Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation $x^3=2x+2$ with initial guesses $x_0=1and{x}_1=2$ .

Answer to Problem 1E

Thus, $x_2=\frac{8}{5}$ and $x_3=1.742268041$ .

Given information:Initial guesses $x_0=1and{x}_1=2$ and equation is $(a)x^3=2x+2$ .

Calculation:

With the initial guesses,  $x_0=1and{x}_1=2$ .

Now, using the formula, we will get the following.

   $\begin{array}{l}{x}_{i+1}=x_i-\frac{f_1\left(x_i\right)\left(x_i-x_{i-1}\right)}{f_1\left(x_i\right)-f_1\left(x_{i-1}\right)}\\ =x_i-\frac{\left(x_i^3-2x_i-2\right)\left(x_i-x_{i-1}\right)}{\left(x_i^3-2x_i-2\right)-\left(x_{i-1}^3-2x_{i-1}-2\right)}\\ =x_i-\frac{\left(x_i^3-2x_i-2\right)\left(x_i-x_{i-1}\right)}{x_i^3-2x_i-x_{i-1}^3+2x_{i-1}}\end{array}$

Let us find the value of  $x_2$ .

   $\begin{array}{cc}{x}_2& =x_1-\frac{\left(x_1^3-2x_1-2\right)\left(x_1-x_0\right)}{x_1^3-2x_1-x_0^3+2x_0}\\ & =2-\frac{\left(2^3-2(2)-2\right)(2-1)}{2^3-2(2)-1^3+2(1)}\end{array}$

   $\begin{array}{l}=2-\frac{2}{5}\hfill \\ =\frac{8}{5}\hfill \end{array}$

Let us find the value of  $x_3$ .

   $\begin{array}{l}{x}_3=x_2-\frac{\left(x_2^3-2x_2-2\right)\left(x_2-x_1\right)}{x_2^3-2x_2-x_1^3+2x_1}\\ =\frac{8}{5}-\frac{\left({\left(\frac{8}{5}\right)}^3-2\left(\frac{8}{5}\right)-2\right)\left(\frac{8}{5}-2\right)}{{\left(\frac{8}{5}\right)}^3-2\left(\frac{8}{5}\right)-2^3+2(2)}\\ \\ =1.742268041\end{array}$

Explanation of Solution

Given information:Initial guesses $x_0=1and{x}_1=2$ and equation is $(a)x^3=2x+2$ .

Calculation:

With the initial guesses,  $x_0=1and{x}_1=2$ .

Now, using the formula, we will get the following.

   $\begin{array}{l}{x}_{i+1}=x_i-\frac{f_1\left(x_i\right)\left(x_i-x_{i-1}\right)}{f_1\left(x_i\right)-f_1\left(x_{i-1}\right)}\\ =x_i-\frac{\left(x_i^3-2x_i-2\right)\left(x_i-x_{i-1}\right)}{\left(x_i^3-2x_i-2\right)-\left(x_{i-1}^3-2x_{i-1}-2\right)}\\ =x_i-\frac{\left(x_i^3-2x_i-2\right)\left(x_i-x_{i-1}\right)}{x_i^3-2x_i-x_{i-1}^3+2x_{i-1}}\end{array}$

Let us find the value of  $x_2$ .

   $\begin{array}{cc}{x}_2& =x_1-\frac{\left(x_1^3-2x_1-2\right)\left(x_1-x_0\right)}{x_1^3-2x_1-x_0^3+2x_0}\\ & =2-\frac{\left(2^3-2(2)-2\right)(2-1)}{2^3-2(2)-1^3+2(1)}\end{array}$

   $\begin{array}{l}=2-\frac{2}{5}\hfill \\ =\frac{8}{5}\hfill \end{array}$

Let us find the value of  $x_3$ .

   $\begin{array}{l}{x}_3=x_2-\frac{\left(x_2^3-2x_2-2\right)\left(x_2-x_1\right)}{x_2^3-2x_2-x_1^3+2x_1}\\ =\frac{8}{5}-\frac{\left({\left(\frac{8}{5}\right)}^3-2\left(\frac{8}{5}\right)-2\right)\left(\frac{8}{5}-2\right)}{{\left(\frac{8}{5}\right)}^3-2\left(\frac{8}{5}\right)-2^3+2(2)}\\ \\ =1.742268041\end{array}$

b.

To determine

To calculate:To apply two steps of the secant method to the equation $e^x+x=7$ with initial guesses $x_0=1and{x}_1=2$ .

Answer to Problem 1E

Thus, $x_2=1.578707248$ and $x_3=1.660160100$

Given information:Initial guesses $x_0=1and{x}_1=2$ and equation is $(b)e^x+x=7$ .

Calculation:

Let us consider the following equation.

   $\begin{array}{l}{e}^x+x=7\hfill \\ e^x+x-7=0\hfill \end{array}$

Or consider the following function

   $f_2(x)=e^x+x-7$

With the initial guesses,  $x_0=1$ and $x_1=2$ .

