Variation of inverse group velocity with angular frequency; for a uniform medium of length $L$, the accumulated group delay dispersion is $L{d}^{2}k/d{\omega }^2$.

Writing the material wavelength as ${\lambda }_m=2\pi /k$ and its phase velocity as $v_p({\lambda }_m)$ gives $v_g=v_p-{\lambda }_{m}d{v}_p/d{\lambda }_m$. This follows by differentiating $\omega =k{v}_p$ and using $kd{\lambda }_m/dk=-{\lambda }_m$. If instead the independent variable is vacuum wavelength ${\lambda }_0=2\pi c/\omega$, then $k=\omega /v_p({\lambda }_0)$ givesSubstituting vacuum wavelengths into the material-wavelength derivative formula gives an erroneous result beyond first order in dispersion. The two coordinates describe the same dispersion relation, but their derivatives differ by the chain rule.

The second derivative of spectral phase with respect to angular frequency; it measures variation of group delay with angular frequency.

On the punctured plane ${\mathbb{R}}^2\setminus \{(0,0)\}$, consider the vector fieldWith $r^2=x^2+y^2$, its partial derivatives satisfyThus the field has zero scalar curl everywhere it is defined. Nevertheless, on any counterclockwise circle ${\gamma }_R(t)=(R\cos t,R\sin t)$ with $R>0$,The radius cancels exactly.

Applying Green theorem to the enclosed disk would be invalid: $G$ is undefined at the origin, inside that disk. Zero curl on the path alone is insufficient.^[ref]

Now take the annulus $A=\{(x,y):1\le r\le 2\}$. Its positive boundary orientation is counterclockwise on the outer circle and clockwise on the inner circle. The complete boundary line integral is thereforeHere Green theorem applies: an open set containing the closed annulus can avoid the origin. The inner contribution remains $-2\pi$ even for an arbitrarily small positive inner radius. Decreasing the hole's area does not decrease this line integral.

The cancellation has a useful proof picture. Cut the annulus along two radial segments into two half-annuli. Each piece has a single simple closed curve as its boundary; along each cut, the two induced boundary orientations oppose, so those line integrals cancel when the two applications of Green theorem are added. Only the outer and inner circles remain.^[ref]

Locally, a choice of angle $\theta$ in polar coordinates has gradient $\nabla \theta =G$. Globally, following a circle once increases that angle by $2\pi$, so it cannot be a single-valued potential of a conservative vector field on the punctured plane. The nonzero closed-path line integral also rules out such a potential by the fundamental theorem for line integrals.

On a simply connected open region, a continuously differentiable planar vector field with zero scalar curl does have a global potential. The hole is what separates the local derivative condition from the global conclusion.^[ref]

Take the vector field $F=(-2y,3x)$. Its scalar curl is the constant $3-(-2)=5$, so Green theorem predicts that its positively oriented boundary line integral is five times the enclosed area. This one field makes both a round and a rectangular calculation transparent.

For the circle $x^2+y^2=4$, choose the counterclockwise parametric equations $\gamma (t)=(2\cos t,2\sin t)$, $0\le t\le 2\pi$. Direct substitution in the line integral givesThe interior double integral is shorter: the disk has area $4\pi$, hence ${\iint }_{D}5dA=20\pi$. Reversing the boundary orientation gives $-20\pi$.

For the rectangle $D=[0,3]\times [0,2]$, traverse its bottom, right, top, and left edges in that order. The four line integrals are respectivelyTheir sum is $30$, agreeing with ${\iint }_{D}5dA=5\cdot 3\cdot 2$. In both examples the path geometry changes the separate contributions, while constant curl makes only area matter to the final answer.

For $r=\sqrt{x_1^2+\cdots +x_d^2}>0$, the multivariable chain rule givesIndeed, $\partial r/\partial x_j=x_j/r$. Thus the gradient points radially outward when $g^{\prime }(r)>0$, inward when it is negative, and vanishes when it is zero.

At the origin the displayed quotient does not apply. If $g$ has a finite right derivative at zero, the radial function is differentiable there exactly when $g^{\prime }(0^{+})=0$, in which case its gradient is zero. Necessity follows by comparing opposite directional difference quotients: their linear parts cannot both equal a common nonzero radial slope. Sufficiency follows from $g(r)-g(0)=o(r)$, the definition of differentiability with zero linear part.

