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.

  • $\begingroup$ 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 ;) $\endgroup$ – 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.


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.