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Entrainment of circadian rhythm synchronizes a circadian clock to a repeating cue, a zeitgeber. One-to-one entrainment of circadian rhythm gives one cycle per environmental cycle with a stable relative circadian phase.^[ref] The stable relation need not put an activity onset exactly at lights-off: a constant offset also expresses phase locking.

For an original local model, let $x_n$ be the activity-onset offset from lights-off on environmental cycle $n$, with a negative value meaning an earlier onset. Choose an intrinsic period $\tau =23.5\mathrm{h}$ and environmental period $T=24\mathrm{h}$. Let $D(x)$ be a stipulated daily delay, measured in hours; positive $D$ moves the next onset later. The recurrence relation isA fixed phase relationship $x_{\ast }$ requires $D(x_{\ast })=T-\tau$: here a half-hour delay balances the half-hour intrinsic advance each cycle. This equality alone describes balance; restoration after a small disturbance also requires fixed point stability for an iteration.

Choose the illustrative local law $D(x)=0.5\mathrm{h}-0.4x$. Thenfor the chosen starting offset $x_0=-3\mathrm{h}$. The error shrinks by a factor of $0.6$ each cycle, proving fixed point stability for an iteration in this model. After five cycles $x_5=-0.23328\mathrm{h}$, about fourteen minutes early. At the fixed point, the imposed delay is $0.5\mathrm{h}$ and the corrected onset interval is $\tau +D(x_{\ast })=24\mathrm{h}$, although the uncoupled period remains $23.5\mathrm{h}$.

The uncorrected comparison has $D=0$ and $x_n=-3\mathrm{h}+n(\tau -T)$, so after five cycles it is $5.5$ hours early. The two traces separate ongoing timekeeping from adjustment to a zeitgeber. The chosen linear delay law is only a local teaching construction: it is not a measured response to light, does not account for wrapping of circadian phase, and does not predict an organism's entrainment range. Nor does it include masking of circadian rhythm. In a real study, tracking several cycles after removal of the cue or after a timed perturbation supplies information that a single activity value under illumination cannot provide.

The Erlang fixed point approximation treats resource blocking as independent and applies the Erlang loss formula to traffic screened by other resources. Its self-consistency equations have a unique fixed point under fixed routing, but can have several under alternative routing.

For a Galton-Watson process with one ancestor, the branching extinction probability is the probability that some generation is empty:The equality uses the absorbing state at zero. The complementary branching survival probability $1-q$ is the probability that every generation is nonempty. The Galton-Watson extinction fixed point computes $q$ from the offspring distribution.

For $p_0=p_1=1/4$, $p_2=1/2$, one ancestor has $q=1/2$. Starting instead from $m$ ancestors with independent descendant families givesAll $m$ families must die out. With three ancestors, eventual extinction has probability $1/8$ and survival has probability $7/8$; by generation two, extinction has probability $(11/32{)}^3=1331/32768\approx 0.040619$. Extra ancestors change the extinction probability without changing the offspring distribution. ^[ref]

For finite offspring mean $\mu$, the extinction classification is:

The critical exception is $p_1=1$: every individual has exactly one child, $F(s)=s$, and $Z_n=1$ forever from one ancestor. Thus $\mu =1$ but $q=0$; a blanket claim that every critical Galton-Watson process becomes extinct would be false. These cases concern the Galton-Watson process with its common offspring distribution and independence assumptions. ^[ref]

This probability answers whether any lineage persists, which the offspring mean alone cannot quantify in the supercritical case. A branching process conditioned on extinction uses the same fixed point to reweight reproduction and describe families known eventually to die out.

A Feigenbaum fixed point is an even analytic unimodal interval map $g$ with $g(0)=1$, a critical maximum of even order $d$, $a=g(1)\in (-1,0)$ andIn the quadratic class $g^{\prime \prime }(0)<0$. Its spatial scaling uses the Feigenbaum constants convention $\alpha =-1/a$; the values depend on the critical order. The central restrictive intervals are successively reduced by $|a|$ under renormalization. This fixed point describes the limiting shape of repeatedly renormalized maps at a period-doubling accumulation.

// Target: dynamical-systems.bigb

For a uniform random permutation $\pi$ of $n\ge 1$ objects, let $M_n$ count its fixed points. Write $M_n={\sum }_{i=1}^n{1}_{\{\pi (i)=i\}}$. Each indicator random variable has expected value $1/n$, so linearity of expectation gives $\mathbb{E}[M_n]=1$ for every $n\ge 1$. This average does not require independence: for $n\ge 2$, fixing two specified objects has probability $1/[n(n-1)]$, greater than the product $1/n^2$ of their individual probabilities.

