Using Newton’s method, consider the map x n + 1 = f(x n ) , where f(x n ) = x n - g(x n ) g'(x n ) , and write Newton’s map x n + 1 = f(x n ) for the equation g ( x ) = x 2 − 4 = 0 . Show that Newton’s map has the fixed points at x&*#x00A0;= ± 2 . Show that the fixed points are superstable. Also, iterate the map numerically from x 0 = 1 and notice the rapid convergence. Concept Introduction: ➢ Obtain Newton’s map from the given function ➢ Determine the fixed points for the given equation and its stability. ➢ Iterate and calculate the numerical values of the function.

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Answer 1

Question

Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map $xn+1 = f(xn)$ , where $f(xn) = xn- g(xn)g'(xn)$ , and write Newton’s map $xn+1 = f(xn)$ for the equation $g(x) = x2−4=0$ . Show that Newton’s map has the fixed points at $x&*#x00A0;= ±2$ . Show that the fixed points are superstable. Also, iterate the map numerically from $x0=1$ and notice the rapid convergence.

Concept Introduction:

➢ Obtain Newton’s map from the given function

➢ Determine the fixed points for the given equation and its stability.

➢ Iterate and calculate the numerical values of the function.

Expert Solution & Answer

Answer to Problem 12E

Solution:

a) The Newton’s map is $f(xn) = (xn)2+42xn$ for the given equation $g(x) = xn2−4$
b) It is shown that the fixed points for the given equation are $x&*#x00A0;= ±2$
c) It is shown that the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) The iteration map with numerical values is shown and $f(x)$ converges rapidly to $f(x) =2$ .

Explanation of Solution

In the given Newton’s method, consider the map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

Here $xn+1$ is the iteration equation for the function.

The function equation is $f(xn)$ and its derivative is $f '(xn)$

a) The Newton map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

As per given, $g(x) = xn2−4=0$

And it’s derivative $g'(x) = 2x$

Newton’s map is calculated by substituting $g(x)$ and $g'(x)$

$f(xn) = xn- g(xn)g'(xn)$

$f(xn) = xn- xn2−42xn$

$f(xn) = 2(xn)2−(xn)2+42xn$

$f(xn) = (xn)2+42xn$

Hence Newton’s map is $f(xn) = (xn)2+42xn$ for the $g(x) = xn2−4$
b) The fixed points of Newton’s map are calculated as,

$x&*#x00A0;= xn+1$

$x&*#x00A0;= xn$

By substituting the value in the equation $x&*#x00A0;= f(xn)$ and $x&*#x00A0;= xn$

$xn+1 = f(xn)$

$x&*#x00A0;= (x*)2+42x*$

$2(x*)2 = (x*)2+4$

$(x*)2 = 4$

$x&*#x00A0;= ±2$

Hence the fixed points for the given equation $x&*#x00A0;= ±2$
c) The stability of the system is determined as below,

Consider the function $f(x) = xn+1$

$f(xn) =xn2+42xn$

The derivative of function $f '(xn)$ is shown below,

$f '(xn) =2xn(2xn)−2(xn2+4)(2xn)2$

$f '(xn) =4(xn)2−2(xn)2−84(xn)2$

$f '(xn) =2(xn)2−84(xn)2$

$f '(xn) =(xn)2−42(xn)2$

Now substituting the fixed points $x&*#x00A0;= ±2$ in $f '(xn)$

$f '(±2) =(xn)2−42(xn)2=(±2)2−42(±2)2=0$

Hence the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) As we iterate the map numerically as shown below, starting from the initial value $x0= 1$ ,

$f(xn) = (xn)2+42xn$

By substituting the initial value

$f(x1) =(x0)2+42(x0) =(1)2+42(1)=2.5$

Similarly $f(x2) =(x1)2+42(x1) =(2.5)2+42(2.5)=2.05$

$f(x3) =(x2)2+42(x2) =(2.05)2+42(2.05)=2.0006$

The iteration of the map and numerical values of the function is given below in tabular form for the initial value of $x0=1$ . From the observed values, $f(x)$ converges rapidly and the exact solution for the given function is $f(x) =2$ .

