Chapter 6, Problem 2P

Determine the highest real root of

$f\left(x\right)=2x^3-11.7x^2+17.7x-5$

(a) Graphically.

(b) Fixed-point iteration method (three iterations, $x_0=3$). Note: Make certain that you develop a solution that converges on the root.

(c) Newton-Raphson method (three iterations, $x_0=3$).

(d) Secant method (three iterations, $x_{-1}=3,x_0=4$).

(e) Modified secant method (three iterations, $x{0}_{}=3,\delta =0.01$).

Compute the approximate percent relative errors for your solutions.

To determine

To calculate: The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ by the use of graphical method.

Answer to Problem 2P

Solution:

The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ is $\text{3}.6$.

Explanation of Solution

Given:

The function, $f\left(x\right)=2x^3-11.7x^2+17.7x-5$.

Formula used:

The roots of the function are the points at which the graph of the function crosses the x-axis.

Calculation:

Consider the function,

$f\left(x\right)=2x^3-11.7x^2+17.7x-5$

Substitute different values of x and find the corresponding values of $f\left(x\right)$ to graph the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$,

For $x=0$,

$\begin{array}{c}f\left(0\right)=2{\left(0\right)}^3-11.7{\left(0\right)}^2+17.7\left(0\right)-5\\ =0-0+0-5\\ =-5\end{array}$

For $x=1$,

$\begin{array}{c}f\left(1\right)=2{\left(1\right)}^3-11.7{\left(1\right)}^2+17.7\left(1\right)-5\\ =2-11.7+17.7-5\\ =19.7-16.7\\ =3\end{array}$

For $x=2$,

$\begin{array}{c}f\left(2\right)=2{\left(2\right)}^3-11.7{\left(2\right)}^2+17.7\left(2\right)-5\\ =16-46.8+35.4-5\\ =51.4-51.8\\ =-0.4\end{array}$

For $x=3$,

$\begin{array}{c}f\left(3\right)=2{\left(3\right)}^3-11.7{\left(3\right)}^2+17.7\left(3\right)-5\\ =54-105.3+53.1-5\\ =107.1-110.3\\ =-3.2\end{array}$

For $x=4$,

$\begin{array}{c}f\left(4\right)=2{\left(4\right)}^3-11.7{\left(4\right)}^2+17.7\left(4\right)-5\\ =128-187.2+70.8-5\\ =198.8-192.2\\ =6.6\end{array}$

Summarize the above values as shown below,

x	$f\left(x\right)=2x^3-11.7x^2+17.7x-5$
0	$-5$
1	3
2	$-0.4$
3	$-3.2$
4	6.6

Plot the above points on the graph and join them as below,

From the above graph, it is observed that the graph of the function crosses the x-axis from three points. That is, approximately $x=0.35,x=1.9\text{ and }x=3.6$.

Hence, the highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ is $\text{3}.6$.

To determine

To calculate: The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ by the use of fixed point iteration method with the initial guess $x_0=3$ and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ is 3.4425.

Explanation of Solution

Given:

The function, $f\left(x\right)=2x^3-11.7x^2+17.7x-5$.

Formula used:

The simple fixed-point iteration formula for the function $x=g\left(x\right)$,

$x_{i+1}=g\left(x_i\right)$

And, formula for approximate error is,

${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$

Calculation:

Consider the function,

$f\left(x\right)=2x^3-11.7x^2+17.7x-5$

The function can be formulated as fixed-point iteration as,

$\begin{array}{c}2x^3-11.7x^2+17.7x-5=0\\ 17.7x=-2x^3+11.7x^2+5\\ x_{i+1}=\frac{-2x_i{}^3+11.7x_i{}^2+5}{17.7}\end{array}$

Use initial guess $x_0=3$, the first iteration is,

$\begin{array}{c}{x}_{0+1}=\frac{-2x_0{}^3+11.7x_0{}^2+5}{17.7}\\ x_1=\frac{-2{\left(3\right)}^3+11.7{\left(3\right)}^2+5}{17.7}\\ =\frac{-54+105.3+5}{17.7}\\ =3.18079\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.18079-3}{3.18079}\right|\times 100\%\\ =\left|\frac{0.18079}{3.18079}\right|\times 100\%\\ =\left|0.05683\right|\times 100\%\\ =5.683\%\end{array}$

Use $x_1=3.18079$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=\frac{-2x_1{}^3+11.7x_1{}^2+5}{17.7}\\ x_2=\frac{-2{\left(3.18079\right)}^3+11.7{\left(3.18079\right)}^2+5}{17.7}\\ =\frac{-64.3628+118.3739+5}{17.7}\\ =3.333959\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.333959-3.18079}{3.333959}\right|\times 100\%\\ =\left|\frac{0.153169}{3.333959}\right|\times 100\%\\ =\left|0.04594\right|\times 100\%\\ =4.594\%\end{array}$

