Minimizing $f(x)=x^{4/3}$ using Newton's method and show as long as initial approximation is not $0$, the algorithm does not converge to $0$.

Question

Consider the problem of minimizing $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x) = x^{4/3}$ . Note that $0$ is the global minimizer of $f$ . (a) Write down the iterations for Newton’s method for this problem. (b) Show that, as long as the initial approximation $x(0)$ is NOT equal to $0$, the algorithm in part (a) does not converge to $0$ (no matter how close to $0$ we start).

So I know Newton's method looks like this: $x(k+1)=x(k) -\dfrac{ f(x(k))}{f'(x(k))}=x(k)-\dfrac{x^{4/3}}{4/3\times x^{1/4}}.$ But I don't know how many iterations there are supposed to be, or what I should choose as my initial approximation, and I don't know how to prove it doesn't converge to $0$. Any help is much appreciated, thanks!

I dont know what tags you should use, but this has nothing to do with probability or statistics. — antkam, Nov 08 '19 at 00:26

score 2 · Accepted Answer · answered Nov 08 '19 at 01:02

Your problem is that you are not minimizing f(x) using Newton's Method, you are solving $f(x)=0$ instead of $f'(x)=0$ using Newton's Method. As you know in order for f(x) to be minimized we need to solve $f'(x)=0$ that is you want to apply Newton's Method to $$f'(x)=(4/3)x^{1/3}$$ It suffices to find the root of $x^{1/3}$

Now we have $$x_{n+1} = x_n - \frac {x_n^{1/3}}{(1/3)x^{-2/3}} = -2x_n$$

Which obviously diverges for every non-zero starting point.

Ooooh I did not think of it that way, thank you!! – mathnerd Nov 08 '19 at 01:34 — mathnerd, Nov 08 '19 at 01:34

Minimizing $f(x)=x^{4/3}$ using Newton's method and show as long as initial approximation is not $0$, the algorithm does not converge to $0$.

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