Chapter 7, Problem 3P

(a)

To determine

To calculate: The positive real root of $f\left(x\right)$ by using Müller’s method if,

$f\left(x\right)=x^3+x^2-4x-4$

Answer to Problem 3P

Solution:

The positive real root of the given equation $f\left(x\right)$ is $2$.

Explanation of Solution

Given Information:

The given equation is,

$f\left(x\right)=x^3+x^2-4x-4$

Use Müller’s method.

Formula used:

The expression for the new roots is,

$x_3=x_2+\frac{-2c}{b+\sqrt{b^2-4ac}}$

The expression for the error is,

$\text{error}=\left|\frac{x_3-x_2}{x_3}\right|\cdot 100$

Calculation:

Recall the equation mentioned in the problem,

$f\left(x\right)=x^3+x^2-4x-4$

Draw the plot of the equation.

From the above plot it is clear that one root is at about $x=2$.

Consider the initial guess be $x_0=1\text{, }{x}_1=1.5$

, and $x_2=2.5$ respectively to determine the positive real root of the equation.

$f\left(x\right)=x^3+x^2-4x-4$

Thus, the values of $f\left(x\right)$ at different initial value are,

$\begin{array}{c}f\left(1\right)=-6\\ f\left(1.5\right)=-4.375\\ f\left(2.5\right)=7.875\end{array}$

Now calculate the $h_0\text{ and }{h}_1$.

$\begin{array}{l}{h}_0=x_1-x_0\\ h_1=x_2-x_1\end{array}$

Substituting the value of $x_0=1\text{, }{x}_1=1.5$

, and $x_2=2.5$.

$\begin{array}{l}{h}_0=0.5\\ h_1=1\end{array}$

Now calculate the ${\delta }_0\text{ and }{\delta }_1$.

$\begin{array}{l}{\delta }_0=\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\\ {\delta }_1=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\end{array}$

Substitute the above calculated values.

$\begin{array}{c}{\delta }_0=\frac{f\left(1.5\right)-f\left(1\right)}{1.5-1}\\ =\frac{-4.375+6}{0.5}\\ =3.25\end{array}$

And,

$\begin{array}{c}{\delta }_1=\frac{f\left(2.5\right)-f\left(1.5\right)}{2.5-1.5}\\ =\frac{7.875+4.375}{1}\\ =12.25\end{array}$

Calculate the value of constants $a,b\text{ and }c$.

$\begin{array}{c}a=\frac{{\delta }_1-{\delta }_0}{h_1+h_0}\\ =\frac{12.25-3.25}{1+0.5}\\ =\frac{9.0}{1.5}\\ =6\end{array}$

For b,

$\begin{array}{c}b=a{h}_1+{\delta }_1\\ =6\times 1+12.5\\ =18.25\end{array}$

For c,

$\begin{array}{c}c=f\left(x_2\right)\\ =7.875\end{array}$

Thus, the new root is calculated as follows.

$x_3=x_2+\frac{-2c}{b+\sqrt{b^2-4ac}}$

Substitute the all values.

$\begin{array}{c}{x}_3=2.5+\frac{-2\times \left(7.875\right)}{18.25+\sqrt{{18.25}^2-\left(4\right)\left(6\right)\left(7.875\right)}}\\ =2.5-0.520616\\ =1.979384\end{array}$

Calculate the error estimate.

$\begin{array}{c}\text{error}=\left|\frac{1.979384-2.5}{1.979384}\right|\left(100\right)\\ =\left|\frac{-0.520616}{2}\right|\left(100\right)\\ =26.03\%\end{array}$

Because the error is large, so required new guesses for $x_0$ is replaced by $x_1$

, $x_1$ is replaced by $x_2$

, and $x_2$ is replaced by $x_3$.

Therefore, for the new iteration, $x_0=1.5$

, $x_1=2.5$

, $x_2=1.98$.

Thus, the values of $f\left(x\right)$ at different initial value are,

$\begin{array}{c}f\left(1.5\right)=-4.375\\ f\left(2.5\right)=7.875\\ f\left(1.98\right)=-0.23721\end{array}$

Now calculate the $h_0\text{ and }{h}_1$.

$\begin{array}{l}{h}_0=x_1-x_0\\ h_1=x_2-x_1\end{array}$

Substituting the value of $x_0=1.5\text{, }{x}_1=2.5$

, and $x_2=1.98$.

$\begin{array}{l}{h}_0=1\\ h_1=-0.52\end{array}$

Now calculate the ${\delta }_0\text{ and }{\delta }_1$.

