Chapter 6, Problem 7P

Locate the first positive root of

$f\left(x\right)=\sin x+\text{cos}\left(1+x^2\right)-1$

Where x is in radians. Use four iterations of the secant method with initial guesses of (a) $x_{i-1}=1.0\text{ and}{x}_i=3.0$: (b) $x_{i-1}=1.5\text{ and }{x}_i=2.5\text{ and }$(c) $x_{i-1}=\text{ 1}\text{.5 and }{x}_i=2.25$ to locate the root. (d) Use the graphical method to explain your results.

To determine

To calculate: Thefirst positive root of the function $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$ by the use of secant method with the initial guess of $x_{i-1}=1.0$ and $x_i=3.0$ up to fourth iteration.

Answer to Problem 7P

Solution:

The first positive root of the function $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$ is $0.39633$.

Explanation of Solution

Given Information:

The function, $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$. And, the initial guessesare $x_{i-1}=1.0$ and $x_i=3.0$.

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,

$f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$

For $i=0$, the initial guess is $x_{-1}=1.0$ and $x_0=3.0$, thus the first iteration is,

$\begin{array}{c}{x}_{0+1}=x_0-\frac{f\left(x_0\right)\left(x_{0-1}-x_0\right)}{f\left(x_{0-1}\right)-f\left(x_0\right)}\\ x_1=3-\frac{\left[\left\{\sin \left(3\right)+\cos \left(1+{\left(3\right)}^2\right)-1\right\}\left(1-3\right)\right]}{\left[\left\{\sin \left(1\right)+\cos \left(1+{\left(1\right)}^2\right)-1\right\}-\left\{\sin \left(3\right)+\cos \left(1+{\left(3\right)}^2\right)-1\right\}\right]}\\ =3-\frac{\left[\left\{-1.69795\right\}\left(-2\right)\right]}{\left[\left\{-0.57468\right\}-\left\{-1.69795\right\}\right]}\\ =-0.02323\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{-0.02323-3}{-0.02323}\right|\times 100\%\\ =\left|\frac{-3.02323}{-0.02323}\right|\times 100\%\\ =13014.33\%\end{array}$

Use $x_1=-0.02323$ and $x_0=3.0$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=x_1-\frac{f\left(x_1\right)\left(x_{1-1}-x_1\right)}{f\left(x_{1-1}\right)-f\left(x_1\right)}\\ x_2=-0.02323-\frac{\left[\left\{\sin \left(-0.02323\right)+\cos \left(1+{\left(-0.02323\right)}^2\right)-1\right\}\left(3+0.02323\right)\right]}{\left[\left\{\sin \left(3\right)+\cos \left(1+{\left(3\right)}^2\right)-1\right\}-\left\{\begin{array}{l}\sin \left(-0.02323\right)+\\ \cos \left(1+{\left(-0.02323\right)}^2\right)-1\end{array}\right\}\right]}\\ =-0.02323-\frac{\left[\left\{-0.48338\right\}\left(3.02323\right)\right]}{\left[\left\{-1.69795\right\}-\left\{-0.48338\right\}\right]}\\ =-1.22643\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{-1.22643-0.02323}{-1.22643}\right|\times 100\%\\ =\left|\frac{-1.24966}{-1.22643}\right|\times 100\%\\ =101.89\%\end{array}$

