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For a real orthogonal matrix $A$, $A^{\mathsf{T}}=A^{-1}$. If $Av=\lambda v$ for a nonzero complex eigenvector, then $A^{\mathsf{T}}v={\lambda }^{-1}v$. Therefore the same eigenvector has eigenvalue $(\lambda +{\lambda }^{-1})/2$ for the symmetric part of a matrix $(A+A^{\mathsf{T}})/2$. Orthogonality implies $|\lambda |=1$, so writing $\lambda =e^{i\theta }$ gives $\cos \theta$. The complete complex orthonormal eigenbasis of an orthogonal matrix shows that all the eigenvalues of its symmetric part lie in $[-1,1]$.

The conjugacy classes of the symmetric group on three letters have sizes $1,3,2$: the identity, the three transpositions and the two 3-cycles. A normal subgroup is a union of whole conjugacy classes and contains the identity. Its candidate orders are therefore $1,4,3,6$. Lagrange's theorem excludes four, leaving precisely $\{e\}$, $A_3$ and $S_3$. The order-three subgroup is the cyclic group of the two 3-cycles and the identity; it is also the kernel of a group homomorphism of the sign map.

The first isomorphism theorem gives the corresponding quotients $S_3$, $C_2$ and the trivial group. Thus these are exactly the possible images of surjective group homomorphisms from $S_3$. Each occurs, by the identity map, the sign map and the constant identity map, respectively.

Suppose $f,g:X\to \mathbb{R}$ are bounded functions on the same nonempty set, and the supremum norm of their difference is at most $\delta$. Then their optimal values differ by at most the same amount:Indeed, $f(x)\le g(x)+\delta \le {\sup }_{X}g+\delta$ for every $x$, so ${\sup }_{X}f\le {\sup }_{X}g+\delta$; exchange $f$ and $g$ for the other direction. Adding a constant shifts a supremum by that constant, a basic compatibility between upper bounds and addition.^[ref]

For a chosen example on $[-1,1]$, take $f(x)=-x^2$ and $g(x)=-x^2+0.2x$. Their supremum error is $0.2$, since $|g(x)-f(x)|=0.2|x|$. Completing the square givesThus $f$ has maximum zero at $x=0$, while $g$ has maximum $0.01$ at $x=0.1$. The change in optimal value, $0.01$, is smaller than the certified bound $0.2$. For the vertical translation $h(x)=f(x)+0.2$, the bound is exact: its supremum error and optimal-value change are both $0.2$.

A stable optimal value does not guarantee a stable maximizing input. On $[-1,1]$, the two linear functions $f_{\epsilon }(x)=\epsilon x$ and $g_{\epsilon }(x)=-\epsilon x$ satisfyFor every positive $\epsilon$, their unique maximizing inputs remain two units apart, at $1$ and $-1$. Both functions approach the flat zero function, which has many maximizing inputs. Consequently an accurate objective function value is not by itself a certificate that the optimizing decision is accurate.

The estimate also gives a convenient consequence of uniform convergence. If $f_n$ converges uniformly to a bounded function $f$ on a nonempty $X$, then ${\sup }_X{f}_n\to {\sup }_{X}f$ for the bounded tail of the sequence. Uniform control handles the changing input that defines the best value. Pointwise convergence alone supplies no such control; the existing examples under uniform convergence illustrate that distinction.

Let $A,B$ be nonempty sets of real numbers bounded above. Their Minkowski sum is $A+B=\{a+b:a\in A,b\in B\}$, allowing every pair from the Cartesian product $A\times B$. ThenThe sum of the two suprema is an upper bound for every permitted sum. For the reverse bound, use the approximation property of a supremum to choose $a>\sup A-\epsilon /2$ and $b>\sup B-\epsilon /2$. Their sum is within $\epsilon$ of the claimed value. This reasoning does not require either supremum to be a maximum.

Separate choices matter. On $X=[0,1]$, choose the functions $f(x)=x$ and $g(x)=1-x$. Each has supremum one, but their sum at a shared input is always one:The maximum of $f$ occurs at $x=1$, while that of $g$ occurs at $x=0$. A common input cannot realize both choices. With separate inputs, choosing $x=1$ and $y=0$ gives $f(x)+g(y)=2$. The allowed pairs of output values are the entire unit square for separate inputs, but only the line segment $a+b=1$ for the shared input.

