# Newton's method

We assume that f(x), f'(x) and f''(x) are continuous in [a,b], and that for some $$\alpha$$ $$\in$$ (a,b) , we have $$f( \alpha )= 0$$ and $$f'( \alpha ) \neq 0$$. We show that if $$x_{0}$$ is chosen close enough to $$\alpha$$, the iterates
$$x_{n+1} = x_{n}- \frac{f(x_{n})}{f'(x_{n})}$$
converge to $$\alpha$$.

I tried to use Taylor's expansion for $$f( \alpha )$$ (centered at $$x_{n}$$), and I got to this expression

$$lim_{n \rightarrow \infty} (\alpha -x_{n+1})= lim_{n \rightarrow \infty} - \frac{1}{2} f''(c) \frac{( \alpha - x_{n} )^{2}}{f'(x_{n})}$$

where $$c \in ( \alpha , x_{n} )$$
and I guess I want the right hand side to be 0 to get to the answer. But I am not sure how to prove this.

Last edited:

Borek
Mentor
Your LaTeX is broken so it is hard to guess what you wrote. But what I am missing from your post is - what is the question? Are you trying to prove why Newton's method works?

And you don't need f'', f' is enough.

I am sorry, I fixed the latex part. (hopefully it's readable now)

By using Taylor's expansion, I got an expression, and by letting n go to infinity, if I could make the RHS become 0, I guess I would have the answer to the problem.
The thing is ,to do that, I am not sure how to use the fact that $$x_{0}$$ is chosen close enough to $$\alpha$$.

Borek
Mentor
I am still not sure what you trying to do and what for.

I am just trying to prove that in this case,

$$lim_{n \rightarrow \infty} (\alpha -x_{n+1})=0$$

so then the iterate $$x_{n+1}$$ converges to $$\alpha$$.

Last edited:
Borek
Mentor
I don't think going through a Taylor expansion is a good idea. Your proof needs f'', but Newton method works even if only f' exists, so your proof will be incomplete.

In the hypothesis, we have that f ''(x) is continuous in [a,b], so it exists.
I guess there must be a reason why they wanted that hypothesis in the problem...

What would be another way to approach this problem?