UNIT 2 • UNEQUAL INTERVAL INTERPOLATION

Lagrange Polynomial
Interpolation Method

Lagrange Interpolation builds a weighted combination of basis polynomials $L_i(x)$ where each basis equals $1$ at its own point $x_i$ and $0$ at all other points—making it ideal for non-uniform data without computing difference tables!

LAGRANGE BASIS FORMULAEXACT POLYNOMIAL WEIGHTS

P(x) = \sum_{i=0}^n y_i \cdot L_i(x) = y_0 L_0(x) + y_1 L_1(x) + \cdots + y_n L_n(x)

L_i(x) = \prod_{j=0, j \neq i}^n \frac{x - x_j}{x_i - x_j}

1. Basis Condition

L_i(x_j) = 1

i=j

, else

0

2. Sum of WeightsAlways sums to 1:

\sum L_i(x) = 1

3. No Grid RestrictionWorks seamlessly for equal or unequal intervals.

Step-by-Step Worked Example

GUIDED STORY

OUR PROBLEM TO SOLVE

Given the unequally spaced dataset:

x	5	6	9	11
y	12	13	14	16

Let's estimate the value at $x = 10$ !

STEP 01 • EVALUATE LAGRANGE BASIS WEIGHTS L_i(10)

Calculate each basis fraction at $x = 10$ :

L_0(10) = \frac{(10-6)(10-9)(10-11)}{(5-6)(5-9)(5-11)} = \frac{-4}{-24} = \frac{1}{6}

L_1(10) = \frac{(10-5)(10-9)(10-11)}{(6-5)(6-9)(6-11)} = \frac{-5}{15} = -\frac{1}{3}

L_2(10) = \frac{(10-5)(10-6)(10-11)}{(9-5)(9-6)(9-11)} = \frac{-20}{-24} = \frac{5}{6}

L_3(10) = \frac{(10-5)(10-6)(10-9)}{(11-5)(11-6)(11-9)} = \frac{20}{60} = \frac{1}{3}

STEP 02 • COMBINE WEIGHTED SUM

Multiply each basis weight by its corresponding $y_i$ value:

P(10) = 12\left(\frac{1}{6}\right) + 13\left(-\frac{1}{3}\right) + 14\left(\frac{5}{6}\right) + 16\left(\frac{1}{3}\right)

P(10) = 2.00000 - 4.33333 + 11.66667 + 5.33333

P(10) = 14.66667 = \frac{44}{3}

FINAL RESULT

Interpolated Value at x = 10: P(10) ≈ 14.6667

\boxed{P(10) = 14.6667 = \frac{44}{3}}

Using Lagrange polynomial interpolation, the estimated value of $y$ at $x = 10$ is 14.6667!

LABORATORY

Interactive Live Calculator & Visualizer

UNEQUAL INTERVAL INTERPOLATION LABORATORY

Lagrange Polynomial Engine

PreviousGauss Backward Interpolation

Unit 2•10 of 28

NextNewton Divided Difference

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 2 • UNEQUAL INTERVAL INTERPOLATION

Lagrange Polynomial
Interpolation Method

LAGRANGE BASIS FORMULAEXACT POLYNOMIAL WEIGHTS

P(x) = \sum_{i=0}^n y_i \cdot L_i(x) = y_0 L_0(x) + y_1 L_1(x) + \cdots + y_n L_n(x)

L_i(x) = \prod_{j=0, j \neq i}^n \frac{x - x_j}{x_i - x_j}

1. Basis Condition

L_i(x_j) = 1

i=j

, else

0

2. Sum of WeightsAlways sums to 1:

\sum L_i(x) = 1

3. No Grid RestrictionWorks seamlessly for equal or unequal intervals.

Step-by-Step Worked Example

GUIDED STORY

OUR PROBLEM TO SOLVE

Given the unequally spaced dataset:

x	5	6	9	11
y	12	13	14	16

Let's estimate the value at $x = 10$ !

STEP 01 • EVALUATE LAGRANGE BASIS WEIGHTS L_i(10)

Calculate each basis fraction at $x = 10$ :

L_0(10) = \frac{(10-6)(10-9)(10-11)}{(5-6)(5-9)(5-11)} = \frac{-4}{-24} = \frac{1}{6}

L_1(10) = \frac{(10-5)(10-9)(10-11)}{(6-5)(6-9)(6-11)} = \frac{-5}{15} = -\frac{1}{3}

L_2(10) = \frac{(10-5)(10-6)(10-11)}{(9-5)(9-6)(9-11)} = \frac{-20}{-24} = \frac{5}{6}

L_3(10) = \frac{(10-5)(10-6)(10-9)}{(11-5)(11-6)(11-9)} = \frac{20}{60} = \frac{1}{3}

STEP 02 • COMBINE WEIGHTED SUM

Multiply each basis weight by its corresponding $y_i$ value:

P(10) = 12\left(\frac{1}{6}\right) + 13\left(-\frac{1}{3}\right) + 14\left(\frac{5}{6}\right) + 16\left(\frac{1}{3}\right)

P(10) = 2.00000 - 4.33333 + 11.66667 + 5.33333

P(10) = 14.66667 = \frac{44}{3}

FINAL RESULT

Interpolated Value at x = 10: P(10) ≈ 14.6667

\boxed{P(10) = 14.6667 = \frac{44}{3}}

Using Lagrange polynomial interpolation, the estimated value of $y$ at $x = 10$ is 14.6667!

LABORATORY

Interactive Live Calculator & Visualizer

UNEQUAL INTERVAL INTERPOLATION LABORATORY

Lagrange Polynomial Engine

PreviousGauss Backward Interpolation

Unit 2•10 of 28

NextNewton Divided Difference

Point #	x_i	y_i	Action
1
2
3
4

Point #	x_i	y_i	Action
1
2
3
4

Lagrange Polynomial Interpolation Method

Step-by-Step Worked Example

Interpolated Value at x = 10: P(10) ≈ 14.6667

Interactive Live Calculator & Visualizer

Lagrange Polynomial Engine

NETZ Engine

Lagrange Polynomial Interpolation Method

Step-by-Step Worked Example

Interpolated Value at x = 10: P(10) ≈ 14.6667

Interactive Live Calculator & Visualizer

Lagrange Polynomial Engine

Lagrange Polynomial
Interpolation Method

Lagrange Polynomial
Interpolation Method