Question

The Lagrange polynomial that
passes through the 3 data points is
given by
15 18 22
y
37 25
f2(x) = L. (x)(24) + L1(x)(37)
+ L2(x)(25)​

Answer 1

The Lagrange polynomial that passes through the $3$ data points is given by

$x & 15 & 18 & 22 \\ y & 24 & 37 & 25$

The value of $L_{1}(x)$ at $x=16$ is

(A) $-0.071430$

(B) $0.50000$

(C) $0.57143$

(D) $4.3333$

Answer:

The correct option is (B) $0.50000$

Step-by-step explanation:

The Lagrange interpolation formula can be used to locate a polynomial known as a Lagrange polynomial that assumes certain values at every location. An Nth degree polynomial approximation to f is called Lagrange's interpolation (x). The only polynomial of lowest degree capable of interpolating a given collection of data is the Lagrange interpolating polynomial.

Given: The Lagrange polynomial passes through the $3$ data points.

To find: The value of $L_{1}(x)$.

According to question,

$L_{i}(x) &=\prod_{\substack{j=0 \\j \neq i}}^{n} \frac{x-x_{j}}{x_{i}-x_{j}} \\L_{1}(x) &=\prod_{\substack{j=0 \\j \neq 1}}^{2} \frac{x-x_{j}}{x_{1}-x_{j}}$

Here,

$j \neq 1}}^{2} \frac{x-x_{j}}{x_{1}-x_{j}}$

Then,

$=\left(\frac{x-x_{0}}{x_{1}-x_{0}}\right)\left(\frac{x-x_{2}}{x_{1}-x_{2}}\right)$

Substitute the values.

$=\left(\frac{16-15}{18-15}\right)\left(\frac{16-22}{18-22}\right)$

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

$=0.50000$

Therefore, the required answer is $0.50000$ and option (B) is the right answer.

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