$\begin{array}{l}{\delta }_0=\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\\ {\delta }_1=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\end{array}$

Substitute the above calculated values.

$\begin{array}{c}{\delta }_0=\frac{f\left(2.5\right)-f\left(1.5\right)}{2.5-1.5}\\ =\frac{7.875+4.375}{1}\\ =12.25\end{array}$

And,

$\begin{array}{c}{\delta }_1=\frac{f\left(1.98\right)-f\left(2.5\right)}{1.98-2.5}\\ =\frac{-0.23721-7.875}{-0.52}\\ =15.60\end{array}$

Calculate the value of constants $a,b\text{ and }c$.

$\begin{array}{c}a=\frac{{\delta }_1-{\delta }_0}{h_1+h_0}\\ =\frac{15.60-12.25}{-0.52+1}\\ =\frac{3.35}{0.48}\\ =6.98\end{array}$

For b,

$\begin{array}{c}b=a{h}_1+{\delta }_1\\ =6\times \left(-0.52\right)+15.60\\ =12.48\end{array}$

For c,

$\begin{array}{c}c=f\left(x_2\right)\\ =-0.23721\end{array}$

Thus, the new root is calculated as follows.

$x_3=x_2+\frac{-2c}{b+\sqrt{b^2-4ac}}$

Substitute the all values.

$\begin{array}{c}{x}_3=1.98+\frac{-2\times \left(-0.23721\right)}{12.78+\sqrt{{12.48}^2-\left(4\right)\left(6.98\right)\left(-0.23721\right)}}\\ =1.98+0.037\\ =2.017\end{array}$

Calculate the error estimate.

$\begin{array}{c}\text{error}=\left|\frac{2.017-1.979384}{2.017}\right|\left(100\right)\\ =\left|\frac{0.037}{2.017}\right|\left(100\right)\\ =1.8\%\end{array}$

Because the error is near about 1, so required new guesses for $x_0$ is replaced by $x_1$

, $x_1$ is replaced by $x_2$

, and $x_2$ is replaced by $x_3$

. And repeat the same process will get the root is $2$.

$i$	$x_3$	error
0	1.979384	26.03
1	$2.017$	1.8
2	2	0.00241

Similarly repeat for all other roots.

To find the exact positive real root use MATLAB.

Write the following code in command window.

>> a=[1 1 -4 -4];

roots(a)

ans =

2.0000

-2.0000

-1.0000

Therefore, the only positive real root of the given equation $f\left(x\right)$ is $2$.

(b)

To determine

To calculate: The positive real root of $f\left(x\right)$ by using Müller’s method if,

$f\left(x\right)=x^3-0.5x^2+4x-2$

Answer to Problem 3P

Solution:

The only positive real root of the given equation $f\left(x\right)$ is $0.5$.

Explanation of Solution

Given Information:

The given equation is,

$f\left(x\right)=x^3-0.5x^2+4x-2$

Use Müller’s method.

Formula used:

The expression for the new roots is,

$x_3=x_2+\frac{-2c}{b+\sqrt{b^2-4ac}}$

The expression for the error is,

$\text{error}=\left|\frac{x_3-x_2}{x_3}\right|\cdot 100$

Calculation:

Recall the equation mentioned in the problem,

$f\left(x\right)=x^3+x^2-4x-4$

Draw the plot of the equation.

From the above plot it is clear that one root is at about $x=0.5$.

Consider the initial guess be $x_0=0.5\text{, }{x}_1=1$

, and $x_2=1.5$ respectively to determine the positive real root of the equation.

$f\left(x\right)=x^3+x^2-4x-4$

Thus, the values of $f\left(x\right)$ at different initial value are,

$\begin{array}{c}f\left(0.5\right)=0\\ f\left(1\right)=2.5\\ f\left(1.5\right)=6.25\end{array}$

Now calculate the $h_0\text{ and }{h}_1$.

$\begin{array}{l}{h}_0=x_1-x_0\\ h_1=x_2-x_1\end{array}$

Substituting the value of $x_0=0.5\text{, }{x}_1=1$

, and $x_2=1.5$.

$\begin{array}{l}{h}_0=0.5\\ h_1=0.5\end{array}$

Now calculate the ${\delta }_0\text{ and }{\delta }_1$.

$\begin{array}{l}{\delta }_0=\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\\ {\delta }_1=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\end{array}$

Substitute the above calculated values.

$\begin{array}{c}{\delta }_0=\frac{f\left(0.5\right)-f\left(1\right)}{0.5-1}\\ =\frac{0-2.5}{-0.5}\\ =5\end{array}$

And,

$\begin{array}{c}{\delta }_1=\frac{f\left(1.5\right)-f\left(1\right)}{1.5-1}\\ =\frac{6.25+2.5}{0.5}\\ =7.5\end{array}$

Calculate the value of constants $a,b\text{ and }c$.