Use $x_2=-1.22643$ and $x_1=-0.02323$, the third iteration is,

$\begin{array}{c}{x}_{2+1}=x_2-\frac{f\left(x_2\right)\left(x_{2-1}-x_2\right)}{f\left(x_{2-1}\right)-f\left(x_2\right)}\\ x_3=-1.22643-\frac{\left[\left\{\sin \left(-1.22643\right)+\cos \left(1+{\left(-1.22643\right)}^2\right)-1\right\}\left(-0.02323+1.22643\right)\right]}{\left[\left\{\begin{array}{l}\sin \left(-0.02323\right)+\\ \cos \left(1+{\left(-0.02323\right)}^2\right)-1\end{array}\right\}-\left\{\begin{array}{l}\sin \left(-1.22643\right)+\\ \cos \left(1+{\left(-1.22643\right)}^2\right)-1\end{array}\right\}\right]}\\ =-1.22643-\frac{\left[\left\{-2.7449\right\}\left(1.2032\right)\right]}{\left[\left\{-0.48338\right\}-\left\{-2.7449\right\}\right]}\\ =0.2339\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{0.2339+1.22643}{0.2339}\right|\times 100\%\\ =\left|\frac{1.46033}{0.2339}\right|\times 100\%\\ =624.34\%\end{array}$

Use $x_3=0.2339$ and $x_2=-1.22643$, the fourth iteration is,

$\begin{array}{c}{x}_{3+1}=x_3-\frac{f\left(x_3\right)\left(x_{3-1}-x_3\right)}{f\left(x_{3-1}\right)-f\left(x_3\right)}\\ x_4=0.2339-\frac{\left[\left\{\sin \left(0.2339\right)+\cos \left(1+{\left(0.2339\right)}^2\right)-1\right\}\left(-1.22643-0.2339\right)\right]}{\left[\left\{\begin{array}{l}\sin \left(-1.22643\right)+\\ \cos \left(1+{\left(-1.22643\right)}^2\right)-1\end{array}\right\}-\left\{\sin \left(0.2339\right)+\cos \left(1+{\left(0.2339\right)}^2\right)-1\right\}\right]}\\ =0.2339-\frac{\left[\left\{-0.27475\right\}\left(-1.46033\right)\right]}{\left[\left\{-2.7449\right\}-\left\{-0.27475\right\}\right]}\\ =0.39633\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{0.39633-0.2339}{0.39633}\right|\times 100\%\\ =\left|\frac{0.16243}{0.39633}\right|\times 100\%\\ =40.98\%\end{array}$

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	$-0.02323$	$13014.33\%$
2	$-1.22643$	$101.89\%$
3	$0.2339$	$624.34\%$
4	$0.39633$	$40.98\%$

Hence, the first positive root is $0.39633$.

To determine

To calculate: The first positive root of the function $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$ by the use of secant method with the initial guess of $x_{i-1}=1.5$ and $x_i=2.5$ up to fourth iteration.

Answer to Problem 7P

Solution:

The first positive root of the function $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$ is $2.53211$.

Explanation of Solution

Given Information:

The function, $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$. And, the initial guesses are $x_{i-1}=1.5$ and $x_i=2.5$.

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,

$f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$

For $i=0$, the initial guess is $x_{-1}=1.5$ and $x_0=2.5$, thus the first iteration is,

$\begin{array}{c}{x}_{0+1}=x_0-\frac{f\left(x_0\right)\left(x_{0-1}-x_0\right)}{f\left(x_{0-1}\right)-f\left(x_0\right)}\\ x_1=2.5-\frac{\left[\left\{\sin \left(2.5\right)+\cos \left(1+{\left(2.5\right)}^2\right)-1\right\}\left(1.5-2.5\right)\right]}{\left[\left\{\sin \left(1.5\right)+\cos \left(1+{\left(1.5\right)}^2\right)-1\right\}-\left\{\sin \left(2.5\right)+\cos \left(1+{\left(2.5\right)}^2\right)-1\right\}\right]}\\ =2.5-\frac{\left[\left\{0.1664\right\}\left(-1\right)\right]}{\left[\left\{-0.9966\right\}-\left\{0.1664\right\}\right]}\\ =2.35692\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{2.35692-2.5}{2.35692}\right|\times 100\%\\ =\left|\frac{-0.14308}{2.35692}\right|\times 100\%\\ =6.07\%\end{array}$

