I need to show that Newton's iteration to compute the root $x^*$ of $$f(x) = |x|^{\frac{3}{7}}$$ does not converge for any starting guess $x_0 \neq 0$.

The first thing I did was to create the Newton's iteration for this $f$:

$$x_{n+1} = x_n + \frac{f(x_n)}{f'(x_n)} = \\x_{n+1} = x_n + \frac{|x|^{\frac{3}{7}}}{\frac{3}{7} \frac{x}{|x|^\frac{11}{7}}} = \\ \cdots = \\ \frac{-4}{3}x_n$$

What I then thought was to try to show for which values of $x_n$, $|\frac{-4}{3}x_n| \leq 1$, and I found that $\frac{-3}{4} \geq x_n$ and $\frac{3}{4} \leq x_n$, but $x_n$ can't be greater than $\frac{3}{4}$ and smaller than $\frac{-3}{4}$ at the same time...

What do you think about what I did?

In the problem's specifications, they tell me to distinguish between when $x_n > 0$ and $x_n < 0$, but I didn't do it.

Please, make sure my calculations are correct, because I usually do some mistakes.


You made a mistake. There shouldn't be any negative sign. If you do the calculation correctly, you will get

$$ x_{n+1} = \phi(x_n) = \frac{10}{3}x_n $$

so that $\phi'(x)=\frac{10}{3}>1, \forall x$.

Edit: remember, when you study the convergence of a fixed point method, it is not the iteration map $\phi$ that you want to be bounded by 1 (in modulus), but rather its derivative.

  • $\begingroup$ I guess. The function $|\cdot |$ is not differentiable in 0. One can capture the derivative outside 0 with a single expression (as you did), or simply consider the two cases separately. $\endgroup$
    – bartgol
    May 9 '16 at 15:32

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