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In the hyperbolic-ball model, the limit set of a Kleinian group is the boundary accumulation set of an orbit $Go$ of an interior point: $\Lambda (G)=\overline{Go}\cap \partial {\mathbb{H}}^3$. The choice of $o$ does not matter, since any two such orbits remain a fixed hyperbolic distance apart. It is closed and $G$-invariant. The complicated/non-elementary hypothesis excludes relatively compact groups and elementary groups with at most two limit points; in particular the limit set is nonempty. Having no finite boundary orbit alone would not exclude a compact rotation group, whose interior orbits are bounded.

If $\Lambda (G)$ were finite and nonempty, the orbit of any of its points would be a finite boundary orbit, contrary to the hypothesis. Thus it is infinite. In fact it cannot be countable. We give the essential minimality/perfectness argument. A sequence of Möbius transformations sending an interior point toward $\xi \in \Lambda$ has a subsequence converging on the boundary away from one exceptional point to the constant $\xi$. This follows by normalizing the representing $2\times 2$ matrices: a diverging sequence has a rank-one projective limit, with one kernel line and one image line. For any boundary point $\eta$, the absence of a finite orbit allows choosing $h\in G$ with $h\eta$ different from that exceptional point; then $g_{j}h\eta \to \xi$. Hence the orbit of every $\eta \in \Lambda$ is dense in $\Lambda$.

If $\xi$ were an isolated limit point, density of every orbit would make every point belong to its orbit: each orbit must hit the open singleton $\{\xi \}$. Every point would then be isolated by equivariance, making the compact set $\Lambda$ discrete and therefore finite, a contradiction. Thus $\Lambda$ is a nonempty perfect compact metric space. Repeatedly choosing two disjoint smaller neighbourhoods meeting this perfect set, with diameters tending to zero, associates distinct nested limits to all infinite binary sequences. The limit set is therefore uncountable.

Jorgensen inequality states that for a non-elementary discrete group generated by lifts $A,B\in \mathrm{SL}(2,\mathbb{C})$,For a concrete failure of discreteness, take $A=\mathrm{diag}(e^{s/2},e^{-s/2})$ and $B=\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)$ with $s>0$ small. The fixed points of $A$ and of $BA{B}^{-1}$ collectively include $0,\infty ,-1$, so the group is non-elementary. Direct multiplication gives both summands equal to $4{\sinh }^2(s/2)$. Their sum is less than one when $s$ is sufficiently small, proving this two-generator group is not discrete.

For a connected-limit-set example, the modular group generated by $z\mapsto z+1$ and $z\mapsto -1/z$ is discrete because its matrices have integer entries. It preserves the upper half-plane; its limit set is the extended real line, a circle. Indeed the orbit of the parabolic fixed point infinity consists of the rational points and infinity, dense in that circle. For a disconnected example, choose four pairwise disjoint round disks and two loxodromic maps pairing the exterior of each negative disk with the interior of its positive disk. The Schottky group generated by these maps is free and discrete by the disjoint-domain ping-pong construction. Successive reduced words give nested disjoint disks with shrinking diameters; their limiting points form a Cantor set. Thus its limit set is disconnected, unlike that of the modular group.

The triangular-grid form is Sperner's lemma: subdivide a triangle into small triangles meeting edge to edge, label its vertices $1,2,3$, and forbid label $i$ on the side opposite corner $i$. The corner labels are thereby forced. An odd number of small triangles, in particular at least one, carry all three labels.

Here is the parity proof. Count incidences of a small triangle with an edge whose endpoint labels are $1,2$. On the outer boundary these edges occur only on the side joining corners $1,2$; traversing that side begins with label one and ends with label two, so the number of changes of label is odd. Interior edges contribute twice to the incidence count. A small triangle with all three labels contributes once; a triangle having only labels one and two contributes twice if it has both, and all other triangles contribute zero. Reduction modulo two therefore proves the assertion.

