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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 $x_1 \neq 0$.

Picture of <span class=$x^1/3$">

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

Bernard
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Ivan_Raki
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    I think this has less to do with the inflection point, for example $y=x^3$ has an inflection point at $x=0$ but Newton's Method applies just fine. I think this has more to do with the fact that $y=x^{1/3}$ is not differentiable at $x=0$. – Ehsaan Mar 17 '20 at 14:04
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    For example solving $\frac{1}{1+\exp(-x)}=\frac12$ should work fine with Newton despite the inflection – Maximilian Janisch Mar 17 '20 at 14:05
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    Even for functions differentiable everywhere, there are examples where the iterative process enters a cycle. Consider the examples from wikipedia under the failure analysis section: https://en.wikipedia.org/wiki/Newton's_method#Bad_starting_points – masiewpao Mar 17 '20 at 14:08
  • @Ehsaan Thank you for the counterexample, sir. If so, in what cases can we not use Newton's method? Thank you for the comment, anyway :) – Ivan_Raki Mar 17 '20 at 14:09
  • @Ehsaan Otherwise from $f'(x_n)=0$ – Ivan_Raki Mar 17 '20 at 14:10
  • Newton's method, when it converges, goes to an attractive fixed point of the function. For your function, $0$ is not an attractive fixed point but $\infty$ is. – GEdgar Mar 17 '20 at 14:51
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    @GEdgar : Not quite sure what you mean by “fixed point” of the function here. As I understand it, a fixed point of a function $f$ is a point $x$ such that $f(x)=x$. If Newton's method converges, it is to a point $x$ such that $f(x)=0$. – John Bentin Mar 17 '20 at 15:06

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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 $\lim_{x\to 0} f'(x) = \infty$.

Matthew Leingang
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