Approximate the root using Newton's method - Convergence of sequence

Question

Let $f(x)=x^3-1$. To approximate the root $x^{\star}=1$, we consider the sequence $(x_n)$ that we get if we apply Newton's method with $x_0>0$. Show that the sequence converges to $1$.

I have done the following:

I used $x_0=0,5$ and applied the method and in that way we see that the sequence converges to $1$.

Is that correct?

Now I think that we could also do the following:

From Newton's method we get $x_{n+1}=x_n-\frac{x_n^3-1}{3x_n^2}=\frac{2x_n^3+1}{3x_n^2}$ and we have to show that this sequence converges to $1$, or not?

Answer 1

Hint: If $x_0 > 1$, show that the sequence is decreasing. If $0 < x_0 < 1$, show that $x_1 > 1$.

Answer 2

I assume you didn't quite understand the other answer, so let me elaborate on that. We have $$x_{n+1}=\frac{2x_n^3+1}{3x_n^2} < x_n \rightarrow1< x_n^3\rightarrow x_n>1$$Now assuming that we have proven that $x_n > 1$ for all $n\ge n_0$, it follows that $x_n$ is decreasing from $n_0$; your sequence being bounded below by $1$ is thus convergent.

To show that $x_n>1$ we proceed by induction as follows: if $x_0 > 1$ we may assume $n_0 = 0$. Now $$x_{n+1}=\frac{2x_n^3+1}{3x_n^2} > 1\rightarrow2x_n^3-3x_n^2+1>0$$Put $t=x_n$. So $2t^3-3t^2+1>0\leftrightarrow2t^2(t-1)>(t-1)(t+1)$. Since $t>1$ by our induction hypotesis we may divide by $(t-1)$, and get $2t^2-t-1>0$ which is again true for $t>1$ as you can check.

If $x_0 < 1$ then we may assume $n_0=1$, for $x_1 = \frac{2x_0^3+1}{3x_ 0^2}>1 \leftrightarrow2x_0^3-3x_0^2+1>0$. As above $2x_0^2(x_0-1)>(x_0-1)(x_0+1)$ and dividing by $x_0-1$ which is negative, we obtain $2x_0^2-x_0-1<0$ which is again satisfied as you can check.