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I've managed to solve the below recurrence,

$a_n = a_{n-1} - b a_{n-1}^{3/2}$

for some constant $b$, with the guess,

$a_n = \alpha n^{-r}$

and solving for $\alpha$ and $r$. (I'm only interested in leading-order asymptotic behaviour as in this example.)

Now I'd like help to solve the related recurrence ,

$a_n = a_{n-1} - b a_{n-1}^{3/2} c^{(n-1)/2}$.

I've tried guesses of varying forms by including $c^{(n-1)/2}$, but I've had no luck, and I'm not really sure how to proceed.

This question is the closest I could find, but is still rather different.

Any help is much appreciated.

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    $\begingroup$ I'm pretty sure $a_n=a_\infty+O\left(c^{n/2}\right)$. I don't have the time to write a full answer at the moment, but I will be back in a day or two. $\endgroup$
    – cvogt8
    Aug 25, 2019 at 7:05
  • $\begingroup$ looking forward to hearing more @cvogt8 ! $\endgroup$
    – tea_pea
    Aug 27, 2019 at 9:40

1 Answer 1

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We consider only a subset of the real convergent sequences spawned by the recurrence relation. $$a_{n+1}=a_n-ba_n^{3/2}c^{n/2}$$ $$b>0 \quad\quad 0<c<1 \quad\quad 0<a_0<b^{-2}$$ There is numerical evidence that $\,a_n\,$ converges for some combinations with $\,b<0\,$ while diverges for others. I have not yet determined the convergence boundary when $\,b<0$, so only $\,b>0\,$ is considered. We prove $\,a_n\in\mathbb{R} \,\,\,\land\,\,\, 0<a_n<b^{-2}\,$ by induction. The base case is given. $$a_n>0 \implies a_{n+1}\in\mathbb{R}\,\,\,\land\,\,\,a_{n+1}<a_n$$ $$a_{n+1}<a_n<b^{-2} \implies a_{n+1}<b^{-2}$$ $$a_{n+1}>0 \iff a_n-ba_n^{3/2}c^{n/2}>0 \iff a_nc^n<b^{-2} \Longleftarrow a_n<b^{-2}$$ $\,a_n\,$ is a strictly decreasing sequence with a lower bound of $\,0$, so it must converge to $\,a_\infty\in\left[0,a_0\right)$.

While I do not use the following substitution in my answer, others may find it helpful: $$a_{n+1}=a_n-ba_n^{3/2}c^{n/2} \implies a_{n+1}c^{n+1}=a_nc^nc\left(1-b\left(a_nc^n\right)^{1/2}\right)$$ $$d_n=a_nc^n \quad\quad d_{n+1}=cd_n-bcd_n^{3/2} \quad\quad d_0=a_0 \quad\quad d_\infty=0 \quad\quad d_{n+1}<d_n$$

Scratch Work:

We assume $\,\exists f:[0, 1]\to\mathbb{R}\,$ such that $\,f\,$ is right-differentiable at $\,0\,$ and the following is satisfied: $$f\left(x\sqrt{c}\right)=f(x)-bxf^{3/2}(x)$$ $$f(1)=a_0$$ We will not attempt to prove the existence of such a function. We use induction to show $\,a_n=f\left(c^{n/2}\right)$. The base case is given. $$f\left(c^{(n+1)/2}\right)=f\left(c^{n/2}\sqrt{c}\right)=f\left(c^{n/2}\right)-bc^{n/2}f^{3/2}\left(c^{n/2}\right)=a_n-bc^{n/2}a_n^{3/2}=a_{n+1}$$ We show $\,f(0)=a_\infty$. Since $\,f\,$ is right-differentiable at $\,0\,$, it is also right-continuous at $\,0$. $$a_\infty=\lim_{n\to\infty}f\left(c^{n/2}\right)=f\left(\lim_{n\to\infty}c^{n/2}\right)=f(0)$$ Now we find the right derivative of $\,f\,$ at $\,0$ in two different ways. $$\sqrt{c}f'\left(x\sqrt{c}\right)=f'(x)-bf^{3/2}(x)-\frac{3}{2}bxf'(x)f^{1/2}(x)$$ $$\sqrt{c}f'(0)=f'(0)-bf^{3/2}(0)$$ $$f'(0)=\frac{bf^{3/2}(0)}{1-\sqrt{c}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}$$ $$f'(0)=\lim_{x\to0^+}\frac{f(x)-f(0)}{x}=\lim_{n\to\infty}\frac{f\left(c^{n/2}\right)-f(0)}{c^{n/2}}=\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}$$ $$\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}$$ I have numerically verified the previous limit in a few cases. Due to our assumption of the existence of a magic function, we cannot trust the above result yet. We use it as guidance for what to rigorously prove.