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Answer 2

Question

Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map $xn+1 = f(xn)$ , where $f(xn) = xn- g(xn)g'(xn)$ , and write Newton’s map $xn+1 = f(xn)$ for the equation $g(x) = x2−4=0$ . Show that Newton’s map has the fixed points at $x&*#x00A0;= ±2$ . Show that the fixed points are superstable. Also, iterate the map numerically from $x0=1$ and notice the rapid convergence.

Concept Introduction:

➢ Obtain Newton’s map from the given function

➢ Determine the fixed points for the given equation and its stability.

➢ Iterate and calculate the numerical values of the function.

Expert Solution & Answer

Answer to Problem 12E

Solution:

a) The Newton’s map is $f(xn) = (xn)2+42xn$ for the given equation $g(x) = xn2−4$
b) It is shown that the fixed points for the given equation are $x&*#x00A0;= ±2$
c) It is shown that the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) The iteration map with numerical values is shown and $f(x)$ converges rapidly to $f(x) =2$ .

Explanation of Solution

In the given Newton’s method, consider the map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

Here $xn+1$ is the iteration equation for the function.

The function equation is $f(xn)$ and its derivative is $f '(xn)$

a) The Newton map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

As per given, $g(x) = xn2−4=0$

And it’s derivative $g'(x) = 2x$

Newton’s map is calculated by substituting $g(x)$ and $g'(x)$

$f(xn) = xn- g(xn)g'(xn)$

$f(xn) = xn- xn2−42xn$

$f(xn) = 2(xn)2−(xn)2+42xn$

$f(xn) = (xn)2+42xn$

Hence Newton’s map is $f(xn) = (xn)2+42xn$ for the $g(x) = xn2−4$
b) The fixed points of Newton’s map are calculated as,

$x&*#x00A0;= xn+1$

$x&*#x00A0;= xn$

By substituting the value in the equation $x&*#x00A0;= f(xn)$ and $x&*#x00A0;= xn$

$xn+1 = f(xn)$

$x&*#x00A0;= (x*)2+42x*$

$2(x*)2 = (x*)2+4$

$(x*)2 = 4$

$x&*#x00A0;= ±2$

Hence the fixed points for the given equation $x&*#x00A0;= ±2$
c) The stability of the system is determined as below,

Consider the function $f(x) = xn+1$

$f(xn) =xn2+42xn$

The derivative of function $f '(xn)$ is shown below,

$f '(xn) =2xn(2xn)−2(xn2+4)(2xn)2$

$f '(xn) =4(xn)2−2(xn)2−84(xn)2$

$f '(xn) =2(xn)2−84(xn)2$

$f '(xn) =(xn)2−42(xn)2$

Now substituting the fixed points $x&*#x00A0;= ±2$ in $f '(xn)$

$f '(±2) =(xn)2−42(xn)2=(±2)2−42(±2)2=0$

Hence the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) As we iterate the map numerically as shown below, starting from the initial value $x0= 1$ ,

$f(xn) = (xn)2+42xn$

By substituting the initial value

$f(x1) =(x0)2+42(x0) =(1)2+42(1)=2.5$

Similarly $f(x2) =(x1)2+42(x1) =(2.5)2+42(2.5)=2.05$

$f(x3) =(x2)2+42(x2) =(2.05)2+42(2.05)=2.0006$

The iteration of the map and numerical values of the function is given below in tabular form for the initial value of $x0=1$ . From the observed values, $f(x)$ converges rapidly and the exact solution for the given function is $f(x) =2$ .

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Answer 3

Question

Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map $xn+1 = f(xn)$ , where $f(xn) = xn- g(xn)g'(xn)$ , and write Newton’s map $xn+1 = f(xn)$ for the equation $g(x) = x2−4=0$ . Show that Newton’s map has the fixed points at $x&*#x00A0;= ±2$ . Show that the fixed points are superstable. Also, iterate the map numerically from $x0=1$ and notice the rapid convergence.

Concept Introduction:

➢ Obtain Newton’s map from the given function

➢ Determine the fixed points for the given equation and its stability.

➢ Iterate and calculate the numerical values of the function.

Expert Solution & Answer

Answer to Problem 12E

Solution:

a) The Newton’s map is $f(xn) = (xn)2+42xn$ for the given equation $g(x) = xn2−4$
b) It is shown that the fixed points for the given equation are $x&*#x00A0;= ±2$
c) It is shown that the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) The iteration map with numerical values is shown and $f(x)$ converges rapidly to $f(x) =2$ .

