Answered: Use Newton’s Method to approximate the…

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

Use Newton’s Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations differ by less than 0.001. Then find the zero(s) using a graphing utility and compare the results. f(x) = −x³ + 2.7x² + 3.55x − 2.422

Expert Solution

Step 1

The given function $f (x) = - x^{3} + 2.7 x^{2} + 3.55 x - 2.422$ .

We have to find the zeros of the function by using Newton's method.

Step 2

Newton's method algorithm:

$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f' (x_{n})}$ .

$f (x) = - x^{3} + 2.7 x^{2} + 3.55 x - 2.422$

Differentiate with respect to $x$ ,

$f' (x) = - 3 x^{2} + 5.4 x + 3.55$

The function values,

x	-2	-1	0	1	3	4
$f (x)$	9.278	-2.272	-2.422	2.828	5.528	-9.022

The zeros lies on the intervals $[- 2, - 1], [0, 1],$ and $[3, 4]$ .

Step 3

The root lies on the interval $[- 2, - 1]$ :

Choose the initial value $x_{0} = - 2$

First iteration:

$f (x_{0}) = f (- 2) = - {(- 2)}^{3} + 2.7 {(- 2)}^{2} + 3.55 (- 2) - 2.422 = 9.278 f' (x_{0}) = f' (- 2) = - 3 {(- 2)}^{2} + 5.4 (- 2) + 3.55 = - 19.25$

$\begin{array}{rcl} x_{1} & = & x_{0} - \frac{f (x_{0})}{f' (x_{0})} \\ = & - 2 - \frac{9.278}{- 19.25} \\ x_{1} & = & - 1.518 \end{array}$

Second iteration:

$f (x_{1}) = f (- 1.518) = - {(- 1.518)}^{3} + 2.7 {(- 1.518)}^{2} + 3.55 (- 1.518) - 2.422 = 1.909 f' (x_{1}) = f' (- 1.518) = - 3 {(- 1.518)}^{2} + 5.4 (- 1.518) + 3.55 = - 11.5605$

$\begin{array}{rcl} x_{2} & = & x_{1} - \frac{f (x_{1})}{f' (x_{1})} \\ = & - 1.518 - \frac{1.909}{- 11.5605} \\ x_{2} & = & - 1.3529 \end{array}$

Third iteration:

$f (x_{2}) = f (- 1.3529) = - {(- 1.3529)}^{3} + 2.7 {(- 1.3529)}^{2} + 3.55 (- 1.3529) - 2.422 = 0.1933 f' (x_{2}) = f' (- 1.3529) = - 3 {(- 1.3529)}^{2} + 5.4 (- 1.3529) + 3.55 = - 9.2466$

$\begin{array}{rcl} x_{3} & = & x_{2} - \frac{f (x_{2})}{f' (x_{2})} \\ = & - 1.3529 - \frac{0.1933}{- 9.2466} \\ x_{3} & = & - 1.332 \end{array}$

Fourth iteration:

$f (x_{3}) = f (- 1.332) = - {(- 1.332)}^{3} + 2.7 {(- 1.332)}^{2} + 3.55 (- 1.332) - 2.422 = 0.0029 f' (x_{3}) = f' (- 1.332) = - 3 {(- 1.332)}^{2} + 5.4 (- 1.332) + 3.55 = - 8.9653$

$\begin{array}{rcl} x_{4} & = & x_{3} - \frac{f (x_{3})}{f' (x_{3})} \\ = & - 1.332 - \frac{0.0029}{- 8.9653} \\ x_{4} & = & - 1.3317 \end{array}$

Fifth iteration:

$f (x_{4}) = f (- 1.3317) = - {(- 1.3317)}^{3} + 2.7 {(- 1.3317)}^{2} + 3.55 (- 1.3317) - 2.422 = 0 f' (x_{4}) = f' (- 1.3317) = - 3 {(- 1.3317)}^{2} + 5.4 (- 1.3317) + 3.55 = - 8.9609$

$\begin{array}{rcl} x_{5} & = & x_{4} - \frac{f (x_{4})}{f' (x_{4})} \\ = & - 1.3317 - \frac{0}{- 8.9609} \\ x_{5} & = & - 1.3317 \end{array}$

After five iterations, the root lies on the interval $[- 2, - 1]$ is $x = - 1.3317$ .

Step 4

The root lies on the interval $[0, 1]$ :

Choose the initial value $x_{0} = \frac{0 + 1}{2} = 0.5$

First iteration:

$f (x_{0}) = f (0.5) = - {(0.5)}^{3} + 2.7 {(0.5)}^{2} + 3.55 (0.5) - 2.422 = - 0.097 f' (x_{0}) = f' (0.5) = - 3 {(0.5)}^{2} + 5.4 (0.5) + 3.55 = 5.5$

$\begin{array}{rcl} x_{1} & = & x_{0} - \frac{f (x_{0})}{f' (x_{0})} \\ = & 0.5 - \frac{- 0.097}{5.5} \\ x_{1} & = & 0.5176 \end{array}$

Second iteration:

$f (x_{1}) = f (0.5176) = - {(0.5176)}^{3} + 2.7 {(0.5176)}^{2} + 3.55 (0.5176) - 2.422 = 0.0004 f' (x_{1}) = f' (0.5176) = - 3 {(0.5176)}^{2} + 5.4 (0.5176) + 3.55 = 5.5414$

$\begin{array}{rcl} x_{2} & = & x_{1} - \frac{f (x_{1})}{f' (x_{1})} \\ = & 0.5176 - \frac{0.0004}{5.5414} \\ x_{2} & = & 0.5176 \end{array}$

After two iterations, the root lies on the interval $[0, 1]$ is $x = 0.5176$ .

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