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Take an open neighbourhood $\tilde{U}$ on which $p$ is one-to-one and a homeomorphism onto its image. By invariance of domain for topological surfaces, that image is open. Shrink $\tilde{U}$ so the image lies in a complex chart $(V,z)$ of $R$, and use $z\circ p$ as a chart of $\tilde{R}$. These charts cover $\tilde{R}$. On any overlap their transition is the transition between the original charts of $R$, restricted to an open set, so is holomorphic with holomorphic inverse. They define the desired complex structure, and in these charts $p$ is locally the identity. Thus $p$ is holomorphic. The openness supplied by invariance of domain is important when the hypothesis is phrased only as a homeomorphism onto the image.

Let $f:\mathbb{C}\to Y$ be holomorphic and let $\pi :\mathbb{D}\to Y$ be the stated covering. Choose a lift of $f(0)$. The covering-space lifting theorem says that a map from a simply connected, locally path-connected domain has a unique lift after fixing this initial point: lift paths from zero, and homotopy lifting makes their endpoint independent of path. Since $\mathbb{C}$ is simply connected, this constructs $\tilde{f}:\mathbb{C}\to \mathbb{D}$ with $\pi \circ \tilde{f}=f$. On a small neighbourhood, the lift is a biholomorphic local inverse of $\pi$ composed with $f$, so it is holomorphic. Its modulus is less than one, and Liouville theorem therefore makes it constant. Hence every holomorphic map $\mathbb{C}\to Y$ is constant.

For a bounded operator on a complex Banach space, the resolvent set is $\rho (T)=\{\lambda :\lambda I-T\text{ is bijective with bounded inverse}\}$, and the resolvent operator is $R_T(\lambda )=(\lambda I-T{)}^{-1}$. The spectrum is its complement $\sigma (T)=\mathbb{C}\setminus \rho (T)$; the point spectrum consists of $\lambda$ with $\ker (\lambda I-T)\ne 0$. For a real Banach space these definitions use its complexification.

If $|\lambda |>\Vert T\Vert$, the norm-convergent Neumann series gives $R_T(\lambda )={\lambda }^{-1}{\sum }_{n\ge 0}(T/\lambda {)}^n$, so the spectrum is bounded. If ${\lambda }_0\in \rho (T)$, factorThe second factor is invertible by a Neumann series whenever $|\lambda -{\lambda }_0|\Vert R_T({\lambda }_0)\Vert <1$. Thus the resolvent set is open and the spectrum is closed and bounded.

The point spectrum need not be closed. On ${\ell }^2$, define $T{e}_n=e_n/n$. Every $1/n$ is an eigenvalue, but zero is not: $Tx=0$ forces every coordinate of $x$ to vanish. No other eigenvalues occur, so ${\sigma }_p(T)=\{1/n:n\ge 1\}$ is not closed.

Finally, if $T$ is self-adjoint and $Tv=\lambda v$, $v\ne 0$, then $⟨Tv,v⟩$ is real and equals $\lambda \Vert v{\Vert }^2$ under the convention linear in the first argument. Therefore every eigenvalue of a self-adjoint operator is real.

Cut the second circle factor at $p,q$. This divides the torus into two annuli whose boundary circles are exactly the two collapsed circles. Collapsing each annulus's two boundaries separately produces a two-sphere; the two spheres then share their two poles. In particular, the two collapsed circles give two distinct vertices, not a single common collapsed point.

For an explicit cellular homology calculation, each suspended circle contributes one edge from the first vertex to the second and one two-cell attached with zero cellular boundary. Thus $C_0\cong {\mathbb{Z}}^2$, $C_1\cong {\mathbb{Z}}^2$, $C_2\cong {\mathbb{Z}}^2$, with no higher cells. Both edges have boundary $v_q-v_p$, so ${\partial }_1$ has rank one and kernel generated by the difference of the edges. Each two-cell's attaching loop traverses its edge in opposite directions, so ${\partial }_2=0$. Consequently the homology of a torus with two parallel circles collapsed isCollapsing either connecting edge as a maximal tree also exhibits the homotopy type $S^1\vee S^2\vee S^2$.

Since $\mathcal{O}=\mathbb{Z}[\theta ]$, reduction gives $\mathcal{O}/p\mathcal{O}\cong {\mathbb{F}}_p[X]/(\bar{g})$. The maximal ideals of this quotient are precisely those generated by the distinct irreducible factors ${\bar{g}}_i$. Their inverse images are thereforethe distinct prime ideals above $p$. In fact $\mathcal{O}/{\mathfrak{p}}_i\cong {\mathbb{F}}_p[X]/({\bar{g}}_i)$ is a field of size $p^{\mathrm{deg}{\bar{g}}_i}$. Every prime containing $p$ arises this way; independence of the choice of lift $g_i$ follows because another lift differs by a multiple of $p$.

To prove the ideal factorization, set $J={\prod }_i{\mathfrak{p}}_i^{e_i}$. A generator of the product either contains a factor $p$, and hence lies in $p\mathcal{O}$, or is the product ${\prod }_i{g}_i(\theta {)}^{e_i}$. The latter also lies in $p\mathcal{O}$, since its polynomial reduces to $\bar{g}$ and $g(\theta )=0$. Thus $J\subset p\mathcal{O}$. The ideal norm is multiplicative for nonzero ideals in the ring of integers; henceEqual finite indices together with the containment imply equality. ThereforeThis proves the Dedekind factorization theorem in the stated monogenic case by containment and norms, rather than assuming the desired factorization.

Let $p_1,\dots ,p_a$ be the primes not exceeding $\sqrt{x}$, where $a=\pi (\sqrt{x})$, and let $\Phi (x,a)$ count integers $1\le n\le x$ divisible by none of these primes. Every surviving integer other than $1$ is prime: a composite integer at most $x$ has a prime factor at most its square root. Conversely, the surviving primes are exactly those between $\sqrt{x}$ and $x$. Thus the Legendre prime-counting formula isBy inclusion-exclusion,where the empty product is one. To cover literally every positive real $x$, replace the subtraction of $1$ by $1_{\{x\ge 1\}}$: for $0<x<1$, both counts are zero and there is no surviving integer $1$.

For $x=48$, the sieving primes are $2,3,5$. The surviving count is $48-(24+16+9)+(8+4+3)-1=13$. Consequently

We consider finite-dimensional continuous complex representations of SU(2). Its maximal torus is $\mathrm{diag}(e^{i\theta },e^{-i\theta })$. The irreducible representations areThey may be realized as homogeneous polynomials of degree $n$ in two variables, with the induced linear substitution action. Their torus weights are $n,n-2,\dots ,-n$, each once. The centre $-I$ acts by $(-1{)}^n$, so precisely the even-$n$ representations descend to SO(3). In physics $n=2j$ labels integral or half-integral spin $j$.

Here is a classification argument. Average any positive definite Hermitian form over the compact group using normalized Haar measure. The averaged form is invariant, so the orthogonal complement of an invariant subspace is invariant. Repeatedly splitting gives complete reducibility. Differentiate an irreducible representation and complexify to the Lie algebra ${\mathfrak{sl}}_2$, with generators $H,E,F$ satisfying $[H,E]=2E$, $[H,F]=-2F$, and $[E,F]=H$. Torus periodicity gives integer $H$-weights. Choose a vector $v$ of maximal weight $n$; then $Ev=0$. Induction on $k$ using the commutators givesThe chain terminates: if $F^{p}v\ne 0$ and $F^{p+1}v=0$, the displayed formula forces $n=p\ge 0$. The vectors $v,Fv,\dots ,F^{n}v$ have distinct weights, span an invariant subrepresentation, and hence span the irreducible representation. Their generator action agrees with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. Invariance under the Lie algebra is equivalent to invariance under the connected group, completing the classification. Conversely, a nonzero invariant subspace of a symmetric power contains a weight vector by projection onto torus weights. Raising it to the highest weight and then lowering generates every monomial, proving that the symmetric power is irreducible.

The character on the torus iswith the endpoints interpreted by continuity. Every representation has a unique decomposition $V\cong {⨁}_{n\ge 0}{m}_n{V}_n$. Its multiplicities can be computed from character inner products,The sine orthogonality directly verifies orthonormality of the irreducible characters. Alternatively, if $w_k$ is the dimension of the weight-$k$ subspace, the weight lists give $w_n=m_n+m_{n+2}+\cdots$, so $m_n=w_n-w_{n+2}$ for $n\ge 0$. This gives a constructive decomposition from torus weights. In particular, multiplying weight characters yields the Clebsch-Gordan decomposition $V_m\otimes V_n\cong V_{m+n}\oplus V_{m+n-2}\oplus \cdots \oplus V_{|m-n|}$. Arbitrary unitary Hilbert-space representations of this compact group similarly decompose into Hilbert direct sums of these finite-dimensional irreducibles; the finite-dimensional statement above is the standard representation-theory setting.

The squared Vandermonde product defining the polynomial discriminant is symmetric in all roots. By the fundamental theorem of symmetric polynomials, it is a polynomial in the elementary symmetric functions, which are the signed coefficients of the monic polynomial. Hence it is a polynomial function of those coefficients over the base field.

For $f=X^3+pX+q$, $\Delta (f)=-\mathrm{Res}(f,f^{\prime })$. Eliminating the two leading rows in the Sylvester determinant reduces the resultant toThusFor $X^3-3X+1$, the only possible rational roots are $\pm 1$, and neither is a root. The cubic is therefore irreducible, so its Galois group is a transitive subgroup of $S_3$. Its discriminant is $81=9^2$. The Vandermonde product changes sign under odd permutations, and its square being a rational square implies that the product itself is rational; hence all Galois permutations are even. The only transitive subgroup of $A_3$ is $A_3$ itself. Consequently, writing $L$ for its splitting field,

Let $d_v$ be the vertex degrees and let $r_{ij}$ count common neighbours of distinct vertices. Count unoriented length-two paths by their centre or endpoints:If no four-cycle occurs, $r_{ij}\le 1$, so $P_2\le \left(\genfrac{}{}{0ex}{}{n}{2}\right)$. By Cauchy-Schwarz inequality, $P_2\ge \frac{1}{2}[(2m{)}^2/n-2m]$. Solving $4m^2-2nm\le n^2(n-1)$ givesIf $m\ge n(n-1)/4$, the same bound and monotonicity of $2m^2/n-m$ in this range give $P_2\ge n(n-1)(n-3)/8$ (the small $n$ cases are immediate). Every pair of common neighbours supplies a four-cycle, counted twice through its two opposite vertex pairs, so four-cycle counting by common neighbours givesLet $N=\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and $\bar{r}=P_2/N\ge (n-3)/4$. Convexity of $r(r-1)/2$ gives $C_4\ge (N/2)\left(\genfrac{}{}{0ex}{}{\bar{r}}{2}\right)$. For $n\ge 7$, this function is increasing on the needed range, givingFor $3\le n\le 6$ the printed right-hand side is nonpositive and the conclusion is trivial. At $n=2$ the printed cycle bound is false: a single edge meets the edge hypothesis but has no four-cycles, while the right-hand side is $5/64>0$. Thus this particular bound needs the intended $n\ge 3$ restriction; the other counting statements above do not have this defect.

On four chosen vertices there are three distinct cycle subgraphs, each present with probability $2^{-4}$ in the stated random graph. Extra edges do not destroy these subgraphs. ThereforeFor positive expectation, Markov inequality giveswhen $0<ϵ<1/2$. If $n<4$, the cycle count is identically zero and the requested event has probability one.

A binary relation $R$ on a set is a well-founded relation if every nonempty subset has an $R$-minimal member, that is, a member with no predecessor inside that subset. Here write $xRy$ when $x\in f(y)$.

Suppose $R$ is well-founded and fix any $g:\mathcal{P}(b)\to b$. Call a partial function $h:D\to b$ consistent if $D$ is downward closed under $R$ and $h(y)=g(\{h(x):x\in f(y)\})$ for $y\in D$. Two consistent partial functions agree on their common domain: otherwise the set of points where they disagree has a minimal member, all of whose predecessors are in both domains and agree, so applying $g$ gives agreement at that member as well, a contradiction. Thus the union of all consistent partial functions is itself a function $h:D\to b$, on a downward-closed domain, and remains consistent. This union is a set, because every such graph is a subset of $a\times b$.

If $D\ne a$, choose a minimal member $y$ of $a\setminus D$. All predecessors of $y$ lie in $D$, so extend $h$ to $y$ by the prescribed value $g(\{h(x):x\in f(y)\})$. The enlarged domain remains downward closed and consistent, contradicting the union's maximality. Hence $D=a$. The same minimal-disagreement argument proves uniqueness. This establishes the required recursion directly, without invoking an unproved well-founded recursion theorem.

Conversely, if $R$ is not well-founded, choose a nonempty $S\subset a$ such that each $y\in S$ has a predecessor in $S$. Let $T$ be all points reachable from $S$ by a finite chain of upward $R$ steps, including $S$. ThenFor a point reached by a nonempty chain, its preceding point is in $T$; for a point of $S$, use its predecessor in $S$. The converse follows by appending the final step to a chain. Choose $b=\{0,1\}$ and $g(B)=1$ if $1\in B$, zero otherwise. Both $h_0\equiv 0$ and $h_1=1_T$ satisfy the required recursion, and they differ because $S\ne \varnothing$. This contradicts recursiveness. Therefore $f$ is recursive exactly when its predecessor relation is well-founded, the characterization of a recursive powerset mapping.

The Hamilton equations are $\dot{q}=p/m$ and $\dot{p}=-V_q$. Applying the chain rule to $H(q,p,\lambda (t))$, their two phase-space contributions cancel, giving $\dot{H}=V_{\lambda }\dot{\lambda }=H_{\lambda }\dot{\lambda }$.

For fixed $\lambda$, a closed periodic orbit of energy $E$ has $p=\pm \sqrt{2m(E-V)}$. Thus its action variable isThe sign matches the direction on each half of the orbit, making the enclosed phase-plane area positive. Since $dq=\dot{q}dt=pdt/m$, differentiating under the integral givesFor slowly varying $\lambda$, the exact energy equation averaged over a nearly frozen cycle gives $d⟨H⟩/dt=⟨V_{\lambda }⟩\dot{\lambda }$ to the retained adiabatic order. Using $E=⟨H⟩$, the two chain-rule terms therefore cancel:Here the angle brackets differentiate the explicit parameter dependence along the frozen cycle, not an independently chosen family of energies. The action is an adiabatic invariant, with errors beyond the slow-cycle approximation.

For $V=\lambda q^{2n}$, scale $q=(E/\lambda {)}^{1/(2n)}s$. ThenThe integral is a positive constant depending only on $n$. Constancy of $I$ gives $E^{n+1}/\lambda$ constant, hence

Use the exponential dispersion family convention $f(y)=\exp \{[y\theta -b(\theta )]/\varphi +c(y,\varphi )\}$. For the Gamma distribution, takeSubstitution reproduces the gamma density exactly. Differentiating its normalization with respect to $\theta$ yields $\mathbb{E}Y=b^{\prime }(\theta )$ and $\mathrm{Var}Y=\varphi b^{\prime \prime }(\theta )$: the first identity follows from zero mean score, and differentiation again gives the variance. ThereforeThe canonical link function is $g(\mu )=\theta =-1/\mu$. Some conventions reverse the sign and call $1/\mu$ the reciprocal link; this only reverses the regression coefficients. We retain the genuine natural parameter $\theta =-1/\mu$.

In the Gamma regression with canonical link, ${\eta }_i=x_i^T\beta <0$ and ${\mu }_i=-1/{\eta }_i$. Differentiating the independent-observation log-likelihood givesHere $d{\mu }_i/d{\eta }_i={\mu }_i^2$, so the negative Hessian equals this information matrix. Fisher scoring or Newton iteration iswith the means and information recalculated at each iteration; damping can preserve $X\beta <0$ and increase the likelihood. Full column rank makes the information positive definite for positive means.

The saturated model has mean $y_i$. Subtracting fitted from saturated log-likelihood yields the unscaled devianceTwice the actual log-likelihood ratio is $D/\varphi$. In particular, the printed parenthetical $\hat{\mu }=X\hat{\beta }$ is incorrect for this nonidentity link: $X\hat{\beta }$ estimates $\eta$, and ${\hat{\mu }}_i=-1/(x_i^T\hat{\beta })$.

A Kleinian group is a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting as orientation-preserving hyperbolic isometries of ${\mathbb{H}}^3$ and as Möbius transformations of its boundary sphere. Its limit set consists of the accumulation points on the boundary of an orbit $Gx$ with $x\in {\mathbb{H}}^3$; it is independent of that interior base point and is closed. A nonidentity parabolic transformation is conjugate to $z\mapsto z+c$, $c\ne 0$. Its iterates on the upper half-space tend to its unique fixed boundary point, infinity in these coordinates. Consequently every parabolic fixed point belongs to the limit set.

The displayed matrix has determinant one, trace two, and is not the identity because $a\ne 0$. Its fixed-point equation is $a(z-w{)}^2=0$, so it has exactly the double fixed point $w$. It is therefore parabolic, and its fixed point is $w$.

The determinant-one Gaussian-integer matrices form a subgroup of ${\mathrm{SL}}_2(\mathbb{C})$. They are discrete because their entries lie in a discrete lattice: a matrix sufficiently near the identity has every entry equal to the corresponding identity entry. Passing through the quotient by $\{\pm I\}$ preserves discreteness, so their Möbius transformations form a Kleinian group.

For $w=n/r$, $n=p+iq$ and $r\ne 0$, set $a=r^2$. This gives the Gaussian-integer parabolic constructionAll entries lie in $\mathbb{Z}[i]$, the determinant is one, and its parabolic fixed point is $n/r=w$. The points with nonzero $p,q,r$ are dense in the finite complex plane, and their closure on the sphere includes infinity. Since the limit set is closed and contains all these parabolic fixed points, the limit set is the whole Riemann sphere.