For any two bounded functions on the same nonempty domain of a function, the useful general estimate isThe inequality can be strict, as the example shows. Equality does not require attained maxima, but it does require that the two large values can be approached at compatible inputs. For example, $f(x)=g(x)=x$ on $(0,1)$ gives equality with value two, although neither function reaches its supremum. This distinction is useful when an objective function combines terms sharing a decision variable: adding the best possible termwise scores gives an upper bound, not necessarily a feasible combined score.

A local derivation makes the acceleration sign transparent. Normalize $a(t_0)=1$ and put $\tau =t_0-t_e$. A Taylor series givesThe radial null geodesic in FLRW spacetime gives $D_C=c{\int }_{t_e}^{t_0}dt/a(t)=c[\tau +H_0{\tau }^2/2+O({\tau }^3)]$. Invert the redshift series, use $d_L=(1+z)D_C$ through second order, and obtain the displayed luminosity distance. Spatial curvature of an FLRW universe first changes this relation at third order. Lower $q_0$ therefore means larger $d_L$ at a fixed sufficiently small positive redshift and common $H_0$.

At $z=0.05$, the quadratic approximation gives $222.4353\mathrm{Mpc}$ for $q_0=-0.55$, within $0.07\%$ of the full flat-model value $222.2891\mathrm{Mpc}$. At $z=0.5$ it gives $2971.1574\mathrm{Mpc}$, about $4.9\%$ too large. A useful sign argument at low redshift is not a precision replacement for the full distance integral in a distant supernova analysis.

Stipulate $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and $c=299792.458\mathrm{km}{\mathrm{s}}^{-1}$, so $c/H_0=4282.7494\mathrm{Mpc}$. A megaparsec is $10^6$ parsecs. Neglect radiation in cosmology. Compare an accelerating flat model $({\Omega }_{m,0},{\Omega }_{\Lambda ,0})=(0.3,0.7)$, an open matter-only model $(0.3,0)$ and the Einstein-de Sitter universe $(1,0)$. Their present deceleration parameters are respectively $-0.55$, $0.15$ and $0.5$. For these models,Here $\chi$ is dimensionless; the radial null geodesic in FLRW spacetime supplies its integral and spatial curvature of an FLRW universe supplies $S_k$. The observer-to-source comoving radial distance is $(c/H_0)\chi$, whereas the transverse distance uses $S_k$. Omitting the latter distinction would bias the open-model comparison.

At $z=0.5$, numerical integration gives $\chi =0.440984335$ for the flat accelerating model and $\chi =0.392227378$ for the open matter-only model. ThusThe accelerating model is $0.21559$ mag fainter than the open matter-only prediction at the same redshift and standardized absolute magnitude, corresponding to $1.10438$ times the luminosity distance and $0.81991$ times the radiative flux. These are model calculations, not observations from either collaboration.

The Milne model has $q=0$, ${\Omega }_{m,0}={\Omega }_{\Lambda ,0}=0$ and ${\Omega }_{k,0}=1$, giving $\chi =\log (1+z)$ and $d_L=(c/H_0)(z+z^2/2)$. The Einstein-de Sitter luminosity-distance relation independently checks the flat matter-only integral. At $z=0.5$, the accelerating model exceeds the Milne model by only $0.12317$ mag, illustrating the dependence of a residual on the chosen reference. Over a longer redshift interval, luminosity distance integrates past deceleration as well as recent acceleration; a residual need not increase forever.

Multiplying $H_0$ by $f>0$ at fixed density fractions divides every $d_L$ by $f$ and shifts every distance modulus by $-5{\log }_{10}f$. The predicted apparent magnitude is unchanged if $M_{B,0}$ simultaneously increases by $5{\log }_{10}f$. This is the absolute magnitude–Hubble constant degeneracy. For example, $70\to 77\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ shifts $\mu$ by $-0.20696$ mag and $M_{B,0}$ by $+0.20696$ mag. Relative curves at the same $H_0$ retain their separation; determining $H_0$ requires an external absolute magnitude calibration.