For $E(a)=a/2-V_0\sqrt{a/(1+a)}$ with $a,V_0>0$, differentiation gives $E^{\prime }(a)=1/2-V_0/[2\sqrt{a}(1+a{)}^{3/2}]$. The positive stationarity equation has a unique solution because its left side increases strictly. Moreover $E^{\prime \prime }(a)=V_0(1+4a)/[4a^{3/2}(1+a{)}^{5/2}]>0$, so it is the global minimum. Substituting the stationarity equation gives the displayed negative energy. The only positive zero is $(\sqrt{1+16V_0^2}-1)/2$; $a=0$ is only a limiting zero, since the limiting constant trial is not a square-integrable function. The Rayleigh-Ritz variational principle therefore proves a negative-energy bound state for every positive well depth. For weak depth, the simpler choice $a=V_0^2$ yields an explicit upper bound $E_0\le -V_0^2/2+\epsilon$ withThis is a variational upper bound, not an equality for the exact weak-depth coefficient.

For integer $n\ge 2$, integration by parts gives $J_n=(n-1)J_{n-2}/2$. The boundary term $-x^{n-1}{e}^{-x^2}/2$ vanishes at both infinities because Gaussian decay dominates every power. The full-line initial values are $J_0=\sqrt{\pi }$ and $J_1=0$, the latter by odd symmetry. HenceUnlike the half-line Gaussian moment recurrence, full-line odd moments vanish and even moments are twice their half-line values. The domain of integration must be retained when transferring a recurrence between the two settings.

For $D,\gamma >0$, the displayed function solves the heat equation $u_t=D{u}_{xx}$ and has initial condition $u(x,0)=A{e}^{-\gamma (x-x_0{)}^2}$. Set $s=1+4\gamma Dt$. Direct differentiation givesThese identities prove the equation. The total integral is $A\sqrt{\pi /\gamma }$, independent of time. The real smooth formula is valid for $t>-1/(4\gamma D)$, including a finite interval of negative time for this particular initial profile. For nonzero $A$, its value at $x_0$ diverges as the lower time endpoint is approached. The superposition principle combines finitely many such profiles with different centers or widths.

For the Gauss-Seidel iteration for a two-variable positive-definite system, define the forward error by $e_x^{(k)}=x^{(k)}-x_{\ast }$ and $e_y^{(k)}=y^{(k)}-y_{\ast }$. Subtract the two exact solution equations from the two update equations:The fresh first coordinate passes the old second-coordinate error through two successive factors. The iteration matrix is thereforeIts eigenvalues are $0$ and $1/12$, as seen from the determinant. The zero eigenvalue explains why the old first-coordinate error disappears completely: the first row overwrites it. After one sweep, every remaining forward error lies on the eigenvector line spanned by $(-3,1{)}^T$.

For this zero start, $e_y^{(0)}=-7/11$, so the exact vectors for $k\ge 1$ areThe Euclidean norm of the forward error decreases by exactly $1/12$ per sweep from $k=1$ onward. The first decrease is different because the initial error has not yet reached that eigenvector line. Each row update moves horizontally or vertically in the coordinate plane; sweep endpoints then approach the solution along one straight line.

The linear-system residual is an independent way to check the calculation. Completing a sweep satisfies the second row exactly, but changing $y$ disturbs the first row:Thus a zero residual in the last updated row alone does not certify a solution; inspect the complete linear-system residual.

For the same matrix, the stale schedule of the Jacobi method gives the iteration matrixHere its eigenvalues are $\pm 1/\sqrt{12}$, whereas the fresh schedule gives $\rho (H)=\rho (B{)}^2$. This comparison belongs to this particular linear system; it is not a universal relation between the two methods' spectral radii.

An invertible matrix and nonzero diagonal entries make the update well defined, but do not guarantee convergence. For the original linear systemthe Gauss-Seidel method updates $x^{(k+1)}=1-2y^{(k)}$ and $y^{(k+1)}=1-2x^{(k+1)}=-1+4y^{(k)}$. Starting at zero gives $(1,-1)$, then $(3,-5)$. The second-coordinate forward error is $e_y^{(k)}=-4^k/3$, and the iteration matrix $H=\left(\begin{array}{cc}0& -2\\ 0& 4\end{array}\right)$ has spectral radius $4$. The linear system has a unique solution while this particular stationary iterative method for a linear system diverges.

For a bound Kepler orbit, specific orbital energy $v^2/2-\mu /r$ is negative. Therefore $\mu >v^{2}r/2$. A projected separation $R_{\perp }\le r$ and a systemic-motion-corrected line-of-sight speed $|v_{\mathrm{los}}|\le v$ give the displayed lower bound. One position and one spectral velocity already constrain the mass in this model; a fitted orbit gives much more. The bound assumes a bound orbit in a point-mass potential; it is not an unrestricted enclosed-mass formula for an arbitrary extended potential.

Choose $D=7.5\mathrm{kpc}$, $\alpha =0.110$ arcseconds, $P=18$ years, $e=0.70$ and $i=55$ degrees. The small-angle conversion gives $a=825\mathrm{AU}$. Taking one model year as $365.25$ days gives $\mu \simeq 2.30\times 10^{26}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ and $K\simeq 1566km/s$.

