## Use Newton’s method with the specified initial approximation x1 to find x3.

1. The problem statement, all variables and given/known data

Use Newton’s method with the specified initial approximation to find , the third approximation to the root of the given equation. (Give your answer to four decimal places.)
x^5+2=0, x_1=-1

2. Relevant equations

3. The attempt at a solution

x5+2=0, x1=-1
y'=5x4
x(n+1)=xn-(x5+2)/(5x4 )

For n=1
x2=-1-((-1)5+2)/(5(-1)4 )=-6/5=-1.2
For n=2

x3=-1.2-((-1.2)5+2)/(5(-1.2)4 )= -1.2-0.4883/10.368=-1.2+0.047=-1.1530

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

Recognitions:
Gold Member
Staff Emeritus
 Quote by phillyolly Please verify my answer. 1. The problem statement, all variables and given/known data Use Newton’s method with the specified initial approximation to find , the third approximation to the root of the given equation. (Give your answer to four decimal places.) x^5+2=0, x_1=-1 2. Relevant equations 3. The attempt at a solution x5+2=0, x1=-1 y'=5x4 x(n+1)=xn-(x5+2)/(5x4 )
This should be
x(n+1)=xn-(xn5+2)/(5xn4 )

 For n=1 x2=-1-((-1)5+2)/(5(-1)4 )=-6/5=-1.2 For n=2 x3=-1.2-((-1.2)5+2)/(5(-1.2)4 )= -1.2-0.4883/10.368=-1.2+0.047=-1.1530
Looks good, but you aren't at "4 decimal places" yet. Continue until you get two consecutive results that are the same to 4 decimal places (and I would recommend doing the calculations to at least 5 decimal places until then).

Hi! What do you mean by
 Quote by HallsofIvy Looks good, but you aren't at "4 decimal places" yet.
?

The task says to do the third approximation, which I found already....
Do you mean I should do the fourth...and the fifth?...

Mentor

## Use Newton’s method with the specified initial approximation x1 to find x3.

I think that HallsOfIvy missed the part about the third approximation, so you're done.