To apply two steps of the secant method to the equation x 3 = 2 x + 2 with initial guesses x 0 = 1 a n d x 1 = 2 .

Question

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Answer 1

Question

Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

a.

Expert Solution

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

b.

Expert Solution

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

c.

Expert Solution

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

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Answer 2

Question

Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

a.

Expert Solution

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

b.

Expert Solution

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

c.

Expert Solution

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

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Answer 3

Question

Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

a.

Expert Solution

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

b.

Expert Solution

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

c.

Expert Solution

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

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Answer 4

Question

Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

a.

Expert Solution

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

b.

Expert Solution

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

c.

Expert Solution

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

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Answer 5

Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

a.

Expert Solution

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

b.

Expert Solution

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

c.

Expert Solution

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Answer 6

Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

a.

Expert Solution

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Answer 7

Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

Answer 8

a.

Expert Solution

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (a) x3=2x+2 .

Calculation:

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f1(xi)(xi−xi−1)f1(xi)−f1(xi−1)=xi−(xi3−2xi−2)(xi−xi−1)(xi3−2xi−2)−(xi−13−2xi−1−2)=xi−(xi3−2xi−2)(xi−xi−1)xi3−2xi−xi−13+2xi−1

Let us find the value of x2 .

x2=x1−(x13−2x1−2)(x1−x0)x13−2x1−x03+2x0=2−(23−2(2)−2)(2−1)23−2(2)−13+2(1)

=2−25=85

Let us find the value of x3 .

x3=x2−(x23−2x2−2)(x2−x1)x23−2x2−x13+2x1=85−((85)3−2(85)−2)(85−2)(85)3−2(85)−23+2(2)=1.742268041

Answer 13

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

b.

Expert Solution

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

Answer 14

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

Answer 15

b.

Expert Solution

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (b) ex+x=7 .

Calculation:

Let us consider the following equation.

ex+x=7ex+x−7=0

Or consider the following function

f2(x)=ex+x−7

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f2(xi)(xi−xi−1)f2(xi)−f2(xi−1)=xi−(exi+xi−7)(xi−xi−1)(exi+xi−7)−(ex4+xi−1−7)=xi−(exi+xi−7)(xi−xi−1)exi+xi−exh−1

Let us find the value of x2 .

x2=x1−(ex1+x1−7)(x1−x0)ex1+x1−ex0−x0=2−(e2+2−7)(2−1)e2+2−e1−1=2−e2−5e2−e1+1=1.578707248

Let us find the value of x3 .

x3=x2−(ex2+x2−7)(x2−x1)ex1+x1−ex0−x0=1.578707248−(e1.580n248+1.578707248−7)(1.578707248−2)e1.5800n48+1.578707248−e2−2=1.660160100

Answer 20

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

c.

Expert Solution

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Answer 21

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Answer 22

c.

Expert Solution

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:Initial guesses x0=1 and x1=2 and equation is (c) ex+sinx=4

Calculation:

Let us consider the following equation.

ex+sin(x)=4ex+sin(x)−4=0

Or consider the following function.

f3(x)=ex+sin(x)−4

With the initial guesses, x0=1 and x1=2 .

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

xi+1=xi−f3(xi)(xi−xi−1)f3(xi)−f3(xi−1)=xi−(ex1+sin(xi)−4)(xi−xi−1)(ex1+sin(xi)−4)−(ex1+1+sin(xi−1)−4)=xi−(exi+sin(xi)−4)(xi−xi−1)exi+sin(xi)−exh−sin(xi−1)

Let us find the value of x2 .

x2=x1−(ex1+sin(x1)−4)(x1−x0)ex1+sin(x1)−ex0−sin(x0)=2−(e2+sin(2)−4)(2−1)e2+sin(2)−e1−sin(1)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.092906580

Let us find the value of x3 .

x3=x2−(ex2+sin(x2)−4)(x2−x1)ex2+sin(x2)−ex1−sin(x1)=1.092906580−(e1.09906550+sin(1.092906580)−4)(1.092906580−2)e1.092906580+sin(1.092906580)−e2−sin(2)=2−e2+sin(2)−4e2−e1+sin(2)−sin(1)=1.119356686

Answer 27

Ch. 1.1 - Use the Intermediate Value Theorem to find an...Ch. 1.1 - Use the Intermediate Value Theorem to find an...Ch. 1.1 - Consider the equations in Exercise 1. Apply two...Ch. 1.1 - Consider the equations in Exercise 2. Apply two...Ch. 1.1 - Consider the equation x4=x3+10 . a. Find an...Ch. 1.1 - Suppose that the Bisection Method with starting...Ch. 1.1 - Prob. 1CP Ch. 1.1 - Use the Bisection Method to find the root to eight...Ch. 1.1 - Use the Bisection Method to locate all solutions...Ch. 1.1 - Prob. 4CP

To apply two steps of the secant method to the equation x 3 = 2 x + 2 with initial guesses x 0 = 1 a n d x 1 = 2 .

Concept explainers

Videos

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1 and x1=2 .

Answer to Problem 1E

Explanation of Solution

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1 and x1=2 .

Answer to Problem 1E

Explanation of Solution

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1 and x1=2 .

Answer to Problem 1E

Explanation of Solution

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