# Find all possible values of $a, p$ for which $x_{n+1} = x_n^2 + (1-2p)x_n + p^2$ converges, given $x_1 = a$ and $n\in\Bbb N$

Given a recurrence relation $$x_{n+1} = x_n^2 + (1-2p)x_n + p^2\\ x_1 = a\\ n\in\Bbb N$$ find all values of $$a, p \in\Bbb R$$ for which $$x_n$$ converges.

First of all I tried to assume that the limit indeed exists. Then it must match a fixed point of the following equation: $$x = x^2 + (1-2p)x + p^2$$ Which solving for $$x$$ gives: $$(x-p)^2 = 0 \iff x = p$$

By this if the recurrence converges then it must follow that: $$\lim_{n\to\infty}x_n = p$$

Lets now take a closer look at the recurrence. I've shown that it must describe a monotonically increasing sequence, namely: $$x_{n+1} = x_n^2 + (1-2p)x_n + p^2 \\ x_{n+1} = x_n^2 - 2px_n + p^2 + x_n \\ x_{n+1} = (x_n - p)^2 + x_n \\ x_{n+1} - x_n = (x_n - p)^2 > 0 \implies \boxed{x_{n+1} > x_n}$$

By this the sequence in monotonically increasing no matter what initial conditions are given.

At this point I'm lost. How do one deduce the constraints for $$a, p$$ for $$x_n$$ to be convergent. Since the sequence is increasing I guess the problem may be reduced to "find the values of $$a, p$$ for which the sequence is bounded". Then the result should follow by monotone convergence theorem. The answer section suggests that: $$0 \le p - a \le 1$$

Which seems true based on the Cobweb plot. But a plot in not a formal proof. Unfortunately I was not able to infer that result. What is a proper way to finish the problem?

Please note this problem is given in the limits section. So even derivatives are not available.

• Perhaps this idea helps (it does not fit to 100%): We would like to show that $lim_{n \to \infty} x_{n+1} - x_{n} = 0$. We know that $lim_{n \to \infty} a^n = 0$ iff $|a| \leq 1$. Hence it is necessary that $|x_n - p| \leq 1$ for all $n$, better $|a-p| \leq 1$ since $x_0 = a$. This leeds to $-1 \leq p-a \leq 1$. This is not what you want, so I made a mistake somewhere, but perhaps it helps ;) – Sqyuli Feb 25 at 16:35

A first simplification is to define $$y_n=x_n-p$$. Then as you already proved $$y_{n+1} -y_n=y_n^2\\ y_1 = a-p\\ n\in\Bbb N$$

the problem is now to find conditions on $$y_1$$ such that $$y_n \to 0$$, i.e to study the recurrence relation $$y_{n+1}=f(y_n)$$ with the fixed (independent of $$a$$ and $$p$$) function $$f(x)=x+x^2$$.

As $$y_{n+1} -y_n=y_n^2$$ the sequence is increasing, and the only possible limits are $$-1$$ and $$0$$ so

• if $$y_1 >0$$ then for all $$n$$, $$y_n>0$$ and the series does not converge.
• if $$y_1 <-1$$ then $$y_2>0$$ and once more for all $$n$$, $$y_n>0$$ and the series does not converge.
• if $$-1 \leq y_1 \leq 0$$ you can show by recurrence that for all $$n$$ $$-1 \leq y_n \leq 0$$, so $$y_n$$ is a bounded monotonous function and thus converges.

You finally obtain the condition $$-1 \leq y_1 \leq 0$$ i.e $$-1 \leq a-p \leq 0$$

Instead of viewing it as a map on $$x_n$$, try examining what happens to $$x_n-p$$.

From the second to last line of your display, it's easy to work out that $$(x_{n+1}-p) = (x_n-p)^2 + (x_n-p).$$

So the sequence $$b_n=x_n-p$$ (which converges exactly when $$x_n$$ does) is iterating under them map $$t\mapsto t^2+t$$. Now you have one function to understand not a class. It sounds like you have the tools necessary to verify this converges if the initial value $$b_1$$ is between -1 and 0.