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Put $e_N(t)=e^{2\pi it/N}$ and use normalized averages ${\mathbb{E}}_x=N^{-1}{\sum }_{x\in \mathbb{Z}/N\mathbb{Z}}$. Our convention for Fourier coefficients on a finite abelian group isThe underlying orthogonality of complex exponentials is ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise. For nonzero $r$ this is the finite geometric series with ratio $e_N(r)\ne 1$; its sum vanishes because that ratio has $N$th power one. The identical orthogonality statement holds when averaging over $r$ instead of $x$.

Using this orthogonality to sum the definition over $r$ proves Fourier inversion on a finite group:Indeed $\mathbb{E}f\overline{g}={\sum }_r\overline{\hat{g}(r)}{\mathbb{E}}_{x}f(x)e_N(-rx)={\sum }_r\hat{f}(r)\overline{\hat{g}(r)}$. This proves the Parseval identity on a finite group and its squared-norm case:For normalized convolution on a finite group, write $(f\ast g)(x)={\mathbb{E}}_{y}f(y)g(x-y)$. Changing variables $x=y+z$ in its transform givesthe convolution theorem on a finite group. Linearity of the transform follows directly from its definition. Translation ${\tau }_{t}f(x)=f(x+t)$ gives $\widehat{{\tau }_{t}f}(r)=e_N(rt)\hat{f}(r)$ by changing variables. Complex conjugation gives $\hat{\overline{f}}(r)=\overline{\hat{f}(-r)}$; for $\tilde{f}(x)=\overline{f(-x)}$, the corresponding identity is $\hat{\tilde{f}}(r)=\overline{\hat{f}(r)}$. Finally $\hat{f}(0)=\mathbb{E}f$ and $|\hat{f}(r)|\le \mathbb{E}|f|$ by the triangle inequality. These identities establish the basic normalized Fourier analysis on a finite abelian group needed below, with all normalizations fixed.

For completeness, pointwise multiplication has the dual convolution formula $\widehat{fg}(r)={\sum }_s\hat{f}(s)\hat{g}(r-s)$: insert the inversion formula for $g$ into the defining average and collect its frequencies. If $c\ne 0$ modulo $N$, the change of variables $u=cx$ gives $\widehat{f(c\cdot )}(r)=\hat{f}(r{c}^{-1})$. These also follow directly from the same additive character orthogonality.

A general feedback shift register of length $d$ over an alphabet $A$ stores $(s_n,\dots ,s_{n+d-1})$, outputs $s_n$, and shifts to $(s_{n+1},\dots ,s_{n+d-1},F(s_n,\dots ,s_{n+d-1}))$ for a fixed function $F:A^d\to A$. When the alphabet is a field $K$, it is a linear-feedback shift register if $F$ is linear over $K$, equivalently $s_{n+d}={\sum }_{j=0}^{d-1}{a}_j{s}_{n+j}$ for fixed coefficients in $K$. The binary case uses the finite field $K={\mathbb{F}}_2$; the examples and counterexamples below are binary.

For a genuinely nonlinear example take length three and $F(a,b,c)=a+bc$. Its values at $(0,1,0)$ and $(0,0,1)$ are both zero, but its value at their sum $(0,1,1)$ is one, violating additivity. Thus this general register is not linear. It even has an invertible state update: from $(b,c,d)$ the old leading bit is $a=d+bc$, so the counterexample does not rely on allowing singular registers.

The ergodic hypothesis states that long-time averages of equilibrium observables along typical trajectories equal their ensemble averages on the accessible phase space with the same conserved quantities. It permits statistical mechanics to predict measured macroscopic quantities from equilibrium distributions rather than tracking a trajectory. In its microcanonical form the measure is on the constant-energy surface. It is an assumption about the relevant dynamics, not a consequence of particles being noninteracting; the calculations below use the equilibrium canonical ensemble at temperature $T$.

Pontryagin maximum principle leaves the initial costate unknown. Terminal conditions determine these unknown constants and, when terminal time is free, the time. In a maximization convention with terminal reward $\Phi$ and running cost $L$, use $H=p\cdot f-L$. A free terminal coordinate gives $p_i(T)={\partial }_i\Phi$; a fixed terminal coordinate instead allows a constraint multiplier. More generally $p(T)-\nabla \Phi$ is normal to the permitted terminal manifold. The free-terminal-time transversality condition is $H(T)+{\Phi }_T=0$. These conditions convert the adjoint and state equations into a boundary-value problem. Reversing the costate sign gives the equivalent minimization convention.

