Lagrange Interpolation Formula

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^{t h} \end{array}

degree polynomial approximation to f(x).

Find the Lagrange Interpolation Formula given below,

lagrange

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 ; x₀ = -2 ; x₁ = 1 ; x₂= 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}) \dots . . (x - x_{n})}{(x_{0} - x_{1}) (x_{0} - x_{2}) \dots . . (x_{0} - x_{n})} \end{array}

\begin{array}{l} y_{0} \end{array}

\begin{array}{l} \frac{(x - x_{0}) (x - x_{2}) \dots . . (x - x_{n})}{(x_{1} - x_{0}) (x_{1} - x_{2}) \dots . . (x_{1} - x_{n})} \end{array}

\begin{array}{l} y_{1} \end{array}

+ …. +

\begin{array}{l} \frac{(x - x_{1}) (x - x_{1}) \dots . . (x - x_{n - 1})}{(x_{n} - x_{0}) (x_{0} - x_{1}) \dots . . (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}