Explanation of Solution

In the given Newton’s method, consider the map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

Here $xn+1$ is the iteration equation for the function.

The function equation is $f(xn)$ and its derivative is $f '(xn)$

a) The Newton map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

As per given, $g(x) = xn2−4=0$

And it’s derivative $g'(x) = 2x$

Newton’s map is calculated by substituting $g(x)$ and $g'(x)$

$f(xn) = xn- g(xn)g'(xn)$

$f(xn) = xn- xn2−42xn$

$f(xn) = 2(xn)2−(xn)2+42xn$

$f(xn) = (xn)2+42xn$

Hence Newton’s map is $f(xn) = (xn)2+42xn$ for the $g(x) = xn2−4$
b) The fixed points of Newton’s map are calculated as,

$x&*#x00A0;= xn+1$

$x&*#x00A0;= xn$

By substituting the value in the equation $x&*#x00A0;= f(xn)$ and $x&*#x00A0;= xn$

$xn+1 = f(xn)$

$x&*#x00A0;= (x*)2+42x*$

$2(x*)2 = (x*)2+4$

$(x*)2 = 4$

$x&*#x00A0;= ±2$

Hence the fixed points for the given equation $x&*#x00A0;= ±2$
c) The stability of the system is determined as below,

Consider the function $f(x) = xn+1$

$f(xn) =xn2+42xn$

The derivative of function $f '(xn)$ is shown below,

$f '(xn) =2xn(2xn)−2(xn2+4)(2xn)2$

$f '(xn) =4(xn)2−2(xn)2−84(xn)2$

$f '(xn) =2(xn)2−84(xn)2$

$f '(xn) =(xn)2−42(xn)2$

Now substituting the fixed points $x&*#x00A0;= ±2$ in $f '(xn)$

$f '(±2) =(xn)2−42(xn)2=(±2)2−42(±2)2=0$

Hence the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) As we iterate the map numerically as shown below, starting from the initial value $x0= 1$ ,

$f(xn) = (xn)2+42xn$

By substituting the initial value

$f(x1) =(x0)2+42(x0) =(1)2+42(1)=2.5$

Similarly $f(x2) =(x1)2+42(x1) =(2.5)2+42(2.5)=2.05$

$f(x3) =(x2)2+42(x2) =(2.05)2+42(2.05)=2.0006$

The iteration of the map and numerical values of the function is given below in tabular form for the initial value of $x0=1$ . From the observed values, $f(x)$ converges rapidly and the exact solution for the given function is $f(x) =2$ .

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Answer 4

Question

Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map $xn+1 = f(xn)$ , where $f(xn) = xn- g(xn)g'(xn)$ , and write Newton’s map $xn+1 = f(xn)$ for the equation $g(x) = x2−4=0$ . Show that Newton’s map has the fixed points at $x&*#x00A0;= ±2$ . Show that the fixed points are superstable. Also, iterate the map numerically from $x0=1$ and notice the rapid convergence.

Concept Introduction:

➢ Obtain Newton’s map from the given function

➢ Determine the fixed points for the given equation and its stability.

➢ Iterate and calculate the numerical values of the function.

Expert Solution & Answer

Answer to Problem 12E

Solution:

a) The Newton’s map is $f(xn) = (xn)2+42xn$ for the given equation $g(x) = xn2−4$
b) It is shown that the fixed points for the given equation are $x&*#x00A0;= ±2$
c) It is shown that the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) The iteration map with numerical values is shown and $f(x)$ converges rapidly to $f(x) =2$ .

Explanation of Solution

In the given Newton’s method, consider the map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

Here $xn+1$ is the iteration equation for the function.

