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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

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

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  • $\begingroup$ What, exactly, do you want proved? That the method converges to x such that f(x)= 0, if it converges, is pretty straight forward since if f(x) did not go to 0, the method would not converge. If you are talking about showing that the method always converges, there is no such proof because that is not always true. $\endgroup$ – user247327 May 8 '18 at 11:42
  • $\begingroup$ @user247327. For sure I totally agree with all your statements. The only thing I tried to do is to formulize the steps using the tangent since he/she wrote I get the big picture, I can see it on the graphs. This was my very modest contribution. Cheers. $\endgroup$ – Claude Leibovici May 8 '18 at 11:47

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