Newtons method to estimate solution to eq.

In summary, the conversation discusses using Newton's method to estimate a solution to an equation. The equation and its derivative are provided and the steps for solving the problem are explained. The final answer obtained using this method is 1.604, which is rounded to three decimal places. The conversation also mentions that the multiple choice answer for the problem may be incorrect.
  • #1
Ashleyz
16
0

Homework Statement


Use Newton's method to estimate the requested solution of the equation. Start with given value of x0 and give x2 as the estimated solution.



Homework Equations

x0 = 2 ( this is the first guess numb. you start with)
the equation you use is... x = x(guuess #) - f(xg)/f'(xg)

f(x) = x^4 -6x +3 and f'(x) = 4(x)^3 -6


The Attempt at a Solution


I have done the entire problem. several times.
when x ='s 2, I got 1. 731

when x ='s 1.731 I got 1.623

when x ='s 1.623 I got 1.604 for my final answer.

perhaps this answer is correct , and the multiple choice answer is wrong.

If you are familiar with Newtons method- then this won't take too long for you.
You just plug and chug.
 
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  • #2
Your answer is right. The next digits are
1.604439
I just punched into maple and got this answer.
 
  • #3
Ashleyz said:

Homework Statement


Use Newton's method to estimate the requested solution of the equation. Start with given value of x0 and give x2 as the estimated solution.



Homework Equations

x0 = 2 ( this is the first guess numb. you start with)
the equation you use is... x = x(guuess #) - f(xg)/f'(xg)

f(x) = x^4 -6x +3 and f'(x) = 4(x)^3 -6


The Attempt at a Solution


I have done the entire problem. several times.
when x ='s 2, I got 1. 731

when x ='s 1.731 I got 1.623

when x ='s 1.623 I got 1.604 for my final answer.

perhaps this answer is correct , and the multiple choice answer is wrong.

If you are familiar with Newtons method- then this won't take too long for you.
You just plug and chug.
That is correct to three decimal places. What are the choices?
 
  • #4
Wouldn't that give you [itex]x_0=2[/itex] , [itex]x_1 \approx 1.731[/itex], [itex]x_2 \approx 1.623[/itex] and [itex]x_3 \approx 1.605[/itex]?

soo if you are asked to find x2...would that not be 1.623?
 
  • #5
my full answer is 1.604938639

when pluging the answers back into the equation and dividing
by the derrivative, I only was tacking my answer to 3 decimal places
keeping in mind to round up the 3rd decimal place.

the choices are 1.600 and 1.604
 
  • #6
gabbagabbahey said:
Wouldn't that give you [itex]x_0=2[/itex] , [itex]x_1 \approx 1.731[/itex], [itex]x_2 \approx 1.623[/itex] and [itex]x_3 \approx 1.605[/itex]?

soo if you are asked to find x2...would that not be 1.623?

no. I think you are confusing x0 with x1. the order goes: x0
x1
x2
ect.
 

Related to Newtons method to estimate solution to eq.

What is Newton's method?

Newton's method, also known as the Newton-Raphson method, is an algorithm used to efficiently estimate the solutions to equations. It involves using an initial guess for the solution and iteratively improving it until it is close enough to the actual solution.

How does Newton's method work?

Newton's method works by using the slope of a tangent line to a point on the curve of the equation to find a better estimate for the solution. This process is repeated until the estimate is accurate enough.

What types of equations can Newton's method be used to solve?

Newton's method can be used to solve a variety of equations, including polynomial equations, exponential equations, and trigonometric equations. It is most commonly used to find the roots of a function, which are the values that make the function equal to zero.

What are the advantages of using Newton's method?

Newton's method is a fast and efficient way to estimate solutions to equations. It also has a high rate of convergence, meaning that it quickly approaches the actual solution. Additionally, it is a versatile method that can be applied to a wide range of equations.

What are the limitations of Newton's method?

One limitation of Newton's method is that it requires an initial guess for the solution, which may not always be easy to obtain. It also may not always converge to the actual solution, particularly if the initial guess is far from the true solution or there are multiple solutions to the equation. In some cases, it may also be computationally expensive to use.

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