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