Use $x_1=2.35692$ and $x_0=2.5$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=x_1-\frac{f\left(x_1\right)\left(x_{1-1}-x_1\right)}{f\left(x_{1-1}\right)-f\left(x_1\right)}\\ x_2=2.35692-\frac{\left[\left\{\sin \left(2.35692\right)+\cos \left(1+{\left(2.35692\right)}^2\right)-1\right\}\left(2.5-2.35692\right)\right]}{\left[\left\{\sin \left(2.5\right)+\cos \left(1+{\left(2.5\right)}^2\right)-1\right\}-\left\{\begin{array}{l}\sin \left(2.35692\right)+\\ \cos \left(1+{\left(2.35692\right)}^2\right)-1\end{array}\right\}\right]}\\ =2.35692-\frac{\left[\left\{0.66986\right\}\left(0.14308\right)\right]}{\left[\left\{0.1664\right\}-\left\{0.66986\right\}\right]}\\ =2.5473\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{2.5473-2.35692}{2.5473}\right|\times 100\%\\ =\left|\frac{0.19038}{2.5473}\right|\times 100\%\\ =7.47\%\end{array}$

Use $x_2=2.5473$ and $x_1=2.35692$, the third iteration is,

$\begin{array}{c}{x}_{2+1}=x_2-\frac{f\left(x_2\right)\left(x_{2-1}-x_2\right)}{f\left(x_{2-1}\right)-f\left(x_2\right)}\\ x_3=2.5473-\frac{\left[\left\{\sin \left(2.5473\right)+\cos \left(1+{\left(2.5473\right)}^2\right)-1\right\}\left(2.35692-2.5473\right)\right]}{\left[\left\{\begin{array}{l}\sin \left(2.35692\right)+\\ \cos \left(1+{\left(2.35692\right)}^2\right)-1\end{array}\right\}-\left\{\sin \left(2.5473\right)+\cos \left(1+{\left(2.5473\right)}^2\right)-1\right\}\right]}\\ =2.5473-\frac{\left[\left\{-0.0829\right\}\left(-0.19038\right)\right]}{\left[\left\{0.66986\right\}-\left\{-0.0829\right\}\right]}\\ =2.52633\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{2.52633-2.5473}{2.52633}\right|\times 100\%\\ =\left|\frac{-0.02097}{2.52633}\right|\times 100\%\\ =0.83\%\end{array}$

Use $x_3=2.52633$ and $x_2=-1.22643$, the fourth iteration is,

$\begin{array}{c}{x}_{3+1}=x_3-\frac{f\left(x_3\right)\left(x_{3-1}-x_3\right)}{f\left(x_{3-1}\right)-f\left(x_3\right)}\\ x_4=2.52633-\frac{\left[\left\{\sin \left(2.52633\right)+\cos \left(1+{\left(2.52633\right)}^2\right)-1\right\}\left(2.5473-2.52633\right)\right]}{\left[\left\{\begin{array}{l}\sin \left(2.5473\right)+\\ \cos \left(1+{\left(2.5473\right)}^2\right)-1\end{array}\right\}-\left\{\sin \left(2.52633\right)+\cos \left(1+{\left(2.52633\right)}^2\right)-1\right\}\right]}\\ =2.52633-\frac{\left[\left\{0.03152\right\}\left(0.02097\right)\right]}{\left[\left\{-0.0829\right\}-\left\{0.03152\right\}\right]}\\ =2.53211\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{2.53211-2.52633}{2.53211}\right|\times 100\%\\ =\left|\frac{0.00578}{2.53211}\right|\times 100\%\\ =0.23\%\end{array}$

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	2.5
1	$2.35692$	$6.07\%$
2	$2.5473$	$7.47\%$
3	$2.52633$	$0.83\%$
4	$2.53211$	$0.23\%$

Hence, the first positive root is $2.53211$.

To determine

To calculate: The first positive root of the function $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$ by the use of secant method with the initial guess of $x_{i-1}=1.5$ and $x_i=2.25$ up to fourth iteration.

Answer to Problem 7P

Solution:

The first positive root of the function $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$ is $1.9446$.

Explanation of Solution

Given Information:

The function, $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$. And, the initial guesses are $x_{i-1}=1.5$ and $x_i=2.25$.

