Question

if f(1) = 1, f(3) =19, f(4) = 49 and f(5) = 101, find the lagrange's interpolation polynomial of f(x)

Answer 1

Answer on Question #54865 – Math – Algorithms | Quantitative Methods

Question

If $f(1) = 1, f(3) = 19, f(4) = 49$ and $f(5) = 101$, find the Lagrange's interpolation polynomial of $f(x)$

Solution

If $f(x_{i}) = y_{i}, i = 0,1,2,\ldots,n$, then Lagrange's interpolation polynomial of $f(x)$ is polynomial of $n$-th degree in form

$$L(x) = \sum_{j=0}^{n} y_j l_j(x)$$

where

$$l_j(x) = \frac{x - x_0}{x_j - x_0} \cdot \frac{x - x_1}{x_j - x_1} \cdots \frac{x - x_{j-1}}{x_j - x_{j-1}} \cdot \frac{x - x_{j+1}}{x_j - x_{j+1}} \cdots \frac{x - x_n}{x_j - x_n}$$

We have $x_0 = 1, x_1 = 3, x_2 = 4, x_3 = 5$ and $y_0 = 1, y_1 = 19, y_2 = 49, y_3 = 101$

$$l_0(x) = \frac{x - 3}{1 - 3} \cdot \frac{x - 4}{1 - 4} \cdot \frac{x - 5}{1 - 5} = -\frac{1}{24}(x - 3)(x - 4)(x - 5) = -\frac{1}{24}(x^3 - 12x^2 + 47x - 60)$$ $$l_1(x) = \frac{x - 1}{3 - 1} \cdot \frac{x - 4}{3 - 4} \cdot \frac{x - 5}{3 - 5} = \frac{1}{4}(x - 1)(x - 4)(x - 5) = \frac{1}{4}(x^3 - 10x^2 + 29x - 20)$$ $$l_2(x) = \frac{x - 1}{4 - 1} \cdot \frac{x - 3}{4 - 3} \cdot \frac{x - 5}{4 - 5} = -\frac{1}{3}(x - 1)(x - 3)(x - 5) = -\frac{1}{3}(x^3 - 9x^2 + 23x - 15)$$ $$l_3(x) = \frac{x - 1}{5 - 1} \cdot \frac{x - 3}{5 - 3} \cdot \frac{x - 4}{5 - 4} = \frac{1}{8}(x - 1)(x - 3)(x - 4) = \frac{1}{8}(x^3 - 8x^2 + 19x - 12)$$

Hence $L(X) = l_0(x) + 19l_1(x) + 49l_2(x) + 101l_3(x) =$

$$= -\frac{1}{24}(x^3 - 12x^2 + 47x - 60) + \frac{19}{4}(x^3 - 10x^2 + 29x - 20) - \frac{49}{3}(x^3 - 9x^2 + 23x - 15) + \frac{101}{8}(x^3 - 8x^2 + 19x - 12) = x^3 - x^2 + 1$$

Answer: $L(x) = x^3 - x^2 + 1$.

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