Derivation of Newton's Fundamental Interpolation Formula

Derivation of Newton's Fundamental Interpolation Formula:

    Let y = f (x) be a function with given values yᵢ = f(xᵢ) for (n+1) points x₀,x₁,x₂,.....,xₙ . Our aim is to construct a polynomial Φ(x) of degree not higher than n satisfying the following conditions 

      Φ(xᵢ) =yᵢ=f(xᵢ) ........(1)

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

Let us take the polynomial Φ(x) in the following form 

Φ(x) =a₀+a₁(x-x₀)+a₂(x-x₀)(x-x₁)+a₃(x-x₀)(x-x₁)(x-x₂) + ......+aₙ(x-x₀)(x-x₁)(x-x₂)....(x-xₙ)

                                                        ........(2)

where a₀,a₁,....aₙ i.e aᵢ's are constants to be determined .

Putting  i=0 in equation (1) ,we get 

Φ(x₀) = y₀ = f(x₀) i.e f(x₀) = Φ(x₀)

Again , putting x= x₀ in equation (2) ,we get 

Φ(x₀) = a₀ ⇒a₀ = f(x₀) 

                  ⇒a₀ = f[x₀] .......(3)

Putting i =1 in and x= x₁ in (2) ,we get 

     f(x₁) = Φ(x₁)

and Φ(x₁) = a₀ + a₁(x₁-x₀)

or, f(x₁) = a₀ +a₁(x₁-x₀)=f(x₀) +a₁(x₁-x₀)

    ⇒a₁   = f(x₁)-f(x₀) / x-x₀ = f[x₀,x₁] .........(4)

Putting i =2 in equation (1) and x= x₂ in equation (2) we get Φ(x₂) = f(x₂)

and Φ(x₂) = a₀+ a₁(x₂-x₀) + a₂(x₂-x₀)(x₂-x₁)

 f(x₂) = a₀ + a₁(x₂-x₀) + a₂(x₂-x₀)(x₂-x₁)

or , f(x₂) = f(x₀) + [f(x₁)-f(x₀)/x-x₀](x-x₀) + 

                                      a₂ (x₂-x₀)(x₂-x₁)

or,a₂(x₂-x₀)(x₂-x₁)=f(x₂)-f(x₀) -[f(x₁)-f(x₀)/x₁-x₀]

                                                           (x₂-x₀)

or, a₂ = f(x₁)-f(x₀) / (x₂-x₀)(x₂-x₁)  -

                                (x₂-x₀)f[x₀,x₁] /(x₂-x₀)(x₂-x₁)

            = f[x₀,x₂] /(x₂-x₁) - f[x₀,x₁] / (x₂-x₁)

            = f[x₀,x₁] - f[x₀,x₁]/(x₂-x₁) = f[x₀,x₁,x₂]

                                                                ........(5)

Similarly , we have 

      a₃ = f[x₀,x₁,x₂,x₃]

      a₄ = f[x₀,x₁,x₂,x₃,x₄]

       .................................

      aₙ = f[x₀,x₁,..........xₙ]

Putting these values of a₀,a₁,a₂,a₃,......aₙ in equation (2) we get 

Φ(x) = f[x₀] + (x-x₀)f[x₀,x₁] + 

                      (x-x₀)(x-x₁)f[x₀,x₁,x₂]+.........+

                 (x-x₀)(x-x₁).....(x-xₙ₋₁)f[x₀,x₁,x₂,...xₙ]

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

Equation(6) is called the Newton's Fundamental Interpolation Formula .It is also called Newton's Divided Difference Interpolation Formula .

Example :

                 Form the interpolation polynomial for the function y = f(x) given by in the table

using Newton's interpolation formula .

    X    -1       1      4       6   

    Y      1       -3     21      127

Solution :

          Given that 

         X        -1       1       4         6

          Y         1       -3      21       127

The Divided Difference Table is 

We know from Newton's Divided Difference Interpopation Formula that

f(x)=f[x₀]+(x-x₀)f[x₀,x₁] +(x-x₀)(x-x₁)f[x₀,x₁,x₂]

                 +(x-x₀)(x-x₁)(x-x₂)f[x₀,x₁,x₂,x₃]   ...(1)

Here x₀= -1 , x₁ = 1 , x₂ = 4 , x₃ = 6 

      f[x₀] = 1 , f[x₀,x₁] = -2 , f[x₀,x₁,x₂] = 2 , 

    f[x₀,x₁,x₂,x₃] =1

Putting the above values in equation (1) we get ,

  f(x) = 1+ (x+1)(-2) + (x+1)(x-1)(2) + 

                     (x+1)(x-1)(x-4)(1)

         =1 -2(x+1) + 2(x+1)(x-1) + (x+1)(x-1)(x-4)

         = x³ -2x² -3x +1

Which is required polynomial .

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Sir Isaac Newton FRS PRS was an English mathematician, physicist, astronomer, theologian, and author who is widely recognised as one of the most influential scientists of all time, and a key figure in the scientific revolution for more.