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,

$f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$

For $i=0$, the initial guess is $x_{-1}=1.5$ and $x_0=2.25$, thus the first iteration is,

$\begin{array}{c}{x}_{0+1}=x_0-\frac{f\left(x_0\right)\left(x_{0-1}-x_0\right)}{f\left(x_{0-1}\right)-f\left(x_0\right)}\\ x_1=2.25-\frac{\left[\left\{\sin \left(2.25\right)+\cos \left(1+{\left(2.25\right)}^2\right)-1\right\}\left(1.5-2.25\right)\right]}{\left[\left\{\sin \left(1.5\right)+\cos \left(1+{\left(1.5\right)}^2\right)-1\right\}-\left\{\sin \left(2.25\right)+\cos \left(1+{\left(2.25\right)}^2\right)-1\right\}\right]}\\ =2.25-\frac{\left[\left\{0.7538\right\}\left(-0.75\right)\right]}{\left[\left\{-0.9966\right\}-\left\{0.7538\right\}\right]}\\ =1.92702\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{1.92702-2.25}{1.92702}\right|\times 100\%\\ =\left|\frac{-0.32298}{1.92702}\right|\times 100\%\\ =16.76\%\end{array}$

Use $x_1=1.92702$ and $x_0=2.25$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=x_1-\frac{f\left(x_1\right)\left(x_{1-1}-x_1\right)}{f\left(x_{1-1}\right)-f\left(x_1\right)}\\ x_2=1.92702-\frac{\left[\left\{\sin \left(1.92702\right)+\cos \left(1+{\left(1.92702\right)}^2\right)-1\right\}\left(2.25-1.92702\right)\right]}{\left[\left\{\sin \left(2.25\right)+\cos \left(1+{\left(2.25\right)}^2\right)-1\right\}-\left\{\begin{array}{l}\sin \left(1.92702\right)+\\ \cos \left(1+{\left(1.92702\right)}^2\right)-1\end{array}\right\}\right]}\\ =1.92702-\frac{\left[\left\{-0.06176\right\}\left(0.32298\right)\right]}{\left[\left\{0.7538\right\}-\left\{-0.06176\right\}\right]}\\ =1.95148\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{1.95148-1.92702}{1.95148}\right|\times 100\%\\ =\left|\frac{0.02446}{1.95148}\right|\times 100\%\\ =1.25\%\end{array}$

Use $x_2=1.95148$ and $x_1=1.92702$, the third iteration is,

$\begin{array}{c}{x}_{2+1}=x_2-\frac{f\left(x_2\right)\left(x_{2-1}-x_2\right)}{f\left(x_{2-1}\right)-f\left(x_2\right)}\\ x_3=1.95148-\frac{\left[\left\{\sin \left(1.95148\right)+\cos \left(1+{\left(1.95148\right)}^2\right)-1\right\}\left(1.92702-1.95148\right)\right]}{\left[\left\{\begin{array}{l}\sin \left(1.92702\right)+\\ \cos \left(1+{\left(1.92702\right)}^2\right)-1\end{array}\right\}-\left\{\sin \left(1.95148\right)+\cos \left(1+{\left(1.95148\right)}^2\right)-1\right\}\right]}\\ =1.95148-\frac{\left[\left\{0.02415\right\}\left(-0.02446\right)\right]}{\left[\left\{-0.06176\right\}-\left\{0.02415\right\}\right]}\\ =1.9446\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{1.9446-1.95148}{1.9446}\right|\times 100\%\\ =\left|\frac{-0.00688}{1.9446}\right|\times 100\%\\ =0.35\%\end{array}$

