## Newton Backword Interpolation Formula

Today in this topic we discous about the **Newton Backword Interpolation Formula **and its example. In previous posts we discus about Newton Forward Interpolation Formula, Langrage Interpolating Formula Derivation, and Lagrange interpolation formula, and we also discuss Secant Method in Numerical Analysis/Techniques 2022 if you did not read yet go and check.

Newton Forward Interpolation is a method of approximating a function’s value for any intermediate value of the independent variable, whereas extrapolation refers to the process of calculating the function’s value outside the defined range.

A Newton polynomial, so named after its creator Isaac Newton[1], is an interpolation polynomial for a given set of data points in the mathematical discipline of numerical analysis. Because the Newton polynomial’s coefficients are determined by applying Newton’s split differences method, it is occasionally referred to as the “Newton’s divided differences interpolation polynomial.”

The precision of polynomial interpolation depends on the distance between the interpolated point and the center of the collection of points’ x values. Naturally, when new points are added at one end, the middle distance from the first data point grows. Therefore, if it is uncertain how many points will be needed to achieve the required accuracy, the interpolation may be performed distant from the middle of the x-values.

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:

∇Y_{r }= Y_{r} – Y_{r+1}

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

When the value of f(x) is needed close to the end of the table, this formula can be helpful. The difference between and interval is known as **u = ( x – an ) / h**, Here an is last term.

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

**Evaluate f(15) using the given data by Newton`s forward interpolation formula.**

x | 10 | 20 | 30 | 40 | 50 |

f(x) | 46 | 66 | 81 | 93 | 101 |

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

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

Output:- Value of f(15) is 96.84