Lagrange Interpolation Formula

Lagrange Interpolation is a way of finding the value of any function at any given point when the function is not given. We use other points on the function to get the value of the function at any required point.

Suppose we have a function y = f(x) in which substituting the values of x gives different values of y. We are given two points (x₁, y₁) and (x₂, y₂) on the curve then the value of y at x = a(constant) is calculated using the Lagrange Interpolation Formula.

Read More:Lagranges Interpolation

Given few real values x₁, x₂, x₃, ..., x_n and y₁, y₂, y₃, ..., y_n and there will be a polynomial P with real coefficients satisfying the conditions P(x_i) = y_i, ∀ i = {1, 2, 3, ..., n} and degree of polynomial P must be less than the count of real values i.e., degree(P) < n.

Lagrange Interpolation Formula for nth Order

The Lagrange Interpolation formula for n^th degree polynomial is given below:

Lagrange Interpolation Formula for the n^th order is,

Lagrange First Order Interpolation Formula

If theDegree of the polynomial is 1 then it is called the First Order Polynomial. Lagrange Interpolation Formula for 1^st order polynomials is,

Lagrange Second Order Interpolation Formula

If the Degree of the polynomial is 2 then it is called Second Order Polynomial. Lagrange Interpolation Formula for 2nd order polynomials is,

Proof of Lagrange Theorem

Let's consider a nth-degree polynomial of the given form,

f(x) = A₀(x - x₁)(x - x₂)(x - x₃)...(x - x_n) + A₁(x - x₀)(x - x₂)(x - x₃)...(x - x_n) + ... + A_n(x - x₀)(x - x₁)(x - x₃)...(x - x_n-1)

Substitute observations x_i to get A_i

Put x = x₀ then we get A₀

f(x₀) = y₀ = A₀(x₀ - x₁)(x₀ - x₂)(x₀ - x₃)...(x₀ - x_n)

A₀ = y₀/(x₀ - x₁)(x₀ - x₂)(x₀ - x₃)...(x₀ - x_n)

By substituting x = x₁ we get A₁

f(x₁) = y₁ = A₁(x₁ - x₀)(x₁ - x₂)(x₁ - x₃)...(x₁ - x_n)

A₁ = y₁/(x₁ - x₀)(x₁ - x₂)(x₁ - x₃)...(x₁ - x_n)

Similarly, by substituting x = x_n we get A_n

f(x_n) = y_n = A_n(x_n - x₀)(x_n - x₁)(x_n - x₂)...(x_n - x_n-1)

A_n = y_n/(x_n - x₀)(x_n - x₁)(x_n - x₂)...(x_n - x_n-1)

If we substitute all values of A_i in function f(x) where i = 1, 2, 3, ...n then we get Lagrange Interpolation Formula as,

Properties of Lagrange Interpolation Formula

Various properties of the Lagrange Interpolation Formula are discussed below,

• This formula is used to find the value of the function at any point even when the function itself is not given.
• It is used even if the points given are not evenly spaced.
• It gives the value of the depennt variable for any independent variable belong to any function and thus is used in Numeracial Analysis for finding the values of the function, etc.

Uses of Lagrange Interpolation Formula

Various uses of the Lagrange Interpolation Formula are discussed below,

• It is used to find the value of the dependent variable at any particular independent variable even if the function itself is not given.
• It is used in image scaling.
• It is used in AI modeling.
• It is used to teach NLPs, etc.

Also Check:

• Interpolation Formula
• Linear Interpolation Formula

Examples Using Lagrange Interpolation Formula

Let's look into a few sample questions on Lagrange Interpolation Formula.

Example 1: Find the value of y at x = 2 for the given set of points (1, 2),(3, 4)

Solution:

Given,

• (x₀, y₀) = (1, 2)
• (x₁, y₁) = (3, 4)

First order Lagrange Interpolation Formula is,

At x = 2
y 
y = (-2/-2) + (4/2)
y = 1 + 2 = 3

The value of y at x = 2 is 3

Example 2: Find the value of y at x = 5 for the given set of points (9, 2), (3, 10)

Solution:

Given,

• (x₀, y₀) = (9, 2)
• (x₁, y₁) = (3, 10)

First order Lagrange Interpolation Formula is,

At x = 5

y = (4/6) + (-40/-6)
y = (2/3) + (20/3)
y = 22/3 = 7.33

The value of y at x = 5 is 7.33

Example 3: Find the value of y at x = 1 for the given set of points (1, 6), (3, 4), (2, 5)

Solution:

Given,

• (x₀, y₀) = (1, 6)
• (x₁, y₁) = (3, 4)
• (x₂, y₂) = (2, 5)

Second Order Lagrange Interpolation Formula is,

At x = 1

y = (12/2) + 0 + 0
y = 6

The value of y at x = 1 is 6

Example 4: Find the value of y at x = 10 for the given set of points (9, 6), (3, 5), (1, 12)

Solution:

Given,

• (x₀, y₀) = (9, 6)
• (x₁, y₁) = (3, 5)
• (x₂, y₂) = (1, 12)

Second Order Lagrange Interpolation Formula is,

At x = 10

y = (63/8) + (-15/4) + (21/4)
y = (63-30 + 42)/8
y = 75/8 = 9.375

The value of y at x = 10 is 9.375

Example 5: Find the value of y at x = 7 for the given set of points (1, 10), (2, 4), (3, 4), (5, 7)

Solution:

Given,

• (x₀, y₀) = (1, 10)
• (x₁, y₁) = (2, 4)
• (x₂, y₂) = (3, 4)
• (x₃, y₃) = (5, 7)

Third Order Lagrange Interpolation Formula is,

At x = 7

y = -50 + 64 - 60 + 35
y = 99 - 110 = -11

The value of y at x = 7 is -11

Example 6: Find the value of y at x = 10 for the given set of points (5, 12), (6, 13), (7, 14), (8, 15)

Solution:

Given,

• (x₀, y₀) = (5, 12)
• (x₁, y₁) = (6, 13)
• (x₂, y₂) = (7, 14)
• (x₃, y₃) = (8, 15)

Third Order Lagrange Interpolation Formula is,

At x = 10,

y = -48 + 195 - 280 + 150
y = 17

The value of y at x = 10 is 17

Example 7: Find the value of y at x = 0 for the given set of points (-2, 5),(1, 7)

Solution:

Given,

• (x₀, y₀) = (-2, 5)
• (x₁, y₁) = (1, 7)

First Order Lagrange Interpolation Formula is,

At x = 0,

y = (5/3) + (14/3)
y = 19/3 = 6.33

The value of y at x = 0 is 6.33