Use $x_2=3.333959$, the third iteration is,

$\begin{array}{c}{x}_{2+1}=\frac{-2x_2{}^3+11.7x_2{}^2+5}{17.7}\\ x_3=\frac{-2{\left(3.333959\right)}^3+11.7{\left(3.333959\right)}^2+5}{17.7}\\ =\frac{-74.1158+130.0488+5}{17.7}\\ =3.4425\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.4425-3.333959}{3.4425}\right|\times 100\%\\ =\left|\frac{0.108541}{3.4425}\right|\times 100\%\\ =\left|0.0315297\right|\times 100\%\\ =3.153\%\end{array}$

Thus, all the iteration can be summarized as below,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	3
1	3.18079	5.683%
2	3.333959	4.594%
3	3.4425	3.153%

Hence, the highest root is 3.4425.

To determine

To calculate: The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ by the use of Newton-Raphson method with the initial guess $x_0=3$ and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ is 3.792837.

Explanation of Solution

Given:

The function, $f\left(x\right)=2x^3-11.7x^2+17.7x-5$.

Formula used:

The Newton-Raphson formula,

$x_{i+1}=x_i-\frac{f\left(x_i\right)}{f^{\prime }\left(x_i\right)}$

And, formula for approximate error is,

${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$

Calculation:

Consider the function,

$f\left(x\right)=2x^3-11.7x^2+17.7x-5$

Differentiate the above function with respect to x,

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(2x^3-11.7x^2+17.7x-5\right)\\ =\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(-11.7x^2\right)+\frac{d}{dx}\left(17.7x\right)+\frac{d}{dx}\left(-5\right)\\ =2\times 3x^2-11.7\times 2x+17.7-0\\ =6x^2-23.4x+17.7\end{array}$

The initial guess is $x_0=3$, thus the first iteration is,

$\begin{array}{c}{x}_{0+1}=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}\\ x_1=3-\frac{\left\{2{\left(3\right)}^3-11.7{\left(3\right)}^2+17.7\left(3\right)-5\right\}}{\left\{6{\left(3\right)}^2-23.4\left(3\right)+17.7\right\}}\\ =3-\left(\frac{-3.2}{1.5}\right)\\ =5.133\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{5.133-3}{5.133}\right|\times 100\%\\ =\left|\frac{2.133}{5.133}\right|\times 100\%\\ =\left|0.41555\right|\times 100\%\\ =41.555\%\end{array}$

Use $x_1=5.133$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=x_1-\frac{f\left(x_1\right)}{f^{\prime }\left(x_1\right)}\\ x_2=5.133-\frac{\left\{2{\left(5.133\right)}^3-11.7{\left(5.133\right)}^2+17.7\left(5.133\right)-5\right\}}{\left\{6{\left(5.133\right)}^2-23.4\left(5.133\right)+17.7\right\}}\\ =5.133-\left(\frac{48.0715}{55.6739}\right)\\ =4.26955\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{4.26955-5.133}{4.26955}\right|\times 100\%\\ =\left|\frac{-0.86345}{4.26955}\right|\times 100\%\\ =\left|-0.20223\right|\times 100\%\\ =20.223\%\end{array}$

Use $x_2=4.26955$, the third iteration is,

$\begin{array}{c}{x}_{2+1}=x_2-\frac{f\left(x_2\right)}{f^{\prime }\left(x_2\right)}\\ x_3=4.26955-\frac{\left\{2{\left(4.26955\right)}^3-11.7{\left(4.26955\right)}^2+17.7\left(4.26955\right)-5\right\}}{\left\{6{\left(4.26955\right)}^2-23.4\left(4.26955\right)+17.7\right\}}\\ =4.26955-\left(\frac{12.9508}{27.16687}\right)\\ =3.792837\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.792837-4.26955}{3.792837}\right|\times 100\%\\ =\left|\frac{-0.476713}{3.792837}\right|\times 100\%\\ =\left|-0.12569\right|\times 100\%\\ =12.569\%\end{array}$

Similarly, all the iteration can be summarized as below,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	3
1	5.133	41.555%
2	4.26955	20.223%
3	3.792837	12.569%

Hence, the highest root is 3.792837.

To determine

To calculate: The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ by the use of Secant method with the initial guess $x_{-1}=3\text{ and }{x}_0=4$ and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ is 3.58629.

Explanation of Solution

Given:

The function, $f\left(x\right)=2x^3-11.7x^2+17.7x-5$.

