I was solving Stewart Calculus Exercise about Newton's Method.
$\mathbf Problem$ Explain why Newton's Method fails when applied to the equation $\sqrt[3]x=0$ with any initial approximation $x_1 \neq 0$.
$x^1/3$">The above picture is a graph of $x^{1/3}$. We can see the concavity is changing in $(0,0)$, which means The Origin is the point of inflection.
If we draw some tangent lines otherwise from The Origin, it gets away and away from the origin and diverges. I got the feeling that we can't use Newton's Method of a function whose root is an inflection point. Am I right? If that is true, how can we prove it?
Topic: We can't use Newton's Method to approximate the root if the function has a root as an inflection point. Prove it or disprove it
Best Answer
Before generalizing, look at the specific problem and see why Newton's method won't converge to the root. If $x_n$ is an “approximation” to the root, $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x - \frac{x_n^{1/3}}{\frac{1}{3} x_n^{-2/3}} = x_n-3x_n = -2x_n $$ So rather than closer to the root, $x_{n+1}$ is twice as far away from the root as $x_n$.
I'm not sure of the essential property that causes this. It might be that $\lim_{x\to 0} f'(x) = \infty$.