Rigorous Approach:

We can represent $\,a_\infty\,$ as a telescoping sum: $$\forall{n} \quad a_\infty=a_n+\sum_{k=n}^\infty{\left(a_{k+1}-a_k\right)}$$ $$a_n-a_\infty=\sum_{k=n}^\infty{\left(a_k-a_{k+1}\right)}=\sum_{k=n}^\infty{ba_k^{3/2}c^{k/2}}$$ Since $\,a_k\,$ is a strictly decreasing sequence, $$\sum_{k=n}^\infty{ba_\infty^{3/2}c^{k/2}}<\sum_{k=n}^\infty{ba_k^{3/2}c^{k/2}}<\sum_{k=n}^\infty{ba_n^{3/2}c^{k/2}}$$ The bounds are geometric series. $$\sum_{k=n}^\infty{ba_\infty^{3/2}c^{k/2}}=ba_\infty^{3/2}c^{n/2}\sum_{k=0}^\infty{c^{k/2}}=\frac{ba_\infty^{3/2}c^{n/2}}{1-\sqrt{c}}$$ $$\sum_{k=n}^\infty{ba_n^{3/2}c^{k/2}}=\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}$$ $$\frac{ba_\infty^{3/2}c^{n/2}}{1-\sqrt{c}}<a_n-a_\infty<\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}$$ $$\frac{ba_\infty^{3/2}}{1-\sqrt{c}}<\frac{a_n-a_\infty}{c^{n/2}}<\frac{ba_n^{3/2}}{1-\sqrt{c}}$$ We use the squeeze theorem to conclude: $$\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}$$ We prove $\,a_\infty\neq0$. Assume $\,a_\infty=0\,$ for contradiction. The above limit gives us something easier to contradict: $$\lim_{n\to\infty}\frac{a_n}{c^{n/2}}=0$$ $$s_n=\frac{a_n}{c^{n/2}} \quad\quad s_0=a_0 \quad\quad s_\infty=0 \quad\quad s_n>0$$ $$s_{n+1}c^{(n+1)/2}=s_nc^{n/2}-bs_n^{3/2}c^{3n/4}c^{n/2}$$ $$s_{n+1}=\frac{s_n}{\sqrt{c}}-bs_n^{3/2}c^{(3n-2)/4}$$ $$s_{n+1}=s_n\left(c^{-1/2}-bs_n^{1/2}c^{(3n-2)/4}\right)$$ To contradict $\,s_\infty=0$, it suffices to show: $$\exists{m}\,\forall{n \geq m}\,\left(s_n\leq1\implies s_{n+1}>s_n\right)$$ This essentially says: For large enough $\,n$, $\,s_n\,$ cannot decrease much further than $\,1$. Hence $\,s_n\,$ cannot converge to $\,0$. $$s_{n+1}>s_n \iff c^{-1/2}-bs_n^{1/2}c^{(3n-2)/4}>1 \iff c^{1/2}+bs_n^{1/2}c^{3n/4}<1$$ $$s_n\leq1 \implies c^{1/2}+bs_n^{1/2}c^{3n/4} \leq c^{1/2}+bc^{3n/4}$$ So it comes down to choosing an $\,m\,$ such that $$\forall{n \geq m} \quad c^{1/2}+bc^{3n/4}<1$$ Since $\,0<c<1$, it is clearly possible to choose such an $\,m$. $$\therefore \quad a_\infty\in\left(0, a_0\right)$$ $$\lim_{n\to\infty}\frac{a_n-a_\infty}{c^{n/2}}=\frac{ba_\infty^{3/2}}{1-\sqrt{c}}\neq0$$ $$a_n=a_\infty+\Theta(c^{n/2})$$ We also have bounds on $\,a_\infty$: $$\forall{n}\quad\left(a_n-\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}<a_\infty<a_n\right)$$ We can use the lower bound to prove $\,a_\infty\neq0$. Assume $\,a_\infty=0\,$ for contradiction. $$a_n-\frac{ba_n^{3/2}c^{n/2}}{1-\sqrt{c}}<0 \iff \sqrt{c}+ba_n^{1/2}c^{n/2}>1$$ Letting $\,n\to\infty$, we get $\,\sqrt{c}\geq1$, which contradicts $\,c<1$.

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