Put $s=j_1+j_2$, with both $j_i>0$. At magnetic quantum number $-s+1$, let $|a⟩$ raise only the first factor of the lowest product state and $|b⟩$ raise only the second. Applying total $J_{+}$ to the lowest product gives the normalized $J=s$ combination $\sqrt{j_1/s}|a⟩+\sqrt{j_2/s}|b⟩$. Its orthogonal combination is annihilated by total $J_{-}$ and therefore has $J=s-1$. This obtains a useful Clebsch-Gordan coefficients without a table.

Apply a total ladder operator to a coupled state and evaluate against an uncoupled product state. The total operator is the sum of the two factor operators, so its one coupled ladder amplitude times the corresponding one of the Clebsch-Gordan coefficients equals the sum of two factor ladder amplitudes times shifted coefficients. An overall common rescaling of the ladder operators cancels from the recurrence. Nonzero coefficients obey $m_1+m_2=m$, and the positive ladder amplitudes require a consistent phase convention within each irreducible summand.

// Target: lie-theory.bigb

For a path-connected, locally path-connected, semilocally simply connected space $X$, base point $x_0$, and subgroup $H\le {\pi }_1(X,x_0)$, take paths starting at $x_0$ modulo the relation $\alpha \sim \beta$ when their endpoints agree and $[\alpha {\beta }^{-1}]\in H$. Projection sends a class to its endpoint. Over a path-connected neighbourhood whose loops die in $X$, classes $[\alpha \eta ]$, with $\eta$ inside that neighbourhood, form a sheet mapping homeomorphically to it. These sheets define a covering space. A loop lifts closed precisely when its class belongs to $H$, so the injective induced fundamental group map has image $H$.

For a semilinear partial differential equation $u_t=\Delta u+F(u)$ with smooth $F$ and smooth fixed boundary data, local boundedness permits positive-time elliptic regularity and parabolic smoothing. If $u_t$ is continuous with values in a Hölder space, elliptic estimates first give continuous $C^{2,\beta }$ spatial regularity for $u$. Time difference quotients then solve a linear parabolic equation with coefficient converging to $F^{\prime }(u)$; smoothing away from the initial time gives $u_t\in C(C^2)$, hence $u\in C^1(C^2)$. It is essential to gain spatial regularity for the time derivative too; $u\in C(C^2)\cap C^1(C^0)$ alone does not imply $C^1(C^2)$.

An argument has logical validity in classical propositional logic when no Boolean valuation makes all premises $\Gamma$ true and the conclusion $A$ false: $\Gamma \models A$. With no premises, $\models A$ says that $A$ is a tautology. See forall x: Calgary, sections 12.4–12.6.

A tautology is a propositional formula true under every Boolean valuation in classical propositional logic, such as the law of excluded middle $p\vee \neg p$. A primitive proposition $p$ is not a tautology, even if it represents a true statement. See forall x: Calgary, section 12.1.

The zero-field Hubbard–Stratonovich transformation of an exponentially coupled Ising model produces $U(m)=A{m}^2-\log [2\cosh (2m)]$, where $A=\tanh (\kappa /2)/J>0$. Since $U^{\prime \prime }(0)=2A-4$ and $U$ tends to positive infinity, a symmetric double well exists exactly when $A<2$. Its nonzero stationary points solve $Am=\tanh (2m)$; the ratio $\tanh (2m)/m$ decreases from 2 to 0 for $m>0$, proving that there is exactly one positive minimum. At $A=2$ the minimum is a single quartic minimum. A small conjugate field adds $-h\tanh (2m)+O(h^2)$ and tilts the wells, while a sufficiently large field can remove the metastable well. A finite barrier permits quantum tunnelling; two classical minima alone do not establish spontaneous symmetry breaking.

For a nearest-neighbor auxiliary-field representation of an Ising model, sharing each local potential between adjacent bonds gives the displayed symmetric integral operator. Its periodic partition function is the trace of its $N$th power, up to Gaussian normalization constants. In the slowly varying continuum limit, a slice of length $ϵ$ has kernel $[M/(2\pi ϵ){]}^{1/2}\exp [-M(m^{\prime }-m{)}^2/(2ϵ)-ϵ\{U(m^{\prime })+U(m)\}/2]$. Expanding its action on a smooth function gives $1-ϵ\hat{H}+O({ϵ}^2)$, with $\hat{H}=-(2M{)}^{-1}{\partial }_m^2+U$. Thus fixed endpoints give an imaginary-time transition kernel and periodic endpoints give a Euclidean path integral trace. The finite-spacing integral operator and the continuum differential operator need not have identical high-energy spectra.

The thermal de Broglie wavelength is a length scale from quantum mechanics entering the classical ideal-gas partition function for nonrelativistic particles of mass $m$ at temperature $T$:

For one free particle in a large three-dimensional volume $V$, the Gaussian integral over momentum givesThe factor $h^{-3}$ counts phase space cells; it is essential for a dimensionless partition function. The thermal wavelength represents a thermal momentum scale rather than a wavelength shared by every particle. ^[ref]

For a single internal state, $n{\lambda }^3\ll 1$ is the dilute classical regime. When $n{\lambda }^3$ becomes comparable to one, Bose-Einstein statistics or Fermi-Dirac statistics matters. Cooling increases $\lambda$ as $T^{-1/2}$, bringing quantum degeneracy within reach even at fixed number density $n$. ^[ref]

For the potential $U=r^{2n}/V^{2n/3}$, rescaling $r=V^{1/3}(kT{)}^{1/(2n)}u$ in the classical one-particle partition function gives the formula above. The parameter-conjugate force is $P=NkT/V$, the internal energy is $(3/2+3/(2n))NkT$, and $C_P-C_V=Nk$. The kinetic term contributes $3NkT/2$, while the potential term contributes $3NkT/(2n)$. For indistinguishable particles the factorial $N!$ changes free energy and entropy but not these force and heat-capacity formulas.

