UNIT 2 • EQUAL INTERVAL INTERPOLATION

Newton Backward
Interpolation Method

Newton Backward Interpolation is used to estimate the value of a function at a given point when 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 by utilizing backward differences.

BACKWARD DIFFERENCE FORMULAEND-POINT ANCHOR

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

Where:

$v = \frac{x - x_n}{h}$ is the backward fractional distance.
$x_n$ is the last value of $x$ in the dataset.
$h$ is the uniform step size between $x$ values ( $h = x_n - x_{n-1}$ ).
$\nabla y_n, \nabla^2 y_n, \nabla^3 y_n, \dots$ are the backward differences along the bottom row.

1. Backward Ratio (v)Distance measured backwards from the final baseline

x_n

2. Equal Spacing (h)Constant gap between consecutive tabulated points.

3. Bottom-Row OperatorsUses backward difference values along the bottom diagonal.

Example of Newton Backward Interpolation

GUIDED STORY TUTORIAL

OUR PROBLEM TO SOLVE

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

x	24	28	32	36	40
y	28.06	30.19	32.75	34.94	40.00

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

STEP 01 • CALCULATE THE BACKWARD DIFFERENCES FOR THE Y VALUES

We construct the difference table by subtracting each $y$ value from its successor. The bottom row values (highlighted in emerald) form our primary backward difference vector $[y_n, \nabla y_n, \nabla^2 y_n, \nabla^3 y_n, \nabla^4 y_n]$ :

x	y	∇y	∇²y	∇³y	∇⁴y
24	28.06	-	-	-	-
28	30.19	2.13	-	-	-
32	32.75	2.56	0.43	-	-
36	34.94	2.19	-0.37	-0.80	-
40 (x_n)	40.00 (y_n)	5.06 (∇y_n)	2.87 (∇²y_n)	3.24 (∇³y_n)	4.04 (∇⁴y_n)

STEP 02 • USE THE FORMULA v = (x - x_n) / h

Given target point $x = 33$ , final base point $x_n = 40$ , and uniform step size $h = 4$ :

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

v = \frac{33 - 40}{4} = \frac{-7}{4} = -1.75

STEP 03 • APPLY THE NEWTON BACKWARD INTERPOLATION FORMULA

Governing polynomial expansion:

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

Substituting values ( $v = -1.75, y_n = 40, \nabla y_n = 5.06, \nabla^2 y_n = 2.87, \nabla^3 y_n = 3.24, \nabla^4 y_n = 4.04$ ):

P(33) = 40 + \frac{-1.75}{1!}(5.06) + \frac{(-1.75)(-0.75)}{2!}(2.87) + \frac{(-1.75)(-0.75)(0.25)}{3!}(3.24) + \frac{(-1.75)(-0.75)(0.25)(1.25)}{4!}(4.04)

P(33) = 40.00000 - 8.85500 + 1.88344 + 0.17719 + 0.06904

P(33) = 33.27467

CONCLUSION

Interpolated Result: y(33) ≈ 33.27466

\boxed{P(33) \approx 33.27466}

Thus, the interpolated value of $y$ at $x = 33$ using Newton Backward Interpolation is approximately 33.27466.

LABORATORY

Interactive Live Calculator & Visualizer

BACKWARD INTERPOLATION LABORATORY

Newton Backward Interactive Engine

PreviousNewton Forward Interpolation

Unit 2•7 of 28

NextGauss Forward Interpolation

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 2 • EQUAL INTERVAL INTERPOLATION

Newton Backward
Interpolation Method

BACKWARD DIFFERENCE FORMULAEND-POINT ANCHOR

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

Where:

$v = \frac{x - x_n}{h}$ is the backward fractional distance.
$x_n$ is the last value of $x$ in the dataset.
$h$ is the uniform step size between $x$ values ( $h = x_n - x_{n-1}$ ).
$\nabla y_n, \nabla^2 y_n, \nabla^3 y_n, \dots$ are the backward differences along the bottom row.

1. Backward Ratio (v)Distance measured backwards from the final baseline

x_n

2. Equal Spacing (h)Constant gap between consecutive tabulated points.

3. Bottom-Row OperatorsUses backward difference values along the bottom diagonal.

Example of Newton Backward Interpolation

GUIDED STORY TUTORIAL

OUR PROBLEM TO SOLVE

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

x	24	28	32	36	40
y	28.06	30.19	32.75	34.94	40.00

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

STEP 01 • CALCULATE THE BACKWARD DIFFERENCES FOR THE Y VALUES

x	y	∇y	∇²y	∇³y	∇⁴y
24	28.06	-	-	-	-
28	30.19	2.13	-	-	-
32	32.75	2.56	0.43	-	-
36	34.94	2.19	-0.37	-0.80	-
40 (x_n)	40.00 (y_n)	5.06 (∇y_n)	2.87 (∇²y_n)	3.24 (∇³y_n)	4.04 (∇⁴y_n)

STEP 02 • USE THE FORMULA v = (x - x_n) / h

Given target point $x = 33$ , final base point $x_n = 40$ , and uniform step size $h = 4$ :

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

v = \frac{33 - 40}{4} = \frac{-7}{4} = -1.75

STEP 03 • APPLY THE NEWTON BACKWARD INTERPOLATION FORMULA

Governing polynomial expansion:

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

Substituting values ( $v = -1.75, y_n = 40, \nabla y_n = 5.06, \nabla^2 y_n = 2.87, \nabla^3 y_n = 3.24, \nabla^4 y_n = 4.04$ ):

P(33) = 40 + \frac{-1.75}{1!}(5.06) + \frac{(-1.75)(-0.75)}{2!}(2.87) + \frac{(-1.75)(-0.75)(0.25)}{3!}(3.24) + \frac{(-1.75)(-0.75)(0.25)(1.25)}{4!}(4.04)

P(33) = 40.00000 - 8.85500 + 1.88344 + 0.17719 + 0.06904

P(33) = 33.27467

CONCLUSION

Interpolated Result: y(33) ≈ 33.27466

\boxed{P(33) \approx 33.27466}

Thus, the interpolated value of $y$ at $x = 33$ using Newton Backward Interpolation is approximately 33.27466.

LABORATORY

Interactive Live Calculator & Visualizer

BACKWARD INTERPOLATION LABORATORY

Newton Backward Interactive Engine

PreviousNewton Forward Interpolation

Unit 2•7 of 28

NextGauss Forward Interpolation

Point #	x_i	y_i	Action
1
2
3
4

Point #	x_i	y_i	Action
1
2
3
4

Newton Backward Interpolation Method

Example of Newton Backward Interpolation

Interpolated Result: y(33) ≈ 33.27466

Interactive Live Calculator & Visualizer

Newton Backward Interactive Engine

NETZ Engine

Newton Backward Interpolation Method

Example of Newton Backward Interpolation

Interpolated Result: y(33) ≈ 33.27466

Interactive Live Calculator & Visualizer

Newton Backward Interactive Engine

Newton Backward
Interpolation Method

Newton Backward
Interpolation Method