$\begin{array}{c}a=\frac{{\delta }_1-{\delta }_0}{h_1+h_0}\\ =\frac{7.5-5}{0.5+0.5}\\ =\frac{2.5}{1.0}\\ =2.5\end{array}$

For b,

$\begin{array}{c}b=a{h}_1+{\delta }_1\\ =2.5\left(0.5\right)+7.5\\ =8.75\end{array}$

For c,

$\begin{array}{c}c=f\left(x_2\right)\\ =6.25\end{array}$

Thus, the new root is calculated as follows.

$x_3=x_2+\frac{-2c}{b+\sqrt{b^2-4ac}}$

Substitute the all values.

$\begin{array}{c}{x}_3=1.5+\frac{-2\times \left(6.25\right)}{8.75+\sqrt{{8.75}^2-\left(4\right)\left(2.5\right)\left(6.25\right)}}\\ =1.5-1.0\\ =0.5\end{array}$

Calculate the error estimate.

$\begin{array}{c}\text{error}=\left|\frac{0.5-1.5}{0.5}\right|\left(100\right)\\ =\left|\frac{-1.0}{0.5}\right|\left(100\right)\\ =200\%\end{array}$

Because the error is large, so required new guesses for $x_0$ is replaced by $x_1$

, $x_1$ is replaced by $x_2$

, and $x_2$ is replaced by $x_3$.

Therefore, for the new iteration, $x_0=1$

, $x_1=1.5$

, $x_2=0.5$.

Thus, the values of $f\left(x\right)$ at different initial value are,

$\begin{array}{c}f\left(1\right)=2.5\\ f\left(1.5\right)=6.25\\ f\left(0.5\right)=0\end{array}$

Now calculate the $h_0\text{ and }{h}_1$.

$\begin{array}{l}{h}_0=x_1-x_0\\ h_1=x_2-x_1\end{array}$

Substituting the value of $x_0=1$

, $x_1=1.5$

, $x_2=0.5$.

$\begin{array}{l}{h}_0=0.5\\ h_1=-1\end{array}$

Now calculate the ${\delta }_0\text{ and }{\delta }_1$.

$\begin{array}{l}{\delta }_0=\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\\ {\delta }_1=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\end{array}$

Substitute the above calculated values.

$\begin{array}{c}{\delta }_0=\frac{f\left(1\right)-f\left(1.5\right)}{1-1.5}\\ =\frac{6.25-2.5}{-0.5}\\ =7.5\end{array}$

And,

$\begin{array}{c}{\delta }_1=\frac{f\left(0.5\right)-f\left(1.5\right)}{0.5-1.5}\\ =\frac{0-6.25}{-1.0}\\ =6.25\end{array}$

Calculate the value of constants $a,b\text{ and }c$.

$\begin{array}{c}a=\frac{{\delta }_1-{\delta }_0}{h_1+h_0}\\ =\frac{6.25-7.5}{-1.0+0.5}\\ =\frac{-2.5}{-0.5}\\ =2.5\end{array}$

For b,

$\begin{array}{c}b=a{h}_1+{\delta }_1\\ =2.5\left(-1\right)+6.25\\ =3.75\end{array}$

For c,

$\begin{array}{c}c=f\left(x_2\right)\\ =0\end{array}$

Thus, the new root is calculated as follows.

$x_3=x_2+\frac{-2c}{b+\sqrt{b^2-4ac}}$

Substitute the all values.

$\begin{array}{c}{x}_3=0.5+\frac{-2\times \left(0\right)}{3.75+\sqrt{{3.75}^2-\left(4\right)\left(2.5\right)\left(0\right)}}\\ =0.5-0\\ =0.5\end{array}$

Calculate the error estimate.

$\begin{array}{c}\text{error}=\left|\frac{0.5-0.5}{0.5}\right|\left(100\right)\\ =0\%\end{array}$

Because the error is zero, the root is $0.5$.

$i$	$x_3$	error
0	$0.5$	200%
1	$0.5$	0%

Similarly repeat for all other roots.

To find the exact positive real root use MATLAB.

Write the following code in command window.

>> a=[1 -0.5 4 -2];

roots(a)

ans =

-0.0000 + 2.0000i

-0.0000 - 2.0000i

0.5000 + 0.0000i

Therefore, the only positive real root of the given equation $f\left(x\right)$ is $0.5$.