Lagrange’s Interpolation Formula
The formula was first published by Waring (1779), rediscovered by Euler in 1783, and published by Lagrange in 1795 (Jeffreys and Jeffreys 1988).
Lagrange interpolating polynomials are implemented in the Wolfram Language as InterpolatingPolynomial[data, var]. They are used, for example, in the construction of Newton-Cotes formulas.
When constructing interpolating polynomials, there is a tradeoff between having a better fit and having a smooth well-behaved fitting function. The more data points that are used in the interpolation, the higher the degree of the resulting polynomial, and therefore the greater oscillation it will exhibit between the data points. Therefore, a high-degree interpolation may be a poor predictor of the function between points, although the accuracy at the data points will be “perfect.”
The Lagrange interpolating polynomials can also be written using what Szegö (1975) called Lagrange’s fundamental interpolating polynomials. Let
Lagrange Interpolation Formula
Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points xj and numbers yj. Lagrange’s interpolation is also an nth
degree polynomial approximation to f(x).
Find the Lagrange Interpolation Formula given below,
What is Lagrange Interpolation Formula?
Given n distinct real values x1,x2,…,xn and n real values y1,y2,…,yn (not necessarily distinct), there is a unique polynomial P with real coefficients satisfying P(xi)=yi
for i ∈ {1, 2, …, n}, such that deg(P) < n. Lagrange interpolation formula for different order of polynomials is given as,
Lagrange First Order Interpolation Formula
Lagrange’s interpolation formula for polynomials of first order can be given as,
Lagrange Second Order Interpolation Formula
Lagrange’s interpolation formula for polynomials of second order can be given as,
Lagrange Interpolation Formula for Nth Degree Polynomial
Lagrange interpolation formula for Nth degree polynomial can be given as,
Let us see how to use the Lagrange interpolation formula in the following solved examples section.
Solved Examples Using Lagrange Interpolation Formula
Example 1: Find the value of y at x = 0 given some set of values (-2, 5), (1, 7), (3, 11), (7, 34)?
Solution:
Given the known values are,
x = 0 ;
= -2 ; = 1 ; = 3 ; = 7 ; = 5 ; = 7 ; = 11 ;
= 34
Using the Lagrange interpolation formula,
y = 21/27 + 4/96 + −77/20 + 51/54
y = 1087/180
Answer: Value of y at (x = 0) = 1087/180
Solution:
Here the intervals are unequal. By Lagrange’s interpolation formula we have
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