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Consider the following equation: $$x_{n+1} = ax_n + b$$

Under which circumstances is there a fixed point solution? Under which circumstances is there a period 2 solution?

So for the first question I just rewrote it to $x=ax+b$ and I got $ x = \dfrac{b}{1-a}$. So my answer was there exists a fixed point for $ a \neq 1$.

I did something similar for the second solution, and I ended up with $x = \dfrac{ab+a}{1-a^2}$, so there exists a period 2 solution for $ a \neq 1$ and $a \neq -1$. (and I believe $\dfrac{b}{1-a} \neq \dfrac{ab+a}{1-a^2}$, but I'm not sure and I wouldn't know how to write that concisely).

Is this correct, or am I missing something here which would make the problem more extensive?

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  • $\begingroup$ You don't just believe $\frac{b}{1 - a} \neq \frac{ab + a}{1 - a^2}$, compute it, which is not hard, and it gives you an extra condition. $\endgroup$ Jun 15, 2013 at 21:11
  • $\begingroup$ Of course, there are many fixed points if $a=1$, $b=0$ and many period-2-points if $a=-1$. To investigate the inequality, multiply both sides with $a^2-1=(a+1)(a-1)$ and see what gives. $\endgroup$ Jun 15, 2013 at 21:11
  • $\begingroup$ @HagenvonEitzen I don't see your 'of course' as straightforward, and I also don't understand your suggestion. Perhaps you can put it in an answer? $\endgroup$
    – iEvenLift
    Jun 15, 2013 at 21:21
  • $\begingroup$ @iEvenLift it's $b\neq a$. $\endgroup$ Jun 15, 2013 at 21:30
  • $\begingroup$ @SecretMath Indeed, I made a tiny mistake. Does this complete the exercise or are there still gaps (as Hagen suggests)? $\endgroup$
    – iEvenLift
    Jun 15, 2013 at 21:33

2 Answers 2

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A fixed point solution exists iff

$$ x = ax + b, $$

has a solution.

If $a \neq 1$, the fixed point solution is given by

$$ x = \frac{b}{1 - a}. $$

If $a = 1$, $x = x + b$ has a solution iff $b = 0$, in which case you have infinitely many solution.

Therefore, the conclusion is $a \neq 1$ or $a = 1, b = 0$.


A period 2 solution exists iff

$$ x \neq ax + b, x = a^2x + b(a + 1), $$

has a solution.

If $a \neq \pm 1$, then

$$ x = \frac{b}{1 - a} $$

which is a fixed point solution.

If $a = 1$, then for a solution to exist $b = 0$, but then the solution is a fixed point solution.

If $a = -1$, then for a solution to exists $b$ can be anything, in which case you have infinitely many solution.

Therefore, the conclusion is $a = -1$.

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1) Under what circumstances does there exist $x\in \mathbb R$ such that $x=ax+b$?

If $a=1$, this just yields $x=x+b$, i.e. either $b=0$ and all $x\in\mathbb R$ are fixed points, or $b\ne 0$ and there are no fixed points. If $a\ne 1$, then $x$ is a fixed point iff $x=\frac b{1-a}$.

2) Under what circumstances does there exist $x\in \mathbb R$ such that $x=a(ax+b)+b$, but $x\ne ax+b$?

The equation simplifies to $(1-a^2)x=(a+1)b$. If $a\ne-1$, we can divide both sides by $a+1$ and obtain $(1-a)x=b$, which is equivalent to $x=ax+b$, i.e. $x$ is already a fixed point. In the remaining case $a=-1$, we simply get $0=0$, i.e. all points have period $2$ (unless they are fixed points). Using the result from (1), we conclude that all $x\ne \frac b2$ have period 2.

In summary: Fixed points exist iff $(a,b)=(1,0)$ or $a\ne 1$. Period-2 points exist iff $a=-1$.

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