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Given the function $f(x)=x-\frac{x^2}2$ and $a_1=1$, $a_n=f(a_{n-1})$. Find the asymptotic expansion of $a_n$ to the $O(1/n^3)$ term as $n\to\infty$.
Attempt (to the $O(1/n)$ term)

Step 1
$$a_n=o(1)$$

As $|f(x)|<x$, $x\in(0,1]$, $\lim\limits_{n\to\infty}a_n=0$.

Step 2
$$a_n=\frac 2n+o(\frac 1n)$$

This step is equivalent to finding the value of $\lim\limits_{n\to\infty}n a_n$.
$$\lim_{n\to\infty}n a_n=\lim_{n\to\infty}\frac n{\frac 1{a_n}}=\lim_{n\to\infty}\frac{n-(n-1)}{\frac 1{a_n}-\frac 1{a_{n-1}}}=\lim_{x\to0}\frac 1{\frac1{x-x^2/2}-\frac1x}=2$$

Step 3 & 4 (only a guess)
$$a_n=\frac 2n-\frac{2\ln n}{n^2}+o(\frac{\ln n}{n^2})$$ $$a_n=\frac 2n-\frac{2\ln n}{n^2}+\frac{\text{a constant}}{n^2}+o(\frac 1{n^2})$$

This is equivalent to finding the limit $\lim\limits_{n\to\infty}\frac{n^2}{\ln n}(a_n-\frac2n)$ and $\lim\limits_{n\to\infty}n^2(a_n-\frac2n+\frac{2\ln n}{n^2})$
I can't go further. I think the condition $a_n=1$ should be used well in Step 3.

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  • $\begingroup$ Your $+$ turns to a $-$. This makes a big difference. Make up your mind. $\endgroup$ – Yves Daoust Aug 31 '18 at 9:24
  • $\begingroup$ @YvesDaoust thanks about pointing out the typo. It is fixed now. :) $\endgroup$ – Kemono Chen Aug 31 '18 at 9:44
  • $\begingroup$ This is very close to MSE question 2861768 "Calculate $\lim_{n\to\infty} n(1-na_n)/\log n$". where there $a_{n+1}=a_n-a_n^2.$ $\endgroup$ – Somos Aug 31 '18 at 15:49
  • $\begingroup$ @Somos Thanks, that helps, but some of the terms the answers give is implicit. I don't know how to find the terms related to $a_1$. $\endgroup$ – Kemono Chen Sep 1 '18 at 1:35
  • $\begingroup$ The dependence on $a_1$ is a numerical constant $c$. Read my answer to the linked question. $\endgroup$ – Somos Sep 1 '18 at 1:47
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One starts with $a_n=o(1)$ which follows from the facts that all $a_n$ are in $[0,1]$, $a_n\leq a_{n-1}$ and $f(x)=x$ has only the zero solution in $[0,1]$. Then the trick is to expand $$\tag1\frac1{a_{n+1}}=\frac1{a_n(1-\frac{a_n}2)}=\frac1{a_n}+\frac12+\frac14a_n+O(a_n^2).$$ This implies that $\frac1{a_{n+1}}-\frac1{a_n}$ tends to $\frac12$. By Cesaro's theorem $\frac1{n-1}\sum_{j=1}^{n-1}(\frac1{a_{j+1}}-\frac1{a_j})=\frac1{n-1}(\frac1{a_n}-\frac1{a_1})$ also tends to $\frac12$ and therefore $a_n=\frac2n+o(\frac1n)$. Inserting into (1) gives $$\frac1{a_{n+1}}-\frac1{a_n}=\frac12+O(\frac1n)$$ and by summing as above we obtain that $\frac1{a_n}=\frac n2+O(\log(n))$. Hence $a_n=\frac2n+O(\frac 1{n^2}\log n)$. Inserting into (1) again gives $$\frac1{a_{n+1}}-\frac1{a_n}=\frac12+\frac1{2n}+O(\frac 1{n^2}\log n).$$ Now the sequence $b_n=\frac1{a_n}-\frac n2-\frac12\log n$ satisfies $b_{n+1}-b_n=O(\frac 1{n^2}\log n).$ (Just use the Taylor expansion of the logarithm). As $\sum_{n=1}^\infty \frac 1{n^2}\log n$ converges, $b_n$ tends to a limit we call $\beta$. Since $\beta-b_n=\sum_{k\geq n}(b_{k+1}-b_k)$, we find that $b_n=\beta+O(\frac1n\log n)$ that is $\frac1{a_n}=\frac n2+\frac12\log n+\beta+O(\frac1n\log(n))$. This means that $$a_n=\frac2n-\frac 2{n^2}\log n-\frac{4\beta}{n^2}+\frac 2{n^3}\log^2 n+O(\frac 1{n^3}\log n)$$ with the above constant $\beta$. This does not solve the problem to the $O(\frac1{n^3})$ term, but the last necessary step is analogous: Inserting into (1) again to refine the asymptotic behavior of the above $b_n$ and sum from $k=n$ to infinity again. I do not want to do that.

The constant $\beta$ can be determined numerically. It is the equivalent of the Euler constant in the present context. I doubt that there is an explicit form of this constant.

Edit: 1.I calculated $\beta=0.38399...$.
2. The recursion $a_{n+1}=\sin(a_n)$ can be treated similarly. Only here, $\frac1{a_{n+1}^2}-\frac1{a_n^2}$ has to be expanded...

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