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Call a set function $f$ with ordinal domain $\gamma$ good if $f(\xi )=G(\xi ,\mathrm{ran}(f{|}_{\xi }))$ for every $\xi <\gamma$. Any two good functions agree on their common domain: otherwise their first disagreement has identical earlier ranges and hence identical values under $G$. This is transfinite induction, not an assumption of the desired recursion.

Good functions exist with every ordinal domain. At a successor stage extend the unique previous function by its prescribed value under $G$. At a limit stage, Axiom schema of replacement collects the unique good functions on all smaller domains into a set, and their compatible union is a good function on the limit domain. Each earlier range is a set, so $G$ is applied only to sets. Transfinite induction now establishes existence at all stages. Define $F(\alpha )$ to be the value at $\alpha$ of the unique good function on $\alpha +1$. This is a definable class function because goodness and the uniqueness assertion are formulas in the defining formula for $G$. Compatibility gives

An initial ordinal is an ordinal not bijective with any smaller ordinal. It is the canonical ordinal representative of a well-orderable cardinal, so cardinal arithmetic can be defined by taking the initial ordinal equinumerous with a disjoint union, Cartesian product or function set. In the usual setting with the axiom of choice, every cardinal has such a representative. Hartogs theorem says that for every set $A$ there is an ordinal $h(A)$ which does not inject into $A$, and there is a least such ordinal. For an initial ordinal $\kappa$ it is the next larger initial ordinal: every smaller ordinal injects into $\kappa$, and if $h(\kappa )$ were equinumerous with a smaller ordinal it too would inject into $\kappa$.

Use the recursion just proved to set ${\omega }_0=\omega$, ${\omega }_{\beta +1}=h({\omega }_{\beta })$ and ${\omega }_{\lambda }={\sup }_{\beta <\lambda }{\omega }_{\beta }$ for nonzero limit $\lambda$. These instructions use the earlier value set alone: at a successor it has a maximum, and at a limit its supremum is taken. They are therefore implemented by an appropriate definable $G$. The limit supremum is initial: if it were bijective with $\gamma$ below it, choose an earlier ${\omega }_{\beta }>\gamma$; then ${\omega }_{\beta }$ would inject into $\gamma$, contradicting initiality. The sequence is strictly increasing and continuous. There are no omitted infinite initial ordinals, since each successor chooses the least next one and each limit is the supremum; more formally, a least omitted initial ordinal cannot lie between successive terms, and the least term above it cannot be a limit term. Terms eventually exceed any prescribed ordinal, since a strictly increasing ordinal sequence satisfies ${\omega }_{\beta }\ge \beta$.

There is a fixed point of a normal ordinal function. Explicitly let ${\alpha }_0=0$, ${\alpha }_{n+1}={\omega }_{{\alpha }_n}$ and $\alpha ={\sup }_{n<\omega }{\alpha }_n$. The sequence is strictly increasing, by induction from $0<{\omega }_0$ and strict increase of the enumeration. It is cofinal in the limit $\alpha$, so continuity gives

Parameterize the second contour by $\zeta =-is$, $s\ge 0$. ThenThe exponential decays when $\mathrm{Im}z>0$, uniformly on compact subsets of that half-plane. Differentiation under the integral there proves holomorphy. Thus the maximal open sector of convergence and holomorphy in this representation is $\boxed{\alpha =0,\beta =\pi }$. The boundary integrals themselves can converge, but that does not enlarge this open analytic sector.

For $0<\arg z<\pi /2$, rotate the original contour clockwise through the fourth quadrant. On a large quarter-circle the integrand is $O(R^{-3})$ times a decaying exponential, so the arc integral tends to zero. The only enclosed pole is ${\zeta }_0=e^{-i\pi /3}$, and its residue is $e^{-2z{\zeta }_0}/(3{\zeta }_0^2)$. The clockwise orientation gives, by the residue theorem,Consequently the desired analytic continuation for $\pi /2\le \arg z<\pi$ isBoth terms are holomorphic in the upper half-plane and the expression agrees with the original function in the overlap, so it is an analytic continuation. The residue term is essential: the pole lies between the two rays.

