Proof of Convergence of Iterates to $\alpha$ via Taylor's Expansion

[math8]

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.

 

[Borek]

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.

 

[math8]

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]

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

 

[math8]

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.

 

[Borek]

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.

 

[math8]

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?