Lagrange Interpolation Method

The Lagrange interpolation method is a polynomial interpolation technique used to construct a polynomial that passes through a given set of points (which need not be equally spaced). This method is particularly useful for finding the value of a function at a specific point given its values at known points.

Formula for Lagrange Interpolation:

P(x) = y_0 \cdot L_0(x) + y_1 \cdot L_1(x) + y_2 \cdot L_2(x) + \cdots + y_n \cdot L_n(x)

L_i(x) = \frac{(x - x_0)(x - x_1) \cdots (x - x_{i-1})(x - x_{i+1}) \cdots (x - x_n)}{(x_i - x_0)(x_i - x_1) \cdots (x_i - x_{i-1})(x_i - x_{i+1}) \cdots (x_i - x_n)}

Where:

$(x_n,y_n)$ are the given data points.
$L_i(x)$ is the Lagrange basis polynomial.
$P(x)$ is the Lagrange interpolating polynomial.

Example of Lagrange Interpolation

Given the following data points:

$x$	$y$
0	5
2	7
3	8
5	10
6	12

We are tasked with finding $y$ where $x = 4$ .

Step 1: Calculate Basis Polynomials $L_n(x)$

$x$	$y$	$L_n(x)$
0	5	$\frac{(4-2)(4-3)(4-5)(4-6)}{(0-2)(0-3)(0-5)(0-6)} = 0.0222$
2	7	$\frac{(4-0)(4-3)(4-5)(4-6)}{(2-0)(2-3)(2-5)(2-6)} = 0.3333$
3	8	$\frac{(4-0)(4-2)(4-5)(4-6)}{(3-0)(3-2)(3-5)(3-6)} = 0.8889$
5	10	$\frac{(4-0)(4-2)(4-3)(4-6)}{(5-0)(5-2)(5-3)(5-6)} = 0.5333$
6	12	$\frac{(4-0)(4-2)(4-3)(4-5)}{(6-0)(6-2)(6-3)(6-5)} = 0.1111$

Step 2: Substitute Values

P(4) = 5(0.0222) + 7(0.3333) + 8(0.8889) + 10(0.5333) + 12(0.1111)

P(4) = 0.1111 + 2.3331 + 7.1112 + 5.3330 + 1.3332 = 16.2216

Conclusion

P(4) \approx 16.22

Thus, the interpolated value of $y$ at $x = 4$ using Lagrange interpolation is approximately $16.22$ .

Lagrange Interpolation Method Calculator

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Lagrange Interpolation Method

Formula for Lagrange Interpolation:

P(x) = y_0 \cdot L_0(x) + y_1 \cdot L_1(x) + y_2 \cdot L_2(x) + \cdots + y_n \cdot L_n(x)

L_i(x) = \frac{(x - x_0)(x - x_1) \cdots (x - x_{i-1})(x - x_{i+1}) \cdots (x - x_n)}{(x_i - x_0)(x_i - x_1) \cdots (x_i - x_{i-1})(x_i - x_{i+1}) \cdots (x_i - x_n)}

Where:

$(x_n,y_n)$ are the given data points.
$L_i(x)$ is the Lagrange basis polynomial.
$P(x)$ is the Lagrange interpolating polynomial.

Example of Lagrange Interpolation

Given the following data points:

$x$	$y$
0	5
2	7
3	8
5	10
6	12

We are tasked with finding $y$ where $x = 4$ .

Step 1: Calculate Basis Polynomials $L_n(x)$

$x$	$y$	$L_n(x)$
0	5	$\frac{(4-2)(4-3)(4-5)(4-6)}{(0-2)(0-3)(0-5)(0-6)} = 0.0222$
2	7	$\frac{(4-0)(4-3)(4-5)(4-6)}{(2-0)(2-3)(2-5)(2-6)} = 0.3333$
3	8	$\frac{(4-0)(4-2)(4-5)(4-6)}{(3-0)(3-2)(3-5)(3-6)} = 0.8889$
5	10	$\frac{(4-0)(4-2)(4-3)(4-6)}{(5-0)(5-2)(5-3)(5-6)} = 0.5333$
6	12	$\frac{(4-0)(4-2)(4-3)(4-5)}{(6-0)(6-2)(6-3)(6-5)} = 0.1111$

Step 2: Substitute Values

P(4) = 5(0.0222) + 7(0.3333) + 8(0.8889) + 10(0.5333) + 12(0.1111)

P(4) = 0.1111 + 2.3331 + 7.1112 + 5.3330 + 1.3332 = 16.2216

Conclusion

P(4) \approx 16.22

Thus, the interpolated value of $y$ at $x = 4$ using Lagrange interpolation is approximately $16.22$ .

Lagrange Interpolation Method Calculator

Plot:

PreviousGauss Backward Interpolation

Unit 2•10 of 28

NextNewton Divided Difference

Lagrange Interpolation Method

Example of Lagrange Interpolation

Step 1: Calculate Basis Polynomials Ln(x)L_n(x)Ln​(x)

Step 2: Substitute Values

Conclusion

Lagrange Interpolation Method Calculator

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NETZ Engine

Lagrange Interpolation Method

Example of Lagrange Interpolation

Step 1: Calculate Basis Polynomials Ln(x)L_n(x)Ln​(x)

Step 2: Substitute Values

Conclusion

Lagrange Interpolation Method Calculator

Plot:

Step 1: Calculate Basis Polynomials $L_n(x)$

Step 1: Calculate Basis Polynomials $L_n(x)$