a and b please **Exercise 5.2.10**: Integrate Newton's divided-difference interpolating polynomial to prove the formulas (a) (5.18) and (b) (5.19).**Figure 5.2(b)** shows the region under the parabola $ P(x) $ interpolating the data points $(-h, 0), (0, 1),$ and $ (h, 0) $. The area is\[\int_{-h}^{h} P(x) \, dx = x - \frac{x^3}{3h^2} = \frac{4}{3}h\]**Figure 5.2(c)** shows the region between the x-axis and the parabola interpolating the data points $(-h, 1), (0, 0),$ and $(h, 0) $. The area is\[\int_{-h}^{h} P(x) \, dx = \frac{1}{3}h\]**(a) (5.18)**: Find the Newton's Divided Difference interpolating polynomial for the points $(-h, 0), (0, 1), (h, 0)$.**(b) (5.19)**: Find the Newton's Divided Difference interpolating polynomial for the points $(-h, 1), (0, 0), (h, 0)$.

Answered: (a) (5.18) Find the Newton's Divided…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

a and b please

**Exercise 5.2.10**: Integrate Newton's divided-difference interpolating polynomial to prove the formulas (a) (5.18) and (b) (5.19).

**Figure 5.2(b)** shows the region under the parabola $ P(x) $ interpolating the data points $(-h, 0), (0, 1),$ and $ (h, 0) $. The area is

\[
\int_{-h}^{h} P(x) \, dx = x - \frac{x^3}{3h^2} = \frac{4}{3}h
\]

**Figure 5.2(c)** shows the region between the x-axis and the parabola interpolating the data points $(-h, 1), (0, 0),$ and $(h, 0) $. The area is

\[
\int_{-h}^{h} P(x) \, dx = \frac{1}{3}h
\]

**(a) (5.18)**: Find the Newton's Divided Difference interpolating polynomial for the points $(-h, 0), (0, 1), (h, 0)$.

**(b) (5.19)**: Find the Newton's Divided Difference interpolating polynomial for the points $(-h, 1), (0, 0), (h, 0)$.

Expert Solution

Step 1: Finding the interpolating polynomial

(a)

$x$ $f (x)$ $1$ st order $2$ nd order

$x_{0} = - h$ $y_{0} = 0$

$[x_{0}, x_{1}] = \frac{y_{1} - y_{0}}{x_{1} - x_{0}} = \frac{1}{h}$

$x_{1} = 0$ $y_{1} = 1$ $[x_{0}, x_{1}, x_{2}] = \frac{[x_{1}, x_{2}] - [x_{0}, x_{1}]}{x_{2} - x_{0}} = \frac{- \frac{2}{h}}{2 h} = - \frac{1}{h^{2}}$

$[x_{1}, x_{2}] = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = - \frac{1}{h}$

$x_{2} = h$ $y_{2} = 0$

$∴ y = f (x) = y_{0} + (x - x_{0}) f [x_{0}, x_{1}] + (x - x_{0}) (x - x_{1}) f [x_{0}, x_{1}, x_{2}] = 0 + (x + h) \frac{1}{h} + (x + h) x (- \frac{1}{h^{2}}) = \frac{(x + h)}{h} (1 - \frac{x}{h}) = \frac{h^{2} - x^{2}}{h^{2}} = 1 - {(\frac{x}{h})}^{2}$