get
lᵢ(x) = (x-x₀)(x-x₁)...(x-xᵢ₋₁)(x-xᵢ₊₁) /
(xᵢ-x₀)(xᵢ-x₁)....(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁).....(xᵢ-xₙ)
.......(6)
putting the values of lᵢ(x) from equation (6) in equation (2) we get ,
Φ(x) =
n
Σ (x-x₀)(x-x₁)...(x-xᵢ₋₁)(x-xᵢ₊₁)....(x-xₙ) y(xᵢ) /
i=1 (xᵢ-x₀)(xᵢ-x₁)...(xᵢ-xᵢ₋₁)(xᵢ-xᵢ₊₁)...(xᵢ-xₙ)
.......(8)
Equation (8) is known as Lagrangian Interpolation Formula.
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About Scientist :
Joseph-Louis Lagrange was an Italian Enlightenment Era mathematician and astronomer. He made significant contributions to the fields of analysis, number theory, and both classical and celestial mechanics.
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