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

In summary, the conversation discusses using Newton's method to find the third approximation of the root of the equation x^5+2=0 with an initial approximation of x_1=-1. The solution involves using the derivative and calculating the next approximation until two consecutive results are the same to four decimal places. The third approximation is found to be -1.1530.
  • #1
phillyolly
157
0
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)=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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  • #2
phillyolly said:
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)=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).
 
  • #3
Hi! What do you mean by
HallsofIvy said:
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?...
 
  • #4
I think that HallsOfIvy missed the part about the third approximation, so you're done.
 

1. What is Newton's method?

Newton's method is an iterative algorithm used to find the roots of a differentiable function. It involves using an initial approximation to approximate the root of the function.

2. How does Newton's method work?

Newton's method works by using the derivative of the function at a given point to approximate the root of the function. It iteratively updates the initial approximation until it converges to the actual root of the function.

3. What is the initial approximation in Newton's method?

The initial approximation, denoted as x1, is the starting point for the iterative process in Newton's method. It is usually chosen close to the actual root of the function for faster convergence.

4. How many iterations are needed to find the root using Newton's method?

The number of iterations needed to find the root using Newton's method depends on the initial approximation and the function itself. Generally, the method converges within a few iterations, but it can take more if the initial approximation is far from the root.

5. What is the significance of finding x3 in Newton's method?

After x3 is found, it can be used as the new initial approximation for the next iteration, which will give a more accurate approximation of the root. This process is repeated until the desired level of accuracy is achieved.

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