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 $n\in\Bbb N$

Question

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.

Answer 1

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 $f(x)=x+x^2$.

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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 $$-1 \leq a-p \leq 0$$

Answer 2

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.