Question

1 The technique of determining an approximate value of f(x) for a non-tabular value of x which lies outside the internal [a, b] is known as............
a. Newton’s second law
b.Newton’s first law
c. Interpolation
d. extapolation

2 The Lagrange’s interpolating polynomial P(x) is given by …………..
a. P(x)=L 0 (x)f 3 +L 1 (x)f 2 +L 2 (x)f 1 +L 3 (x)f 0
b. p(x)=L 1 (x)f 0 +L 2 (x)f 1 +L 3 (x)f 2 +L 4 (x)f 3
c. P(x)=L 0 (x)f 0 +L 1 (x)f 1 +L 2 (x)f 2 +L 3 (x)f 3
d. P(x)=L 0 (x)f 0 +L 1 (x)f 2 +L 2 (x)f 2 +L 3 (x)f 3

Answer 1

Answer on Question #55516 – Math - Discrete Math

Question 1. The technique of determining an approximate value of $f(x)$ for a non-tabular value of $x$ which lies outside the interval $[a, b]$ is known as...

a. Newton's second law

b. Newton's first law

c. Interpolation

d. extrapolation

Solution

The correct answer is "d. Extrapolation".

Question 2. The Lagrange's interpolating polynomial $P(x)$ is given by ...

a. $P(x) = L \cdot 0(x)f \cdot 3 + L \cdot 1(x)f \cdot 2 + L \cdot 2(x)f \cdot 1 + L \cdot 3(x)f \cdot 0$

b. $p(x) = L \cdot 1(x)f \cdot 0 + L \cdot 2(x)f \cdot 1 + L \cdot 3(x)f \cdot 2 + L \cdot 4(x)f \cdot 3$

c. $P(x) = L \cdot 0(x)f \cdot 0 + L \cdot 1(x)f \cdot 1 + L \cdot 2(x)f \cdot 2 + L \cdot 3(x)f \cdot 3$

d. $P(x) = L \cdot 0(x)f \cdot 0 + L \cdot 1(x)f \cdot 2 + L \cdot 2(x)f \cdot 2 + L \cdot 3(x)f \cdot 3$

Solution

The correct answer is "c. $P(x) = L \cdot 0(x)f \cdot 0 + L \cdot 1(x)f \cdot 1 + L \cdot 2(x)f \cdot 2 + L \cdot 3(x)f \cdot 3$".

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