## Newton Forward Interpolation Formula

While extrapolation refers to the process of calculating the value of the function outside the specified range, Newton Forward Interpolation is the technique of estimating the value of a function for any intermediate value of the independent variable**.**

The degree of a **Newton interpolating polynomial** can be increased by including additional terms and points without removing any already present ones, just like with other difference formulas. The simplicity of Newton’s form is that any additional points are always added at one end. Newton’s forward formula can add additional points to the right, and Newton’s backward formula can add additional points to the left.

The distance between the interpolated point and the center of the set of points’ x values determines how accurate polynomial interpolation is. Naturally, the middle distance from the first data point increases as new points are added at one end. Therefore, the middle of the x-values may be far from where the interpolation is done if it is unknown how many points will be required for the desired **accuracy.**

**Also read:**

Forward Differences: The differences y1 – y0, y2 – y1, y3 – y2,……, yn – yn-1 are referred to as the first forward differences, respectively, when they are denoted by dy0, dy1, dy2,……, dyn-1. The following are the first forward differences:

ΔY_{r }= Y_{r+1} – Y_{r}

**NEWTON’S GREGORY FORWARD INTERPOLATION FORMULA :**

Newton Forward Interpolation is particularly helpful for extrapolating f(x) values close to the start of the given set of values. u = (x – a) / h, where a is the first term, and ‘h’ is referred to as the interval of difference.

let’s take an example for a better understanding of Newton Forward Interpolation

**Evaluate f(1.2) using the given data by Newton`s forward interpolation formula.**

x | 0 | 1 | 2 | 3 | 4 |

f(x) | 1 | 1.5 | 1.2 | 3.1 | 4.3 |

To solve the problem first make a finite table by using the given data

Putting all the values in **Newton`s forward interpolation formula.**

**u = x-x_{0}/h =1.2 – 0 = 1.2**

**(Where h =**

**x**)_{1}-x_{0}Output:- Value of f(1.2) is 1.656

** **

In this post, we see the forward interpolation formula and its example and how to solve the problem in the next post we discuss the back work interpolation formula.

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