Dividing by the nominal solar mass parameter, $(GM{)}_{\odot }^{\mathrm{N}}=1.3271244\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, gives $1.73\times 10^6$ nominal solar masses. The nominal parameter is an exact conversion convention, rather than an assertion that the Sun's measured mass is exact.^[ref]

The chosen periapsis radius is $r_p=a(1-e)=247.5\mathrm{AU}$. Its corresponding Schwarzschild radius, $r_s=2\mu /c^2$, is about $0.0342\mathrm{AU}$. Thus this illustrative orbit remains roughly $7200$ Schwarzschild radii from the centre at closest approach. Inferring a very concentrated mass from such an orbit is different from directly resolving its event horizon. Every number here follows from the chosen orbit, not from a fit to the actual Galactic-centre measurements.

Angular astrometry determines the combination displayed by substituting $a=D\alpha$ into Kepler's third law. Keeping the fitted shape, orientation and orbital period fixed, the rescalingleaves the entire angular time series unchanged. A longer physical orbit around a larger mass can trace precisely the same angular path. More angular samples improve the fitted combination; by themselves they cannot break this scaling symmetry of the model.

The radial-velocity semi-amplitude isFor $\sin i\ne 0$, combining it with the angular fit gives the orbital parallax distanceThe spectral velocity scale supplies the missing physical ruler. For a face-on orbit, $\sin i=0$, this radial signal vanishes and that method cannot supply a distance. In a partial or noisy data set, fitted orientation and systemic motion also correlate with the inferred scale.

Let $z$ increase away from the observer, with $x,y$ spanning the sky plane. A rotation $R=R_z(\Omega )R_x(i)R_z(\omega )$ sends the orbital-plane coordinates to $\mathbf{r}=R{\mathbf{r}}^{\prime }$. Here $i$ is orbital inclination, $\Omega$ the longitude of ascending node, and $\omega$ the argument of periapsis. For distance $D\gg a$, the small-angle model gives angular positions ${\theta }_x=x/D$, ${\theta }_y=y/D$ in radians. Doppler spectroscopy instead supplies the line-of-sight velocity $v_{\mathrm{los}}=\dot{z}+v_{\mathrm{sys}}$, with a fitted systemic offset $v_{\mathrm{sys}}$.

Define $\alpha =a/D$, the angular scale of the deprojected fitted orbit. It is not generally the semi-major axis of the apparent ellipse on the sky. Projection also need not send a true focus to a focus of the apparent ellipse. Fitting the rotated orbit preserves that distinction; measuring a bounding ellipse alone loses it.

For a complex irreducible representation $D$ of a finite group $G$, the sum $A={\sum }_{g}D(g^2)$ commutes with every $D(h)$: conjugation sends its summands to $D((hg{h}^{-1}{)}^2)$, merely permuting the summation index. Schur's lemma gives $A=cI$. Taking the trace yields $c\dim D={\sum }_g\chi (g^2)=|G|\nu (D)$, proving the identity. Here $\nu (D)$ is the Frobenius-Schur indicator. For a reducible representation the sum is still central in its image, but need not be one scalar on all irreducible constituents.

For an interface between two lossless, isotropic, nonmagnetic media with unequal refractive index values $n_1,n_2$, Brewster's angle is the incidence angle at which the reflected amplitude in transverse magnetic polarization vanishes. The Fresnel equations and Snell's law give $\tan {\theta }_B=n_2/n_1$; the reflected and transmitted rays are perpendicular there. The usual formula assumes equal magnetic permeabilities.

For the regular isothermal Lane-Emden equation, the first fixed-mass pressure turning point occurs at ${\xi }_{\mathrm{crit}}\simeq 6.451$. Thus the boundary radius there is $R_{\mathrm{crit}}\simeq 1.820c_s/\sqrt{G{\rho }_c}$, where $c_s^2=K$ is the squared isothermal sound speed. The approximate choice ${\xi }_{\mathrm{crit}}=6.5$ gives coefficient $1.83$. Comparing with the Jeans length requires specifying the reference density and remembering that the latter is a wavelength rather than a boundary radius.

For $N$ identical masses between fixed walls with equal springs, the dimensionless stiffness matrix has diagonal entries $2$ and adjacent entries $-1$. Its normalized eigenvectors and eigenvalues areExtend each vector by $a_0=a_{N+1}=0$. The sine addition identity gives $2a_i-a_{i-1}-a_{i+1}={\lambda }_j{a}_i$, including the fixed-end rows. The finite sine orthogonality identity supplies the normalization and makes distinct vectors orthogonal. The normal mode angular frequencies are ${\omega }_j=\sqrt{k{\lambda }_j/m}$ for mass $m$ and spring constant $k$. This complements the fixed-end spring-chain determinant and its repeated-root limit with the spatial mode shapes.

For the uniform noninteracting electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.