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Let $g : I \rightarrow \mathbb{R}$ be a $C^1$ map such that $g'(x) \ne 0$ for any $x$ in $I$. Assume that there exists $r \in I$ such that $g(r)=0$. Prove that for $\eta \in I$ sufficiently close to $r$ then the fixed point iteration $$x_{k+1}=x_k-\frac{g(x_k)}{g'(x_k)},\,x_0=\eta$$ satisfies $\lim_{k\rightarrow\infty}x_k=r$.

I noticed that this is Newton's method for finding the roots of $g$, but I couldn't manage yet to prove the desired conclusion.

Carl Christian
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George R.
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  • Retagged: there is no differential equation and no initial-value problem here. –  Jun 28 '18 at 13:27

1 Answers1

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Just write $$ x_{k+1}-r=x_k-r-\frac{g(r)+g'(\tilde x_k)(x_k-r)}{g'(x_k)} $$ so that $$ |x_{k+1}-r|\le\frac{M-m}{m}|x_k-r| $$ where $M=\max_{x\in[r-c,r+c]}|g'(x)|$ and $m=\min_{x\in[r-c,r+c]}|g'(x)|$. Obviously, $M-m\to 0$ for $c\to 0$.

Lutz Lehmann
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