[motivation]
### Linear Stability: What It Achieves
The simplest stability criterion for a fixed point $p$ of a smooth map $f$ is *linear*: compute the multiplier $\lambda = f'(p)$ and check whether $|\lambda| < 1$ (stable) or $|\lambda| > 1$ (unstable). The logic is that near $p$, the map is well-approximated by its linearisation $x \mapsto p + \lambda(x - p)$, which contracts distances when $|\lambda| < 1$ and expands them when $|\lambda| > 1$. For the vast majority of fixed points — those with $|\lambda| \neq 1$ — this is the complete story, and no further analysis is needed.
### The Failure at the Critical [Boundary](/page/Boundary)
At $|\lambda| = 1$, the linear approximation preserves distances: $|f(x) - p| \approx |x - p|$. The linearisation is neutral, and it cannot distinguish between attraction, repulsion, or any intermediate behaviour. Yet the actual nonlinear map $f$ need not be neutral — the higher-order terms that the linearisation discards are precisely the terms that decide the outcome.
The two concrete cases are $\lambda = +1$ and $\lambda = -1$. When $\lambda = +1$, the linearisation is the identity $x \mapsto x$, so the displacement $f(x) - x$ is entirely governed by the nonlinear remainder. When $\lambda = -1$, the linearisation flips each orbit point to the opposite side of $p$, creating an oscillation $x_0, f(x_0), x_2 \approx x_0, \ldots$ whose amplitude is again controlled by the nonlinear terms.
### Two Maps, Same Multiplier, Different Stability
To see that higher-order information is essential, consider two maps with the same critical multiplier $\lambda = +1$ at $p = 0$. The map $f(x) = x + x^2$ satisfies $f'(0) = 1$ and $f''(0) = 2 > 0$. Near $x = 0$, the displacement $f(x) - x = x^2 > 0$ for all $x \neq 0$, so every orbit moves to the right: points to the left of $0$ are pushed toward $0$, while points to the right are pushed away. The fixed point is *semi-stable* — attracting from one side and repelling from the other.
Now consider $g(x) = x - x^3$, which also satisfies $g'(0) = 1$ but has $g''(0) = 0$ and $g'''(0) = -6 < 0$. Here $g(x) - x = -x^3$, which is positive for $x < 0$ and negative for $x > 0$. Points on both sides are pushed toward $0$, and the fixed point is *asymptotically stable*.
The same multiplier produces qualitatively opposite outcomes. The resolution lies in the first nonvanishing higher-order derivative: the sign of $f''(p)$ when it is nonzero, and the sign of $f'''(p)$ when $f''(p) = 0$. Developing this idea rigorously — and extending it to the oscillatory case $\lambda = -1$ via the Schwarzian derivative — is the purpose of this page.
[/motivation]