Formula used:

The iterative equation of secant method is,

$x_{i+1}=x_i-\frac{f\left(x_i\right)\left(x_{i-1}-x_i\right)}{f\left(x_{i-1}\right)-f\left(x_i\right)}$

And, formula for approximate error is,

${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$

Calculation:

Consider the function,

$f\left(x\right)=2x^3-11.7x^2+17.7x-5$

The initial guess is $x_{-1}=3\text{ and }{x}_0=4$, thus the first iteration is,

$\begin{array}{c}{x}_{0+1}=x_0-\frac{f\left(x_0\right)\left(x_{0-1}-x_0\right)}{f\left(x_{0-1}\right)-f\left(x_0\right)}\\ x_1=4-\frac{\left[\left\{2{\left(4\right)}^3-11.7{\left(4\right)}^2+17.7\left(4\right)-5\right\}\left(3-4\right)\right]}{\left[\left\{2{\left(3\right)}^3-11.7{\left(3\right)}^2+17.7\left(3\right)-5\right\}-\left\{2{\left(4\right)}^3-11.7{\left(4\right)}^2+17.7\left(4\right)-5\right\}\right]}\\ =4-\frac{\left[\left\{6.6\right\}\left(-1\right)\right]}{\left[\left\{-3.2\right\}-\left\{6.6\right\}\right]}\\ =3.3265\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.3265-4}{3.3265}\right|\times 100\%\\ =\left|\frac{-0.6735}{3.3265}\right|\times 100\%\\ =\left|-0.2025\right|\times 100\%\\ =20.25\%\end{array}$

Use $x_1=3.3265$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=x_1-\frac{f\left(x_1\right)\left(x_{1-1}-x_1\right)}{f\left(x_{1-1}\right)-f\left(x_1\right)}\\ x_2=3.3265-\frac{\left[\left\{2{\left(3.3265\right)}^3-11.7{\left(3.3265\right)}^2+17.7\left(3.3265\right)-5\right\}\left(4-3.3265\right)\right]}{\left[\left\{2{\left(4\right)}^3-11.7{\left(4\right)}^2+17.7\left(4\right)-5\right\}-\left\{\begin{array}{l}2{\left(3.3265\right)}^3-11.7{\left(3.3265\right)}^2\\ +17.7\left(3.3265\right)-5\end{array}\right\}\right]}\\ =3.3265-\frac{\left[\left\{-1.969\right\}\left(0.6735\right)\right]}{\left[\left\{6.6\right\}-\left\{-1.969\right\}\right]}\\ =3.4812\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.4812-3.3265}{3.4812}\right|\times 100\%\\ =\left|\frac{0.1574}{3.4812}\right|\times 100\%\\ =\left|0.04443\right|\times 100\%\\ =4.443\%\end{array}$

Use $x_2=3.4812$, the third iteration is,

$\begin{array}{c}{x}_{2+1}=x_2-\frac{f\left(x_2\right)\left(x_{2-1}-x_2\right)}{f\left(x_{2-1}\right)-f\left(x_2\right)}\\ x_3=3.4812-\frac{\left[\left\{2{\left(3.4812\right)}^3-11.7{\left(3.4812\right)}^2+17.7\left(3.4812\right)-5\right\}\left(3.3265-3.4812\right)\right]}{\left[\left\{\begin{array}{l}2{\left(3.3265\right)}^3-11.7{\left(3.3265\right)}^2\\ +17.7\left(3.3265\right)-5\end{array}\right\}-\left\{\begin{array}{l}2{\left(3.4812\right)}^3-11.7{\left(3.4812\right)}^2+\\ 17.7\left(3.4812\right)-5\end{array}\right\}\right]}\\ =3.4812-\frac{\left[\left\{-0.7965\right\}\left(-0.1547\right)\right]}{\left[\left\{-1.969\right\}-\left\{-0.7965\right\}\right]}\\ =3.58629\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.58629-3.4812}{3.58629}\right|\times 100\%\\ =\left|\frac{0.10509}{3.58629}\right|\times 100\%\\ =\left|0.02930\right|\times 100\%\\ =2.93\%\end{array}$

Similarly, all the iteration can be summarized as below,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	4
1	3.3265	20.25%
2	3.4812	4.443%
3	3.58629	2.93%

Hence, the highest root is 3.58629.

To determine

To calculate: The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ by the use of modified Secant method with the initial guess $x_0=3\text{ and }\delta =0.01$ and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function $f\left(x\right)=2x^3-11.7x^2+17.7x-5$ is 3.7429.

Explanation of Solution

Given:

The function, $f\left(x\right)=2x^3-11.7x^2+17.7x-5$.

