Assume that a sequence $\{x_i\}_{i=0}^\infty$ of Newton's method $x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$ is convergent and set $x_\infty = \lim_{i\to\infty} x_i$. Furthermore assume that $f'(x_\infty) \neq 0$. How is it possible to show that $x_\infty$ is an solution to $f(x)=0$?

My solution for this were assuming that $x_{\infty+1}$=$x_\infty$ so $$x_{\infty+1}=x_\infty-\frac{f(x_\infty)}{f'(x_\infty)}$$ $$0=-\frac{f(x_\infty)}{f'(x_\infty)}$$ $$0=-f(x_\infty)$$ $$f(x_\infty)=0$$

  • $\begingroup$ Newton's method only works for certain differentiable functions $f$ and when $x_0$ is a close enough "guess" of what $x_\infty$ might be. See: en.wikipedia.org/wiki/Newton%27s_method $\endgroup$ – B. Núñez Nov 13 '19 at 1:10
  • $\begingroup$ I'm taking Calculus Beta right now and this is an assignment problem. In the problem description it is derived from the tangent line equation that $x_i+1=x_i-\frac{f(x_i)}{f'(x_i)}$. Furthermore the question is that we assume a sequence is convergent and $f'(x_\infty) \neq 0$ and for $x_\infty$ it is defined as $\lim_{i\to\infty} x_i $, how do we show that $x_\infty$ is an solution (the root) to $f(x)=0$. There aren't any other details about this sub-problem in the assignment. $\endgroup$ – Rakozay Nov 13 '19 at 1:53
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    $\begingroup$ One might be able to guess from your use of the term "Newton method" in your question that you mean $x_{i+1}=x_i-\frac{f(x_i}{f'(x_i)}$. Yet, I would think you should put this info in your question, to make it clear. You dont need any $x_{\infty+1}$. If $x_i\to x_\infty$ then also $x_{i+1}\to x_\infty$, so $x_\infty=x_\infty-\frac{f(x_\infty)}{f'(x_\infty)}$. Welcome to MSE ! $\endgroup$ – Mirko Nov 13 '19 at 1:55
  • $\begingroup$ Incredible! Makes much more sense and thanks @Mirko! $\endgroup$ – Rakozay Nov 13 '19 at 2:05
  • $\begingroup$ You are welcome ! $\endgroup$ – Mirko Nov 13 '19 at 2:06

The following proof works if we assume that $\{x_i\}_{i\geq 0}$ is a convergent sequence such that $x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$ for all $i\geq 0$ (this assumption wasn't on your initial post, hence my comment) and $f$ and $f'$ are both continuous in some interval containing $x_\infty$ (i.e. $f$ is continuously differentiable in $x_\infty$).

Your intuition that "$x_{\infty}=x_{\infty+1}$" is good, but doesn't yet constitute a proof. We can use the fact that $\displaystyle \lim_{i\to\infty} x_i=\lim_{i\to\infty} x_{i+1}$ (see here; this stems from the fact that $\{x_{i+1}\}_{i\geq 0}$ is a subsequence of $\{x_i\}_{i\geq 0}$) to deduce the following: $$\lim_{i\to \infty} x_{i+1}=\lim_{i\to \infty}\left(x_i-\frac{f(x_{i})}{f'(x_i)}\right) ~~\implies~~x_{\infty}=x_{\infty}-\frac{f(x_{\infty})}{f'(x_{\infty})},$$ by sum and fraction of limits and the convergence of $\{f(x_i)\}_{i\geq 0}$ and $\{f'(x_i)\}_{i\geq 0}$ if $f,f'$ are continuous and $f'(x_{\infty})\neq 0$. Substracting $x_{\infty}$ on both sides and multiplying by $-f'(x_{\infty})$, we obtain: $$-\frac{f(x_{\infty})}{f'(x_{\infty})}=0 ~~~~\text{and thus}~~~~ f(x_{\infty})=0.$$ That is, $x_{\infty}$ is a solution for $x$ in $f(x)=0$.

I'd like to remark (as in my comment) that the Newton-Rhapson method does not always converge to a root; this proof required fairly strong assumptions. Here's an interesting discussion about this.

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