Lagmental Vicfred

The Logistic Map Changes Stability at r Equals Three by Vicfred

The nonzero fixed point of r x times one minus x loses stability when its multiplier crosses minus one. The example is deliberately concrete; it is a test of the statement, not a substitute for it.

The data

A discrete dynamical system iterates \(x_{n+1}=F(x_n)\), while a flow solves \(\dot x=V(x)\). Fixed points, periodic orbits, and invariant sets organize long-term behavior.

$$ F_r(x)=rx(1-x),\qquad x^\ast=1-\frac1r $$

The definition determines which expressions are legal; only then does the identity become meaningful. An equality in \(\mathcal A\) may change ambient meaning, so I keep \(\mathsf D\) separate from \(\mathsf C\).

$$ F_r'(x^\ast)=2-r $$

Derivation

The middle display is intentionally dense: it is where signs, bounds, multiplicities, or normalising factors are most likely to be lost.

$$ \begin{cases}1<r<3&\Longrightarrow|2-r|<1\ \text{and }x^\ast\text{ attracts},\\r>3&\Longrightarrow|2-r|>1\ \text{and }x^\ast\text{ repels initially}.\end{cases} $$

Invariant content

The invariant statement is the one that does not depend on a convenient choice of coordinates, representatives, basis, or enumeration.

$$ \begin{aligned} \mathsf{D}\;&:\quad F_r(x)=rx(1-x),\qquad x^\ast=1-\frac1r,\\[5pt] \mathsf{C}\;&:\quad F_r'(x^\ast)=2-r. \end{aligned} $$

Scope

Sensitive dependence is not the same as randomness. A deterministic system may be chaotic while remaining exactly specified by its initial condition.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] F_r'(x^\ast)=2-r \end{gathered}} $$

The final box is a summary, not a new assumption; the proof still lives in the definitions and the intervening calculation. The source keeps each scope delimiter visible for later inspection.

This article was posted on Sun 10 March 2019. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.