Newtons method to estimate solution to eq.

Ashleyz
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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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Your answer is right. The next digits are
1.604439
I just punched into maple and got this answer.
 
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?
 
Wouldn't that give you x_0=2 , x_1 \approx 1.731, x_2 \approx 1.623 and x_3 \approx 1.605?

soo if you are asked to find x2...would that not be 1.623?
 
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
 
gabbagabbahey said:
Wouldn't that give you x_0=2 , x_1 \approx 1.731, x_2 \approx 1.623 and x_3 \approx 1.605?

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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