Newton Forward Interpolation Method

Newton Forward Interpolation is used to estimate the value of a function at a given point when the data points are tabulated at equal intervals. This method is particularly useful when you want to interpolate a value near the beginning of the data set. It utilizes forward differences to form the interpolation polynomial.

The formula for Newton Forward Interpolation is:

P(x)=y0+vΔy0+v(v1)2!Δ2y0+v(v1)(v2)3!Δ3y0+v(v1)(v2)(v3)4!Δ4y0+ P(x) = y_0 + v \cdot \Delta y_0 + \frac{v(v-1)}{2!} \cdot \Delta^2 y_0 + \frac{v(v-1)(v-2)}{3!} \cdot \Delta^3 y_0 + \frac{v(v-1)(v-2)(v-3)}{4!} \cdot \Delta^4 y_0 + \cdots

where:

  • v=xx0hv = \frac{x - x_0}{h}
  • x0x_0 is the first value of xx in the data.
  • hh is the uniform difference between the xx values (where h=x1x0h = x_1 - x_0).
  • Δy0,Δ2y0,\Delta y_0, \Delta^2 y_0, \dotsare the forward differences.

This method is efficient for interpolating at or near the beginning of the data set.

Example of Newton Forward Interpolation

Let's say we are given the following data points:


xxyy
12
25
310
417

We are tasked with finding yy where x=2.5x = 2.5


Step 1: Calculate the forward differences for the yy values.


xxyyΔy{\Delta y} Δ2y{\Delta^2 y} Δ3y{\Delta^3 y}
12320
2552
3107
417

Step 2: Using the formula v=xx0hv = \frac{x - x_0}{h} values.


Given x=2.5 and x0=1 with h=1\text{Given } x = 2.5 \text{ and } x_0 = 1 \text{ with } h = 1
v=xx0hv = \frac{x - x_0}{h}
v=2.511v = \frac{2.5 - 1}{1}
v=1.5v = 1.5

Step 3: Apply the Newton Forward Interpolation formula:


P(x)=y0+vΔy0+v(v1)2!Δ2y0+v(v1)(v2)3!Δ3y0P(x) = y_0 + v \cdot \Delta y_0 + \frac{v(v-1)}{2!} \cdot \Delta^2 y_{0} + \frac{v(v-1)(v-2)}{3!} \cdot \Delta^3 y_0

Substitute the known values:

P(2.5)=2+1.53+1.5(1.51)2!2+1.5(.51)(1.52)3!(0) P(2.5) = 2 + 1.5 \cdot 3 + \frac{1.5(1.5-1)}{2!} \cdot 2 + \frac{1.5(.5-1)(1.5-2)}{3!} \cdot (0)

Step-by-step:

  1.  extFirstterm=2\ ext{First term = } 2
  2.  extSecondterm:1.53=4.5\ ext{Second term: } 1.5 \cdot 3 = 4.5
  3.  extThirdterm:\ rac1.5(1.51)2!2=0.75 \ ext{Third term: } \ rac{1.5(1.5-1)}{2!} \cdot 2 = 0.75
  4.  extFourthterm:\ rac1.5(.51)(1.52)3!(0)=0 \ ext{Fourth term: } \ rac{1.5(.5-1)(1.5-2)}{3!} \cdot (0) = 0

Adding them together:

P(33)=2+(4.5)+(0.75)+(0)P(33)= 2+(4.5)+(0.75)+(0)

Thus, the interpolated value of yy at x=2.5x =2.5 is approximately 7.257.25

Newton Forward Interpolation Calculator

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