[Math] Trouble Understanding Newton-Raphson Iterative Method

approximationnumerical methods

Currently i am reading this page which discusses the newton-raphson method of approximating the roots of an equation. It says given a function $f$ over the reals $x$, and its derivative $f$,we begin with a first guess $x_{0}$ for a root of the function $f$. Provided the function is reasonably well behaved a better approximation $x_{1}$ is
$$x_{1} = x_{0} – \frac{f(x_{0})}{f \prime(x_{0})}.$$

First question: What does it mean to say that the function is reasonably well behaved?What's according to them is "Reasonably well behaved?"

The process is repeated as $$x_{n+1} = x_{n} – \frac{f(x_{n})}{f \prime(x_{n})}$$ until a sufficiently accurate value is reached.

For example i tried solving the equation $x^3 – 3x – 4 = 0$ where $f(1) < 0$ and $f(3) > 0$ so there's a solution to the equation between 1 and 3.We shall take this to be 2,by bisection and hence $x_{0} = 2$.A better approximation $x_{1}$ is given by the above formula.

Second question:"until a sufficiently accurate value is reached." When do i get to the "sufficiently accurate" value?How do i know that i've reached a point where i stop using this formula?

Best Answer

Suppose you want to compute $\sqrt{2}$, so you use $f(x) = x^2-2$ and start with $x_1=1$. Say you want 50 places. Here we go:

x[ 1] = 1.00000000000000000000000000000000000000000000000000 
x[ 2] = 1.50000000000000000000000000000000000000000000000000 
x[ 3] = 1.41666666666666666666666666666666666666666666666667 
x[ 4] = 1.41421568627450980392156862745098039215686274509804 
x[ 5] = 1.41421356237468991062629557889013491011655962211574 
x[ 6] = 1.41421356237309504880168962350253024361498192577620 
x[ 7] = 1.41421356237309504880168872420969807856967187537723 
x[ 8] = 1.41421356237309504880168872420969807856967187537695 
x[ 9] = 1.41421356237309504880168872420969807856967187537695 
x[10] = 1.41421356237309504880168872420969807856967187537695 
x[11] = 1.41421356237309504880168872420969807856967187537695 
x[12] = 1.41421356237309504880168872420969807856967187537695 
x[13] = 1.41421356237309504880168872420969807856967187537695 
x[14] = 1.41421356237309504880168872420969807856967187537695 
x[15] = 1.41421356237309504880168872420969807856967187537695

So presumably we can stop at $x_8$.

Related Question