Thread: Newton's Method to approximate the real root of f(x)=2x^7-1 to 3 decimal places.

1. Newton's Method to approximate the real root of f(x)=2x^7-1 to 3 decimal places.

Hi so I have been having some issues with Newton's Method. I understand what you have to do, however, I was solving this one equation, yet no matter what I did I could not arrive at the correct answer.

The question was to use Newton's method to approximate the real root of f(x)=2x^7-1 to 3 decimal places. I see x1=1 and when I do it I get 0.928571429. What I am supposed to get is 0.907343 as my x2. What do I do?

Also this one problem I cannot figure out at all. Use Newton's Method to find the value of squareroot7 until two consecutive terms of the sequence are within 0.0001. (Hint squareroot7 is a root of what function?) I'm very confused on this problem, any help is appreciated.

Edit I have fixed my error in the first problem, I was too stupid to not keep going. However, the second confuses me still, I've tried multiple methods to solve it but cannot come up with anything.

2. Originally Posted by LANavjeet
Hi so I have been having some issues with Newton's Method. I understand what you have to do, however, I was solving this one equation, yet no matter what I did I could not arrive at the correct answer.

The question was to use Newton's method to approximate the real root of f(x)=2x^7-1 to 3 decimal places. I see x1=1 and when I do it I get 0.928571429. What I am supposed to get is 0.907343 as my x2. What do I do?

Also this one problem I cannot figure out at all. Use Newton's Method to find the value of squareroot7 until two consecutive terms of the sequence are within 0.0001. (Hint squareroot7 is a root of what function?) I'm very confused on this problem, any help is appreciated.
How do you know that? What is x2?

If x2 is 2nd approximation - then your calculation is correct.

After five iteration the answer will converge.

Hint for the second problem: What are the roots of y = x^2 - b ('b' is a real constant)?

3. Let us not forget that this method relies on a steep slope. If you are too near horizontal, you are out of luck.

Did you use your hint? $\sqrt{7}$ is a solution to what equation? Use your best algebra.