# Newtons Method Help

• Aug 4th 2008, 04:43 AM
NoAsherelol
Newtons Method Help
Using newtons method, solve the equation, x^3-4x-12=0 accuratey to six decimal places?

• Aug 4th 2008, 06:03 AM
CaptainBlack
Quote:

Originally Posted by NoAsherelol
Using newtons method, solve the equation, x^3-4x-12=0 accuratey to six decimal places?

Newton's method uses the recursion:

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

to find a root of f(x)=0, so here we put:

$f(x)=x^3-4x-12$

$f'(x)=3x^2-4$

Then:

$x_{n+1}=x_n-\frac{x_n^3-4x_n-12}{3x^2-4}$

Now we need a guess to get us started, we observe that $f(x)$ is negative at $x=2$ and positive at $x=3$ (and nearer to zero at $x=3$) so we take $x_0=3$ as our starting value.

You will find this converges to the required accuracy after about 3 itterations.

RonL