# Does the infinite iteration sequence of Newtons method always converge to a root?

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 $$x_{\infty+1}=x_\infty-\frac{f(x_\infty)}{f'(x_\infty)}$$ $$0=-\frac{f(x_\infty)}{f'(x_\infty)}$$ $$0=-f(x_\infty)$$ $$f(x_\infty)=0$$

• 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 – B. Núñez Nov 13 '19 at 1:10
• 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. – Rakozay Nov 13 '19 at 1:53
• 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 ! – Mirko Nov 13 '19 at 1:55
• Incredible! Makes much more sense and thanks @Mirko! – Rakozay Nov 13 '19 at 2:05
• You are welcome ! – 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 $$f(x)=0$$.