Hey, I've been stuck on a question for a while now, please help! (Worried)

Using the Newton Raphson method, which I'm not too clear on, I need to solve: x^3 + x -1 = 0, to 3 d.p

and also

cos x = x^2

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- Dec 16th 2009, 07:56 AMtchimpNewton Raphson, help please!
Hey, I've been stuck on a question for a while now, please help! (Worried)

Using the Newton Raphson method, which I'm not too clear on, I need to solve: x^3 + x -1 = 0, to 3 d.p

and also

cos x = x^2 - Dec 16th 2009, 03:22 PMkjchauhan
Newton-Raphson :

$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

Solve then by taking $\displaystyle f(x)=x^3+x-1$ and $\displaystyle f'(x)=3x^2+1$ and find the initial value $\displaystyle x_2=\frac{x_1+x_0}{2}$, where approx root lies between $\displaystyle x_0$ and $\displaystyle x_1$

and then

$\displaystyle x_{2}=x_1-\frac{f(x_1)}{f'(x_1)}$

$\displaystyle x_{3}=x_2-\frac{f(x_2)}{f'(x_2)}$

etc..