Derivation of Newton's Forward Difference Interpolation Formula

Newton's Forward Difference Interpolation Formula:

     Let y = f(x) be a function of x and let us suppose that yᵢ = f(xᵢ) ...(1) for i = 1,2,3,.....,n satisfying the condition xᵢ = x₀+ih where 

'h' is the interval of difference .

       Now our aim is to constuct a function Φ(x) of degree not higher than n such that
  Φ(xᵢ) = yᵢ   ............(2)

          Since Φ(x) is a polynomial of degree n
 then we can write

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₋₁)...(3)

Let us find the value of a₀,a₁,a₂.....aₙ satisfying the equations (2) and (3)

From equation (2) , we get Φ(x₀) = y₀
 From equation (3) , we get Φ(x₀) = a₀

        so , a₀ = y₀ ...........(4)

From equation (2) we get Φ(x₁) = y₁
From equation (3) we get Φ(x₁) = a₀+a₁(x₁-x₀)

Therefore  y₁ = a₀ + a₁(x₁-x₀)

             ⇒y₁ = y₀ + a₁h    as a₀=y₀ and h = x₁-x₀

              ⇒a₁ = y₁-y₀ / h

              ⇒ a₁ = Δy₀ / h .........(5)

From equation (2) we get Φ(x₂) = y₂ and from equation (3) we get

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

Therefore

     y₂ = a₀ + a₁(x₂-x₀) + a₂(x₂-x₀)(x₂-x₁)s

⇒y₂ = y₀ + (y₁-y₀)(2h)/h + a₂(2h)(h)

⇒ y₂ = y₀ + 2(y₁-y₀) + a₂(2h²)

⇒ a₂ = y₂-2y₁+y₀ / 2!h²

⇒ a₂ = Δ² y₀/2!h² ..........(4)

From equation (2) we get Φ(x₃) = y₃

From equation (3) we get

    Φ(x₃) = a₀ + a₁(x₃-x₀) + a₂(x₃-x₀)(x₃-x₁)
                                     + a₃(x₃-x₀)(x₃-x₁)(x₃-x₂)

⇒y₃ =y₀+(y₁-y₀)(3h)/h+
                                     (y₂-2y₁+y₀)(3h)(2h)/2
                                     + a₃(3h)(2h)(h)

⇒ y₃ = y₀ + 3(y₁-y₀) + 3y₂-6y₁ + 3y₀ + 6a₃h³

⇒ a₃ = y₃ + 3y₁ -3y₂-y₀ /6h³ = Δ³y₀ /1*2*3*h³

                                                 = Δ³y₀/3!h³

Therefore a₃ = Δ³y₀ /3!h³ ...............(7)

Continuing in the process ,we get

   a₄ = Δ⁴y₀/4!h⁴  , a₅ = Δ⁵ y₀ /5! h⁵ ,........

           aₙ  = Δⁿy₀ / n!hⁿ    ............(8)

Now , substituting the values of a₀, a₁,a₂...aₙ
in equation (3) ,we get

Φ(x) = y₀ + Δy₀(x-x₀)/h + Δ²y₀(x-x₀)(x-x₁)/2!h²

                 + Δ³ y₀ (x-x₀)(x-x₁)(x-x₂)/3!h³ + ......

            + Δⁿy₀(x-x₀)(x-x₁)......(x-xₙ₋₁)/n!hⁿ ....(9)

The equation (9) is known as Newton's Forward Difference  Interpolation
Formula .

      But for practical purpose this formula is modified by using the equation

                x = x₀ + sh  ..........(10)

Then we have  x - x₀ = sh    , so    s = x - x₀ / h

      x-x₁ = x₀+sh - (x₀+h) = (s-1)h

      x-x₂ = x₀+sh - (x₀+2h) = (s-2)h

      x-x₃ = x₀+sh - (x₀ +3h) = (s-3)h

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

  x- xₙ₋₁ = x₀ +sh - {x₀+ (n-1)h} = [s-(n-1)]h
                                                              ...........(11)

Substituting these values of (x-x₀),(x-x₁),
 (x-x₂),.......(x-xₙ) in equation (9) , we get

Φ(x) = Φ(x₀+sh)

         = y₀ + sΔy₀ + s(s-1)Δ²y₀ /2!

            + s(s-1)(s-2)Δ³y₀/3! + .........

             + s(s-1)(s-2).....{s-(n-1)}Δⁿy₀/n! ......(12)

This equation (12) is the Modified Newton's Forward Difference Interpolation Formula.

For Example :

         Evaluate f(1.2) using the following table .

        x     0     1      2      3      4

 
     f(x)   1    1.5    2.2   3.1   4.2

Solution : 

          The finite difference table is 

    x          f(x)        Δf(x)         Δ²f(x)

    0            1
                               0.5
    1          1.5                           0.2

                               0.7
     2         2.2                           0.2
                               0.9
     3         3.1                           0.2
                               1.1
      4        4.2

Here x = 1.2  ,let x₀ = 1.0

  Then s = x - x₀ /h = 1.2-1 / 1 = 0.2

Now by Newton's Forward Difference Interpolation Formula   , we get

y(x) = y₀ + sΔy₀ + s(s-1) Δ²y₀ / 2!

Therefore

         y(0.2) = 1.5 + 0.2 * 0.7 + 0.2(0.2-1) *0.2 /2

                    = 1.624 

which is required solution .



for more