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.
Best Answer
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.