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Consider the sequence $\{a_{n}\}$ satisfying $a_{1}\in(0,1)$,such $$a_{n+1}=a_{n}(1-a_{n})$$ question:

Find a sequence $\{x_{n}\}$ such that $$\lim_{n\to \infty}x_{n}\left(1-\dfrac{n(1-na_{n})}{\ln{n}}\right)=1$$

I have prove this $$\lim_{n\to \infty}\dfrac{n}{\ln{n}}(1-na_{n})=1$$ But I can't find this $x_{n}$ Thank you

such as:How prove this $\displaystyle\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\frac{n}{\ln{n}}\right)=-1$

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math110
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    What about $x_{n}=\frac{1}{\left(1-\dfrac{n(1-na_{n})}{\ln{n}}\right)}$? – P.. Jan 08 '14 at 06:59
  • No,I think we can find $x_{n}=g(n)$,such http://math.stackexchange.com/questions/485408/how-prove-this-displaystyle-lim-n-to-infty-fracn-ln-lnn-left1-a?rq=1 – math110 Jan 08 '14 at 07:40
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    Could you please share your proof that $\frac{n}{\ln n}(1-n a_n) \to 1$? – Antonio Vargas Jan 11 '14 at 06:21
  • $a_n=f(n)$, doesn't it? Then $x_n=g(n)$ in the comment of @P.. Please reedit your question to make clear what is your request. – vesszabo Jan 19 '14 at 19:01
  • No, $a_n=f(n, a_1)$; that is, its terms depend on the starting point. The premise of the question, I think, is that they actually don't depend on the starting point, at least asymptotically. That is, there is a function $x(n)$ such that $1 - n(1-n a_n)/\log n \sim 1/x(n)$. – mjqxxxx Jan 21 '14 at 22:56

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This is a question of determining the behavior of the solutions to $$ a_{n+1}=a_{n}(1-a_{n}) $$ as an asymptotic expansion as $n\rightarrow\infty$. I'll reiterate how to derive the result you already have, then find the next term. First, note that $$ \Delta a_n\equiv a_{n+1}-a_{n}=-a_n^2; $$ this is analogous to the differential equation $a'(n)=-a(n)^2$, whose general solution is $$ a(n)=\frac{1}{n+C},$$ suggesting that $a_n \sim 1/n$ for large $n$. Defining an auxiliary sequence $b_n$ via $$ a_n=\frac{1}{n}\left(1+b_n\right), $$ and substituting this into the equation for $\Delta a_n$, we find (after some algebra) the form of $\Delta b_n$ to be $$ \Delta b_{n}=-\frac{1}{n^2}-\frac{1}{n}\left(1+\frac{2}{n}\right)b_n - \frac{1}{n}\left(1+\frac{1}{n}\right)b_n^2. $$ The leading-order terms are analogous to $b'(n)=-1/n^2-b(n)/n$, whose solution is $b(n)=(A/n -\log n/n)$, which behaves as $-\log n/n$ for large $n$. Indeed, taking $b_n\sim -\log n/n$ makes both sides of the above equation asymptotic to $\log n/n^2$. We repeat the procedure of defining a new variable: $$ b_n=-\frac{\log n}{n} + c_n. $$ Note that $$ 1-\frac{n(1-na_n)}{\log n}=1+\frac{nb_n}{\log n}=\frac{nc_n}{\log n}, $$ so the dominant behavior of $c_n$ is all we need. Plugging in the definition, we get $$ \Delta c_n = \Delta\left(\frac{\log n}{n}\right)-\frac{1}{n^2}+\frac{1}{n}\left(1+\frac{2}{n}\right)\left(\frac{\log n}{n}-c_n\right)-\frac{1}{n}\left(1+\frac{1}{n}\right)\left(\frac{\log n}{n}-c_n\right)^2. $$ Note that $$ \Delta\left(\frac{\log n}{n}\right)=\frac{\log (n+1)}{n+1} - \frac{\log n}{n}=\frac{n\log(n+1)-(n+1)\log n}{n(n+1)} \\ = -\frac{\log n}{n^2} + \frac{1}{n^2} + \frac{\log n}{n^3} - \frac{3}{2n^3} + \ldots; $$ the first two terms cancel the $-1/n^2$ and $\log n/n^2$ terms in the remainder of the right-side, as expected. We are left with $$ \Delta c_n = \left(\frac{\log n}{n^3}+\ldots\right) + \left(-\frac{c_n}{n}+\frac{2\log n}{n^3}+\ldots\right) +\left(-\frac{\log^2 n}{n^3}+\ldots\right). $$ The leading-order equation is just $\Delta c_n \sim -c_n / n$, with solution $c_n \sim K/n$. Plugging everything back in, we find that $$ 1-\frac{n(1-na_n)}{\log n} \sim \frac{K}{\log n}, $$ and so $$ \lim_{n\rightarrow\infty}\left(\log n\right)\left(1-\frac{n(1-na_n)}{\log n}\right)=K(a_0); $$ i.e., the limit exists for each starting value $a_0\in(0,1)$. However, despite how your question is written, the value of the limit is not universal; it depends on the starting value. Empirically, the form of the dependence looks something like $$ K(a_0)\approx -\frac{\log^{4/5} a_1}{2.9a_1}, $$ where $a_1=a_0(1-a_0)$.

mjqxxxx
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  • How did you reach empirically the dependence at the end? – Did Jan 22 '14 at 21:40
  • I just played around with it. It's not valid near the endpoints ($0$ and $1$), but is pretty good through the middle. – mjqxxxx Jan 22 '14 at 22:47
  • "played around with it" meaning what? Numerical simulations? – Did Jan 23 '14 at 06:20
  • When you are simplifying to $b'=-1/n^2-b/n$, why don't you keep $b^2/n$? How do you know that $b^2/n$ is not bigger than $1/n^2$? – user10676 Jan 23 '14 at 10:55
  • @Did: yes, numerical simulations. – mjqxxxx Jan 23 '14 at 18:02
  • @user10676: you're assuming that $b \ll 1$, so $b^2/n \ll b/n$. you don't know how it compares to $1/n^2$. – mjqxxxx Jan 23 '14 at 18:03
  • Can you provide an outline of the simulation procedure you used? – Did Jan 23 '14 at 18:05
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    I just looked at different values of $a_0$, calculated $a_n$ out to about $n=10^6$, and checked that $(\log n)(1-n(1-na_n / \log n))$ appeared to be converging to a constant. – mjqxxxx Jan 23 '14 at 19:38
  • Would you mind proving the steps after the establishing the equation for $c_n$, particularly the dropping of the terms such as $\frac{c_n}{n^2}$ and the quadratic term of $c_n$? – Hans Dec 01 '22 at 06:51