The first fit is grouped-binomial logistic regression: independently, $Y_i\sim \mathrm{Bin}(N_i,{\pi }_i)$, with the observed group sizes $N_i$ andThe reference group is under 50 with a nonmalignant initial tumour. The weights specify binomial numbers of trials, not six equally weighted individual observations. There is no age-by-malignancy interaction. Each age coefficient compares that age group with the reference age at fixed malignancy; $\exp {\beta }_3$ is the common malignancy odds ratio.

The maximum-likelihood estimates are obtained by maximizing the binomial likelihood. The reported standard errors come from the inverse fitted Fisher information. Each displayed Wald test tests a coefficient equal to zero, using $z=\hat{\beta }/\mathrm{se}(\hat{\beta })$ and the approximate standard normal distribution; its two-sided probability is $2\Phi (-|z|)$. At the 5% level the middle age coefficient and the malignancy coefficient are significant, with negative signs; the oldest age coefficient is not, although its 9.17% probability is weaker evidence in the same direction. The oldest group has only 19 observations and its larger standard error does not establish absence of an age effect. The intercept tests reference log-odds zero, equivalently reference survival probability $1/2$, rather than an age contrast. The reference fitted probability is about $0.888$; the fitted malignancy odds ratio is about $0.481$.

The binomial deviance is twice the saturated-minus-fitted log-likelihood. The null fit has one parameter, hence $6-1=5$ residual degrees of freedom; the first fit has four parameters, hence $6-4=2$. The improvement $12.65585-0.49409=12.16176$ is an approximate likelihood-ratio test on three degrees of freedom, with probability about $0.0069$. The residual deviance $0.49409$, compared with ${\chi }_2^2$, gives about $0.781$, so there is no evidence of lack of fit. These are asymptotic comparisons, especially approximate for the small oldest strata. The Akaike information criterion is $-2\ell +2p$, here with $p=4$.

Pooling the two older age groups gives a grouped-binomial logistic regression with a common coefficient for age at least 50 and three parameters in total. The similar older-age estimates and the sparse oldest group motivate examining this restriction, but the significance of either coefficient separately is not a test of their equality. The actual analysis of deviance for pooling compares the nested fits: the deviance increases by only $0.784-0.49409\simeq 0.28991$ for one lost parameter, giving a ${\chi }_1^2$ probability about $0.590$. The pooled model also has smaller Akaike information criterion, $28.723$ rather than $30.433$. Prefer the pooled model for its simpler adequate fit. Its common age and malignancy coefficients are negative and significant at 5% in the reported Wald tests.

There is a literal parenthesis error in the printed final R fragment: the first sum is not closed before subtracting the second sum. As printed, the expression is incomplete. Closing that sum gives twice the saturated-minus-fitted log-likelihood, including the binomial constants which cancel. Its intended final value is therefore the second fit's binomial deviance, $\boxed{0.784\text{ approximately}}$.

A frieze group is a discrete subgroup of Euclidean isometries whose translation subgroup is generated by one nonzero vector and has finite index. A plane crystallographic group is a discrete subgroup of Euclidean isometries with a rank-two translation lattice of finite index; equivalently its action has a compact fundamental region.

Here are three frieze groups, with explicit generators. The translation $T(x,y)=(x+1,y)$ alone gives $\mathbb{Z}$. Adding the vertical reflection in mathematics $S(x,y)=(-x,y)$ gives $\mathbb{Z}⋊C_2$, the infinite dihedral group, since $STS=T^{-1}$. Instead adding the horizontal reflection in mathematics $H(x,y)=(x,-y)$ gives $\mathbb{Z}\times C_2$, since $HT=TH$. Their translation parts are exactly $⟨T⟩$ and their point parts have respectively one, two and two elements.

