Newton's Divided Difference Interpolation Method

Newton's Divided Difference formula is used for polynomial interpolation. It provides a way to construct a polynomial that passes through a given set of points. The formula is particularly useful when the points are not equally spaced.

f(x)=y_0+(x-x_0)f[x_0,x_1]+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+(x-x_0)(x-x_1)(x-x_2)f[x_0,x_1,x_2,x_3]

Where:

$f(x)$ is the interpolating polynomial ,which gives the value at $x$
$y_0$ is the initial value corresponding to $x_0$ given.
$f[x_0,x_1,....,x_n]$ are the divided differences. $f[x_0,x_1...x_n] = \ rac{ f[x_1...x_n] -f[x_0,x_1...x_{n-1}]}{ x_n - x_0}$
The terms $(x - x_0)(x - x_1) \dots$ are Basis Polynomials,Here, 𝑥 is the point at which you want to evaluate the interpolating polynomial. $x_ n$ is one of the known data points

Example of Newton's Divided Difference Interpolation

Given the following data points:

x	$y= f(x_0)$
0	1
1	3
3	49
4	129
7	813

We are tasked with finding $f(x)$ where $x = 0.3$ .

Step 1: Calculate the divided differences for the $y$ values.

x	$y=f(x_0)$	$1^{st} order$	$2^{st} order$	$3^{st} order$	$4^{st} order$
0	1	$\ rac{3-1}{1-0}=2$	$\ rac{23-2}{3-0}=7$	$\ rac{19-7}{4-0}=3$	$\ rac{3-3}{7-0}=0$
1	3	$\ rac{49-3}{3-1}=23$	$\ rac{80-23}{4-1}=19$	$\ rac{37-19}{7-1}=3$
3	49	$\ rac{129-49}{4-3}=80$	$\ rac{228-80}{7-3}=37$
4	129	$\ rac{813-129}{7-4}=228$
7	813

Step 2: Apply the Newtons Divided Interpolation formula:

f(x)=y_0+(x-x_0)f[x_0,x_1]+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+(x-x_0)(x-x_1)(x-x_2)f[x_0,x_1,x_2,x_3] \cdots

Step 3: Substituting the values into the interpolation formula:

f(0.3) = 1 + (0.3-0)2 + (0.3)(0.3-1)7 + (0.3)(0.3-1)(0.3-3)3 + 0

Step 4: After solving the equation, the interpolated value of $f(x)$ at $x = 0.3$ is approximately:

P(0.3) \approx 1.831

This is the final result of the Newtons Divided Interpolation method.

Newton's Divided Difference Interpolation Calculator

Plot:

Difference Table

x