The Supernova Cosmology Project measured the brightness-redshift relation of Type Ia supernovae to constrain matter density and the contribution represented by a cosmological constant. Its high-redshift sample was fitted together with nearby events after standardizing peak brightness using light-curve information.^[ref]

Its importance includes an independent observing and analysis effort alongside the High-Z Supernova Search Team. Independence of teams does not eliminate all shared uncertainty: both rely on related supernova physics, nearby calibration and assumptions connecting luminosity distance with an expansion model.

The High-Z Supernova Search Team used distant Type Ia supernovae to compare the observed relation between luminosity distance and redshift with models of cosmic expansion. Its analysis combined distant events with nearby calibrators and tested how light-curve fitting and sample choices affected the inferred acceleration.^[ref]

The physical inference is that sources at a specified redshift were farther away than predicted by relevant nonaccelerating comparison models. Calibration, extinction and population evolution are therefore central to interpreting the brightness difference, rather than secondary corrections to an otherwise direct acceleration measurement.

The current-phase relation of a superconducting weak link gives its pair electric current $I_s$ as a function of the Gauge-invariant Josephson phase $\varphi$, at specified temperature and other conditions. For the ideal junction used below, $I_s=I_c\sin \varphi$ with $I_c>0$. More general periodic relations can contain higher Fourier harmonics; their maximum magnitude defines the corresponding superconducting critical current.

An Abrikosov vortex has a core where the superconducting order parameter is suppressed, surrounding circulating electric currents, and a winding of its quantum phase. A singly wound vortex carries one superconducting flux quantum in the appropriate bulk flux-quantization limit. The exact contour statement is fluxoid and flux quantization in a superconductor; the enclosed magnetic flux alone need not equal a flux quantum on a contour with appreciable electric current.

Super-Kamiokande used $535$ days ($33.0$ kiloton-years) and $22.5$ fiducial kilotons of water. Inner photomultiplier times and charges reconstructed interactions; the outer anticoincidence detector vetoed entering particles and marked exits. Contained single-ring events used likelihood classification: electron particle showers blur rings; muons give sharper edges. The $1.33\mathrm{GeV}$ visible-energy sample division is not a universal detection threshold. Simulation estimated classification efficiency above about $98\%$, distinct from total detection efficiency; stopping muons and decay electrons checked classification.^[ref]

Downward neutrinos travel about $15\mathrm{km}$; upward ones about $13000\mathrm{km}$. The multi-GeV muon-like up/down asymmetry was $-0.296\pm 0.048\pm 0.010$; electron-like events lacked a comparable effect. Calibration constrained directional bias. This was disappearance evidence; tau neutrino appearance was untagged and sterile conversion remained possible.^[ref]

Writing $U$ and $D$ for the selected upward and downward populations, the asymmetry is $A=(U-D)/(U+D)$: a negative value means fewer upward events. An absolute source-normalization error alone rescales both populations and leaves this ratio unchanged. A direction-dependent response error need not cancel.

Why compare inferred neutrino fluxes rather than raw event counts? For channel $j$, exposure time $T$, target population $N_j$, cross section ${\sigma }_{j\alpha }$, and detection efficiency ${ϵ}_{j\alpha }$ including reconstruction and selection, a forward model isHere $B_j$ is expected background; $N_j$ on the left counts events, while $N_j^{\mathrm{target}}$ counts targets. Distinct cross sections, target numbers, thresholds and detection efficiencies prevent a direct CC/NC count ratio from being an electron neutrino fraction. Channel populations themselves can be correlated estimates from overlapping event patterns.

An original reduced linear algebra model isolates the inference. Assume specified spectral shapes and divide each background-corrected channel rate by its own calibrated electron neutrino response. Put ${\varphi }_x={\varphi }_{\mu }+{\varphi }_{\tau }$ and let $r$ be the spectrum- and response-weighted non-electron/electron ES sensitivity ratio. ThenThe equal $\mu$ and $\tau$ responses determine their sum, not their separate neutrino fluxes. Changing the assumed spectra generally changes the integrated coefficients; with arbitrary unknown spectra this small matrix is insufficient.