For plane crystallographic groups, let $T_1(x,y)=(x+1,y)$ and $T_2(x,y)=(x,y+1)$. These generate ${\mathbb{Z}}^2$. Adding the half-turn $R_2(x,y)=(-x,-y)$ gives ${\mathbb{Z}}^2⋊C_2$, with inversion on both translation generators. Adding the quarter-turn $R_4(x,y)=(-y,x)$ gives ${\mathbb{Z}}^2⋊C_4$, with $R_4{T}_1{R}_4^{-1}=T_2$ and $R_4{T}_2{R}_4^{-1}=T_1^{-1}$. These are the translation, half-turn and square-rotation wallpaper groups. Integer translation vectors and a finite point part make each group a discrete subgroup; the unit square meets every orbit, establishing a cocompact group action.

For the final claim, suppose a Euclidean isometry $R$ has order five. An orientation-reversing planar isometry has finite order only when it is a reflection in mathematics, so $R$ is a rotation through $\theta =2\pi j/5$, $j=1,2,3,4$, about some centre. Write $A$ for its linear rotation part. If $t_v\in G$ is a nonzero translation, conjugation gives $R{t}_v{R}^{-1}=t_{Av}$. Thus $G$ contains translations by $v$ and by $(A+A^{-1})v=cv$, where $c=2\cos \theta$.

For a primitive fifth root $\zeta$, dividing $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4=0$ by ${\zeta }^2$ gives $c^2+c-1=0$. Hence $c$ is irrational. The pigeonhole principle, applied to $N+1$ fractional parts of $0,c,\dots ,Nc$ in $N$ equal intervals, gives integers $m,n$ with $0<|nc-m|\le 1/N$. Consequently $G$ contains nonidentity translations by $(nc-m)v$ tending to the identity. This contradicts the discrete subgroup hypothesis. There is no element of order five. This argument needs only the given nonzero translation, not a rank-two lattice.

Assume dilute nondegenerate nonrelativistic particles, a common temperature, chemical equilibrium, charge neutrality $n_p=n_e$, and photons with zero chemical potential. The reaction balance gives ${\mu }_p+{\mu }_e={\mu }_H$. Dividing the two charged-species equilibrium densities by the neutral density cancels their chemical-potential factors and leavesFor ground-state hydrogen, $g_p=g_e=2$ and $g_H=4$, and $m_H\simeq m_p$ in the thermal prefactor. The rest-mass difference is $I/c^2$. These approximations give the Saha ionization equationExcited levels or different degeneracy conventions would change the prefactor; the stated assumptions give the printed normalization.

Write the right side as $S(T)$. Since $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the hydrogen ionization fraction obeys $X_e^2/(1-X_e)=S(T)/n_B$. With $n_B=\eta n_{\gamma }$, the given blackbody photon density then yieldsThere are vastly more photons than baryons. Even when the typical photon energy is below the ionization energy, the high-energy tail can contain enough ionizing photons to keep nearly every baryon ionized. Neutral hydrogen forms only when that tail becomes sufficiently depleted relative to the very small baryon-to-photon ratio. This entropy/number-density effect explains why cosmological recombination occurs at a temperature far below $I$, rather than simply when $kT$ first falls below the binding energy. The Saha ionization equation describes the equilibrium estimate; late kinetic freeze-out is a separate limitation of that approximation.

Use coordinates $(i,j)\in \{0,\dots ,7{\}}^2$, with the starting corner $(0,0)$. The bishop stays on the 32 squares where $i+j$ is an even number. Its moves form a random walk on a graph, with an edge between distinct squares whenever $|i-i^{\prime }|=|j-j^{\prime }|$. If $d(i,j)$ is the number of reachable squares, the transition probability to each neighbor is $1/d(i,j)$.

The graph is connected: from $(i,j)$ move along its antidiagonal to $((i+j)/2,(i+j)/2)$, then along the main diagonal to the corner. The midpoint coordinates are integers because $i+j$ is an even number, and this square always lies on the board. Thus this is the full communicating state space of the bishop random walk.

