Lagrange Interpolation Formula

Lagrange Interpolation Formula finds a polynomial called Lagrange Polynomial that takes on certain values at an arbitrary point. It is an n^th degree polynomial expression to the function f(x). The interpolation method is used to find the new data points within the range of a discrete set of known data points.

Lagrange Interpolation Formula

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. The Lagrange Interpolation formula for different orders i.e., n^th order is given as,

Lagrange Interpolation Formula for n^th order is-

$f(x)=\frac{(x-x_1)(x-x_2)...(x-x_n)}{(x_0-x_1)(x_0-x_2)...(x_0-x_n)}\times y_0+\frac{(x-x_0)(x-x_2)...(x-x_n)}{(x_1-x_0)(x_1-x_2)...(x_1-x_n)}\times y_1+...+\frac{(x-x_0)(x-x_1)...(x-x_n-1)}{(x_n-x_0)(x_n-x_1)...(x_n-x_n-1)}\times y_n$

If theDegree of the polynomial is 1 then,

Lagrange Interpolation Formula for 1^st order polynomial is-

$f(x)=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1$

Similarly for 2^nd Order polynomial, the Lagrange Interpolation formula is-

$f(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\times y_0+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\times y_1+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\times y_2$

Proof of Lagrange Theorem

Let’s consider a nth degree polynomial of given form
$f(x)=A_0(x-x_1)(x-x_2)(x-x_3)...(x-x_n)+A_1(x-x_0)(x-x_2)(x-x_3)...(x-x_n)+...+A_n(x-x_1)(x-x_2)(x-x_3)...(x-x_{n-1})$
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)
In similar way 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.
$f(x)=\frac{(x-x_1)(x-x_2)...(x-x_n)}{(x_0-x_1)(x_0-x_2)...(x_0-x_n)}\times y_0+\frac{(x-x_0)(x-x_2)...(x-x_n)}{(x_1-x_0)(x_1-x_2)...(x_1-x_n)}\times y_1+...+\frac{(x-x_0)(x-x_1)...(x-x_n-1)}{(x_n-x_0)(x_n-x_1)...(x_n-x_n-1)}\times y_n$

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

Sample Problems

Question 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)
x = 2
As per the 1^st order Lagrange Interpolation Formula,
$y=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1$
$=\frac{(2-3)}{(1-3)}\times 2+\frac{(2-1)}{(3-1)}\times 4$
= (-2/-2) + (4/2)
= 1 + 2
y = 3

Question 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)
x = 5
As per the 1^st order Lagrange Interpolation Formula,
$y=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1 =\frac{(5-3)}{(9-3)}\times 2+\frac{(5-9)}{(3-9)}\times 10$
= (4/6) + (-40/-6)
= (2/3) + (20/3)
= 22/3
y = 7.33

Question 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)
x = 1
As per the 2^nd order Lagrange Interpolation Formula
$y=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\times y_0+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\times y_1+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\times y_2 =\frac{(1-3)(1-2)}{(1-3)(1-2)}\times 6+\frac{(1-1)(1-2)}{(3-1)(3-2)}\times 4+\frac{(1-1)(1-3)}{(2-1)(2-3)}\times 5 =\frac{(-2)(-1)}{(-2)(-1)}\times 6+\frac{(0)(-1)}{(2)(1)}\times 4+\frac{(0)(-2)}{(1)(-1)}\times 5$
= (12/2) + 0 + 0
y = 6

Question 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)
x = 10
As per the 2^nd order Lagrange Interpolation Formula,
$y=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\times y_0+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\times y_1+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\times y_2 =\frac{(10-3)(10-1)}{(9-3)(9-1)}\times 6+\frac{(10-9)(10-1)}{(3-9)(3-1)}\times 5+\frac{(10-9)(10-3)}{(1-9)(1-3)}\times 12 =\frac{(7)(9)}{(6)(8)}\times 6+\frac{(1)(9)}{(-6)(2)}\times 5+\frac{(1)(7)}{(-8)(-2)}\times 12$
= (63/8) + (-15/4) + (21/4)
= (63-30 + 42)/8
= 75/8
y = 9.375

Question 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)
x = 7
As per the 3^rd order Lagrange Interpolation Formula,
$y=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0+\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1+\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3 =\frac{(7-2)(7-3)(7-5)}{(1-2)(1-3)(1-5)}\times 10+\frac{(7-1)(7-3)(7-5)}{(2-1)(2-3)(2-5)}\times 4+\frac{(7-1)(7-2)(7-5)}{(3-1)(3-2)(3-5)}\times 4+\frac{(7-1)(7-2)(7-3)}{(5-1)(5-2)(5-3)}\times 7 =\frac{(5)(4)(2)}{(-1)(-2)(-4)}\times 10+\frac{(6)(4)(2)}{(1)(-1)(-3)}\times 4+\frac{(6)(5)(2)}{(2)(1)(-2)}\times 4+\frac{(6)(5)(4)}{(4)(3)(2)}\times 7$
= -50 + 64 – 60 + 35
= 99 – 110
y = -11

Question 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)
x = 10
As per the 3^rd order Lagrange Interpolation Formula
$y=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0+\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1+\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3 =\frac{(10-6)(10-7)(10-8)}{(5-6)(5-7)(5-8)}\times 12+\frac{(10-5)(10-7)(10-8)}{(6-5)(6-7)(6-8)}\times 13+\frac{(10-5)(10-6)(10-8)}{(7-5)(7-6)(7-8)}\times 14+\frac{(10-5)(10-6)(10-7)}{(8-5)(8-6)(8-7)}\times 15 =\frac{(4)(3)(2)}{(-1)(-2)(-3)}\times 12+\frac{(5)(3)(2)}{(1)(-1)(-2)}\times 13+\frac{(5)(4)(2)}{(2)(1)(-1)}\times 14+\frac{(5)(4)(3)}{(3)(2)(1)}\times 15$
= -48 + 195 – 280 + 150
y = 17

Question 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)
x = 0
As per the 1^st order Lagrange Interpolation Formula
$y=\frac{(x-x_1)}{(x_0-x_1)}\times y_0+\frac{(x-x_0)}{(x_1-x_0)}\times y_1 =\frac{(0-1)}{(-2-1)}\times 5+\frac{(0+2)}{(1+2)}\times 7$
= (5/3) + (14/3)
= 19/3
y = 6.33

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Last Updated : 24 Feb, 2022

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