# Proof of the Newton Raphson method

I know only a little about this method. I get about the big picture. You draw a tangent from an appropriate point on function f, find the spot where it meets the x-axis, and draw another tangent from the point which has the same x coordinate on the function f. the meet converges toward the root. I get the big picture, I can see it on the graphs, but I can't prove it. Can you please elaborate a proof for me? I'm in a hurry, plz

Consider that you look for the zero of $f(x)$. At a starting point $x_0$, make a Taylor series limited to first order $$f(x)=f(x_0)+f'(x_0)(x-x_0)+ O\left((x-x_0)^2\right)$$ Ignore the higher order terms and solve for $x$ and let us call $x_1$ the solution of $$0=f(x_0)+f'(x_0)(x-x_0)\implies x_1=x_0-\frac{f(x_0)}{f'(x_0)}$$ Repeating $$f(x)=f(x_1)+f'(x_1)(x-x_1)+ O\left((x-x_1)^2\right)$$ Ignore the higher order terms and solve for $x$ and let us call $x_2$ the solution of $$0=f(x_1)+f'(x_1)(x-x_1)\implies x_2=x_1-\frac{f(x_1)}{f'(x_1)}$$ Repeat again and again to get the famous $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$