The function equation is $f(xn)$ and its derivative is $f '(xn)$

a) The Newton map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

As per given, $g(x) = xn2−4=0$

And it’s derivative $g'(x) = 2x$

Newton’s map is calculated by substituting $g(x)$ and $g'(x)$

$f(xn) = xn- g(xn)g'(xn)$

$f(xn) = xn- xn2−42xn$

$f(xn) = 2(xn)2−(xn)2+42xn$

$f(xn) = (xn)2+42xn$

Hence Newton’s map is $f(xn) = (xn)2+42xn$ for the $g(x) = xn2−4$
b) The fixed points of Newton’s map are calculated as,

$x&*#x00A0;= xn+1$

$x&*#x00A0;= xn$

By substituting the value in the equation $x&*#x00A0;= f(xn)$ and $x&*#x00A0;= xn$

$xn+1 = f(xn)$

$x&*#x00A0;= (x*)2+42x*$

$2(x*)2 = (x*)2+4$

$(x*)2 = 4$

$x&*#x00A0;= ±2$

Hence the fixed points for the given equation $x&*#x00A0;= ±2$
c) The stability of the system is determined as below,

Consider the function $f(x) = xn+1$

$f(xn) =xn2+42xn$

The derivative of function $f '(xn)$ is shown below,

$f '(xn) =2xn(2xn)−2(xn2+4)(2xn)2$

$f '(xn) =4(xn)2−2(xn)2−84(xn)2$

$f '(xn) =2(xn)2−84(xn)2$

$f '(xn) =(xn)2−42(xn)2$

Now substituting the fixed points $x&*#x00A0;= ±2$ in $f '(xn)$

$f '(±2) =(xn)2−42(xn)2=(±2)2−42(±2)2=0$

Hence the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) As we iterate the map numerically as shown below, starting from the initial value $x0= 1$ ,

$f(xn) = (xn)2+42xn$

By substituting the initial value

$f(x1) =(x0)2+42(x0) =(1)2+42(1)=2.5$

Similarly $f(x2) =(x1)2+42(x1) =(2.5)2+42(2.5)=2.05$

$f(x3) =(x2)2+42(x2) =(2.05)2+42(2.05)=2.0006$

The iteration of the map and numerical values of the function is given below in tabular form for the initial value of $x0=1$ . From the observed values, $f(x)$ converges rapidly and the exact solution for the given function is $f(x) =2$ .

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Answer 5

Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map $xn+1 = f(xn)$ , where $f(xn) = xn- g(xn)g'(xn)$ , and write Newton’s map $xn+1 = f(xn)$ for the equation $g(x) = x2−4=0$ . Show that Newton’s map has the fixed points at $x&*#x00A0;= ±2$ . Show that the fixed points are superstable. Also, iterate the map numerically from $x0=1$ and notice the rapid convergence.

Concept Introduction:

➢ Obtain Newton’s map from the given function

➢ Determine the fixed points for the given equation and its stability.

➢ Iterate and calculate the numerical values of the function.

Expert Solution & Answer

Answer to Problem 12E

Solution:

a) The Newton’s map is $f(xn) = (xn)2+42xn$ for the given equation $g(x) = xn2−4$
b) It is shown that the fixed points for the given equation are $x&*#x00A0;= ±2$
c) It is shown that the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) The iteration map with numerical values is shown and $f(x)$ converges rapidly to $f(x) =2$ .

Explanation of Solution

In the given Newton’s method, consider the map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

Here $xn+1$ is the iteration equation for the function.

The function equation is $f(xn)$ and its derivative is $f '(xn)$

a) The Newton map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

As per given, $g(x) = xn2−4=0$

And it’s derivative $g'(x) = 2x$

Newton’s map is calculated by substituting $g(x)$ and $g'(x)$

$f(xn) = xn- g(xn)g'(xn)$

$f(xn) = xn- xn2−42xn$

$f(xn) = 2(xn)2−(xn)2+42xn$

$f(xn) = (xn)2+42xn$

Hence Newton’s map is $f(xn) = (xn)2+42xn$ for the $g(x) = xn2−4$
b) The fixed points of Newton’s map are calculated as,

$x&*#x00A0;= xn+1$

$x&*#x00A0;= xn$

By substituting the value in the equation $x&*#x00A0;= f(xn)$ and $x&*#x00A0;= xn$

$xn+1 = f(xn)$

$x&*#x00A0;= (x*)2+42x*$

$2(x*)2 = (x*)2+4$

$(x*)2 = 4$

$x&*#x00A0;= ±2$

Hence the fixed points for the given equation $x&*#x00A0;= ±2$
c) The stability of the system is determined as below,

