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$0 \le x_0 \le \frac{1}{2}$ , and $x_{n+1}=x_n-\dfrac{4x_n^3}{n+1}$

When I take $x_0=\sqrt{\frac{1}{12}}$, it converges very very slow. I can see it is monotonic decreasing but don't know how to find its limit.

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Let's study $$x_{n+1}^{-2} - x_n^{-2} = \left( x_n - \frac{4x_n^3}{n+1} \right)^{-2} - x_n^{-2} = x_n^{-2} \left( \left( 1- \frac{4x_n^2}{n+1}\right)^{-2} -1\right)$$

So $$x_{n+1}^{-2} - x_n^{-2} = x_n^{-2} \left( \frac{8x_n^2}{n+1} + o\left(\frac{x_n^2}{n+1} \right) \right) = \frac{8}{n+1} + o\left( \frac{1}{n+1} \right)$$

So you get that $$x_{n+1}^{-2} - x_n^{-2} \sim \frac{8}{n+1}$$

The series $\sum \frac{8}{n+1}$ diverges, so we can sum to obtain $$x_{n+1}^{-2} \sim \sum_{k=1}^n \frac{8}{k+1}$$

You obtain finally $$x_{n+1} \sim \frac{1}{\sqrt{\sum_{k=1}^n \frac{8}{k+1}}}$$

In particular, the limit is indeed $0$.

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  • $\begingroup$ That's just amazing! $\endgroup$ – Zhenduo Cao Sep 30 '18 at 19:16
  • $\begingroup$ Yes, the technic to study $u_{n+1}^{\alpha} - u_n^{\alpha}$ is very powerful ! $\endgroup$ – TheSilverDoe Sep 30 '18 at 19:20
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If $x_0=0$ or $x_0=\frac12$, then $x_1=0$ and so $x_n=0$ for all $n\ge1$.


If $0\lt x_n\lt\frac12$, then $$ \begin{align} x_{n+1} &=x_n-\frac{4x_n^3}{n+1}\\ &=x_n\left(1-\frac{4x_n^2}{n+1}\right)\\[3pt] &\in(0,x_n) \end{align} $$ This means that $x_n$ is decreasing and bounded below; therefore, it converges.


Suppose the limit is $a$. Then, because $x_n$ is decreasing and bounded below by $0$, $$ \begin{align} x_0-x_m &=\sum_{n=0}^{m-1}(x_n-x_{n+1})\\ &=\sum_{n=0}^{m-1}\frac{4x_n^3}{n+1}\\ &\ge\sum_{n=0}^{m-1}\frac{4a^3}{n+1}\\ &=4a^3H_m \end{align} $$ Thus, $$ a^3\le\frac{x_0}{4H_m}\to0. $$

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  • $\begingroup$ That's quite clear $\endgroup$ – Zhenduo Cao Oct 1 '18 at 20:39
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Notice that $x_{n+1}=\left(1-\frac{x_n^2}{n+1}\right)x_n$ and, therefore, $$x_{n+1}=x_0\prod_{k=0}^{n-1}\left(1-\frac{4x_k^2}{n+1}\right)$$

Now, recall this standard result on infinite products:

Let $0\le u_n<1$; then $\lim\limits_{n\to\infty} \prod\limits_{k=0}^n (1-u_k)\ne 0$ if and only if $\sum\limits_{k=0}^\infty u_k<\infty$.

If $\inf_n x_n=\beta>0$ then we'd have $\sum_{k=0}^\infty\frac{4x_k^2}{n+1}\ge 4\beta^2\sum_{k=1}^\infty \frac1n=\infty$. But by the previous theorem, this conclusion implies $\lim_{n\to\infty} x_n=0$.

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  • $\begingroup$ Thanks. Really powerful result. $\endgroup$ – Zhenduo Cao Oct 1 '18 at 20:47

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