- #1

- 160

- 0

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

[tex]x_{n+1} = x_{n}- \frac{f(x_{n})}{f'(x_{n})}[/tex]

converge to [tex]\alpha[/tex].

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

[tex]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})}[/tex]

where [tex] c \in ( \alpha , x_{n} )[/tex]

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.

[tex]x_{n+1} = x_{n}- \frac{f(x_{n})}{f'(x_{n})}[/tex]

converge to [tex]\alpha[/tex].

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

[tex]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})}[/tex]

where [tex] c \in ( \alpha , x_{n} )[/tex]

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: