1. Newton raphson method

i need to solve this equation with newton raphson method

x^3-3x+1
thank you

2. By Bolzano's Theorem there is a solution in $\displaystyle (0,1)$ . Choose $\displaystyle x_0=0$ and construct the corresponding sequence.

3. Surely you are not expecting people to do this problem for you. Show what you have done. Of course, Newton-Raphson is an iterative method- for each approximation, $\displaystyle x_n$, to the solution of f(x)= 0, $\displaystyle x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)}$ should be a better approximation and, hopefully, the sequence will converge to a solution.

4. Originally Posted by HallsofIvy
Surely you are not expecting people to do this problem for you. Show what you have done. Of course, Newton-Raphson is an iterative method- for each approximation, $\displaystyle x_n$, to the solution of f(x)= 0, $\displaystyle x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)}$ should be a better approximation and, hopefully, the sequence will converge to a solution.

maybe i should find the location of the root by the method of tabulation ?

5. Look at FernandoRevilla's post!

6. for an equation like this ... $\displaystyle x^3 -8x-4$

newton raphson iterative scheme is given by

$\displaystyle x_n_+_1 = x_n - f(x_n)/f'(x_n)$

for the given equation f(x)= $\displaystyle x^3-8x-4$

first we find the location of the root by the method of tabulation . .

the table for f(x) is

x 0 1 2 3 4
f(x) -4 -13 -12 -1 28

the positive root is near x = 3

we take $\displaystyle x_0 = 3$ in newton raphson iterative scheme

$\displaystyle x_n_+_1 =x_n - x_n^3-8x_n-4/3x_n^2-8$

we get , $\displaystyle x_1 = 3 - 27-24-4/27-8 = 3.0526$

similarly , $\displaystyle x_2 = 3.05138$

and

$\displaystyle x_3= 3.05138$

thus , the positive root is 3.0514 , correct to five significant digits