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Sulaiman Khan
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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. This method is particularly useful for finding the value of a function at a specific point, given its values at known points.

The formula for Lagrange interpolation method Interpolation is:

P(x)=y0⋅L0(x)+y1⋅L1(x)+y2⋅L2(x)+⋯yn⋅Ln(x) 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)
Li(x)=(x−x0)(x−x1)⋯(x−xi−1)(x−xi+1)⋯(x−xn)(xi−x0)(xi−x1)⋯(xi−xi−1)(xi−xi+1)⋯(xi−xn)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:

  • (xn,yn)(x_n,y_n) are the given data points.
  • Li(x)L_i(x) is the Lagrange basis polynomial,
    Li(x)=∏j=0,j ≠ inx−xjxi−xjL_i(x) = \prod_{j=0, j \,\neq \, i}^{n} \frac{x - x_j}{x_i - x_j}
  • nn is the number of data points.
  • P(x)P(x) is the The Lagrange polynomial.

Example of Newton Backward Interpolation

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


xxyy
05
27
38
510
612

We are tasked with finding yy where x=4x = 4


Step 1: Calculate the Ln(x)L_n (x) for the yy values.


xxyyLn(x)L_n(x)
05\ rac(4−2)(4−3)(4−5)(4−6)(0−2)(0−3)(0−5)(0−6)=0.0222\ rac{(4-2)(4-3)(4-5)(4-6)}{(0-2)(0-3)(0-5)(0-6)} = 0.0222
27\ rac(4−0)(4−3)(4−5)(4−6)(2−0)(2−3)(2−5)(2−6)=0.3333\ rac{(4-0)(4-3)(4-5)(4-6)}{(2-0)(2-3)(2-5)(2-6)} = 0.3333
38\ rac(4−0)(4−2)(4−5)(4−6)(3−0)(3−2)(3−5)(3−6)=0.8889\ rac{(4-0)(4-2)(4-5)(4-6)}{(3-0)(3-2)(3-5)(3-6)} = 0.8889
510\ rac(4−0)(4−2)(4−3)(4−6)(5−0)(5−2)(5−3)(5−6)=0.5333\ rac{(4-0)(4-2)(4-3)(4-6)}{(5-0)(5-2)(5-3)(5-6)} = 0.5333
612\ rac(4−0)(4−2)(4−3)(4−5)(6−0)(6−2)(6−3)(6−5)=0.1111\ rac{(4-0)(4-2)(4-3)(4-5)}{(6-0)(6-2)(6-3)(6-5)} = 0.1111

Step 3: Apply the Newton Backward Interpolation formula:


P(x)=y0⋅L0(x)+y1⋅L1(x)+y2⋅L2(x)+⋯yn⋅Ln(x) 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)

Substituting the values:

P(4)=5⋅0.0222+7⋅0.3333+8⋅0888910⋅0.5333+12⋅0.1111P(4) = 5 \cdot 0.0222 + 7 \cdot 0.3333 + 8 \cdot 08889 10 \cdot 0.5333 + 12 \cdot 0.1111

Substituting the values:

P(4)=0.1111−2.3333+7.1111+5.3333+−1.3333P(4) = 0.1111 - 2.3333 + 7.1111+ 5.3333 + - 1.3333

Adding them together:

P(4)=8.8889P(4) = 8.8889

Thus, the interpolated value of yy at x=4x = 4 is approximately 8.8889 8.8889

Lagrange Interpolation Method Calculator

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