Use $x_3=1.9446$ and $x_2=1.95148$, the fourth iteration is,

$\begin{array}{c}{x}_{3+1}=x_3-\frac{f\left(x_3\right)\left(x_{3-1}-x_3\right)}{f\left(x_{3-1}\right)-f\left(x_3\right)}\\ x_4=1.9446-\frac{\left[\left\{\sin \left(1.9446\right)+\cos \left(1+{\left(1.9446\right)}^2\right)-1\right\}\left(1.95148-1.9446\right)\right]}{\left[\left\{\begin{array}{l}\sin \left(1.95148\right)+\\ \cos \left(1+{\left(1.95148\right)}^2\right)-1\end{array}\right\}-\left\{\sin \left(1.9446\right)+\cos \left(1+{\left(1.9446\right)}^2\right)-1\right\}\right]}\\ =1.9446-\frac{\left[\left\{-0.00002961\right\}\left(0.00688\right)\right]}{\left[\left\{0.02415\right\}-\left\{-0.00002961\right\}\right]}\\ =1.9446\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{1.9446-1.9446}{1.9446}\right|\times 100\%\\ =\left|\frac{0}{1.9446}\right|\times 100\%\\ =0.00\%\end{array}$

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	$2.25$
1	$1.92702$	16.76%
2	$1.95148$	$1.25\%$
3	$1.9446$	$0.35\%$
4	$1.9446$	$0.00\%$

Hence, the first positive root is $1.9446$.

To determine

The explanation of the result obtained by the secant method of the function $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$ with the initial guesses $\left(x_{i-1}=1.0\text{ and }{x}_i=3.0\right)$, $\left(x_{i-1}=1.5\text{ and }{x}_i=2.5\right)$ and $\left(x_{i-1}=1.5\text{ and }{x}_i=2.25\right)$ by the graphical method.

Answer to Problem 7P

Solution:

For the initial guesses $x_{i-1}=1.0\text{ and }{x}_i=3.0$, the first iteration gives a negative value due to the improper choice of initial guess.

For the initial guesses $x_{i-1}=1.5\text{ and }{x}_i=2.5$, the root is converged towards the second positive root instead of first positive root.

For the initial guesses $x_{i-1}=1.5\text{ and }{x}_i=2.25$, the root is converged towards the first positive root $x=1.9446$.

Explanation of Solution

Given Information:

The function, $f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$.

Consider the function,

$f\left(x\right)=\sin x+\cos \left(1+x^2\right)-1$

From part (a), the result obtained with the initial guesses $x_{i-1}=1.0\text{ and }{x}_i=3.0$ is,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	3
1	$-0.02323$	$13014.33\%$
2	$-1.22643$	$101.89\%$
3	$0.2339$	$624.34\%$
4	$0.39633$	$40.98\%$

Use MATLAB to draw the graph of the function as below,

Code:

function f = g(x)

% f is assigned a value of function f(x).

f=sin(x)+cos(1+(x^2))-1;

%function is defined.

end

Output:

The graph obtained is,

From the graph, it is observed that the first iteration gives a negative value due to the improper choice of initial guess.

From part (b), the result obtained with the initial guesses $x_{i-1}=1.5\text{ and }{x}_i=2.5$ is,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	2.5
1	$2.35692$	$6.07\%$
2	$2.5473$	$7.47\%$
3	$2.52633$	$0.83\%$
4	$2.53211$	$0.23\%$

Use MATLAB to draw the graph of the function as below,

Code:

function f = g(x)

% f is assigned a value of function f(x).

f=sin(x)+cos(1+(x^2))-1;

%function is defined.

end

Output:

The graph obtained is,

From the graph, it is observed that the root is converges towards the second positive root with these initial guesses instead of first positive root.

From part (c), the result obtained with the initial guesses $x_{i-1}=1.5\text{ and }{x}_i=2.25$ is,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	$2.25$
1	$1.92702$	16.76%
2	$1.95148$	$1.25\%$
3	$1.9446$	$0.35\%$
4	$1.9446$	$0.00\%$

Use MATLAB to draw the graph of the function as below,

Code:

function f = g(x)

% f is assigned a value of function f(x).

f=sin(x)+cos(1+(x^2))-1;

%function is defined.

Output:

The graph obtained is,

From the graph, it is observed that the root is converges towards the first positive root with these initial guesses rapidly. Hence, these guesses are the proper selection to locate the first positive root by secant method.