To obtain the continuous Brouwer fixed-point theorem on the triangle, use barycentric coordinates and a continuous map $F$ from the triangle to itself. If a grid vertex is not already fixed, label it by an index $i$ with $x_i>F_i(x)$, which exists since both coordinate sums are one. A point on the side $x_i=0$ cannot receive label $i$, so the discrete lemma applies. On successively finer grids take a fully labelled small triangle. Compactness gives a subsequence converging to one point $x$; all three vertices converge to it because the mesh tends to zero. Passing to the limit in the three label inequalities gives $x_i\ge F_i(x)$ for every $i$. Equal coordinate sums force equality throughout, hence $F(x)=x$.

A closed disk is homeomorphic to a triangle: radial coordinates from an interior point, scaled by the distance to the boundary in each direction, give such a homeomorphism. Conjugation therefore gives the same fixed-point assertion on the disk. If $r$ were a continuous retraction from the unit disk to its boundary, define $F(x)=-r(x)$. A fixed point would have $|x|=1$, hence $r(x)=x$ and $x=-x$, impossible on the unit circle. Thus no continuous disk-to-boundary retraction exists. The continuous conclusion has been deduced from the grid parity argument, not used to prove it.

Let $S_m$ be the simple symmetric random walk started at the origin. Odd-time returns are impossible by parity. A return in $2n$ steps requires equal numbers of positive and negative steps along each axis. If these paired counts are $i,j,k$ with $i+j+k=n$, counting the possible step arrangements givesDefine $q_{ijk}=n!/(3^{n}i!j!k!)$. These are multinomial distribution probabilities and sum to one. Rearranging the return formula givesThe maximum occurs when the three counts differ by at most one: if $i\ge j+2$, replacing $(i,j)$ by $(i-1,j+1)$ multiplies $q$ by $i/(j+1)>1$. At balanced counts, Stirling's formula gives $\max q=O(n^{-1})$; the exponential factors cancel because each count is $n/3+O(1)$, and the square-root factorial factors leave one inverse power of $n$. The same formula gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim (\pi n{)}^{-1/2}$. HenceThis is the multinomial collision proof of three-dimensional walk transience. The sum is the expected total number of visits to the origin. If the probability of returning after departure were one, the Strong Markov property at successive returns would make every successive return finite almost surely, giving infinitely many visits and contradicting that finite expectation. Thus the origin is a transient state. Translation invariance gives the same conclusion at every lattice point: the three-dimensional walk is transient.

An $n$-node Gaussian quadrature formula chooses nodes and weights for an integral with a positive weight so that every polynomial of degree at most $2n-1$ is integrated exactly. Its nodes are the zeros of the degree-$n$ orthogonal polynomial for that weight.

For the even weight $w(x)=1-x^2$, the monic orthogonal quadratic has the form $p_2(x)=x^2-d$. Orthogonality to $x$ follows from oddness; orthogonality to one givesThe nodes are $\pm 1/\sqrt{5}$. The equal weights sum to $4/3$, hence each is $2/3$. The two-node Gaussian quadrature with quadratic weight is thereforeTo verify cubic exactness, divide any cubic as $P=p_{2}Q+R$ with $\mathrm{deg}Q,\mathrm{deg}R\le 1$. Orthogonality removes $p_{2}Q$ from the integral and its nodal values vanish; linear interpolation integrates $R$ exactly. This also proves the Gaussian property rather than merely matching two moments.

For static fields, the Maxwell equations give $\nabla \times B={\mu }_{0}J$ and $\nabla \cdot B=0$. Write $B=\nabla \times A$. A gauge transformation $A\mapsto A+\nabla \chi$ leaves $B$ unchanged; choose $\Delta \chi =-\nabla \cdot A$ to impose the Coulomb gauge. ThenFor a localized conserved current with the potential chosen to vanish at infinity, the supplied fundamental solution gives the magnetic vector potentialIts divergence vanishes by integration by parts and $\nabla \cdot J=0$. For a filamentary loop the current integral becomes $I{\oint }_{L}dr$, soFor $R=|x|$ much larger than the fixed loop size, expand $|x-r{|}^{-1}=R^{-1}+(x\cdot r)R^{-3}+O(R^{-3})$. The leading constant term integrates to zero because $\oint dr=0$. ThereforeThe vector triple-product identity gives $x\times (r\times dr)=r(x\cdot dr)-(x\cdot r)dr$. Integrating the derivative of $(x\cdot r)r$ around the closed loop shows $\oint r(x\cdot dr)=-\oint (x\cdot r)dr$. ConsequentlyEquivalently, $A={\mu }_{0}m\times x/(4\pi R^3)+O(R^{-3})$, where $m=(I/2)\oint r\times dr$ is the magnetic dipole moment. The leading displayed potential is of order $R^{-2}$; the remainder is smaller by a factor $R^{-1}$.

The operator commutator is $[A,B]=AB-BA$, acting on a common domain where both products are defined. Use the canonical commutation relation $[x_i,p_j]=i\hslash {\delta }_{ij}$, with positions mutually commuting and momenta mutually commuting. On smooth test functions the only nonzero terms in the expansion areThus the orbital angular momentum commutation relations giveThe product rule for a operator commutator givesbecause the last operator expression is symmetric in $i,k$, whereas ${ϵ}_{iwk}$ is antisymmetric. Hence $\boxed{[L^2,L_w]=0}$ for every component.

Write $r=\sqrt{x^2+y^2+z^2}$ and $P=x+y+z$. The differential angular momentum operator $L=-i\hslash x\times \nabla$ annihilates every radial function, so $L_i(P{e}^{-r})=e^{-r}{L}_{i}P$ and the same holds for its square. For example $L_{x}x=0$, $L_y^{2}x={\hslash }^{2}x$ and $L_z^{2}x={\hslash }^{2}x$. Cyclic symmetry gives $L^{2}y=2{\hslash }^{2}y$ and $L^{2}z=2{\hslash }^{2}z$ as well. ThereforeThe wavefunction is an $l=1$ orbital angular momentum state; its nontrivial radial factor does not change its angular eigenvalue. The calculation holds away from the origin and defines the same angular-operator identity in the square-integrable state space.

Take characteristic coordinates $\xi =ct-x$ and $\eta =ct+x$. Then $c^{-2}{y}_{tt}-y_{xx}=4y_{\xi \eta }$. The wave equation is therefore $y_{\xi \eta }=0$, whose general twice-differentiable solution isThe first fixed-end boundary gives $f(s)=-g(s)$. The second gives $g(s+L)=g(s-L)$, hence $g(s+2L)=g(s)$; $f$ has the same period. Thus $y=g(ct+x)-g(ct-x)$ with a $2L$-periodic function $g$.

Let $A=g^{\prime }(ct+x)$ and $B=g^{\prime }(ct-x)$. Then $y_t=c(A-B)$ and $y_x=A+B$, so the quadratic energy density reduces to $\frac{1}{2}[(A-B{)}^2+(A+B{)}^2]=A^2+B^2$. Changing $x$ to $-x$ in the second integral provesThe right side integrates a periodic function over one full period and is independent of its starting point. More explicitly its time derivative is $c[g^{\prime }(ct+L{)}^2-g^{\prime }(ct-L{)}^2]=0$. This is conservation of the fixed-end wave energy.

The integral is finite and nonnegative because $|f|$ is continuous on a compact interval. If it vanishes and $f(x_{\ast })\ne 0$, continuity gives a relative interval of positive length on which $|f|\ge |f(x_{\ast })|/2$, a contradiction. Thus the integral is zero only for $f=0$. For a real scalar $a$, $\Vert af\Vert =|a|\Vert f\Vert$, and the pointwise triangle inequality $|f+g|\le |f|+|g|$ integrates to $\Vert f+g\Vert \le \Vert f\Vert +\Vert g\Vert$. These are all the norm axioms.

