i need to solve this equation with newton raphson method

thank youQuote:

x^3-3x+1

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- Apr 4th 2011, 02:23 AMthomaztrizNewton raphson method
i need to solve this equation with newton raphson method

Quote:

x^3-3x+1

- Apr 4th 2011, 02:48 AMFernandoRevilla
By Bolzano's Theorem there is a solution in $\displaystyle (0,1)$ . Choose $\displaystyle x_0=0$ and construct the corresponding sequence.

- Apr 4th 2011, 03:15 AMHallsofIvy
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.

- Apr 4th 2011, 04:54 AMthomaztriz
- Apr 4th 2011, 05:44 AMHallsofIvy
Look at FernandoRevilla's post!

- Apr 4th 2011, 10:44 AMthomaztriz
http://i51.tinypic.com/2cdigq0.jpg

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