Now, using the formula, we will get the following.

   $\begin{array}{l}{x}_{i+1}=x_i-\frac{f_2\left(x_i\right)\left(x_i-x_{i-1}\right)}{f_2\left(x_i\right)-f_2\left(x_{i-1}\right)}\\ \begin{array}{l}=x_i-\frac{\left(e^{x_i}+x_i-7\right)\left(x_i-x_{i-1}\right)}{\left(e^{x_i}+x_i-7\right)-\left(e^{x_4}+x_{i-1}-7\right)}\hfill \\ =x_i-\frac{\left(e^{x_i}+x_i-7\right)\left(x_i-x_{i-1}\right)}{e^{x_i}+x_i-e^{x_{h-1}}}\hfill \end{array}\end{array}$

Let us find the value of  $x_2$ .

   $\begin{array}{cc}{x}_2& =x_1-\frac{\left(e^{x_1}+x_1-7\right)\left(x_1-x_0\right)}{e^{x_1}+x_1-e^{x_0}-x_0}\\ & =2-\frac{\left(e^2+2-7\right)(2-1)}{e^2+2-e^1-1}\\ & \begin{array}{l}=2-\frac{e^2-5}{e^2-e^1+1}\\ \end{array}\\ & =1.578707248\end{array}$

Let us find the value of  $x_3$ .

   $\begin{array}{l}{x}_3=x_2-\frac{\left(e^{x_2}+x_2-7\right)\left(x_2-x_1\right)}{e^{x_1}+x_1-e^{x_0}-x_0}\\ =1.578707248-\frac{\left(e^{1.580n248}+1.578707248-7\right)(1.578707248-2)}{e^{1.5800n48}+1.578707248-e^2-2}\\ \\ \\ =1.660160100\end{array}$

c.

To calculate:To apply two steps of the secant method to the equation $e^x+\sin x=4$ with initial guesses $x_0=1and{x}_1=2$ .

Thus, $x_2=1.092906580$ and $x_3=1.119356686$

Given information:Initial guesses $x_0=1and{x}_1=2$ and equation is $(c)e^x+\sin x=4$

Calculation:

Let us consider the following equation.

   $\begin{array}{l}{e}^x+\sin (x)=4\hfill \\ e^x+\sin (x)-4=0\hfill \end{array}$

Or consider the following function.

   $f_3(x)=e^x+\sin (x)-4$

With the initial guesses,  $x_0=1_{}$ and $x_1=2$ .

Now, using the formula, we will get the following.

   $\begin{array}{l}{x}_{i+1}=x_i-\frac{f_3\left(x_i\right)\left(x_i-x_{i-1}\right)}{f_3\left(x_i\right)-f_3\left(x_{i-1}\right)}\\ =x_i-\frac{\left(e^{x_1}+\sin \left(x_i\right)-4\right)\left(x_i-x_{i-1}\right)}{\left(e^{x_1}+\sin \left(x_i\right)-4\right)-\left(e^{x_{1+1}}+\sin \left(x_{i-1}\right)-4\right)}\\ =x_i-\frac{\left(e^{x_i}+\sin \left(x_i\right)-4\right)\left(x_i-x_{i-1}\right)}{e^{x_i}+\sin \left(x_i\right)-e^{x_h}-\sin \left(x_{i-1}\right)}\end{array}$

Let us find the value of  $x_2$ .

   $\begin{array}{l}{x}_2=x_1-\frac{\left(e^{x_1}+\sin \left(x_1\right)-4\right)\left(x_1-x_0\right)}{e^{x_1}+\sin \left(x_1\right)-e^{x_0}-\sin \left(x_0\right)}\\ =2-\frac{\left(e^2+\sin (2)-4\right)(2-1)}{e^2+\sin (2)-e^1-\sin (1)}\\ =2-\frac{e^2+\sin (2)-4}{e^2-e^1+\sin (2)-\sin (1)}\\ \\ =1.092906580\end{array}$

Let us find the value of  $x_3$ .

   $\begin{array}{l}{x}_3=x_2-\frac{\left(e^{x_2}+\sin \left(x_2\right)-4\right)\left(x_2-x_1\right)}{e^{x_2}+\sin \left(x_2\right)-e^{x_1}-\sin \left(x_1\right)}\\ =1.092906580-\frac{\left(e^{1.09906550}+\sin (1.092906580)-4\right)(1.092906580-2)}{e^{1.092906580}+\sin (1.092906580)-e^2-\sin (2)}\\ =2-\frac{e^2+\sin (2)-4}{e^2-e^1+\sin (2)-\sin (1)}\\ \\ =1.119356686\end{array}$

Explanation of Solution

Given information:Initial guesses $x_0=1and{x}_1=2$ and equation is $(b)e^x+x=7$ .

Calculation:

Let us consider the following equation.

