Newton Backward Interpolation Method

Newton Backward 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 end of the data set. It utilizes backward differences to form the interpolation polynomial.

The formula for Newton Backward Interpolation is:

P(x) = y_n + v \cdot \Delta y_n + \frac{v(v+1)}{2!} \cdot \Delta^2 y_{n-2} + \frac{v(v+1)(v+2)}{3!} \cdot \Delta^3 y_{n-3} + \frac{v(v+1)(v+2)(v+3)}{4!} \cdot \Delta^4 y_{n-4} + \cdots

where:

$v = \frac{x - x_n}{h}$
$x_n$ is the last value of $x$ in the data.
$h$ is the uniform difference between the $x$ values (where $h = x_n - x_{n-1}$ ).
$\Delta y_n, \Delta^2 y_n, \dots$ are the backward differences.

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

Example of Newton Backward Interpolation

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

$x$	$y$
24	28.06
28	30.19
32	32.75
36	34.94
40	40

We are tasked with finding $y$ where $x = 33$

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

$x$	$y$	${\Delta y}$	${\Delta^2 y}$	${\Delta^3 y}$	${\Delta^4 y}$
24	28.06
28	30.19	2.13
32	32.75	2.56	0.43
36	34.94	2.19	-0.37	-0.8
40	40	5.06	2.87	3.24	4.04

Step 2: use the formula $v = \frac{x - x_n}{h}$ values.

\text{Given } x = 33 \text{ and } x_n = 40 \text{ with } h = 4,

v = \frac{x - x_n}{h}

v = \frac{33 - 40}{4}

v = -1.75

Step 3: Apply the Newton Backward Interpolation formula:

P(x) = y_n + v \cdot \Delta y_n + \frac{v(v+1)}{2!} \cdot \Delta^2 y_{n-2} + \frac{v(v+1)(v+2)}{3!} \cdot \Delta^3 y_{n-3} + \frac{v(v+1)(v+2)(v+3)}{4!} \cdot \Delta^4 y_{n-4} + \cdots

Substituting the values:

P(40) = 40 + \frac{(-1.75 \cdot (-1.75+1))}{1!} \cdot 5.0600 + \frac{(-1.75 \cdot (-1.75+1) \cdot (-1.75+2))}{2!} \cdot 2.8700 + \frac{(-1.75 \cdot (-1.75+1) \cdot (-1.75+2) \cdot (-1.75+3))}{3!} \cdot 3.2400 + \frac{(-1.75 \cdot (-1.75+1) \cdot (-1.75+2) \cdot (-1.75+3) \cdot (-1.75+4))}{4!} \cdot 4.0400

Step-by-step:

First term: $40$
Second term: $\frac{(−1.75)}{1!}∗5.0600 = −8.85500$
Third term: $\frac{(−1.75∗(−1.75+1))}{2!} ∗2.8700 = 0.17718$
Fourth term: $\frac{(−1.75∗(−1.75+1)∗(−1.75+2))}{3!} ∗3.2400 = 0.17718$
Fifth term: $\frac{(−1.75∗(−1.75+1)∗(−1.75+2)∗(−1.75+3))}{4!} ∗4.0400 =0.06904$

Adding them together:

P(33)= 40+(−8.855000000000004)+(1.883437500000003)+(0.1771875000000003)+(0.06904296875000004)

Thus, the interpolated value of $y$ at $x = 33$ is approximately $33.27466$

Newton Backward Interpolation Calculator

Newton Backward Interpolation Method

Example of Newton Backward Interpolation

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

Step-by-step:

Newton Backward Interpolation Calculator

Plot: