6
$\begingroup$

Let a sequence, $\{x_n\}$ such that: $x_{n+1}=x_n-x_n^3$ and $0<x_1<1$.
1) Prove $\mathop {\lim }\limits_{n \to \infty } {x_n} = 0$
2) Calculate $\mathop {\lim }\limits_{n \to \infty } n{x_n}^2$

So, section (1) is very easy. I didn't really bother to write it down - just show the sequence is monotonically decreasing and bounded bellow by zero.

Section (2) is the real fun, I do familiar with the Lemma says: "If $a_n$ limit is $0$ and $b_n$ is bounded then the limit of $a_nb_n$ is also zero" - But I don't think it can work here.

I tried separating the limit using limits-arithmetic into two limits, but then I got:
$$\mathop {\lim }\limits_{n \to \infty } n{x_n} \cdot \mathop {\lim }\limits_{n \to \infty } {x_n}$$

$\endgroup$
  • $\begingroup$ is it $n \to \infty$ $\endgroup$ – Suraj M S Jan 25 '14 at 18:34
  • $\begingroup$ it's a typo. will be corrected. Thanks $\endgroup$ – SuperStamp Jan 25 '14 at 18:40
6
$\begingroup$
  1. $\lim_{x \to \infty } {x_n}=0$. Because the sequence is monotone and bounded by $0$ and $1$. Last statement is proved by mathematical induction. Passing to the limit in the recurrence relation is obtained limit $0$. Dividing the recurrence relation by $x_n$ and then passing to the limit we find $\lim_{x \to \infty}\frac{x_{n+1}}{{x_n}} = 1$

  2. Applied Cesaro-Stolz's Lemma:

$$\lim_{x \to \infty } n{x_n}^2 = \lim_{x \to \infty }\frac{n}{(\frac{1}{x_n})^2}=\lim_{x \to \infty }\frac{n+1-n}{(\frac{1}{x_{n+1}})^2-(\frac{1}{x_n})^2}= \lim_{x \to \infty }\frac{(x_n)^2(x_{n+1})^2}{(x_n)^2-(x_{n+1})^2} =$$$$=\lim_{x \to \infty }\frac{(x_n)^4(1-(x_{n})^2)^2}{(x_n)^4(1+\frac{x_{n+1}}{x_n})} =\frac{1}{2}$$

$\endgroup$
  • $\begingroup$ How can you assume that ${nx_n^2} = {n \over \left({{1 \over x_n}}\right)^2}$? $\endgroup$ – SuperStamp Jan 25 '14 at 19:03
  • 1
    $\begingroup$ It is just a rewriting: $a^2 = \frac{1}{\left(\frac{1}{a}\right)^2}$ $\endgroup$ – Clement C. Jan 25 '14 at 19:07
  • $\begingroup$ 1. is not justified and in 2. it is assumed that $\lim\limits_{k\to\infty}\frac{x_{k+1}}{x_k}=1$, but that is not justified. $\endgroup$ – robjohn Jan 26 '14 at 15:12
  • $\begingroup$ also $1 + \frac{1}{n}$ is monotically decreasing and bounded below by $0$, but does not go to $0$ $\endgroup$ – Ant Jan 26 '14 at 20:26
  • $\begingroup$ Passing to the limit in the recurrence relation is obtained limit $0$. $\endgroup$ – medicu Jan 26 '14 at 20:29
3
$\begingroup$

Suppose that $0\lt x_k\lt1$, then $0\lt x_k^2\lt1$ and since $x_{k+1}=x_k(1-x_k^2)$, we know that $0\lt x_{k+1}\lt1$. Furthermore, since $x_k^3\gt0$, we have that $x_{k+1}=x_k-x_k^3\lt x_k$.

Thus, $x_k$ is decreasing and bounded below by $0$. Therefore, $\lim\limits_{k\to\infty}x_k$ exists and we have $$ \lim_{k\to\infty}x_k=\lim_{k\to\infty}x_{k+1}=\lim_{k\to\infty}x_k-\left(\lim_{k\to\infty}x_k\right)^3\tag{1} $$ Equation $(1)$ implies that $\lim\limits_{k\to\infty}x_k=0$.


Since $x_{k+1}=x_k-x_k^3$, dividing by $x_k$ and taking limits, we get that $$ \begin{align} \lim_{k\to\infty}\frac{x_{k+1}}{x_k} &=\lim_{k\to\infty}\left(1-x_k^2\right)\\ &=1\tag{2} \end{align} $$ Next notice that $$ \begin{align} \frac1{x_{k+1}^2}-\frac1{x_k^2} &=\frac{x_k^2-x_{k+1}^2}{x_k^2x_{k+1}^2}\\ &=\frac{(x_k-x_{k+1})(x_k+x_{k+1})}{x_k^2x_{k+1}^2}\\ &=\frac{x_k^3(x_k+x_{k+1})}{x_k^2x_{k+1}^2}\\ &=\frac{x_k^2}{x_{k+1}^2}+\frac{x_k}{x_{k+1}}\tag{3} \end{align} $$ Combining $(2)$ and $(3)$ yields $$ \lim_{k\to\infty}\frac1{x_{k+1}^2}-\frac1{x_k^2}=2\tag{4} $$ Apply $\displaystyle\lim_{k\to\infty}a_k=b\implies\lim_{n\to\infty}\frac1n\sum_{k=1}^na_k=b$ to $(4)$ to get $$ \lim_{n\to\infty}\frac1{nx_{n+1}^2}=2\tag{5} $$ which leads to $$ \lim_{n\to\infty}nx_n^2=\frac12\tag{6} $$


Note that the justification for $(5)$ is saying that if a sequence converges, then its Cesaro means converge to the same limit.

$\endgroup$
1
$\begingroup$

For what it's worth:

Since $x_n > 0$ for all $n$, one can safely divide both sides of the recurrence relation by $x_n$ to get that $$ \frac{x_{n+1}}{x_n} = 1 - x_n^2 $$ and thus $$ n x_n^2 = n\left( 1 - \frac{x_{n+1}}{x_n} \right) $$

Now, since $x_n\searrow 0$, $x_n^3=o(x_n)$ and $x_{n+1}\operatorname*{\sim}_{n\to\infty}x_n$; and in particular $\frac{x_{n+1}}{x_n} \xrightarrow[n\to\infty]{} 1$. Suppose you can get a second order expansion of the form $\frac{x_{n+1}}{x_n}\operatorname*{=}_{n\to\infty} 1 + \varepsilon_n + o(\varepsilon_n)$ for some "simple" $\varepsilon_n$ (I would bet on $Cn^{-\alpha}$ for some $\alpha > 0$). Then, $$ n x_n^2 \operatorname*{\sim}_{n\to\infty}n\varepsilon _n $$ and depending on $\varepsilon_n$, that'd give you the limit...

$\endgroup$
  • $\begingroup$ I'm not that familiar with o-notation, tough I should $\endgroup$ – SuperStamp Jan 25 '14 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.