Prove that $x_{n+1} = \frac{1}{3}(2x_n + \frac{a}{x_n^2})$ is decreasing

Question

Prove that $x_{n+1} = \frac{1}{3}(2x_n + \frac{a}{x_n^2})$ is decreasing where $x_1$ $> 0$.

I have been asked the above question and the working out given to me skipped some steps in between.

It gets to a point where I show $x_{n+1}$ - $x_n$ = $\frac{a - x_n^3}{3x_n^2}$ = $(a^\frac{1}{3}-x_n)(\frac{a^\frac{2}{3}+a^\frac{1}{3}x_n+x^2_n}{3x_n^2})$

This makes sense to me until the worked solution says:

$(a^\frac{1}{3}-x_n)^2>0$ -> $a^\frac{2}{3} + x^2_n$ > $2a^\frac{1}{3}$$x_n$ > $a^\frac{1}{3}x_n$ < $0$.

My question is, how does he arrive to the conclusion that it is less than 0 given no condition for $a$ since it is arbitrary and can be positive or negative. How does this tie into my factorisation $(a^\frac{1}{3}-x_n)(\frac{a^\frac{2}{3}+a^\frac{1}{3}x_n+x^2_n}{3x_n^2})$?

Answer 1

The sequence might only decrease if you ignore $x_1$ and $a>0$. For example, when $x_1=0.1, a=1,$ 1 → 2 increases, then decreases.

Also, in this case, you have $\operatorname{sgn}(a^{\frac 1 3}x_n)=1, \forall n$.

When $x_1=0.1, a=-1,$ 1 → 2 decreases, then increases.

x = 0.1
a = 1
for i in range(20):
    print(i+1,x)
    x = (2*x + a/(x**2))/3

Answer 2

Another way to look at the problem $$x_{n+1} = \frac{1}{3}\left(2x_n + \frac{a}{x_n^2}\right)\implies x_{n+1} = x_n -\frac{x_n^3-a}{3x_n^2}$$ This means that Newton method is used for solving the equation $$f(x)=x^3-a$$ Suppose that we start with $x_1=b >0$

$$f(b)\times f''(b)=6\,b\,(b^3-a)$$

If $(b^3-a)>0$, by Darboux theorem, the solution will be reached without any overshoot and the sequence will decrease.

Try with $a=8$ and $x_1=1$ which generates the sequence $$\{1.00000,3.33333,2.46222,2.08134,2.00314,2.00000\}$$

Try with $a=8$ and $x_1=3$ which generates the sequence $$\{3.00000,2.29630,2.03659,2.00065,2.00000\}$$

Answer 3

Claim. If $a > 0$ and $a^{1/3} \ne x_1 > 0$, then $x_2 > x_3 > x_4 > \cdots$, i.e., $x_{n+1} > x_{n+2}$ for all $n\ge 1$.

Proof.

We have, for all $n \ge 1$, $$x_{n+1} - a^{1/3} = \frac{(2x_n + a^{1/3})(x_n - a^{1/3})^2}{3x_n^2}.$$ By the Mathematical Induction, we can prove that $x_n \ne a^{1/3}$ for all $n \ge 1$.

We have, for all $n\ge 1$, $$x_{n+1} - x_{n+2} = \frac{(8x_n^3 + a)(x_n^3 - a)^2}{9a_n^2 (2x_n^3 + a)^2} > 0.$$ The claim is proved.

Answer 4

Let set $\mathbb{A}$ contain all values generated by this series. We need to prove $x_n-x_{n+1}>0,x_1>0$ By inserting your formula for $x_{n+1}$,we get $$x_n-\bigg(\frac{2x_n}{3}+\frac{a}{3x^2_n}\bigg)=\frac{1}{3}\bigg(x_n-\frac{a}{x_n^2}\bigg)$$

Thus we basically need to prove that $$\bigg(x_n-\frac{a}{x_n^2}\bigg)>0\implies x_n^3>a \forall x\in \mathbb{A}$$

But this is not true always. It seems your question is missing many details. It is highly possible that $x_n^3<a\implies x_n<\sqrt[3]a$.

If your question is asking if it's always decreasing $\forall a>0,n>0,x_1>0$, then it's totally false. Examples are infinite, take any number $a$ more than $x_1^3$ and you'll be atleast sure to get $x_2>x_1$.Take $x_1=1,a=3$ and you'll get $x_2=\frac{5}{3}$

Now if your question doesn't care about $n$ and takes considers the series to run till infinity

Then as the series increases when $x_n<\sqrt[3]a$, at some value $i,i\in \mathbb{Z^+}$, $x_i≥\sqrt[3]a$. After that, it will be a decreasing series. In the previous example,$x_2>\sqrt[3]3$ so $x_3=\frac{331}{225}<x_2$

Hope what I presented is clear