For four objects, exact enumeration gives counts $9,8,6,0,1$ for $0,1,2,3,4$ fixed points among the $24$ equally likely permutations. Their weighted arithmetic mean is $(0\cdot 9+1\cdot 8+2\cdot 6+4\cdot 1)/24=1$. Exactly three fixed points are impossible: the last object must then also be fixed.

Choose the $m$ fixed objects and derange the remaining $n-m$. Their count is $\left(\genfrac{}{}{0ex}{}{n}{m}\right)D_{n-m}$ among $n!$ equally likely permutations. The inclusion-exclusion principle for $D_{n-m}$ proves the displayed law, valid for $0\le m\le n$ and zero otherwise. With $D_0=1$, it includes the all-fixed case and the impossibility of exactly $n-1$ fixed objects. For fixed $m$, the law tends to $e^{-1}/m!$, the Poisson distribution of mean one.

A finite group of Euclidean isometries has a common fixed point: average any orbit. Each isometry is an affine map and therefore preserves this finite affine average; applying a group element merely permutes the orbit terms. To see that an isometry is affine, subtract its value at zero and use the polarization identity to show that the resulting map preserves inner products. Images of a standard orthonormal basis then determine a linear orthogonal map.

For a strictly increasing ordinal class function continuous at limits, start with ${\alpha }_0=0$ and iterate ${\alpha }_{n+1}=f({\alpha }_n)$. If the sequence stabilizes, its stable value is a fixed point. Otherwise its supremum $\alpha$ is a limit and continuity gives $f(\alpha )={\sup }_{n}f({\alpha }_n)={\sup }_n{\alpha }_{n+1}=\alpha$. Starting the iteration above any prescribed ordinal gives arbitrarily large fixed points.

For two contraction mappings of a nonempty complete metric space, their fixed points obey $d(x_f,x_g)\le q_{f}d(x_f,x_g)+d_{\infty }(f,g)$. Rearranging proves the displayed bound whenever the uniform metric is finite. It establishes continuity of the fixed-point map at each $f$ without assuming a contraction constant shared by all maps near $f$.

To solve $\Phi (p)=p$, let $J$ be a bounded approximation to $(D\Phi (p_0)-I{)}^{-1}$ and set $A(p)=p-J(\Phi (p)-p)$. Its derivative is $DA(p)=I-J(D\Phi (p)-I)$. Every fixed point of $\Phi$ is fixed by $A$; the converse holds when $J$ is injective. A good inverse approximation makes $A$ a contraction mapping even if $\Phi$ itself has an expanding direction.

// Target: analysis.bigb

For a Galton-Watson process with one ancestor and offspring probability generating function $F$, the branching extinction probability $q$ is the smallest fixed point of $F$ in $[0,1]$. A direct way to find it is to compute extinction by generation $n$:Zero is an absorbing state, so extinction by generation $n$ is exactly the event $Z_n=0$. ^[ref]

For $F(s)=1/4+s/4+s^2/2$, the equation isBoth $1/2$ and $1$ are fixed points, but iteration from zero selects $q=1/2$. The first values are $q_1=1/4$, $q_2=11/32$, $q_3=809/2048\approx 0.395020$, and $q_6\approx 0.462186$. Iteration started at $1$ would stay at $1$ forever and would give the wrong extinction answer for this offspring distribution.

The vertical steps in the figure apply the offspring probability generating function; horizontal steps return to the diagonal to start the next iteration. This gives a useful numerical calculation even when solving $F(s)=s$ explicitly is difficult. For the displayed law, every finite iteration is a lower bound for the eventual branching extinction probability; at generation six, about $0.037814$ additional extinction probability remains beyond the current empty families.

Why the smallest fixed point? The ancestor's children's descendant families are independent; given $k$ children, all are empty $n$ generations later with probability $q_n^k$. Averaging gives $q_{n+1}={\sum }_k{p}_k{q}_n^k=F(q_n)$. The increasing extinction events give the limit $q$, and continuity of the probability generating function yields $F(q)=q$. For any other fixed point $r\in [0,1]$, $q_0\le r$ and monotonicity of $F$ imply $q_n\le r$ at every step; hence $q\le r$. ^[ref]

At the normalized quadratic fixed point $g$, one expanding eigenvalue $\delta >1$ of $D\mathcal{T}(g)$ and a contracting complement give a stable manifold of finite codimension one. A generic one-parameter family crosses that manifold transversely at its cascade accumulation. Its unstable coordinate is initially proportional to parameter distance and multiplies by $\delta$ per renormalization, yielding $s_{\infty }-s_n\sim C{\delta }^{-n}$. Stable components disappear, while the fixed point's orientation-reversing scale $a$ yields signed spatial ratios $a^{-1}=-\alpha$.

// Target: dynamical-systems.bigb