Question

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

a. P(x)=x 3 −x 2 +1
b. P(x)=x 3 –3x 2 −5x–4
c. P(x)=x 3 –3x 2 +5x–6
d. P(x)=2x 2 –3x+5x–6
4 The first divided difference of f with respect to
x i
and
x i+1
denoted by
f[x i ,x i+1 ]
is defined as
a. f[x i ,x i+1 ]=f[x i+1 ]−f[x i ]x i+1 −x i
b. f[x i ,x i+1 ]=x i+1 −x i f[x i+1 ]−f[x i ]
c. f[x i ,x i+1 ]=f[x i ]−f[x i+1 ]x i+1 −x i
d. f[x i ,x i+1 ]=f[x i ]−f[x i+1 ]f[x i+1 ]−x i

Answer 1

Answer on Question #55518 – Math – Algorithms | Quantitative Methods

Question 1

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

a. $P(x) = x \cdot 3 - x \cdot 2 + 1$

b. $P(x) = x \cdot 3 - 3x \cdot 2 - 5x - 4$

c. $P(x) = x \cdot 3 - 3x \cdot 2 + 5x - 6$

d. $P(x) = 2x \cdot 2 - 3x + 5x - 6$

Solution

The Lagrange’s interpolation polynomial has the form

$$L(x) = \sum_{i=0}^{n} y_i l_i(x),$$

where

$$l_i(x) = \prod_{j=0, j\neq i}^{n} \frac{x - x_j}{x_i - x_j} = \frac{x - x_0}{x_i - x_0} \cdots \frac{x - x_{i-1}}{x_i - x_{i-1}} \cdot \frac{x - x_{i+1}}{x_i - x_{i+1}} \cdots \frac{x - x_n}{x_i - x_n}.$$

Thus, we obtain

$$\begin{aligned} P(x) = & \frac{(x - 3)(x - 4)(x - 5)}{(1 - 3)(1 - 4)(1 - 5)} f(1) + \frac{(x - 1)(x - 4)(x - 5)}{(3 - 1)(3 - 4)(3 - 5)} f(3) + \frac{(x - 1)(x - 3)(x - 5)}{(4 - 1)(4 - 3)(4 - 5)} f(4) + \frac{(x - 1)(x - 3)(x - 4)}{(5 - 1)(5 - 3)(5 - 4)} f(5) = \\ & = \frac{-1}{24}(x - 3)(x - 4)(x - 5) + \frac{19}{4}(x - 1)(x - 4)(x - 5) + \frac{-49}{3}(x - 1)(x - 3)(x - 5) + \frac{101}{8}(x - 1)(x - 3)(x - 4) = \\ & = x^3 - x^2 + 1. \end{aligned}$$

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

Question 2

The first divided difference of $f$ with respect to $x$ $i$ and $x$ $i+1$ denoted by $f[x \cdot i, x \cdot i+1]$ is defined as

a. $f[x \cdot i, x \cdot i+1] = f[x \cdot i+1] - f[x \cdot i] \cdot x \cdot i+1 - x i$

b. $f[x \cdot i, x \cdot i+1] = x \cdot i+1 - x \cdot i f[x \cdot i+1] - f[x \cdot i]$

c. $f[x \cdot i, x \cdot i+1] = f[x \cdot i] - f[x \cdot i+1] \cdot x \cdot i+1 - x i$

d. $f[x \cdot i, x \cdot i+1] = f[x \cdot i] - f[x \cdot i+1] \cdot f[x \cdot i+1] - x i$

Solution

The first divided difference of $f$ with respect to $x_i$ and $x_{i+1}$ denoted by $f[x_i, x_{i+1}]$ is defined as

$$f[x_i, x_{i+1}] = \frac{f(x_{i+1}) - f(x_i)}{x_{i+1} - x_i}.$$

Answer: a. $f[x_i, x_{i+1}] = \frac{f(x_{i+1}) - f(x_i)}{x_{i+1} - x_i}$.

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