Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points
\(\begin{array}{l}x_{j}\end{array} \)
and numbers
\(\begin{array}{l}y_{j}\end{array} \)
. Lagrange’s interpolation is also an
\(\begin{array}{l}n^{th}\end{array} \)
degree polynomial approximation to
f(
x).
Find the Lagrange Interpolation Formula given below,
Solved Example
Question:Â
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 ; x0 = -2 ; x1 = 1 ; x2Â = 3 ; x3 = 7 ; y0 = 5 ; y1 = 7 ; y2 = 11 ; y3 = 34
Using the interpolation formula,
y = \(\begin{array}{l}\frac{(x-x_{1})(x-x_{2})…..(x-x_{n})}{(x_{0}-x_{1})(x_{0}-x_{2})…..(x_{0}-x_{n})}\end{array} \)
\(\begin{array}{l}y_{0}\end{array} \)
+
\(\begin{array}{l}\frac{(x-x_{0})(x-x_{2})…..(x-x_{n})}{(x_{1}-x_{0})(x_{1}-x_{2})…..(x_{1}-x_{n})}\end{array} \)
\(\begin{array}{l}y_{1}\end{array} \)
+ …. +
\(\begin{array}{l}\frac{(x-x_{1})(x-x_{1})…..(x-x_{n-1})}{(x_{n}-x_{0})(x_{0}-x_{1})…..(x_{n}-x_{n-1})}\end{array} \)
\(\begin{array}{l}y_{n}\end{array} \)
y =
\(\begin{array}{l}\frac{(0-1)(0-3)(0-7)}{(-2-1)(-2-3)(-2-7)}\times 5\end{array} \)
 +
\(\begin{array}{l}\frac{(0+2)(0-3)(0-7)}{(1+2)(1-3)(1-7)}\times 7\end{array} \)
 +
\(\begin{array}{l}\frac{(0+2)(0-1)(0-7)}{(3+2)(3-1)(3-7)}\times 11\end{array} \)
 +
\(\begin{array}{l}\frac{(0+2)(0-1)(0-3)}{(7+2)(7-1)(7-3)}\times 34\end{array} \)
y = \(\begin{array}{l}\frac{21}{27}\end{array} \)
+ \(\begin{array}{l}\frac{49}{6}\end{array} \)
+ \(\begin{array}{l}\frac{-77}{20}\end{array} \)
+ \(\begin{array}{l}\frac{51}{54}\end{array} \)
y = \(\begin{array}{l}\frac{1087}{180}\end{array} \)
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