Why does the Newton-Raphson method for solving equations always work?
What I know:
I know that if $f(x) = 0$, then the Newton-Raphson method may be applied. It states that $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ where the solution to $f(x) = 0$ is $x_\infty$
I also know that you can select a point on the graph of $f(x)$ (your first estimate $x_1$), and Newton-Raphson takes the gradient of that point, and where that gradient meets the $x$-axis, and the x value is where $x_2$ is.
I know how it works, but why the $$\frac{f(x)}{f'(x)} ?$$
It seems that the curve is divided by the gradient, but what does this give?
I would just like an explanation of why this works.
What I can give you is a derivation of the method.
Assume some $x^*$ is some zero of $f$. Bcause $f$ is differentiable, sufficiently close at $x^*$ it looks like a straight line. So when you are not exactly in $x^*$ but close to it, then you can just take this line and look where this line intersects the $x$-axis instead of the more complicated function $f$. This is because the zeros of a linear function are very easy to determine. E.g. for $y=mx+n$ the zero is simply at
$$x =-\frac nm.$$
So say we are in $x_i$, close to $x^* $. At this point the curve looks like the line
$$y=f'(x_i)(x-x_i)+f(x_i).$$
This is the unique linear function through $f(x_i)$ with the appropriate slope to be a tangent to $f$. Now what is the unqiue zero $x$ of this line? Set $y$ to zero and rearragne and you will find
$$x=x_i-\frac{f(x_i)}{f'(x_i)}$$
in analogy to the zero for a general linear function $y=mx+n$. So when we call this new zero $x_{i+1}$, then starting from an estimation of a zero $x_i$, we found a better one.
This is essentially what the method does. It take an approximation and makes a better one. When your approximation is bad, so will the result be bad too (maybe).
Note:
Newtons method does not always work. There are examples where the sequence of $x_i$ either diverges or oscilltes, either because there is no zero or your initial estimation was bad.