Question

Use Newton's method to find the positive fourth root of 3 by solving the equation x4 – 3 = 0. Start with x1 = 1 and find x2.

Answer 1

Newton's method:
x1=x0-f(x0)/f'(x0)
here
f(x)=x^4-3
f'(x)=4x^3
let x0=1
x1=1-f(1)/f'(1)=3/2
x2=97/72
x3=115403137/87616608
x4=236297297271008837816738085152257/179546943199700984864483416264832
and so on.
x4^4=3.000011742...
the precision approximately doubles after every iteration.

Answer 2

I appreciate the answer, but it doesn't really help me to understand it

Answer 3

I thought the formula was Xn+1=Xn-(f(Xn)/f'(Xn))

Answer 4

The standard Newton's method is based on
x1=x0-f(x0)/f'(x0)
where x1 is a new approximation, and x0 is an estimate.
f(x)=x^4-3=0 is the equation to be solved.
f'(x) is the derivative (calculus!) of f(x).
It is essentially a special form of linearization that we learn in Calculus.

Answer 5

This problem is for my calculus class :)

Answer 6

Would you prefer an explanation on how it is done, or why it is done this way?

Answer 7

Newtons method is just an improved trial and error method :)

Answer 8

How it is done... so I am basically substituting an x value (1) for the first root to equation 1-(x^4-3)/4x^3? Is this right?

Answer 9

exactly.
After that, you substitute the result you get on the left-hand side to the one on the right. It is an iterative process, using the previous result to get a better result.

Answer 10

So with x0=1, we calculate x1=3/2.
Substitute x1=3/2 on the right hand side to calculate another value for x.
This iterative process goes on until the desired accuracy is obtained.
Note that sometimes it may not converge to the desired root, which is one problem of Newton's method.

Answer 11

Thank you Amis64, long time no see!

Answer 12

Can you show me how you did x2 then I think I can get the other two. You say substitute x1 to the right side but Im not getting what you got.

Answer 13

x0=1
x1=1-(1^4-3)/(4*1^3)=3/2
x2=3/2-((3/2)^4-3)/(4*(3/2)^3)=97/72
x3=97/72 - ((97/72)^4-3)/(4*(97/72)^3)=115403137/87616608=1.317137...
and so on.

Answer 14

Oh I'm supposed to substitute x1 for all of x in x2! Thats why I was getting a different answer then you

Answer 15

You are awesome thank you so much I get it now!

Answer 16

Pleasure to be of help!