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Let $u = (u_n)_{n \geq 0}$ be a sequence such that $u_0 = 0$ and $$u_n = u_{n-1} + f(u_{n-1})$$ with $f$ a decreasing bijection from $\mathbb{R}$ to $(0,1)$ (hence $f$ vanishes at infinity). We assume moreover that $f$ is smooth.

My question is about the asymptotic behavior of $u$. The sequence $u_n$ is increasing. Let's denote by $\ell \in \mathbb{R} \cup \{\infty\}$ its limit. At the limit, we have $\ell = \ell + f(\ell)$, which is not possible for $\ell$ finite since $f > 0$. Hence, $$u_n \underset{n\rightarrow \infty}{\longrightarrow} \infty. $$

Question: What is the speed of divergence of the sequence $u$? How does it relate to the decay of $f$?

A typical example for $f$ is, defining the sigmoid function $\sigma(x) = \frac{1}{1+e^{-x}}$, for $$f(x) = \sigma(-x).$$

Any advice or answer for this case is also welcome.

Thibaut
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Goulifet
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    For the sigmoid case, you can expect that $u_n\sim \ln n$ – Stefan Lafon Apr 09 '25 at 16:09
  • See these two questions for the case $f(x)\sim_{x\to\infty} e^{-x}$ (which is equivalent to the sigmoid case) where we indeed get $u_n\sim \ln n$. My answer below provides a more general “recipe”. – Thibaut May 25 '25 at 00:20
  • You can try ansatz $u_n\sim Cn^{\alpha_1}(\log n)^{\alpha_2}(\log\log n)^{\alpha_3}...$ with $\alpha_1<1$ , this class of functions is unbounded yet having vanishing derivative. It's ansatz so it can be false but I feel like it works if $f(x)$ is not "exotic". – Quý Nhân May 25 '25 at 02:41

2 Answers2

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This kind of questions can often be treated by introducing an integral. Specifically, observe that since $f$ is positive and decreasing, we have $$\frac{u_{n+1}-u_n}{f(u_n)} = 1\leq \int_{u_n}^{u_{n+1}} \frac{dt}{f(t)}\leq \frac{u_{n+1}-u_n}{f(u_{n+1})}=1\cdot\frac{f(u_n)}{f\big(u_n+f(u_n)\big)}.$$ By summation, the lower estimate immediately implies that $G(u_n)\geq n$, where $G$ is the antiderivative of $1/f$ that vanishes at $0$. Under the additional assumption that $\lim_{x\to\infty} f(x)/f\big(x+f(x)\big)=1$ (this is the case for the sigmoid example), the lower and upper estimates combined imply that $G(u_n)\sim n$. In the sigmoid example this yields $u_n= \log n + o(1)$.

Addendum: the assumption $\lim_{x\to\infty} f(x)/f\big(x+f(x)\big)=1$ is satisfied under “natural” assumptions. For example, assume that $\lim_{x\to+\infty} f'(x)=0$. Then, for each $\varepsilon<0$ we have for $x$ large enough $$\varepsilon\cdot f(x)\leq \int_{x}^{x+f(x)} f'(t)dt=f\big(x+f(x)\big)-f(x)\leq 0.$$ The conclusion follows after dividing by $f(x)$.

Thibaut
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If $g$ is a smooth function such that $g(n)=u_n$, then $g'(n)\approx g(n+1)-g(n)=f(u_n)=f(g(n))$. So solving the differential equation $$y'=f(y)$$ should give you a reasonable approximation or at least a good starting point for heuristics.

Hagen von Eitzen
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