Choose illustrative inputs, not historical measurements: $r=0.15$, ${\varphi }_e=2$ and ${\varphi }_x=3$, with every neutrino flux in the same arbitrary unit. The forward prediction is $(y_{\mathrm{CC}},y_{\mathrm{NC}},y_{\mathrm{ES}})=(2,5,2.45)$. CC and NC independently invert to ${\varphi }_e=y_{\mathrm{CC}}=2$, ${\varphi }_x=y_{\mathrm{NC}}-y_{\mathrm{CC}}=3$; substituting into ES returns $2+0.15(3)=2.45$. Alternatively, CC and ES give ${\varphi }_x=(y_{\mathrm{ES}}-y_{\mathrm{CC}})/r=3$. Thus the third channel tests the response model and the inferred mixture.

CC alone admits any nonnegative ${\varphi }_x$. CC and ES lose the second independent constraint when $r=0$; for small positive $r$, an ES-minus-CC error is amplified by $1/r$. With chosen independent errors ${\sigma }_{\mathrm{ES}}={\sigma }_{\mathrm{CC}}=0.1$, their propagated standard deviation is $\sqrt{0.1^2+0.1^2}/0.15\simeq 0.943$. If CC and NC instead have those independent errors, the difference has standard deviation $\sqrt{0.1^2+0.1^2}\simeq 0.141$. Actual analyses must retain covariances, including those introduced by statistical channel separation. Complementarity supplies information; calibration establishes what information each channel carries.

In 2001, $240.95$ live days, reconstructed radius below $5.50\mathrm{m}$ and effective electron kinetic energy above $6.75\mathrm{MeV}$ gave a CC neutrino flux of $1.75\times 10^6{\mathrm{cm}}^{-2}{\mathrm{s}}^{-1}$ for the assumed boron-8 spectrum. Source calibrations measured $1.4\%$ instrumental-cut signal loss. SNO's own CC–ES difference was $1.6\sigma$; comparison with Super-Kamiokande's precise ES flux gave $3.3\sigma$ evidence for non-electron active neutrinos. This was not an NC measurement.^[ref]

In 2002 the pure-heavy water analysis used $306.4$ live days and a $5\mathrm{MeV}$ reconstructed kinetic-energy threshold. NC neutrons captured on deuterons emitted $6.25\mathrm{MeV}$ gamma rays, producing secondary electron light. The NC reaction threshold was $2.2\mathrm{MeV}$, a different quantity. Californium-252 calibrated $29.9\%$ capture efficiency, but only $14.4\%$ detection efficiency after fiducial and energy cuts. Assays constrained backgrounds from radioactive decay and photodisintegration.^[ref]

Energy, radius and direction relative to the Sun statistically separated channels. Assuming the standard boron-8 spectral shape, the NC total active neutrino flux was $5.09_{-0.43}^{+0.44}{(stat.)}_{-0.43}^{+0.46}(syst.)\times 10^6{\mathrm{cm}}^{-2}{\mathrm{s}}^{-1}$, consistent with stellar structure predictions. The non-electron active component differed from zero by $5.3\sigma$. This established active flavour conversion, without individually tagging muon neutrinos and tau neutrinos. It did not exclude every sterile admixture.^[ref]

A steady spherical subsonic isothermal breeze with isothermal sound speed $a$ obeys $(u-a^2/u)u^{\prime }=2a^2/r-GM/r^2$. Since $0<u<a$, its finite-radius speed maximum lies at $r_M=GM/(2a^2)$. Spherical mass conservation then gives the displayed density, where ${\mathcal{M}}_M=u(r_M)/a\in (0,1)$ is the peak Mach number. This free parameter is selected by a boundary condition, not by $T$, $M$ and $\dot{M}$ alone. Therefore ${\rho }_M>\dot{M}{a}^3/(\pi G^2{M}^2)$, while a peak Mach number of order unity gives that expression as a characteristic scale.

A direct demonstration of the free parameter comes from the first integral $u^2/2-a^2\ln (u/a)=2a^2\ln (r/r_M)+GM/r+C$. Its additive constant remains free for a breeze; imposing the regular transonic branch would be a different condition.