For continuous real coefficients with $p>0$ and $q,r\ge 0$ on a bounded interval, use the clamped second-order Sobolev space and the displayed bilinear form. A forcing in $L^2$ defines $\ell (v)=\int fv$. The weak solution satisfies $a(u,v)=\ell (v)$ for every clamped test vector. Compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution; density extends that equation to the energy space. Continuous coefficients alone do not justify differentiating them classically. The clamped interval derivative norm bound gives coercivity, so Lax-Milgram theorem supplies a unique solution and the quadratic energy has that unique minimum.

For $v\in H_0^2(-1,1)$, integrate its weak derivative from the left endpoint, where $v=v^{\prime }=0$. The Cauchy-Schwarz inequality gives $\Vert v^{\prime }{\Vert }_2\le 2\Vert v^{\prime \prime }{\Vert }_2$ and $\Vert v{\Vert }_2\le 2\Vert v^{\prime }{\Vert }_2$. Adding the three squared derivative norms gives the displayed bound. The constant is sufficient, rather than asserted sharp. Thus the second-derivative norm is equivalent to the full Sobolev norm on this clamped space.

Claims inflation changes the monetary size of a claim over calendar time. With a fixed annual rate $r$, a time-zero size $c_0$ becomes the displayed end-of-year amount. Historical claim amounts must be converted to a common price year before their empirical monetary averages are combined. When the per-claim amounts are known, dividing total amount by the year's per-claim amount recovers the claim count.

An inner NFW profile has $M_i(r_i)=A{r}_i^2$, where $A=2\pi {\rho }_s{r}_s$. If the final total circular speed is $V$, circular-shell adiabatic contraction gives $r_i=(V^2/(GA){)}^{1/3}{r}^{2/3}$ and $M_D^{\prime }=A^{1/3}(V^2/G{)}^{2/3}{r}^{4/3}$. Thus ${\rho }_D^{\prime }=A^{1/3}(V^2/G{)}^{2/3}{r}^{-5/3}/(3\pi )$. The total density producing the flat curve is $V^2/(4\pi G{r}^2)$, so the dark fraction tends to zero as $r^{1/3}$ despite its increase relative to the original cusp.

A synchronous circular Kepler orbit has an orbital period equal to its central body's sidereal rotation period $T$. For the ideal spherical planet of mass $M$, Kepler's third law givesThe satellite's own mass cancels in the test-particle approximation. The radius must exceed the planet's physical radius for an external orbit.

Matching periods alone does not keep the satellite above a fixed surface point. That stronger condition requires a prograde equatorial circular orbit, giving a planet-stationary orbit; the terrestrial case is a geostationary orbit. An inclined synchronous circular orbit moves north and south relative to the equator despite its matching period. This distinction is useful when choosing between repeated ground coverage and an apparently fixed satellite.

For two nonempty planar circles with centers $c_i$ and radii $r_i\ge 0$, the greatest Euclidean distance between their points is the displayed expression. The triangle inequality gives the upper bound. When the centers differ, choose the two boundary points outward along the line of centers to attain it; when they coincide, choose opposite radial directions. For any third nonempty circle and a point $u$ on it, $|z-w|\le |z-u|+|u-w|$ proves $m(C_1,C_2)\le m(C_1,C_3)+m(C_2,C_3)$. This is not a metric, since $m(C,C)=2r$ for a positive-radius circle.

For real $\alpha \ne 0,\gamma$ and complex $\beta \ne 0$, completing the square in $\alpha |z{|}^2-\beta \bar{z}-\bar{\beta }z+\gamma =0$ gives center $\beta /\alpha$ and radius $\sqrt{|\beta {|}^2-\alpha \gamma }/|\alpha |$, when the radicand is nonnegative. The displayed map is reflection in the line through $0$ and $\beta$: it preserves the complex modulus, fixes the center, and squares to the identity. Hence it maps the circle onto itself, including a radius-zero degeneration.

A chord joins two points of a circle by a straight segment. If the circle has radius $r$ and the chord midpoint lies a distance $d$ from its center, the perpendicular bisector and Pythagorean theorem give the displayed length. The length exceeds that of an inscribed equilateral triangle precisely when $d<r/2$.

Half of a circle, with endpoints at opposite ends of a diameter. For radius $R>0$, the upper semicircle in the complex plane is parametrized by $z=R{e}^{it}$, $0\le t\le \pi$, going counterclockwise from $R$ to $-R$.

For integers $1\le r\le n$, if a simple graph has the same chromatic polynomial as the Turán graph $T_r(n)$, it has the same number of vertices and edges, and the same chromatic number $r$. It is therefore $K_{r+1}$-free and attains equality in Turán's theorem. The equality case forces it to be isomorphic to $T_r(n)$.