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 data points are not equally spaced.

Formula for Newton's Divided Difference:

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.
$y_0$ is the initial value corresponding to $x_0$ .
$f[x_0,x_1,\dots,x_n]$ are the divided differences.

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 Divided Differences

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

Step 2 & 3: Substitute Values into 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

Conclusion

P(0.3) \approx 1.831

Thus, the interpolated value of $f(x)$ at $x = 0.3$ is approximately $1.831$ .

Newton's Divided Difference Interpolation Calculator

Plot:

Difference Table

x

PreviousLagrange Interpolation

Unit 2•11 of 28

NextTrapezoidal Rule

Newton's Divided Difference Interpolation Method

Formula for Newton's Divided Difference:

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.
$y_0$ is the initial value corresponding to $x_0$ .
$f[x_0,x_1,\dots,x_n]$ are the divided differences.

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 Divided Differences

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

Step 2 & 3: Substitute Values into 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

Conclusion

P(0.3) \approx 1.831

Thus, the interpolated value of $f(x)$ at $x = 0.3$ is approximately $1.831$ .

Newton's Divided Difference Interpolation Calculator

Plot:

Difference Table

x

PreviousLagrange Interpolation

Unit 2•11 of 28

NextTrapezoidal Rule

Newton's Divided Difference Interpolation Method

Example of Newton's Divided Difference Interpolation

Step 1: Calculate Divided Differences

Step 2 & 3: Substitute Values into Formula

Conclusion

Newton's Divided Difference Interpolation Calculator

Plot:

Difference Table

NETZ Engine

Newton's Divided Difference Interpolation Method

Example of Newton's Divided Difference Interpolation

Step 1: Calculate Divided Differences

Step 2 & 3: Substitute Values into Formula

Conclusion

Newton's Divided Difference Interpolation Calculator

Plot:

Difference Table