Taylor's Theorem:
f(x)=f(a)+f'(a)(x-a)+f"(a)(x-a)2/2!+f"'(a)(x-a)3/3!+...
Newton's Method uses the iterative formula to solve for x such that f(x)=0:
xn+1=xn-f(xn)/f'(xn)
If we use the first two terms of Taylor's expansion we get:
f(x)≈f(a)+f'(a)(x-a)
When f(x)=0 x=x0 is an approximate root of f(x), i.e., f(x0)≈0.
In the approximate Taylor expansion a=x0 when f(a)≈0, so:
if f(x0)+f'(x0)(x-x0)=0 (approx), f'(x0)(x-x0)=-f(x0), x-x0=-f(x0)/f'(x0), x=x0-f(x0)/f'(x0).
The value of x given by this expression is just another approximation for the root, and we can simply call this x1 and apply the formula again.
This leads us to Newton's Iterative Method: xn+1=xn-f(xn)/f'(xn).