UNIT 2 • UNEQUAL INTERVAL INTERPOLATION

Newton's Divided
Difference Method

When your data points are unequally spaced, standard forward/backward difference tables break down. Newton's Divided Difference Method solves this by dividing each difference by its corresponding interval span!

DIVIDED DIFFERENCE FORMULAUNEQUAL SPACING READY

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

1. First Divided Diff

f[x_0,x_1] = \frac{y_1 - y_0}{x_1 - x_0}

2. Second Divided Diff

f[x_0,x_1,x_2] = \frac{f[x_1,x_2] - f[x_0,x_1]}{x_2 - x_0}

3. Polynomial TermsMultiplies differences by progressive factors

(x - x_0)(x - x_1)\dots

Step-by-Step Worked Example

GUIDED STORY

OUR PROBLEM TO SOLVE

Given the unequally spaced dataset:

x	0	1	3	4	7
y	1	3	49	129	813

Let's estimate the value of $f(x)$ at $x = 0.3$ !

STEP 01 • CONSTRUCT DIVIDED DIFFERENCE TABLE

x	y	1st Order	2nd Order	3rd Order	4th Order
0 (x_0)	1 (y_0)	2 (f[x_0,x_1])	7 (f[x_0,x_1,x_2])	3 (f[x_0..x_3])	0
1	3	23	19	3	-
3	49	80	37	-	-
4	129	228	-	-	-
7	813	-	-	-	-

STEP 02 • FORMULA & NUMERICAL SUBSTITUTION

Newton Divided Difference 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]

Substituting values ( $x = 0.3, y_0 = 1, f[x_0,x_1] = 2, f[x_0,x_1,x_2] = 7, f[x_0..x_3] = 3$ ):

f(0.3) = 1 + (0.3)(2) + (0.3)(-0.7)(7) + (0.3)(-0.7)(-2.7)(3)

f(0.3) = 1.00000 + 0.60000 - 1.47000 + 1.70100

f(0.3) = 1.83100

FINAL RESULT

Interpolated Value at x = 0.3: f(0.3) = 1.831

\boxed{f(0.3) = 1.831}

Using Newton's divided differences for unequal spacing, the estimated functional value at $x = 0.3$ is 1.831!

LABORATORY

Interactive Live Calculator & Visualizer

UNEQUAL INTERVAL INTERPOLATION LABORATORY

Newton Divided Difference Engine

PreviousLagrange Interpolation

Unit 2•11 of 28

NextTrapezoidal Rule

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm