Chapter 10.1, Problem 12E

Interpretation Introduction

Interpretation:

Using Newton’s method, consider the map ${\text{x}}_{n+1}{\text{ = f(x}}_n)$, where ${\text{f(x}}_{\text{n}}{\text{) = x}}_{\text{n}}\text{- }\frac{{\text{g(x}}_{\text{n}}\text{)}}{{\text{g'(x}}_{\text{n}}\text{)}}$, and write Newton’s map ${\text{x}}_{n+1}{\text{ = f(x}}_n)$ for the equation $g(x){\text{ = x}}^2-4=0$. Show that Newton’s map has the fixed points at $\text{x\&*\#x00A0;= }\pm \text{2}$. Show that the fixed points are superstable. Also, iterate the map numerically from ${\text{x}}_0=1$ 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.

Answer to Problem 12E

Solution:

• a) The Newton’s map is ${\text{f(x}}_{\text{n}}\text{) = }\frac{{\left({\text{x}}_{\text{n}}\right)}^2+4}{{\text{2x}}_{\text{n}}}$ for the given equation ${\text{g(x) = x}}_n{}^2-4$

• b) It is shown that the fixed points for the given equation are $\text{x\&*\#x00A0;= ±2}$

• c) It is shown that the fixed points $\text{x\&*\#x00A0;= ±2}$ are super stable points.

• d) The iteration map with numerical values is shown and $\text{f(x)}$ converges rapidly to $\text{f(x) =}2$.

Explanation of Solution

In the given Newton’s method, consider the map

${\text{x}}_{n+1}{\text{ = f(x}}_n)$,

where ${\text{f(x}}_{\text{n}}{\text{) = x}}_{\text{n}}\text{- }\frac{{\text{g(x}}_{\text{n}}\text{)}}{{\text{g'(x}}_{\text{n}}\text{)}}$

Here ${\text{x}}_{n+1}$ is the iteration equation for the function.

The function equation is ${\text{f(x}}_n)$ and its derivative is ${\text{f '(x}}_n)$

• a) The Newton map

  ${\text{x}}_{n+1}{\text{ = f(x}}_n)$,

  where ${\text{f(x}}_{\text{n}}{\text{) = x}}_{\text{n}}\text{- }\frac{{\text{g(x}}_{\text{n}}\text{)}}{{\text{g'(x}}_{\text{n}}\text{)}}$

  As per given, ${\text{g(x) = x}}_n{}^2-4=0$

  And it’s derivative $\text{g'(x) = 2x}$

  Newton’s map is calculated by substituting $\text{g(x)}$ and $\text{g'(x)}$

  ${\text{f(x}}_{\text{n}}{\text{) = x}}_{\text{n}}\text{- }\frac{{\text{g(x}}_{\text{n}}\text{)}}{{\text{g'(x}}_{\text{n}}\text{)}}$

  ${\text{f(x}}_{\text{n}}{\text{) = x}}_{\text{n}}\text{- }\frac{{\text{x}}_{\text{n}}{}^{\text{2}}-4}{{\text{2x}}_{\text{n}}}$

  ${\text{f(x}}_{\text{n}}\text{) = }\frac{\text{2}{\left({\text{x}}_{\text{n}}\right)}^2-{\left({\text{x}}_{\text{n}}\right)}^{\text{2}}+4}{{\text{2x}}_{\text{n}}}$

  ${\text{f(x}}_{\text{n}}\text{) = }\frac{{\left({\text{x}}_{\text{n}}\right)}^2+4}{{\text{2x}}_{\text{n}}}$

  Hence Newton’s map is ${\text{f(x}}_{\text{n}}\text{) = }\frac{{\left({\text{x}}_{\text{n}}\right)}^2+4}{{\text{2x}}_{\text{n}}}$ for the ${\text{g(x) = x}}_n{}^2-4$

• b) The fixed points of Newton’s map are calculated as,

  ${\text{x\&*\#x00A0;= x}}_{n+1}$

  ${\text{x\&*\#x00A0;= x}}_n$

  By substituting the value in the equation ${\text{x\&*\#x00A0;= f(x}}_n)$ and ${\text{x\&*\#x00A0;= x}}_{\text{n}}$

  ${\text{x}}_{n+1}{\text{ = f(x}}_n)$

  $\text{x\&*\#x00A0;= }\frac{{\left(\text{x*}\right)}^2+4}{\text{2x*}}$

  $\text{2}{\left(\text{x*}\right)}^2\text{ = }{\left(\text{x*}\right)}^2+4$

  ${\left(\text{x*}\right)}^2\text{ = }4$

  $\text{x\&*\#x00A0;= }\pm 2$

  Hence the fixed points for the given equation $\text{x\&*\#x00A0;= ±2}$

• c) The stability of the system is determined as below,

  Consider the function ${\text{f(x) = x}}_{\text{n+1}}$

  ${\text{f(x}}_{\text{n}}\text{) =}\frac{{\text{x}}_{\text{n}}{}^{\text{2}}+4}{{\text{2x}}_{\text{n}}}$

  The derivative of function ${\text{f '(x}}_{\text{n}}\text{)}$ is shown below,

  ${\text{f '(x}}_{\text{n}}\text{) =}\frac{{\text{2x}}_{\text{n}}\left({\text{2x}}_{\text{n}}\right)-2\left({\text{x}}_{\text{n}}{}^{\text{2}}+4\right)}{{\left({\text{2x}}_{\text{n}}\right)}^2}$

  ${\text{f '(x}}_{\text{n}}\text{) =}\frac{\text{4}{\left({\text{x}}_{\text{n}}\right)}^2-2{\left({\text{x}}_{\text{n}}\right)}^2-8}{4{\left({\text{x}}_{\text{n}}\right)}^2}$

  ${\text{f '(x}}_{\text{n}}\text{) =}\frac{2{\left({\text{x}}_{\text{n}}\right)}^2-8}{4{\left({\text{x}}_{\text{n}}\right)}^2}$

  ${\text{f '(x}}_{\text{n}}\text{) =}\frac{{\left({\text{x}}_{\text{n}}\right)}^2-4}{2{\left({\text{x}}_{\text{n}}\right)}^2}$

  Now substituting the fixed points $\text{x\&*\#x00A0;= ±2}$ in ${\text{f '(x}}_{\text{n}}\text{)}$

  $\text{f '(}\pm \text{2) =}\frac{{\left({\text{x}}_{\text{n}}\right)}^2-4}{2{\left({\text{x}}_{\text{n}}\right)}^2}=\frac{{\left(\pm \text{2}\right)}^2-4}{2{\left(\pm \text{2}\right)}^2}=0$

  Hence the fixed points $\text{x\&*\#x00A0;= ±2}$ are super stable points.

• d) As we iterate the map numerically as shown below, starting from the initial value ${\text{x}}_0\text{= 1}$,

  ${\text{f(x}}_{\text{n}}\text{) = }\frac{{\left({\text{x}}_{\text{n}}\right)}^2+4}{{\text{2x}}_{\text{n}}}$

  By substituting the initial value

  ${\text{f(x}}_1\text{) =}\frac{{\left({\text{x}}_0\right)}^2+4}{\text{2}\left({\text{x}}_0\right)}\text{ =}\frac{{\left(\text{1}\right)}^2+4}{\text{2}\left(1\right)}=2.5$

  Similarly ${\text{f(x}}_2\text{) =}\frac{{\left({\text{x}}_1\right)}^2+4}{\text{2}\left({\text{x}}_1\right)}\text{ =}\frac{{\left(\text{2}\text{.5}\right)}^2+4}{\text{2}\left(2.5\right)}=2.05$

  ${\text{f(x}}_3\text{) =}\frac{{\left({\text{x}}_2\right)}^2+4}{\text{2}\left({\text{x}}_2\right)}\text{ =}\frac{{\left(\text{2}\text{.05}\right)}^2+4}{\text{2}\left(2.05\right)}=2.0006$

  The iteration of the map and numerical values of the function is given below in tabular form for the initial value of ${\text{x}}_0=1$. From the observed values, $\text{f(x)}$ converges rapidly and the exact solution for the given function is $\text{f(x) =}2$.