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

by phillyolly
Tags: approximation, initial, method, newton’s
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phillyolly
#1
Jul14-10, 10:15 PM
P: 157
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 )

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
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HallsofIvy
#2
Jul15-10, 06:52 AM
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Quote Quote by phillyolly View Post
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).
phillyolly
#3
Jul15-10, 11:34 AM
P: 157
Hi! What do you mean by
Quote Quote by HallsofIvy View Post
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
#4
Jul15-10, 12:34 PM
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P: 21,214
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.


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