2
$\begingroup$

I have a recurrence of the form

$u_0=0, u_1=50, u_n=-a_{n-1}+u_{n-1}+50$, where $a_{n-1}$ is a probabilistic amount which I can not describe in a simple formula. But I can set bounds for the original recurrence in the following way:

$u_n\le-0.001 u_{n-1}^2+u_{n-1}+50$, and

$u_n\ge-0.001 u_{n-1}^2+0.9 u_{n-1}+45$

Plotting the two functions which correspond to the right side of each inequality; i.e.

$f_1(x)=-0.001 x^2+x+50$ and

$f_2(x)=-0.001 x^2+0.9 x+45$

figure

I get the attached figure, which shows $f_1$ in blue and $f_2$ in red, and function $f_3(x)=x$ in black.

Now, my question is:

Say that $f_1$ intersects with $f_3$ in $(x_1,y_1)$, and $f_2$ intersects with $f_3$ in $(x_2,y_2)$

Is it correct to say that (based on Banach fixed point theorem) $u_n$ is convergent, and its limit $l$ satisfies $y_2 \le l \le y_1$ ?

Update:

recently, I was able to find another lower bound of the form $u_n \ge f_3(u_{n-1}) = 0.001 u_{n-1}^2+0.005 u_{n-1}+50$

Now upper and lower bounds intersect like the attached shape Does that change my problem?

$\endgroup$
  • $\begingroup$ Welcome to Math.SE! I have tried to improve the readability of your question by improving the $\rm \LaTeX$ code of the question. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. I've edited away the $\times$ sign since it can be confused with the variable $x$ in your question. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 27 at 15:51
  • $\begingroup$ @gnusupporter-8964民主女神-地下教會 Thank you very much. $\endgroup$ – Angie Apr 27 at 16:10
1
$\begingroup$

Without additional restrictions imposed on the sequence $\{a_n\}$, the sequence $\{u_n\}$ is not necessarily convergent. For instance, a sequence $0, 50, 90, 125, 150, 170, 171, 170, 171,\dots $ satisfies $f_2(u_{n-1})\le u_n \le f_1(u_{n-1})$ for each $n\ge 1$, but does not converge.

Let’s investigate the behavior of the sequence $\{u_n\}$. Condition $f_3(u_n)\le f_1(u_n)$ implies $0\le u_n\le 497.5$. We have $$[y_2, y_1]=[50\sqrt{19}-50,100\sqrt{5}]~\simeq [168, 224].$$ If $u_n\le y_1$ then $f_1(u_n)<y_1$ so $u_m<y_1$ for all $m\ge n$. If $u_n<y_2$ then $u_{n+1}>u_n$, so $u_{m+1}>u_{m}$ for all $m\ge n$ or $u_m\ge y_2$ for some $m\ge n$. In both cases $u_m<y_1$ for all $m\ge n$. In the first case, moreover, $\{u_m:m\ge n\}$ is a monotonic bounded sequence, which, therefore, has a limit $u\le y_2$. If $u<y_2$ then $f_2(u)>u$ so, by continuity of the function $f_2$, there exists $n$ such that $u_{n+1}\ge f_2(u_n)>u$, a contradiction. So $u=y_2$.

$\endgroup$
  • $\begingroup$ Thank you very much, and sorry for my weak mathematics background, recently I was able to find another lower bound of the form $u_n \ge f_3(u_{n-1}) = 0.001 u_{n-1}^2+0.005 u_{n-1}+50$, now upper bound and lower bound intersect in one point, does that change anything? $\endgroup$ – Angie May 2 at 8:12
  • $\begingroup$ @Angie Now the situation looks even worse because the graph suggests that the domain $D$ between the upper and lower bound is bigger. And if $D$ contains a square $S$ with two opposite vertices (s,s) and $t$ placed at the line $x=y$ then we can reach in a sequence $s\le u_n\le t$ for some $n$ then we can continue it with arbitrary values between $s$ and $t$. But if our case even this simple observation is not needed because the sequence from the answer still satisfies the given conditions but does not converge. $\endgroup$ – Alex Ravsky May 2 at 20:31
  • $\begingroup$ Thanks a lot, I choose the second lower boundary which intersects with the upper one, just to say that $u_n$ is surely bounded. but can I describe the behavior of $u_n$ relative to the intersection points with y=x? In order to understand the direction of $u-n$ I computed $u_{n+1}-u_n$ where $f_2(u_n)-u_n \le u_{n+1}-u_n \le f_1(u_n)-u_n$ this amount will change its sign outside the area bounded by intersection points, so, is it correct to say that for some n $y_2 \le u_n \le y_1$? $\endgroup$ – Angie May 3 at 6:49
  • $\begingroup$ @ Almost correct. I updated my answer. $\endgroup$ – Alex Ravsky May 4 at 7:24
  • $\begingroup$ I couldn't figure out why $\{u_m : m \ge n\}$ is monotonic , It is a key point, but not clear for me. $\endgroup$ – Angie May 4 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.