# Find the fixed points

We have $x_{n+1} = ax_n +b$ with $x_0$ given. We have to find the fixed points of this function, and decide for which values of $a$ they are stable.

So I looked it up and found that a fixed point is a point for which $f(x) = x$, so basically intersections of a certain function with the line $y=x$.

My question is; How can we use this for the above function? Does $x_{n+1}$ have to equal $x_n$? And how would we able to find these fixed points?

Your system needs to remain at the same point after an additional iteration. I.e. $x_{n}=x_{n+1}$.

$x_{n+1} = ax_{n}+b$ hence becomes $x_{n}=ax_{n}+b$, which yields $x_{n}-ax_{n}=b$
then
$x_{n}(1-a)=b$
and finally
$x_{n}=\frac{b}{1-a}$.

For $b=0$ we get $x_{n+1} = ax_n$, which is feasible for any $a$, with the fixed point coming to lie at 0. Otherwise this is feasible for any $a\neq 1$ (which would make the system $x_{n+1}=x_{n}+b$, which makes visible how b will for ever increase the system value).

This also explains the result, which means in simple terms that at the fixed point the factor a diminishes the value of x as much as b adds to it.

Again, this relates to the stability. A BIBO stable system will remain bounded on a bounded input b (or returns to zero for asymptotic stability). This is achieved by limiting the eigenvalues of the discrete system (here a) to be strictly smaller than 1 in magnitude: $|a|<1$.

• Its feasible if $b=0$ and any $a$, or if $b\neq0$ and $a\neq 1$
Commented Sep 6, 2014 at 7:37
• Thank you @Adam, I've added the case $b=0$ to the solution. Commented Dec 7, 2014 at 14:53

Fixed points are where the $n+1$th step is the same as the $n$th step, ie $x=ax+b$.

• So $\dfrac{b}{1-a}$? Commented Jun 15, 2013 at 15:23
• Stable for $|a|<1$. Commented Jun 15, 2013 at 15:24

$$x_{n}=ax_{n}+b$$

$$(1-a)x_n=b$$

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

It's worth noting that in general, by the fundamental theorem of algebra, there will be exactly $d$ fixed points in a $d$ degree polynomial (some of them may be complex).