# Approximate the limit of quadratic recurrence by applying Banach fixed-point theorem to its bounds

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

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

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 Does that change my problem?

• 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. – GNUSupporter 8964民主女神 地下教會 Apr 27 '19 at 15:51
• @gnusupporter-8964民主女神-地下教會 Thank you very much. – Angie Apr 27 '19 at 16:10

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 $$f_1(u_n) so $$u_m for all $$m\ge n$$. If $$u_n then $$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 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 $$u then $$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$$.
• 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? – Angie May 2 '19 at 8:12
• @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. – Alex Ravsky May 2 '19 at 20:31
• 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 $y_2 \le u_n \le y_1$? – Angie May 3 '19 at 6:49
• I couldn't figure out why $\{u_m : m \ge n\}$ is monotonic , It is a key point, but not clear for me. – Angie May 4 '19 at 10:11