Existence And Uniqueness Of Interpolating Polynomial

          Existence And Uniqueness Of Interpolating Polynomial

We now examine the existence and uniqueness of a polynomial that interpolates a function f(x) at a given set of distinct nodes x₀,x₁,x₂….xₙ . Note that the interpolation means if i≠j then xᵢ≠xⱼ .

Suppose Pₙ(x) given by is the polynomial interpolating f at a set of n+1 distinct points x₀,x₁,x₂……xₙ . Then we have

  Pₙ(x) = fᵢ= f(xᵢ) ; i=  0,1,2….n

        a₀+a₁x₀+…..+aₙx₀ⁿ = f₀ 
⇒   a₀ +  a₁x₁ + ….+aₙx₁ⁿ = f₁
       ………………………………..       }...(2.2.1)
    a₀ + a₁xₙ + …….+aₙxₙⁿ = fₙ 

This is a system of n+1 linear equations in n+1 unknowns : a₀,a₁,a₂…..aₙ; hence the system will have a unique solution if the determinant

Δ =  DET(f₀,f₁,......fₙ) ≠ 0 ..........(2.2.2)

  Indeed , the value of the determinant Δ is not zero since

      Δ = Π (xᵢ - xⱼ)   0≤j≤i≤n

        xᵢ ≠ xⱼ for i≠ j as the points x₀,x₁,....xₙ are distinct .

        Therefore an unique interpolating polynomial exists whose co_ efficients are the solution of the system of linear equations (2.2.1) for a given function and a given set of nodes aᵢ and functional values fᵢ .

Note :

            If the number of interpolating points are n+1 , then the degree of the interpolating polynomial ≤ n .

  The Alternative Method : 

→         The uniqueness of the interpolating polynomial can be demonstrated in the following alternative way . 

             Let qₙ(x) be a polynomial of degree ≤ n which also interpolates f(x) at the points x₀,x₁,x₂,....xₙ . Let

                        Φ(x) = pₙ(x) - qₙ(x) 

Then Φ(x) is a polynomial of degree ≤ n 

However , Φ(xⱼ) = pₙ(xⱼ) - qₙ(xⱼ) = fⱼ - fⱼ = 0

                                             for j = 0,1,2......n

      Thus Φ(x) which is   a polynomial of degree≤ n has n+1 zeros ,x₀,x₁,x₂,.....xₙ . 

     This can happen only if Φ(x) = 0 

    i.e  if pₙ(x) = qₙ(x) .