The invariant distribution is proportional to degree. Its normalizing constant can be counted by summing the ordered pairs along each diagonal. The diagonals of one orientation have lengths $8,6,6,4,4,2,2$, contributing $144$; the other have lengths $1,3,5,7,7,5,3,1$, contributing $136$. ThereforeFor any neighboring states $s,t$, $\pi (s)P(s,t)=1/280=\pi (t)P(t,s)$, proving the detailed balance equations and hence that the chain is a reversible Markov chain.

The corner has seven neighbors. Kac's lemma gives its expected first positive return time:

For distinct Gaussian quadrature nodes, let ${\ell }_i$ be the Lagrange interpolation polynomial with ${\ell }_i(x_j)={\delta }_{ij}$. Since ${\ell }_i^2$ has degree $2n-2$, exactness givesThe last inequality follows from $w>0$ on the interval and the nonzero polynomial ${\ell }_i$ being nonzero on a subinterval. It does not require prior knowledge of the signs of the weights.

For $n=2$, the nodes are the roots of the supplied orthogonal polynomial $p_2$:Indeed exactness for $1$ gives $b_1+b_2={\int }_{-1}^1(1+x^2)dx=8/3$, and exactness for $x$ gives equality of the weights.

With consistent flux and loop orientations, Faraday's law is $V=-d\Phi /dt$. The magnetic force on a current-carrying wire is $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$.

Let $z(t)$ be the top edge height and $v=-\dot{z}$ the downward speed. While $0<z<l$, only the top horizontal edge lies in the field, and the linked magnetic flux is $Bwz$. Therefore the induced electromotive force has magnitude $Bwv$, and Ohm's law gives current magnitude $I=Bwv/R$. Its force opposes the downward motion and has upward magnitude $IBw=B^2{w}^{2}v/R$. Forces on the vertical sides cancel horizontally. HenceFor a release from rest, $v=v_{\mathrm{terminal}}(1-e^{-B^2{w}^{2}t/(mR)})$. The terminal speed is the formal limit while partial overlap is maintained; after the entire loop leaves the field this magnetic braking term vanishes.

Put $D=1+i\hslash t/m$ and write the Gaussian wave packet as $\psi ={\pi }^{-1/4}{D}^{-1/2}{e}^{-x^2/(2D)}$. Differentiation givesHence $\boxed{i\hslash {\psi }_t=-({\hslash }^2/2m){\psi }_{xx}}$, the free Time-dependent Schrödinger equation.

With $\tau =\hslash t/m$, its probability density isThe stated Gaussian integral shows its integral is one. Symmetry gives $⟨x⟩=0$; differentiating that Gaussian integral with respect to its coefficient gives $⟨x^2⟩=(1+{\tau }^2)/2$. Thus the position standard deviation isIt is not a stationary state: its probability density changes with time as the packet spreads. A stationary state would have only an overall time-dependent phase and a fixed position density.

At a regular constrained stationary point, $\nabla g\ne 0$ and the directional derivative of $f$ vanishes along the tangent to $g=0$. Therefore $\nabla f=\lambda \nabla g$ for a Lagrange multiplier $\lambda$. Solve this equation together with the constraint, then compare candidate values and any exceptional nonregular points. Proportionality of the two gradients givesIf $\nabla g=0$ at an exceptional point, this determinant vanishes trivially, although the regular multiplier method need not identify every such point.

For the given ellipse, maximize $r^2=x^2+y^2$. The constraint is $r^2+xy=4$, and $xy\ge -r^2/2$, so $4\ge r^2/2$. Equality requires $x=-y$ and the constraint then gives $(x,y)=(2,-2)$ or $(-2,2)$. ThusThe constraint gradient vanishes only at the origin, which is not on this ellipse, so the regularity condition holds everywhere relevant.