For unattenuated isotropic radiation from a point source, the radiation pressure force on each absorbing particle is $\sigma L/(4\pi c{r}^2)$. Dividing by its mass $\bar{m}$ changes the effective inverse-square inward acceleration to $[GM-\sigma L/(4\pi \bar{m}c)]/r^2$. The wind equation therefore places the maximum of a subsonic isothermal breeze at the displayed radius, provided the bracket is positive. If it is nonpositive, the positive-temperature subsonic outflow decreases in speed throughout its domain and has no interior speed maximum. A cross section per particle requires that particle's mass; an opacity per unit mass instead absorbs this conversion.

The trivial subgroup contains only the identity element of a group. It is a subgroup of every group and is proper whenever the group has more than one element. Thus a group of prime group order has no proper nontrivial subgroups, rather than having no proper subgroups at all.

A nonempty subset $H$ of a group $G$ is a subgroup if and only if $g{h}^{-1}\in H$ for every $g,h\in H$. Necessity follows from closure and the existence of inverse elements in a subgroup.

Conversely, choose $a\in H$. Setting $g=h=a$ gives the identity element $e=a{a}^{-1}\in H$. Taking $g=e$ then gives $h^{-1}\in H$ for every $h\in H$. Finally, using $h^{-1}$ as the second element gives $g(h^{-1}{)}^{-1}=gh\in H$, proving closure. The inherited multiplication is associative, so these properties establish the group axioms on $H$.

For $0<x<1$ and zero endpoint values, write $L=w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]$ with $p=x^{-2}$, $w=x^2$ and $q=-a{x}^2$. The change of variable $z=x^3/3$ has $dz=x^{2}dx$ and sends $y^{\prime }=x^2{Y}_z$, $y^{\prime \prime }=2x{Y}_z+x^4{Y}_{zz}$. Thus it is an isometry from the weighted inner product to the ordinary integral on $0<z<1/3$, and the apparent singular operator becomes $-Y_{zz}-aY$. Its Dirichlet eigenvalues are $9n^2{\pi }^2-a$ and its normalized eigenfunctions are $\sqrt{6}\sin (n\pi x^3)$, $n\ge 1$. The singular endpoint is handled through this transformed regular domain; a zero eigenvalue occurs exactly when $a=9n^2{\pi }^2$.

For $-y^{\prime \prime }=\lambda y$, positive eigenvalues have eigenfunctions $\sin (q_{n}x)$, where $\tan q_n=\alpha q_n$ and $q_n>0$. Their squared norms are $N_n=1/2-\sin (2q_n)/(4q_n)$. A zero eigenvalue occurs precisely at $\alpha =1$, with eigenfunction $x$. A negative eigenvalue occurs precisely at $0<\alpha <1$: it is $-{\kappa }^2$ with eigenfunction $\sinh (\kappa x)$, where $\tanh \kappa =\alpha \kappa$. There is exactly one such $\kappa$, since $\tanh \kappa /\kappa$ decreases strictly from $1$ to $0$. The integration by parts identityand the Cauchy-Schwarz inequality show positivity for $\alpha >1$ and nonnegativity at equality. The complete expansion of a square-integrable function therefore includes the extra term $3x{\int }_0^{1}tf(t)dt$ when $\alpha =1$, or $\sinh (\kappa x){\int }_0^{1}f(t)\sinh (\kappa t)dt/[\sinh (2\kappa )/(4\kappa )-1/2]$ when $\alpha <1$, in addition to ${\sum }_n[{\int }_0^{1}f(t)\sin (q_{n}t)dt/N_n]\sin (q_{n}x)$. Omitting that extra mode makes the system incomplete. In particular, $f=x$ at $\alpha =1$ is orthogonal to every positive sine mode and is an immediate counterexample to a sine-only expansion.

Strong induction does not prove more than ordinary mathematical induction. Put $Q(n)={\bigwedge }_{k=n_0}^{n}P(k)$. Ordinary mathematical induction on $Q$ retains the earlier cases and obtains the next one, establishing the same conclusion. Conversely, an ordinary inductive step can ignore the extra cases allowed in a strong induction hypothesis.^[ref]

The well-ordering principle gives another view. A least counterexample would have all smaller cases true, so the strong induction step would establish that allegedly false case. The starting case excludes the beginning.