Aspire Faculty ID #17163 · Topic: AMU MCA 2020 · Just now
AMU MCA 2020

Let $f(x)$ be continuous whose values are known at $-2,-1,1,2$. If the Lagrange interpolation formula $f(x)=L_1f(-2)+L_2f(-1)+L_3f(1)+L_4f(2)$ is used to approximate $f(0)$, then $L_3$ is

Solution

Lagrange basis for node $x_3=1$:

$L_3(0)=\frac{(0+2)(0+1)(0-2)}{(1+2)(1+1)(1-2)}$

$=\frac{(2)(1)(-2)}{(3)(2)(-1)} =\frac{-4}{-6}=\frac{2}{3}$

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