search

mandli / intro-numerical-methods

Jupyter notebooks and other materials developed for the Columbia course APMA 4300
Creative Commons Attribution 4.0 International
278 stars 398 forks source link

A probable mistake in differentiation at Asymptotic Convergence of Newton's Method #3

Closed yeejoh closed 9 years ago

yeejoh commented 9 years ago

At 5 Root Finding and Optimization, Asymptotic Convergence of Newton's Method

... Let $g(x) = x - \frac{f(x)}{f'(x)}$, then ... What about $g'(x^*)$ though: $$g'(x) = 1 - \frac{f'(x)}{f'(x)} + \frac{f(x)}{f''(x)}$$ which simplifies when evaluated at $x = x^*$ to $$g'(x^*) = \frac{f(x^*)}{f''(x^*)} = 0$$ ...

Since the Quotient Rule, $$\left ( \frac{u}{v} \right )' = \frac{u' v - v' u}{v^2}$$,

I think it should be $$g'(x) = 1 - \frac{f'(x) f'(x) - f(x) f''(x)}{f'^2(x)} = 1 - 1 + \frac{f(x) f''(x)}{f'^2(x)}$$.

Nonetheless, $$g'(x^*) = \frac{f(x^*) f''(x^*)}{f'^2(x^*)} = 0$$, $$g''(x^*) = \frac{f''(x^*)}{f'(x^*)}$$, so it won't effect the following result.

mandli commented 9 years ago

There is a mistake as you said, there is a missing extra $1/f'(x)$ term. Thanks for pointing this out!