Newton's Method and Secant Method Convergence proof

Asked Oct 02 '13 at 00:33

Active Oct 02 '13 at 00:33

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Show that if the iterates in Newton's method converge to a point $r$ for which $f'(r)\ne0$, then $f(r)=0$. Also establish the same assertion for the secant method.

Hint: For the secant method, the mean-value theorem is useful. This is the case $n=0$ in Taylor's Theorem.

Not sure how to approach this problem for either method. Thanks for any help!

asked Oct 02 '13 at 00:33

Drake

What are you stuck on, or what have you tried? If you write down the equation for Newton's method and use your assumptions, the result is immediate. – Tyler Oct 02 '13 at 00:40
I'm stuck at $x_{n+1}=r-\frac{f(r)}{f'(r)}$ – Drake Oct 02 '13 at 00:51
looks like you forgot to take the limit on the left side as well ;) – Tyler Oct 02 '13 at 00:51
$lim x_{n+1} = r$ ? I would assume – Drake Oct 02 '13 at 00:56
Right. $x_{n+1}$ and $x_{n}$ would both have to converge to $r$ (if they converge at all... but we are given that they do!) – Tyler Oct 02 '13 at 00:58
Ah I see, thanks! Any advice for secant method? – Drake Oct 02 '13 at 00:59

Newton's Method and Secant Method Convergence proof

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