The given sequence satisfiesHence it converges to the zero function in this integral norm. Although $f_n(0)=1$ for all $n$, an isolated endpoint contributes no integral. The pointwise limit is discontinuous, and the sequence does not converge in the uniform norm; neither fact obstructs convergence in the stated norm.

The Eisenstein criterion says that a primitive integer polynomial $P(x)=a_m{x}^m+\cdots +a_0$ is irreducible over $\mathbb{Q}$ if some prime number $p$ divides every $a_j$ for $j<m$, does not divide $a_m$, and $p^2$ does not divide $a_0$. By Gauss lemma for polynomials, this gives irreducibility in $\mathbb{Z}[x]$ as well.

For prime $n=p$, translate the variable in $P_p(x)=(x^p-1)/(x-1)$:Its leading coefficient is one, all other coefficients are divisible by $p$, and its constant coefficient is exactly $p$. The Eisenstein criterion proves this translated polynomial irreducible. Since $x\mapsto x+1$ is an invertible substitution in $\mathbb{Z}[x]$, $P_p$ is irreducible too.

If $n=ab$ with $a,b>1$, thenBoth factors have positive degree and integer coefficients, proving reducibility. Consequently $\boxed{1+x+\cdots +x^{n-1}\text{ is irreducible exactly when }n\text{ is prime}.}$

Start with a feasible network flow, meaning capacity bounds and flow conservation at every vertex other than source and sink. Its residual network has, for an arc with capacity $c$ and current flow $f$, a forward residual capacity $c-f$ and a reverse residual capacity $f$. A reverse arc allows earlier flow to be cancelled. The Ford-Fulkerson algorithm repeatedly finds a residual source-to-sink augmenting path, takes the smallest residual capacity $\delta$ on it, and augments by $\delta$, increasing forward flows and decreasing reverse flows as appropriate. Feasibility is preserved and the total flow increases by $\delta$.

For rational capacities and initial flows, choose one common denominator $D$ and scale all quantities by $D$. Every residual capacity and augmentation is then integral; each augmentation increases the integer-valued flow by at least one. The flow value is bounded by the finite sum of capacities out of the source, so termination occurs after finitely many augmentations. When no residual path remains, let $U$ be the vertices reachable from the source. Every original arc leaving $U$ is saturated and every original arc entering $U$ carries zero flow, or its reverse arc would extend reachability. Conservation then makes the flow value equal to the capacity of the cut $(U,U^c)$. Every feasible flow is bounded by that cut capacity. Thus the terminating flow is globally optimal, proving the needed max-flow min-cut theorem conclusion.

For the given directed graph, begin at zero. One valid sequence of augmentations is:

Each amount is the current path bottleneck. The resulting nonzero arc flows areevery other arc has zero flow. Directly summing incoming and outgoing flows verifies conservation, and every displayed flow is at most its printed capacity. A minimum-cut certificate isIts only outgoing arcs are $SD$, $EK$, $HJ$ and $BT$, with capacities $10,6,8,6$. ThereforeThe matching feasible flow and cut prove optimality independently of the augmenting-path choices.

Use the standard complex inner product $⟨u,v⟩=u^{\ast }v$, conjugate-linear in its first argument. A unitary matrix satisfies $U^{\ast }U=I$, and a Hermitian matrix satisfies $H^{\ast }=H$, where $\ast$ is conjugate transpose.

If $Hv=\lambda v$ with $v\ne 0$, thenPositivity of $⟨v,v⟩$ proves $\boxed{\lambda \in \mathbb{R}}$. If $Uv=\lambda v$, preservation of the norm gives $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\lambda {|}^2\Vert v{\Vert }^2$, so $\boxed{|\lambda |=1}$.

For eigenvectors $Hv=\lambda v$ and $Hw=\mu w$, the Hermitian matrix identity gives $⟨v,Hw⟩=⟨Hv,w⟩$. The already established reality of the eigenvalues therefore implies $(\mu -\lambda )⟨v,w⟩=0$. Thus distinct eigenvalues have orthogonal eigenvectors.