   $\begin{array}{l}{e}^x+x=7\hfill \\ e^x+x-7=0\hfill \end{array}$

Or consider the following function

   $f_2(x)=e^x+x-7$

With the initial guesses,  $x_0=1$ and $x_1=2$ .

Now, using the formula, we will get the following.

   $\begin{array}{l}{x}_{i+1}=x_i-\frac{f_2\left(x_i\right)\left(x_i-x_{i-1}\right)}{f_2\left(x_i\right)-f_2\left(x_{i-1}\right)}\\ \begin{array}{l}=x_i-\frac{\left(e^{x_i}+x_i-7\right)\left(x_i-x_{i-1}\right)}{\left(e^{x_i}+x_i-7\right)-\left(e^{x_4}+x_{i-1}-7\right)}\hfill \\ =x_i-\frac{\left(e^{x_i}+x_i-7\right)\left(x_i-x_{i-1}\right)}{e^{x_i}+x_i-e^{x_{h-1}}}\hfill \end{array}\end{array}$

Let us find the value of  $x_2$ .

   $\begin{array}{cc}{x}_2& =x_1-\frac{\left(e^{x_1}+x_1-7\right)\left(x_1-x_0\right)}{e^{x_1}+x_1-e^{x_0}-x_0}\\ & =2-\frac{\left(e^2+2-7\right)(2-1)}{e^2+2-e^1-1}\\ & \begin{array}{l}=2-\frac{e^2-5}{e^2-e^1+1}\\ \end{array}\\ & =1.578707248\end{array}$

Let us find the value of  $x_3$ .

   $\begin{array}{l}{x}_3=x_2-\frac{\left(e^{x_2}+x_2-7\right)\left(x_2-x_1\right)}{e^{x_1}+x_1-e^{x_0}-x_0}\\ =1.578707248-\frac{\left(e^{1.580n248}+1.578707248-7\right)(1.578707248-2)}{e^{1.5800n48}+1.578707248-e^2-2}\\ \\ \\ =1.660160100\end{array}$

c.

To determine

To calculate:To apply two steps of the secant method to the equation $e^x+\sin x=4$ with initial guesses $x_0=1and{x}_1=2$ .

Answer to Problem 1E

Thus, $x_2=1.092906580$ and $x_3=1.119356686$

Explanation of Solution

Given information:Initial guesses $x_0=1and{x}_1=2$ and equation is $(c)e^x+\sin x=4$

Calculation:

Let us consider the following equation.

   $\begin{array}{l}{e}^x+\sin (x)=4\hfill \\ e^x+\sin (x)-4=0\hfill \end{array}$

Or consider the following function.

   $f_3(x)=e^x+\sin (x)-4$

With the initial guesses,  $x_0=1_{}$ and $x_1=2$ .

Now, using the formula, we will get the following.

   $\begin{array}{l}{x}_{i+1}=x_i-\frac{f_3\left(x_i\right)\left(x_i-x_{i-1}\right)}{f_3\left(x_i\right)-f_3\left(x_{i-1}\right)}\\ =x_i-\frac{\left(e^{x_1}+\sin \left(x_i\right)-4\right)\left(x_i-x_{i-1}\right)}{\left(e^{x_1}+\sin \left(x_i\right)-4\right)-\left(e^{x_{1+1}}+\sin \left(x_{i-1}\right)-4\right)}\\ =x_i-\frac{\left(e^{x_i}+\sin \left(x_i\right)-4\right)\left(x_i-x_{i-1}\right)}{e^{x_i}+\sin \left(x_i\right)-e^{x_h}-\sin \left(x_{i-1}\right)}\end{array}$

Let us find the value of  $x_2$ .

   $\begin{array}{l}{x}_2=x_1-\frac{\left(e^{x_1}+\sin \left(x_1\right)-4\right)\left(x_1-x_0\right)}{e^{x_1}+\sin \left(x_1\right)-e^{x_0}-\sin \left(x_0\right)}\\ =2-\frac{\left(e^2+\sin (2)-4\right)(2-1)}{e^2+\sin (2)-e^1-\sin (1)}\\ =2-\frac{e^2+\sin (2)-4}{e^2-e^1+\sin (2)-\sin (1)}\\ \\ =1.092906580\end{array}$

Let us find the value of  $x_3$ .

   $\begin{array}{l}{x}_3=x_2-\frac{\left(e^{x_2}+\sin \left(x_2\right)-4\right)\left(x_2-x_1\right)}{e^{x_2}+\sin \left(x_2\right)-e^{x_1}-\sin \left(x_1\right)}\\ =1.092906580-\frac{\left(e^{1.09906550}+\sin (1.092906580)-4\right)(1.092906580-2)}{e^{1.092906580}+\sin (1.092906580)-e^2-\sin (2)}\\ =2-\frac{e^2+\sin (2)-4}{e^2-e^1+\sin (2)-\sin (1)}\\ \\ =1.119356686\end{array}$