A model of three-dimensional hyperbolic space is ${\mathbb{H}}^3=\{(z,t):z\in \mathbb{C},t>0\}$ with metric $d{s}^2=(|dz{|}^2+d{t}^2)/t^2$, the complete simply connected metric of curvature $-1$. The translation length of an isometry $g$ is $\ell (g)={\inf }_{P}d(P,gP)$. A hyperbolic isometry has positive translation length attained on its invariant complete geodesic, the axis of a hyperbolic-space isometry. It joins the two fixed ideal points and is translated by distance $\ell (g)$; a three-dimensional axial isometry can also rotate around it.

For the boundary map $z\mapsto kz$, the hyperbolic extension of a Möbius transformation is $(z,t)\mapsto (kz,|k|t)$. Every path from $P$ to $gP$ has length at least the absolute change of $\log t$, namely $|\log |k||$, because the metric length dominates $|dt|/t$. On the vertical line $z=0$ a vertical segment attains that bound. HenceIf $|k|\ne 1$ the vertical line is the axis. If $|k|=1$ the map is a rotation with zero translation length, including the identity when $k=1$; it is not a positive-length hyperbolic isometry.

Set ${\theta }_k=k\pi /(M+1)$ and extend the proposed vector by $v_0=v_{M+1}=0$. For $v_m=i^m\sin (m{\theta }_k)$, the angle-addition identity givesHence the Toeplitz antisymmetric tridiagonal matrix has eigenvaluesThe real sine vectors are eigenvectors of the real symmetric tridiagonal matrix with both off-diagonals one, with distinct eigenvalues $2\cos {\theta }_k$; thus they are orthogonal and form a basis. Multiplication by the unitary diagonal matrix $\mathrm{diag}(i,i^2,\dots ,i^M)$ preserves orthogonality. Their squared norms are $(M+1)/2$, so multiplying by $\sqrt{2/(M+1)}$ gives a common orthonormal basis for all $a,b$, even when some eigenvalues coincide.

Let $K$ have superdiagonal one, subdiagonal minus one and diagonal zero. The prescribed endpoint values remove the exterior unknowns, and the scheme isIn the common orthonormal eigenbasis, the two eigenvalues are $1\mp i(\mu /2)\cos {\theta }_k$. Every eigenvalue of $B$ is nonzero for real $\mu$, and the amplification matrix $Q=B^{-1}C$ has eigenvaluesThus $Q$ is a unitary matrix and $\Vert u^{n+1}{\Vert }_2=\Vert u^n{\Vert }_2$ exactly. The same holds for the mesh-weighted norm. This matrix argument proves $\boxed{\text{stability for every real }\mu \text{, hence every physical }\mu \ge 0}$, without a Fourier stability calculation.

The printed mesh ratio has a consistency error: advection requires $\mu =\Delta t/\Delta x$, while the PDF prints $\Delta t/(\Delta x{)}^2$. Indeed the centered spatial difference is $2\Delta x{u}_x+O((\Delta x{)}^3)$, so division of the scheme by $\Delta t$ gives leading right-hand coefficient $\mu \Delta x/\Delta t$. The printed choice makes this $1/\Delta x$ rather than one. The stability conclusion is valid for the algebraic scheme with either definition, but does not fix this inconsistency. The continuous first-order advection problem also ordinarily specifies only the inflow endpoint, here $x=1$; zero data at both endpoints restrict admissible initial data. For example the initially boundary-zero profile $\sin \pi x$ develops nonzero outflow $u(0,t)=\sin \pi t$ before inflow changes reach that boundary. The matrix norm calculation concerns the stated two-endpoint discretization and does not establish well-posedness for arbitrary continuous initial data.

The left group action on the homogeneous polynomials is ${\rho }_n(g)P(z)=P(g^{-1}z)$, since ${\rho }_n(g){\rho }_n(h)={\rho }_n(gh)$. Every SU(2) matrix is conjugate to $\mathrm{diag}(e^{i\theta },e^{-i\theta })$ with $0\le \theta \le \pi$. In the monomial basis $z_1^{n-j}{z}_2^j$, its eigenvalues are $e^{-i(n-2j)\theta }$. Consequently the character of the homogeneous polynomial representation of SU2 isThe quotient is interpreted by continuity at the endpoints: ${\chi }_n(0)=n+1$ and ${\chi }_n(\pi )=(-1{)}^n(n+1)$. The introductory action and character requests are independent of parts (i)–(iii), so this root Solution slot has been restored.

The angles $\psi$ and $\varphi$ are cyclic. Their conjugate momentum components are $p_{\psi }=I_3(\dot{\psi }+\dot{\varphi }\cos \theta )$ and $p_{\varphi }=I_1\dot{\varphi }{\sin }^2\theta +p_{\psi }\cos \theta$, so both are conserved; consequently ${\omega }_3=p_{\psi }/I_3$ is constant. With $q=I_3{\omega }_3=p_{\varphi }>0$,Eliminating the cyclic velocities gives the nutation effective potentialapart from the constant axial spin energy. ThusThe endpoint $\theta =\pi$ is excluded for these nonzero conserved momenta: the effective potential diverges there. The apparent Euler-angle singularity at $\theta =0$ is interpreted by its smooth small-tilt limit.

