Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation.

Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation.

x5 − x − 6 = 0,    x1 = 2

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Newton's method is numerical method for solving nonlinear equations of type `f(x)=0.` If you have `n`-th approximation you can calculate `(n+1)`-st approximation using the formula:

`x_(n+1)=x_n-(f(x_n))/(f'(x_n)).`

For further explenation see the link below.

Lets calculate derivation: `f'(x)=5x^4-1.` So you have:

`x_2 = 2 - (24)/(79) = 1.6962`

`x_3 = 1.6962 - (6.34439)/(40.38836) = 1.53912` 

Your final solution would be something like `x_n = 1.49612.` Of course that's only one of 5 solutions to this equation. Other 4 solutions are complex numbers. If you want to get complex solutions, your initial approximation must be complex.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial