Model the four land regions by vertices $A,B,C,D$ and the seven bridges by distinct edges: two $AB$, two $AC$, and one each of $AD,BD,CD$. Their vertex degrees are $(5,3,3,3)$. Leonhard Euler's paper records these connections in §6 and the obstruction in §9; §20 states the zero-or-two odd-region rule. Four odd vertex degrees exclude an Eulerian trail, even if returning to the starting vertex is optional.^[ref]

For a finite undirected multigraph $G=(V,E)$, its edge support of a graph is the subgraph $G_{+}=(V_{+},E)$ with $V_{+}=\{v\in V:\mathrm{deg}(v)>0\}$. It removes exactly the isolated vertices, retaining every edge, including self-loops and parallel edges. When $E\ne \varnothing$, connectivity of this subgraph is the relevant condition for an Eulerian trail or Euler circuit.

An inviscid core is the part of a high-Reynolds number flow outside thin wall or shear layers, where viscosity is negligible at the retained asymptotic order. It obeys the Euler equations for an inviscid fluid. Its tangential velocity need not satisfy no slip at a material wall; adjoining viscous boundary layers supply that condition and can influence the core through displacement and pressure matching.

A clamped filament touching a laterally sliding, moment-free wall has fixed projected gap $d$. For an inextensible small-amplitude elastica, parameterize its tip angle by $k=\sin ({\theta }_{\mathrm{tip}}/2)$. Then $\ell =\sqrt{B/F}K(k)$ and $d=\sqrt{B/F}[2E(k)-K(k)]$, where $K,E$ are complete elliptic integrals. Their small-$k$ expansions give the displayed decrease of wall force with excess contour length. It concerns the weakly buckled branch, before overhang or additional wall contact changes the boundary conditions.

A connected open subset of the Euclidean plane. For a bounded polygonal example, the Dirichlet Laplacian is the self-adjoint operator associated with the Dirichlet energy on $H_0^1(\Omega )$; its eigenfunctions are smooth in the interior but need not be smooth at reentrant boundary corners.

A complex lattice is a discrete rank-two subgroup of the complex plane, generated by two real-linearly independent periods. An oriented basis can be chosen with $\mathrm{Im}({\omega }_1/{\omega }_2)>0$. Scaling by ${\omega }_2^{-1}$ then gives $\mathbb{Z}\tau +\mathbb{Z}$ for a point in the complex upper half-plane.

For an integer basis of a Euclidean lattice, the algorithm uses integer column operations and swaps to obtain a size-reduced basis with Gram-Schmidt coefficients $|{\mu }_{ij}|\le 1/2$ and the condition $\delta \Vert b_i^{\ast }{\Vert }^2\le \Vert b_{i+1}^{\ast }{\Vert }^2+{\mu }_{i+1,i}^2\Vert b_i^{\ast }{\Vert }^2$. It preserves the lattice and exposes short vectors and near-relations. It is useful for improving proven Diophantine bounds by encoding scaled logarithms in integer coordinates. A certified lower bound for every nonzero lattice vector is ${\min }_i\Vert b_i^{\ast }\Vert$: in a nonzero integer linear combination, project onto the last nonzero Gram-Schmidt direction. Better-reduced bases improve this lower bound. Rounding and logarithm errors must still be bounded before turning it into a certified lower bound for a logarithmic form.

Repeated division with remainder that determines successive digits of a quotient. To generate the fractional digits in base $b$, multiply the current remainder by $b$, divide by the denominator, and keep the resulting remainder for the next step.

For an integer $a$ and positive integer $b$, the remainder in Euclidean division is the unique integer $r$ in $0\le r<b$ for which $a=bq+r$ with an integer $q$. It is not the full quotient $a/b$.

A proposed optical bench test uses a steady illuminated diffuse patch, a relay made from two optical lens elements, interchangeable centred circular stops, and a calibrated optical power meter large enough to collect the whole output bundle. Start with small cone angles so an ordinary relay approximates the area-angle tradeoff. Measure the patch's image diameter and the accepted angular extent at both conjugate planes; the transverse magnification gives the spatial scale, but does not by itself measure etendue.

Map the local radiance with a baffled photodiode on translation and rotation stages, using two small aligned holes to select a narrow known solid angle. Keep the sampled position fixed while scanning direction, calibrate the sampling footprint and angular response, and repeat at several positions. Numerically integrate $n^2\cos \theta dAd\Omega$ over the occupied samples, and integrate $(L/n^2)dG$ to predict the collected radiant flux. This checks the filled-cone assumption instead of inferring it from two outer dimensions.

Compare the relay without a stop and with a stop of $0.6$ times the illuminated diameter in the same plane. The uniform-bundle prediction is a radiant flux ratio of $0.36$, with unchanged local radiance in the surviving samples. Cover the input for a dark reading, monitor source drift, repeat the open-stop reading, check detector linearity with calibrated optical attenuators, and enlarge the collecting detector or scan it to check for missed light rays. Lens losses, angular response, illumination gradients, alignment errors and diffraction can spoil the ideal prediction; their effects must be measured separately. This is a procedure to try, with predicted readouts, not purported experimental data.

An optical concentrator redirects light from an entrance area $A_{\mathrm{in}}$ to a smaller receiver area $A_{\mathrm{out}}$. Its geometric concentration ratio is $C=A_{\mathrm{in}}/A_{\mathrm{out}}$. Specify the admitted angular distribution and losses before interpreting $C$ as an irradiance gain: if a fraction $\eta$ of incident radiant flux reaches the receiver, the ratio of mean receiver to mean entrance irradiance is $\eta C$.

Insert a centred opaque circular stop of diameter $d=6\mathrm{mm}$ directly in the plane of the $D=10\mathrm{mm}$ bundle. It transmits the same $30^{\circ }$ directions at every surviving position, but discards the surrounding annulus. ThereforeWith the uniform radiance of etendue of a circular aperture and cone, the transmitted radiant flux is $2.2207\mathrm{mW}$ and the blocked fraction is $64\%$. Surviving light rays retain their radiance; the stop has not concentrated them. Lossless propagation after the stop preserves the etendue of this surviving subset, not that of the original incident bundle. A stop placed elsewhere can select positions and directions together, so its physical area ratio alone need not give the transmitted radiant flux fraction.

When both endpoint bundles have the filled circular-cone geometry of etendue of a circular aperture and cone, conservation of etendue givesFor the same refractive index, quadrupling the area halves the sine of the cone half-angle. In the $10\mathrm{mm}$, $30^{\circ }$ example, an ideal output of diameter $20\mathrm{mm}$ must have $\sin {\theta }_2=1/4$, so ${\theta }_2\approx 14.4775^{\circ }$. The exact result differs from simply halving $30^{\circ }$. Conversely, reducing the diameter to $5\mathrm{mm}$ requires a complete forward hemisphere, ${\theta }_2=90^{\circ }$, as a limiting case.

Equal bounding area and solid angle do not alone prove equal etendue: an actual output can have unoccupied directions or position-direction correlations. The filled-cone equation is an exact model of the specified bundles and a capacity constraint on an optical instrument, not a guarantee of an aberration-free implementation.

Consider a planar circular aperture of diameter $D$ in constant refractive index $n$. Suppose every position in its area $A=\pi D^2/4$ is filled with directions inside the same rotationally symmetric cone about the unit normal, with half-angle ${\theta }_{\max }$. Using spherical coordinates, the exact etendue iswhere $\mathrm{NA}=n\sin {\theta }_{\max }$ is the appropriate side's numerical aperture. The ordinary cone solid angle is $2\pi (1-\cos {\theta }_{\max })$; its cosine-weighted value is $\pi {\sin }^2{\theta }_{\max }$. A full forward hemisphere therefore gives $G=n^{2}A\pi$, rather than $2\pi n^{2}A$. ^[ref]

For an original model example, choose $n=1$, $D=10\mathrm{mm}$, and ${\theta }_{\max }=30^{\circ }$. ThenIf the radiance is uniform across this filled bundle and equals $L=100\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, its radiant flux is $\Phi =LG/n^2\approx 6.1685\mathrm{mW}$. Neither the geometry nor the chosen radiance is an experimental measurement.

Lossless reversible propagation preserves the etendue of the same bundle of light rays. Here the model uses stationary isotropic media, specular reflection and refraction, with every selected light ray reaching the output and with no optical gain or scattering of light. An aperture that blocks some incident light rays changes which bundle is counted; the full incident and transmitted etendue need not be equal. ^[ref]

The underlying phase space has four coordinates. In a planar section with constant refractive index, let $(x,y)$ locate a light ray and setThe Jacobian determinant from spherical coordinates givesThus etendue is a four-dimensional volume, preserved by the optical counterpart of Liouville theorem in Hamiltonian mechanics. Individual position and direction extents may change, and correlations between them matter. The Lagrange optical invariant is a related paraxial approximation invariant for pairs of light rays, rather than the full four-dimensional measure.

Consider the nonstiff scalar initial value problem $y^{\prime }=-y$, $y(0)=1$, on $0\le t\le 8$. Its exact solution is the exponential decay $y(t)=e^{-t}$. Use step-doubling error estimation with the explicit Euler method: from the same numerical starting value $y_n$, form one coarse step $C_n$ and two half steps giving $F_n$:Accept $y_{n+1}=F_n$ when the absolute and relative error tolerances test $E_n=|{\hat{e}}_n|/s_n\le 1$ passes, with $s_n=10^{-5}+10^{-3}\max (|y_n|,|F_n|)$. There is no Richardson extrapolation in the accepted value. Both approximations have order of a numerical method one, while this local error estimator has step exponent $r=2$. Consequently the adaptive step-size controller uses a square root, not a cube root:Start with $h_0=1$. A rejected trial retains both $t_n$ and $y_n$, caps $q_n$ at one, and retries. A zero estimate uses $q_n=5$. Before every trial, cap $h_n$ at $8-t_n$. Stop with a failure if floating-point arithmetic gives $t_n+h_n=t_n$, or after 100000 trials. These rules specify the example completely; they are not a prescription for every ordinary differential equation solver.

The first two trials, $h=1$ and $h=0.2$, have $E\approx 247.525$ and $E\approx 9.90099$ and are rejected. The next trial has $h\approx 0.0572049$ and $E\approx 0.81$ and is accepted. For positive decreasing numerical values, $s_n=10^{-5}+10^{-3}{y}_n$, so the local error estimator givesWhen the relative term dominates, this suggests nearly constant accepted steps, approximately $2(0.9)\sqrt{10^{-3}}$. When $y_n$ falls below the crossover size $10^{-5}/10^{-3}=0.01$, the absolute term increasingly permits longer steps. The sharp final decrease in $h$ only lands on $t=8$; it does not indicate a sudden loss of smoothness.

The actual signed local defect from the same numerical input isHere $O$ is Big O notation, obtained from Taylor's theorem for the exponential function. Thus ${\hat{e}}_n$ agrees with the leading term as $h_n\to 0$, but is not identical to ${\ell }_n$. At $h=1$, $y_n=1$, the estimate is $0.25$ while the actual local defect is $e^{-1}-1/4\approx 0.117879$. Their closeness improves for the smaller accepted steps. Had we added ${\hat{e}}_n$ using Richardson extrapolation, the accepted formula would instead be $y_n(1-h_n+h_n^2/2)$, of order of a numerical method two; the same raw difference would still scale as $h_n^2$ and would not estimate that improved value's leading $h_n^3$ defect.

The global error $g_n=y_n-e^{-t_n}$ obeys an exact recurrence relation involving the actual local defects:Each previous global error contracts, yet the local defects keep adding. In the plotted run, 101 steps are accepted and two trials are rejected. Every accepted $E_n$ is at most approximately $0.81$, while the largest $|g_{n+1}|/s_n$ is approximately $38.6927$. At $t=8$, the numerical value is approximately $0.00027184218$, compared with the exact $e^{-8}\approx 0.00033546263$, giving absolute error approximately $6.36204\times 10^{-5}$. The right-hand plot deliberately scales actual global error by the same $s_n$ used for local acceptance; it is not a new global error tolerance test.

Tightening both absolute and relative error tolerances by a factor ten gives 315 accepted steps and endpoint absolute error approximately $2.14922\times 10^{-5}$; tightening both by a factor one hundred gives 991 accepted steps and endpoint absolute error approximately $6.94171\times 10^{-6}$. This is a convergence comparison for this particular initial value problem. The endpoint error decreases by about a factor three per tenfold tightening, rather than by ten: a first-order Euler method accumulates defects across many steps, while its local error estimator varies quadratically with $h$. A variable-step global ODE error bound needs propagation information as well as accepted local tests.

The absolute stability boundary remains a separate constraint. On the Dahlquist test equation $y^{\prime }=-\lambda y$ with $\lambda >0$, the accepted two-half-step formula has stability function $R(-\lambda h)=(1-\lambda h/2{)}^2$. It strictly damps perturbations only for $0<\lambda h<4$. Its local error estimator is ${\lambda }^2{h}^2{y}_n/4$, which can pass an absolute-tolerance test at tiny $y_n$ even outside this damping interval. In a stiff two-mode linear system, a fast mode can therefore constrain the step long after its visible amplitude has become small. An adaptive step-size controller does not turn an explicit Euler method into an A-stable method. Official solver documentation likewise distinguishes explicit Runge-Kutta methods for nonstiff problems from implicit Runge-Kutta methods and backward differentiation formulas for stiff equations.^[ref]

A local error estimator approximates a numerical step's defect from the exact solution starting at the same input. Differences between formulas estimate error only after their asymptotic orders and constants are accounted for. Step-doubling error estimation, an embedded Runge-Kutta pair and predictor-corrector error estimation are common constructions. They do not by themselves bound accumulated global error.

Let ${\Phi }_h(t_n,y_n)$ be the exact endpoint of the ordinary differential equation started at the numerical value $y_n$, and let $Y_{n+1}$ be the candidate numerical endpoint. The signed local defect being estimated is ${\ell }_{n+1}={\Phi }_h(t_n,y_n)-Y_{n+1}$. In contrast, actual global error compares $Y_{n+1}$ with the solution of the original initial value problem, including all previous steps. The local truncation error convention inserts original exact data instead. These starting-value distinctions matter even when the same formula is used for every step. A practical local error estimator also relies on a smooth Taylor expansion; a small difference can reflect accidental cancellation rather than a rigorous upper bound.

Absolute tolerances provide a nonzero error scale near zero, while relative tolerances follow component magnitude. A max-norm test ${\max }_i|{\hat{e}}_i|/s_i\le 1$ controls every scaled estimated component locally. A root-mean-square test has a different componentwise implication. Neither test automatically certifies the complete trajectory's global error.

Write $Y_i$ for the candidate accepted component and require $s_i>0$. The absolute and relative error tolerances scale in the title produces the dimensionless maximum norm testAbsolute tolerances have the same units as the corresponding solution components; relative tolerances are dimensionless. Choose separate absolute scales for components measured in different units or with different significant magnitudes. Near zero, a purely relative test can demand a zero allowable error; a positive absolute scale avoids that difficulty. The crossover is $\max (|y_i|,|Y_i|)\approx {\mathrm{atol}}_i/{\mathrm{rtol}}_i$ when ${\mathrm{rtol}}_i>0$. Official solver documentation specifies the analogous blend $\mathrm{atol}+\mathrm{rtol}|y|$ and allows component-specific absolute tolerances; the old/new maximum above is the convention used in this example.^[ref]

A root mean square version replaces the maximum norm by $E_{\mathrm{rms}}=\sqrt{m^{-1}{\sum }_i(|{\hat{e}}_i|/s_i{)}^2}$. It permits one scaled component as large as $\sqrt{m}$ when all others vanish, so passing that test does not mean every component is at most one. State the aggregation rule along with the absolute and relative error tolerances. Neither rule is a bound on actual global error.

If a scaled local error estimator behaves as $E\propto h^r$ with approximately unchanged error scale, the displayed update targets a fixed local level. A safety factor $0<\eta <1$ aims below the acceptance boundary. For an embedded Runge-Kutta pair with order of a numerical method $p$ for the accepted output and $p-1$ for the companion formula, the difference generally has $r=p$; for the actual unscaled local truncation error of an order-$p$ formula, $r=p+1$. The exponent belongs to the quantity being controlled. Dividing the defect by $h$ changes that exponent. This follows by solving $E(h_{\mathrm{new}})\approx E(h)(h_{\mathrm{new}}/h{)}^r$ for the next step.^[ref]

Rejecting a trial means retrying from the same accepted time and state, with a smaller step. Advancing time or retaining the rejected output would change the initial value problem being approximated. Minimum and maximum change factors prevent extreme updates; a zero estimate requires finite capped growth. After acceptance, cap the next step at the remaining interval length. Detect a step that cannot advance time in floating-point arithmetic, and report failure rather than an endless retry. An adaptive step-size controller can use earlier estimates to smooth step sequences, but the memoryless rule here is sufficient to explain acceptance and rejection.^[ref]

For the $D_3$ action $\rho (z_1,z_2)=({\zeta }^{-1}{z}_1,\zeta z_2)$, $m(z_1,z_2)=(z_2,z_1)$ and common phase multiplication, the fifth-order normal form has first component $z_1[\alpha +a|z_1{|}^2+b|z_2{|}^2+c|z_1{|}^4+d|z_1{|}^2|z_2{|}^2+e|z_2{|}^4]+f{\bar{z}}_1^2{z}_2^3$ and second component obtained by swapping indices. Here all coefficients may be complex. Its three primary symmetry types are a one-component rotating wave and the two standing-wave lines $z_1=z_2$ and $z_1=-z_2$, together with their group conjugates. The two standing amplitudes split at fifth order through $\mathrm{Re}f$.