Formula used:

The iteration formula for modified secant method is,

$x_{i+1}=x_i-\frac{\delta x_{i}f\left(x_i\right)}{f\left(x_i+\delta x_i\right)-f\left(x_i\right)}$

And, formula for approximate error is,

${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$

Calculation:

Consider the function,

$f\left(x\right)=2x^3-11.7x^2+17.7x-5$

Use initial guess of $x_0=3$ and $\delta =0.01$, the first iteration is,

$\begin{array}{c}{x}_{0+1}=x_0-\frac{\delta x_{0}f\left(x_0\right)}{f\left(x_0+\delta x_0\right)-f\left(x_0\right)}\\ x_1=3-\frac{0.01\times 3\times f\left(3\right)}{f\left(3+0.01\times 3\right)-f\left(3\right)}\\ =3-\frac{0.03\left\{2{\left(3\right)}^3-11.7{\left(3\right)}^2+17.7\left(3\right)-5\right\}}{\left\{2{\left(3.03\right)}^3-11.7{\left(3.03\right)}^2+17.7\left(3.03\right)-5\right\}-\left\{2{\left(3\right)}^3-11.7{\left(3\right)}^2+17.7\left(3\right)-5\right\}}\\ =3-\frac{0.03\times \left\{-3.2\right\}}{\left\{-3.149276\right\}-\left\{-3.2\right\}}\end{array}$

Simplify furthermore,

$\begin{array}{c}{x}_1=3-\frac{\left(-0.096\right)}{\left(0.050724\right)}\\ =3-\left(-1.89259\right)\\ =3+1.89259\\ =4.89259\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{4.89259-3}{4.89259}\right|\times 100\%\\ =\left|\frac{1.89259}{4.89259}\right|\times 100\%\\ =38.68\%\end{array}$

Use $x_1=4.89259$ and $\delta =0.01$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=x_1-\frac{\delta x_{1}f\left(x_1\right)}{f\left(x_1+\delta x_1\right)-f\left(x_1\right)}\\ x_2=4.89259-\frac{0.01\times 4.89259\times f\left(4.89259\right)}{f\left(4.89259+0.01\times 4.89259\right)-f\left(4.89259\right)}\\ =4.89259-\frac{0.049\left\{2{\left(4.89259\right)}^3-11.7{\left(4.89259\right)}^2+17.7\left(4.89259\right)-5\right\}}{\left\{\begin{array}{l}2{\left(4.9415\right)}^3-11.7{\left(4.9415\right)}^2\\ +17.7\left(4.9415\right)-5\end{array}\right\}-\left\{\begin{array}{l}2{\left(4.89259\right)}^3-11.7{\left(4.89259\right)}^2\\ +17.7\left(4.89259\right)-5\end{array}\right\}}\\ =4.89259-\frac{0.049\times \left\{35.763\right\}}{\left\{38.096\right\}-\left\{35.763\right\}}\end{array}$

Simplify furthermore,

$\begin{array}{c}{x}_2=4.89259-\frac{0.049\times \left\{35.763\right\}}{\left\{38.096\right\}-\left\{35.763\right\}}\\ =4.89259-\frac{1.7524}{2.333}\\ =4.89259-0.75114\\ =4.14145\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{4.14145-4.89259}{4.14145}\right|\times 100\%\\ =\left|\frac{-0.75114}{4.14145}\right|\times 100\%\\ =18.14\%\end{array}$

Use $x_2=4.14145$ and $\delta =0.01$, the third iteration is,

$\begin{array}{c}{x}_{2+1}=x_2-\frac{\delta x_{2}f\left(x_2\right)}{f\left(x_2+\delta x_2\right)-f\left(x_2\right)}\\ x_3=4.14145-\frac{0.01\times 4.14145\times f\left(4.14145\right)}{f\left(4.14145+0.01\times 4.14145\right)-f\left(4.14145\right)}\\ =4.14145-\frac{0.041415\left\{2{\left(4.14145\right)}^3-11.7{\left(4.14145\right)}^2+17.7\left(4.14145\right)-5\right\}}{\left\{\begin{array}{l}2{\left(4.183\right)}^3-11.7{\left(4.183\right)}^2\\ +17.7\left(4.183\right)-5\end{array}\right\}-\left\{\begin{array}{l}2{\left(4.14145\right)}^3-11.7{\left(4.14145\right)}^2\\ +17.7\left(4.14145\right)-5\end{array}\right\}}\\ =4.14145-\frac{0.041415\times \left\{9.6949\right\}}{\left\{10.7025\right\}-\left\{9.6949\right\}}\end{array}$

Simplify furthermore,

$\begin{array}{c}{x}_3=4.14145-\frac{0.041415\times \left\{9.6949\right\}}{\left\{10.7025\right\}-\left\{9.6949\right\}}\\ =4.14145-\frac{0.4015}{1.0076}\\ =4.14145-0.3985\\ =3.7429\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{3.7429-4.14145}{3.7429}\right|\times 100\%\\ =\left|\frac{-0.39855}{3.7429}\right|\times 100\%\\ =10.65\%\end{array}$

Similarly, all the iteration can be summarized as below,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	3
1	4.89259	38.68%
2	4.14145	18.14%
3	3.7429	10.65%

Hence, the highest root is 3.7429.