Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points $x_{j}$ and numbers $y_{j}$. Lagrange’s interpolation is also an $N^{th}$ degree polynomial approximation to f(x).
Find the Lagrange Interpolation Formula given below,
Question: 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 ; $x_{0}$ = -2 ; $x_{1}$ = 1 ; $x_{2}$= 3 ; $x_{3}$ = 7 ; $y_{0}$ = 5 ; $y_{1}$ = 7 ; $y_{2}$ = 11 ; $y_{3}$ = 34
Using the interpolation formula,
y = $\frac{(x-x_{1})(x-x_{2})…..(x-x_{n})}{(x_{0}-x_{1})(x_{0}-x_{2})…..(x_{0}-x_{n})}$ $y_{0}$ + $\frac{(x-x_{0})(x-x_{2})…..(x-x_{n})}{(x_{1}-x_{0})(x_{1}-x_{2})…..(x_{1}-x_{n})}$ $y_{1}$ + …. + $\frac{(x-x_{1})(x-x_{1})…..(x-x_{n-1})}{(x_{n}-x_{0})(x_{0}-x_{1})…..(x_{n}-x_{n-1})}$ $y_{n}$
y = $\frac{(0-1)(0-3)(0-7)}{(-2-1)(-2-3)(-2-7)}$ $\times$ 5 + $\frac{(0+2)(0-3)(0-7)}{(1+2)(1-3)(1-7)}$ $\times$ 7 + $\frac{(0+2)(0-1)(0-7)}{(3+2)(3-1)(3-7)}$ $\times$ 11 + $\frac{(0+2)(0-1)(0-3)}{(7+2)(7-1)(7-3)}$ $\times$ 34
y = $\frac{21}{27}$ + $\frac{49}{6}$ + $\frac{-77}{20}$ + $\frac{51}{54}$
y = $\frac{1087}{180}$
No comments:
Post a Comment