Rouché's theorem states that if $f,g$ are holomorphic on and inside a simple closed contour and $|g|<|f|$ on it, then $f$ and $f+g$ have the same number of zeros inside, counted with multiplicity. On $|z|=1$, the power series for sine givesApply Rouché's theorem with $f=2z$ and $g=-\sin z$. There is one zero inside and none on the boundary. Since $z=0$ is a zero and $(2z-\sin z{)}^{\prime }{|}_{z=0}=1$, it is simple. Exactly one complex number qualifies: $z=0$.

Uniform convergence means that for every $\epsilon >0$ there is $N$ such that $|s_n(x)-s(x)|<\epsilon$ for every $x\in \mathbb{R}$ and every $n\ge N$. Equivalently, ${\sup }_x|s_n(x)-s(x)|\to 0$. The nonzero constant functions $s_n=1/n$ converge uniformly to zero. In contrast, the continuous moving bumps $s_n(x)=\max (0,1-|x-n|)$ converge pointwise to zero but have supremum one for every $n$.

On the compact interval,Thus uniform convergence permits the stated interchange with integration.

Choose $\boxed{s(x)=1/(1+|x|)}$. It is continuous, nonnegative and tends to zero at infinity, but its improper integral diverges logarithmically. This even choice also makes the prescribed cutoff continuous at both $n$ and $-n$: beyond either endpoint it remains at $s(n)$ for another unit of distance and then decreases as $s(n)/(|x|-n{)}^2$.

Inside $[-n,n]$, $s_n=s$. Outside, both functions lie between zero and $1/(n+1)$, so ${\sup }_x|s_n-s|\le 1/(n+1)\to 0$. Nevertheless each cutoff has finite integral:Here the value of the minimum at $t=0$ is its continuous limiting value one. This example shows that uniform convergence on an unbounded domain does not by itself ensure convergence of the integrals over that domain.

Let $T$ consist of all torsion elements of a module in $M$. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$. Since $R$ is an integral domain, $rs\ne 0$. Also $r(am)=0$ for every scalar $a\in R$. Thus $T$ is the torsion submodule.

If $r(m+T)=0$ in $M/T$ with $r\ne 0$, then $rm\in T$, so some $s\ne 0$ satisfies $srm=0$. The domain property again gives $sr\ne 0$, hence $m\in T$. Therefore $M/T$ is a torsion-free module.

Finally, any submodule $N$ with the stated properties satisfies $N\subseteq T$. Conversely, if $m\in T$, its coset in $M/N$ is torsion and therefore zero, so $m\in N$. Hence $\boxed{N=T}$ is the unique choice.

Interpret a beneficial trade as weakly improving both participants and strictly improving at least one; otherwise the identity trade would always remain available. Put $g_j=\nabla U_j(y_j)$. For a globally strictly increasing concave function, every component of its gradient is strictly positive: concavity makes a coordinate derivative nonincreasing along that coordinate, and monotonicity makes it nonnegative. If it vanished at a finite point, it would vanish farther along the coordinate, contradicting strict increase on the whole real line.

To construct a pairwise improvement from nonparallel utility gradients, suppose two positive gradients $g,h$ are not proportional. Strict Cauchy-Schwarz inequality givesChoose $t$ between these bounds and put $\delta =g-th$. Then $g\cdot \delta >0$ and $h\cdot \delta <0$. For sufficiently small $\epsilon >0$, the feasible transfer $y_i\mapsto y_i+\epsilon \delta$, $y_j\mapsto y_j-\epsilon \delta$ strictly improves both utilities by differentiability. This contradicts the absence of beneficial pairwise trades.

Hence all gradients are proportional. Choose $v=g_1$; positivity makes each proportionality scalar positive:Now take $\boxed{a_j=1/{\lambda }_j}$. The concave supporting-tangent inequality gives, for any other allocation $(z_j)$ with the same total endowment,Thus the final allocations globally maximize the social planner's weighted objective. They need not be unique when utilities are not strictly concave, but the required positive supporting weights exist.