Consider the function $f(x) = xn+1$

$f(xn) =xn2+42xn$

The derivative of function $f '(xn)$ is shown below,

$f '(xn) =2xn(2xn)−2(xn2+4)(2xn)2$

$f '(xn) =4(xn)2−2(xn)2−84(xn)2$

$f '(xn) =2(xn)2−84(xn)2$

$f '(xn) =(xn)2−42(xn)2$

Now substituting the fixed points $x&*#x00A0;= ±2$ in $f '(xn)$

$f '(±2) =(xn)2−42(xn)2=(±2)2−42(±2)2=0$

Hence the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) As we iterate the map numerically as shown below, starting from the initial value $x0= 1$ ,

$f(xn) = (xn)2+42xn$

By substituting the initial value

$f(x1) =(x0)2+42(x0) =(1)2+42(1)=2.5$

Similarly $f(x2) =(x1)2+42(x1) =(2.5)2+42(2.5)=2.05$

$f(x3) =(x2)2+42(x2) =(2.05)2+42(2.05)=2.0006$

The iteration of the map and numerical values of the function is given below in tabular form for the initial value of $x0=1$ . From the observed values, $f(x)$ converges rapidly and the exact solution for the given function is $f(x) =2$ .

Answer 6

Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map $xn+1 = f(xn)$ , where $f(xn) = xn- g(xn)g'(xn)$ , and write Newton’s map $xn+1 = f(xn)$ for the equation $g(x) = x2−4=0$ . Show that Newton’s map has the fixed points at $x&*#x00A0;= ±2$ . Show that the fixed points are superstable. Also, iterate the map numerically from $x0=1$ and notice the rapid convergence.

Concept Introduction:

➢ Obtain Newton’s map from the given function

➢ Determine the fixed points for the given equation and its stability.

➢ Iterate and calculate the numerical values of the function.

Expert Solution & Answer

Answer to Problem 12E

Solution:

a) The Newton’s map is $f(xn) = (xn)2+42xn$ for the given equation $g(x) = xn2−4$
b) It is shown that the fixed points for the given equation are $x&*#x00A0;= ±2$
c) It is shown that the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) The iteration map with numerical values is shown and $f(x)$ converges rapidly to $f(x) =2$ .

Explanation of Solution

In the given Newton’s method, consider the map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

Here $xn+1$ is the iteration equation for the function.

The function equation is $f(xn)$ and its derivative is $f '(xn)$

a) The Newton map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

As per given, $g(x) = xn2−4=0$

And it’s derivative $g'(x) = 2x$

Newton’s map is calculated by substituting $g(x)$ and $g'(x)$

$f(xn) = xn- g(xn)g'(xn)$

$f(xn) = xn- xn2−42xn$

$f(xn) = 2(xn)2−(xn)2+42xn$

$f(xn) = (xn)2+42xn$

Hence Newton’s map is $f(xn) = (xn)2+42xn$ for the $g(x) = xn2−4$
b) The fixed points of Newton’s map are calculated as,

$x&*#x00A0;= xn+1$

$x&*#x00A0;= xn$

By substituting the value in the equation $x&*#x00A0;= f(xn)$ and $x&*#x00A0;= xn$

$xn+1 = f(xn)$

$x&*#x00A0;= (x*)2+42x*$

$2(x*)2 = (x*)2+4$

$(x*)2 = 4$

$x&*#x00A0;= ±2$

Hence the fixed points for the given equation $x&*#x00A0;= ±2$
c) The stability of the system is determined as below,

Consider the function $f(x) = xn+1$

$f(xn) =xn2+42xn$

The derivative of function $f '(xn)$ is shown below,

$f '(xn) =2xn(2xn)−2(xn2+4)(2xn)2$

$f '(xn) =4(xn)2−2(xn)2−84(xn)2$

$f '(xn) =2(xn)2−84(xn)2$

$f '(xn) =(xn)2−42(xn)2$

Now substituting the fixed points $x&*#x00A0;= ±2$ in $f '(xn)$

$f '(±2) =(xn)2−42(xn)2=(±2)2−42(±2)2=0$

Hence the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) As we iterate the map numerically as shown below, starting from the initial value $x0= 1$ ,

$f(xn) = (xn)2+42xn$

By substituting the initial value

$f(x1) =(x0)2+42(x0) =(1)2+42(1)=2.5$

Similarly $f(x2) =(x1)2+42(x1) =(2.5)2+42(2.5)=2.05$

$f(x3) =(x2)2+42(x2) =(2.05)2+42(2.05)=2.0006$

The iteration of the map and numerical values of the function is given below in tabular form for the initial value of $x0=1$ . From the observed values, $f(x)$ converges rapidly and the exact solution for the given function is $f(x) =2$ .

