Fixed Points of Iterated Maps

Take any number, repeatedly apply the same function to it — square it, average it with something, whatever — and watch where it goes. Some starting points settle into a stable value, some run off to infinity, some bounce around forever. The single most useful question to ask about any such process is: what are its fixed points, the values $x^*$ that satisfy $f(x^*) = x^*$, and does nearby stuff move toward them or away?

Finding fixed points

A fixed point of $f$ is a value the function maps to itself. Algebraically, you just solve $f(x) = x$. For $f(x) = x^2 - 2$, this means solving $x^2 - 2 = x$, or $x^2 - x - 2 = 0$, which factors as $(x-2)(x+1) = 0$, giving two fixed points: $x^* = 2$ and $x^* = -1$. In plain English: if you start the iteration exactly at $x=2$, applying $f$ gives $2^2 - 2 = 2$ again — you're stuck there forever. Same for $x=-1$: $(-1)^2 - 2 = -1$.

Stability: does nearby stuff get pulled in or pushed out

Finding a fixed point tells you where the process can rest — not whether it will actually end up there if you start nearby instead of exactly on it. That's a question about the slope of $f$ at the fixed point. If $|f'(x^*)| < 1$, the map shrinks distances near $x^*$, so nearby points get pulled closer with every iteration — the fixed point is attracting. If $|f'(x^*)| > 1$, the map stretches distances, pushing nearby points away — the fixed point is repelling, reachable only if you land on it exactly.

Worked example: classifying both fixed points of $x^2-2$

$f'(x) = 2x$, so $f'(2) = 4$ and $f'(-1) = -2$. Both have magnitude greater than 1, so both fixed points are repelling — this particular map has no stable rest point at all for real starting values in general position, and iterating it from almost any starting value sends you bouncing away rather than settling down (in fact this specific map is chaotic on the interval $[-2,2]$). Contrast this with $f(x) = \tfrac12(x + 2/x)$ from Newton's method for $\sqrt2$: its fixed point at $x^*=\sqrt2$ has $f'(\sqrt2) = 0$, strongly attracting, which is exactly why that iteration converges so fast.

Worked example: a linear map you can check by hand

Let $f(x) = 0.5x + 3$. Solve $x = 0.5x + 3 \Rightarrow 0.5x = 3 \Rightarrow x^* = 6$. Here $f'(x) = 0.5$ everywhere, so $|f'(x^*)| = 0.5 < 1$: attracting. Starting at $x_0 = 0$: $x_1 = 3$, $x_2 = 4.5$, $x_3 = 5.25$, $x_4 = 5.625$ — visibly closing in on 6, each gap exactly half the previous gap, matching the slope of 0.5 precisely.

Drag the curve's parameters and compare it against the diagonal line $y=x$ (imagine it overlaid): every point where the curve crosses that diagonal is a fixed point, and the curve's steepness there — flatter or steeper than the diagonal — tells you whether it attracts or repels.

What this means in practice

Fixed-point thinking is the backbone of any interview question about repeated processes: does compounding a rule converge, does a pricing iteration stabilize, does a feedback loop in a trading system settle down or blow up. The two-step habit — solve $f(x^*)=x^*$, then check $|f'(x^*)|$ against 1 — answers "what happens if I do this forever" without ever running the process for more than a couple of terms by hand.