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Suppose we have a sequence of the (recursive) form $a_{n+1}=a_n - \frac{a_n^2}{2}$. And $a_0$ is set to be $<1$. Hopefully this sequence goes to zero. Is there any way to estimate the asymptotic behaviour of $a_n$? Is it $a_n= \mathcal{O}(1/n)$?

What I have tried is that we can rewrite the recursive formula to obtain: $$a_{n+1}= -\frac{1}{2}(a_n^2-2a_n+1-1)= \frac{-1}{2}(a_n-1)^2 + \frac{1}{2},$$ then subtract 1 on both sides we have: $$a_{n+1}-1 = \frac{-1}{2}(a_n-1)^2 - \frac{1}{2}.$$

Define $1-a_n$ as $b_n$ we can have $$b_{n+1} = \frac{1}{2}b_n^2 + \frac{1}{2}$$

Further denote $\frac{1}{2}b_n$ as $c_n$, we have $$c_{n+1} = c_n^2 + \frac{1}{4}.$$

SouthChinaSeaPupil
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  • Check this: https://math.stackexchange.com/q/3338103/42969 – Martin R Oct 07 '24 at 09:29
  • See also https://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n with a general treatment of $a_{n+1}=a_n+ca_n^d+...$, $d\ge 1$ and topics linked from there. – Lutz Lehmann Oct 07 '24 at 11:05

1 Answers1

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You loosely compare this with the class of Bernoulli equations to find a suitable variable change. Here $u_n=\frac1{a_n}$. Then use series expansion using that $a_n$ is very small for large $n$: $$ u_{n+1}=u_n\left(1-\frac{a_n}2\right)^{-1}=u_n\left(1+\frac{a_n}2+\frac{a_n^2}4+...\right)=u_n+\frac12+O(a_n) $$ so that asymptotically $u_n\asymp \frac n2$ and thus $a_n\asymp\frac2n$.

Lutz Lehmann
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