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

[phillyolly]

Please verify my answer.

Homework Statement

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

Homework Equations

The Attempt at a Solution

x⁵+2=0, x₁=-1
y'=5x⁴
x_(n+1)=x_n-(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

 

[HallsofIvy]

Please verify my answer.

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

This should be
x_(n+1)=x_n-(x_n⁵+2)/(5x_n⁴ )

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).

 

[phillyolly]

Hi! What do you mean by

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?...

 

[Mark44]

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