Nonlinear Dynamics and Chaos
Nonlinear Dynamics and Chaos
2nd Edition
ISBN: 9780813349107
Author: Steven H. Strogatz
Publisher: PERSEUS D
Question
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Chapter 10.1, Problem 12E
Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map xn+1 = f(xn), where f(xn) = xng(xn)g'(xn), and write Newton’s map xn+1 = f(xn) for the equation g(x) = x24=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
Check Mark

Answer to Problem 12E

Solution:

  • a) The Newton’s map is f(xn) = (xn)2+42xn for the given equation g(x) = xn24

  • 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) = xng(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) = xng(xn)g'(xn)

    As per given, g(x) = xn24=0

    And it’s derivative g'(x) = 2x

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

    f(xn) = xng(xn)g'(xn)

    f(xn) = xnxn242xn

    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) = xn24

  • 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)22(xn)284(xn)2

    f '(xn) =2(xn)284(xn)2

    f '(xn) =(xn)242(xn)2

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

    f '(±2) =(xn)242(xn)2=(±2)242(±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.

    Nonlinear Dynamics and Chaos, Chapter 10.1, Problem 12E

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