Answer 7

Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map $xn+1 = f(xn)$ , where $f(xn) = xn- g(xn)g'(xn)$ , and write Newton’s map $xn+1 = f(xn)$ for the equation $g(x) = x2−4=0$ . Show that Newton’s map has the fixed points at $x&*#x00A0;= ±2$ . Show that the fixed points are superstable. Also, iterate the map numerically from $x0=1$ and notice the rapid convergence.

Concept Introduction:

➢ Obtain Newton’s map from the given function

➢ Determine the fixed points for the given equation and its stability.

➢ Iterate and calculate the numerical values of the function.

Answer 8

Expert Solution & Answer

Answer to Problem 12E

Solution:

a) The Newton’s map is $f(xn) = (xn)2+42xn$ for the given equation $g(x) = xn2−4$
b) It is shown that the fixed points for the given equation are $x&*#x00A0;= ±2$
c) It is shown that the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) The iteration map with numerical values is shown and $f(x)$ converges rapidly to $f(x) =2$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

In the given Newton’s method, consider the map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

Here $xn+1$ is the iteration equation for the function.

The function equation is $f(xn)$ and its derivative is $f '(xn)$

a) The Newton map

$xn+1 = f(xn)$ ,

where $f(xn) = xn- g(xn)g'(xn)$

As per given, $g(x) = xn2−4=0$

And it’s derivative $g'(x) = 2x$

Newton’s map is calculated by substituting $g(x)$ and $g'(x)$

$f(xn) = xn- g(xn)g'(xn)$

$f(xn) = xn- xn2−42xn$

$f(xn) = 2(xn)2−(xn)2+42xn$

$f(xn) = (xn)2+42xn$

Hence Newton’s map is $f(xn) = (xn)2+42xn$ for the $g(x) = xn2−4$
b) The fixed points of Newton’s map are calculated as,

$x&*#x00A0;= xn+1$

$x&*#x00A0;= xn$

By substituting the value in the equation $x&*#x00A0;= f(xn)$ and $x&*#x00A0;= xn$

$xn+1 = f(xn)$

$x&*#x00A0;= (x*)2+42x*$

$2(x*)2 = (x*)2+4$

$(x*)2 = 4$

$x&*#x00A0;= ±2$

Hence the fixed points for the given equation $x&*#x00A0;= ±2$
c) The stability of the system is determined as below,

Consider the function $f(x) = xn+1$

$f(xn) =xn2+42xn$

The derivative of function $f '(xn)$ is shown below,

$f '(xn) =2xn(2xn)−2(xn2+4)(2xn)2$

$f '(xn) =4(xn)2−2(xn)2−84(xn)2$

$f '(xn) =2(xn)2−84(xn)2$

$f '(xn) =(xn)2−42(xn)2$

Now substituting the fixed points $x&*#x00A0;= ±2$ in $f '(xn)$

$f '(±2) =(xn)2−42(xn)2=(±2)2−42(±2)2=0$

Hence the fixed points $x&*#x00A0;= ±2$ are super stable points.
d) As we iterate the map numerically as shown below, starting from the initial value $x0= 1$ ,

$f(xn) = (xn)2+42xn$

By substituting the initial value

$f(x1) =(x0)2+42(x0) =(1)2+42(1)=2.5$

Similarly $f(x2) =(x1)2+42(x1) =(2.5)2+42(2.5)=2.05$

$f(x3) =(x2)2+42(x2) =(2.05)2+42(2.05)=2.0006$

The iteration of the map and numerical values of the function is given below in tabular form for the initial value of $x0=1$ . From the observed values, $f(x)$ converges rapidly and the exact solution for the given function is $f(x) =2$ .

Answer 12

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

Answer to Problem 12E

Explanation of Solution

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