Lubrication theory exploits a film thickness $b$ much smaller than its horizontal variation scale $\ell$. Incompressibility then gives vertical velocity of order $(b/\ell )U$, vertical viscous derivatives dominate horizontal ones, and inertia is negligible when $\rho U{b}^2/(\mu \ell )\ll 1$. The leading equations reduce to $\mu u_{yy}=p_x$ and vertical hydrostatic balance $p_y=-\rho g$. Boundary conditions determine the local horizontal velocity profile; integrating it across the film and enforcing mass conservation produces an equation for flux or surface shape. This reduction turns the thin-film problem into a one-dimensional pressure problem while retaining the dominant viscous stresses.

In SI units, Gauss's law relates the outward electric flux through a closed surface to its enclosed charge:The outward unit normal specifies the sign of the electric flux.

The modified transition matrix $\tilde{P}$ describes the original Markov chain until it hits $B$, and then keeps it at the first state hit. Each state of $B$ is therefore an absorbing state, andFor a nonnegative superharmonic function for a Markov chain $g$ satisfying the boundary condition, part (c) and positivity of the transition matrix give by induction $g\ge {\tilde{P}}^{n}g$. Also $g\ge 1_B$. ConsequentlyLet $n\to \infty$ to obtain $g_i\ge h_i$. Since $h$ satisfies the conditions by parts (a),(b), $h$ is their pointwise minimal solution. This extends the hitting probability is the minimal nonnegative harmonic extension principle to the superharmonic inequality. Nonnegative sums justify the argument on a countable state space as well as a finite one.

Now suppose the Markov chain is irreducible and recurrent, and let $g$ satisfy condition (a). In such a chain every state is hit almost surely from every other state. To see this directly, recurrence gives infinitely many visits to a starting state $i$. Irreducibility supplies a finite path from $i$ to any target $j$ with positive probability $\alpha$, chosen with no return to $i$ before reaching $j$. At each successive return to $i$, the Strong Markov property gives another trial with conditional success probability at least $\alpha$. The chance of avoiding $j$ through $N$ such trials is at most $(1-\alpha {)}^N$, which tends to zero.

Apply the absorbing modification at the singleton $\{j\}$. Part (c) holds for arbitrary $g$ satisfying (a), soSince $g_j\ge 0$ and the probability tends to one, $g_i\ge g_j$. Interchanging $i,j$ yields equality. Thus every nonnegative superharmonic function is constant on an irreducible recurrent Markov chain.

Conversely, if an irreducible Markov chain is not recurrent, choose a state $j$ with return probability $r_j={\mathbb{P}}_j({\tau }_j^{+}<\infty )<1$. Its hitting probability $h$ for $\{j\}$ satisfies condition (a), and $h_j=1$. The first-step formula for a return after time zero gives $(Ph{)}_j=r_j<1$. Therefore $h$ cannot be constant: a constant function with value one at $j$ would have $(Ph{)}_j=1$. This supplies a nonconstant nonnegative superharmonic solution. HenceThe admissible constant solutions are precisely the nonnegative constants. This is recurrence characterized by constant nonnegative superharmonic functions.

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

Use the steady inviscid, hydrostatic, slowly varying channel approximation, without a hydraulic jump. Conservation of volume flux per unit width gives $Q=u(x)h(x)=u_1{h}_1$. The free-surface Bernoulli equation isDividing by $h_1$ yieldsThe parameter $F$ here is the squared upstream Froude number. For nonzero discharge, regarding $D$ as a function of $h$, its derivative is $dD/dh=-1+Q^2/(g{h}^3)$. It has its unique maximum at $h_c=(Q^2/g{)}^{1/3}=h_1{F}^{1/3}$, where the local Froude number is one. Substitution gives $D^{\ast }=h_1(1-3F^{1/3}/2+F/2)$. This maximum is the hydraulic control over a smooth hump.

Let $v\ne 0$ be an eigenvector of the Hermitian matrix $A$, with eigenvalue $\lambda$. Multiplying $Av=\lambda v$ by $v^{†}$ givesThe denominator is the positive real number $\Vert v{\Vert }^2$, whileThus the numerator is real, and every eigenvalue of a Hermitian matrix is real. This argument does not require distinct eigenvalues.

For a complex matrix, its Hermitian conjugate is obtained by transposing and taking entrywise complex conjugates:The defining conditions areFor a square matrix, either one of the two inverse identities in the unitary matrix condition implies the other. A Hermitian matrix may have complex entries, but its diagonal entries are real.