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I am exploring a recursively defined sequence involving non-integer exponents, and I aim to find or confirm the limit of $\left\{b_n\right\} \text { as } n \rightarrow \infty \text {. } $ The sequence is defined as follows:

Let $x_1, x_2, \ldots, x_t$ be initial fixed positive real numbers, and define the sequence $\{b_n\}$ by:

  • For $n \leq t$, $b_n = x_n$.
  • For $n > t$, $$ b_n = b_{n-1}^p + b_{n-2}^p + \cdots + b_{n-k}^p, $$ where $p$ is a real number with $0 < p < 1$.

Progress and Approach:

Given the concavity of the function $f(x) = x^p$ for $x > 0$ and $0 < p < 1$, we observe that $f(x) \leq x$ for $x \geq 1$. This suggests that each term in the sequence contributes progressively less as it grows larger, potentially indicating boundedness or stabilization of $\{b_n\}$.

Assuming the sequence might converge to a limit $L$, we can expect $L$ to satisfy the steady-state condition: $$ L = k L^p. $$ Solving this leads to: $$ L = k^{1/(1-p)}. $$ This suggests a fixed point, which we hypothesize might act as an attractor for the sequence.

To further validate this, I considered the function $f : [0, \infty)^k \to [0, \infty)$ defined by: $$ f(x_1, \ldots, x_k) = x_1^p + \ldots + x_k^p. $$ Given that the derivative $\frac{d}{dx}(x^p) = p x^{p-1}$ is less than 1 for all $x > 1$ and $p < 1$, I aimed to examine if $f$ acts as a contraction mapping under a suitable metric. Defining the metric as: $$ d((x_1, \ldots, x_j), (y_1, \ldots, y_j)) = \max |x_i - y_i|, $$ one needs to show that: $$ |f(x_1, \ldots, x_j) - f(y_1, \ldots, y_j)| \leq C \max |x_i - y_i|, $$ where $C < 1$. This would imply that $f$ gradually brings terms closer together, suggesting convergence towards the fixed point $L$.

Initially, setting $M = \max(x_1, \ldots, x_t)$, it leads to: $$ b_{t+1} \leq k M^p $$ and following the recurrence: $$ b_{t+2} \leq k \cdot (k M^p)^p = k^{1+p} M^{p^2} $$ indicating each term might be bounded by a decreasing sequence as $M^{p^n}$ approaches zero for large $n$, given $p^n$’s exponential decay.

Considering these, how can I rigorously prove the hypothesized convergence towards $L = k^{1/(1-p)}$, or assess how the sequence will converge/bound instead? I would greatly appreciate any insights, corrections, or suggestions on methods to rigorously confirm whether $\{b_n\}$ converges and to what limit.

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    $\begingroup$ A small note: you distinguish $t$ from $k$, but that doesn't really increase generality, since we must have $t \ge k$ for the sequence to be well-defined, and if $t > k$ then the first $t - k$ terms of the sequence have no bearing on the rest. $\endgroup$
    – crb233
    May 13 at 16:49
  • $\begingroup$ @crb233 That's correct, but I am afraid I don't understand what conclusions you end up with this. Does it mean $t$ is dependent on $k$ or what exactly? $\endgroup$
    – Snowball
    May 13 at 17:11
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    $\begingroup$ We can simplify the recurrent formula as follows. We assume that $t\ge k$ to define the sequence well, as was noted by crb233. Let $n>t$. Then $$b_n = b_{n-1}^p + b_{n-2}^p + \cdots + b_{n-k}^p,$$ $$b_{n+1} = b_{n}^p + b_{n-1}^p + \cdots + b_{n-k+1}^p=b_{n}^p+b_n-b_{n-k}^p.$$ $\endgroup$ May 13 at 19:55
  • $\begingroup$ @AlexRavsky Yes, that's true. But I am afraid I do not know to understand how to find a limit for this recurrence after having that. $\endgroup$
    – Snowball
    May 14 at 1:51
  • $\begingroup$ @crb233 If there is any progress you think deserves attention, I would be very glad to see them. $\endgroup$
    – Snowball
    May 14 at 12:52

1 Answer 1

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Following crb233's comment we assume that $t\ge k$. Put $K=k^{1/(1-p)}$, $$L_{-}=\liminf_{n\to\infty} b_n:=\lim_{n\to\infty}\inf_{k\ge n} b_k\mbox{ and } L_+=\limsup_{n\to\infty} b_n:=\lim_{n\to\infty}\sup_{k\ge n} b_k$$

Let $c_1>0$ be any number and $(c_n)_{n\in\mathbb N}$ be the sequence of real numbers such that $c_{n+1}=c_n^p$ for each natural $n\ge 2$. It is easy to see that $\lim_{n\to\infty} c_n=1$. Since the numbers $b_1,b_2,\dots,b_t$ are positive and for each natural $n>t$ we have $b_n\ge b_{n-1}^p$, we obtain that $L_-\ge 1$. Moreover, the recurrence for the sequence $(b_n)_{n\in\mathbb N}$ implies that $L_-\ge k(L_-)^p$, so $L_-\ge K$.

Now let $M=\max\{b_1,\dots,b_k,K\}$. Then $kM^p\le M$, so $b_n\le M$ for each natural $n$. Moreover, the recurrence for the sequence $(b_n)_{n\in\mathbb N}$ implies that $L_+\le k(L_+)^p$, so $L_+\le K$.

Thus $K=L_-=L_+=L=\lim_{n\to\infty} b_n$ which proves the desired answer.

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    $\begingroup$ Nice one, thanks for the detailed solution~ $\endgroup$
    – Snowball
    May 17 at 8:20

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