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Choose the incident photon along the $x$ direction. Conservation of four-momentum preserves the squared total energy invariant $\mathcal{S}=E_{\mathrm{tot}}^2-c^2|{\mathbf{p}}_{\mathrm{tot}}{|}^2$. Initially,In the centre-of-momentum frame let the outgoing photon energy be $e_{\ast }>0$; the larger particle has opposite momentum magnitude $e_{\ast }/c$ and energy $\sqrt{M^2{c}^4+e_{\ast }^2}$. ThusConversely, whenever $\mathcal{S}>M^2{c}^4$, the positive value $e_{\ast }=(\mathcal{S}-M^2{c}^4)/(2\sqrt{\mathcal{S}})$ constructs a final state with the required invariant, which can be boosted to the laboratory frame. The limiting photon excitation threshold with a soft final photon is thereforeStrictly, an outgoing physical photon requires $E>E_{\mathrm{th}}$; the printed minimum is the threshold infimum approached as the final photon becomes soft.

Take the electric dipole moment along the polar axis, $\mathbf{p}=p\hat{\mathbf{z}}$. The electric field is minus the gradient of the electric potential; in spherical polar coordinates,Since $\mathbf{p}=p(\cos \theta {\hat{\mathbf{e}}}_r-\sin \theta {\hat{\mathbf{e}}}_{\theta })$ and $\mathbf{p}\cdot {\hat{\mathbf{e}}}_r=p\cos \theta$, the radial and angular components combine asThe origin is excluded because the point-dipole potential is singular there.

For a regular smooth constraint $g=0$, meaning $\nabla g\ne 0$ at the point, constrained stationarity requires $\nabla f=\lambda \nabla g$. Solve these Lagrange multiplier equations together with the constraint, then compare candidate values and possible boundary or singular cases.

Let $r>0$ and $h>0$ be cylinder radius and height. Its volume is $V=\pi r^{2}h$ and the surface-area constraint is $2\pi r^2+2\pi rh=A$. The multiplier equations areThe second gives $\lambda =r/2$, and substituting into the first gives $h=2r$. Consequently $6\pi r^2=A$. To check global maximality, eliminate $h$ and obtain $V(r)=Ar/2-\pi r^3$ on $0<r<\sqrt{A/(2\pi )}$; it tends to zero at both ends and has exactly this one stationary point. The largest volume is

Use three compactness facts: a closed subset of a compact space is compact; a continuous image of a compact space is compact; and a compact subset of a Hausdorff space is closed. For a closed $C\subseteq X$, these facts show that $f(C)$ is closed in $Y$. Therefore the bijection $f$ is a closed map, and $(f^{-1}{)}^{-1}(C)=f(C)$ proves continuity of its inverse. Thus $f$ is a homeomorphism. This proves the compact-to-Hausdorff continuous bijection theorem without requiring $X$ itself to be Hausdorff.

Let $q:[0,1]\to [0,1]/\sim$ be the quotient map and $h(t)=(\cos 2\pi t,\sin 2\pi t)$. The map $h$ is continuous and identifies precisely the equal parameters and the endpoint pair. Hence it induces a bijection $\bar{h}$ from the quotient space to the unit circle, with $h=\bar{h}\circ q$; the definition of the quotient topology makes $\bar{h}$ continuous. The quotient is compact as a continuous image of $[0,1]$, while the circle is Hausdorff as a subspace of ${\mathbb{R}}^2$. Applying the preceding theorem gives

The sequence has uniform convergence to $f$ when, for every $\epsilon >0$, there is an $N$ independent of $x$ such that $|f_n(x)-f(x)|<\epsilon$ for every $n\ge N$ and every $x\in [a,b]$. Fix $x$ and choose a single $N$ with $\Vert f_N-f{\Vert }_{\infty }<\epsilon /3$. By continuity of $f_N$, sufficiently small $|y-x|$ within the interval gives $|f_N(y)-f_N(x)|<\epsilon /3$. The triangle inequality then givesThis proves continuity at every point, including endpoints with the relative topology.

For the moving points, use the just-proved continuity of $f$:This is uniform convergence at moving evaluation points: uniform control is what allows the evaluation point to vary with $n$.

A complex number is an algebraic number over $\mathbb{Q}$ when it is a root of a nonzero rational polynomial. Its minimal polynomial of an algebraic element is the unique monic polynomial in $\mathbb{Q}[X]$ of smallest degree annihilating it; polynomial division shows that it divides every rational polynomial annihilating the element, and it is irreducible.

Consider the surjective evaluation ring homomorphism ${\mathrm{ev}}_{\alpha }:\mathbb{Z}[X]\to \mathbb{Z}[\alpha ]$. Certainly $(f)\subseteq \ker {\mathrm{ev}}_{\alpha }$. Because the nonconstant $f$ is irreducible over $\mathbb{Z}$, its content is a unit: otherwise its content and its nonconstant primitive part give a factorization into nonunits. By the Gauss lemma for polynomials, $f$ is irreducible in $\mathbb{Q}[X]$ and is a scalar multiple of the minimal polynomial.

If $g\in \mathbb{Z}[X]$ vanishes at $\alpha$, then $f$ divides $g$ in $\mathbb{Q}[X]$. Primitive divisibility in Gauss lemma for polynomials implies division in $\mathbb{Z}[X]$ as well. Explicitly, write the rational quotient as $(a/b)h$ with $h$ primitive integral and $\gcd (a,b)=1$; the primitive product $fh$ shows that integrality of $(a/b)fh$ forces $b=1$. Thus the kernel is precisely $(f)$, and the first isomorphism theorem for rings yieldsNo monicity of $f$ is needed; this is an isomorphism of rings, not an assertion that either ring is a field.

All specified nearest-neighbour probabilities are positive, making the birth-death chain irreducible. The row sums give $p_0+q_0=1$ and $p_i+q_i=1$ for $i\ge 1$. Write $T_j$ for the hitting time of $j$, and on $\{0,\dots ,N\}$ let $h_i={\mathrm{Pr}}_i(T_N<T_0)$. The first-step equations giveDefine ${\rho }_0=1$ and ${\rho }_k={\prod }_{r=1}^k{q}_r/p_r$. The successive increments are proportional to ${\rho }_k$, and normalization at $N$ yieldsThe chain cannot remain forever in a fixed finite set avoiding zero: from every state in that set there is a uniformly positive chance to leave or hit zero within finitely many steps, so iterating the Markov property makes indefinite confinement have probability zero. Thus avoiding zero is the limit event of hitting every upper boundary before zero. Sending $N\to \infty$ shows that the probability of never hitting zero from one vanishes exactly when ${\sum }_k{\rho }_k=\infty$. At zero, return is immediate with probability $q_0$ and otherwise begins an excursion from one; irreducibility then givesFor positive recurrence define $w_0=1$ and $w_n={\prod }_{r=1}^n{p}_{r-1}/q_r$. First assume recurrence and let $\tau =T_0^{+}$ be the first return from zero. For each fixed $i$, the expected number $m_i$ of visits to $i$ before $\tau$ is finite: each visit to $i\ge 1$ has a positive chance of following the successive downward steps to zero before another visit, so its visit count has a geometric tail. Also $m_0=1$.

Each finite excursion starts and ends below every edge $(i,i+1)$, so its upward and downward crossing counts of that edge are equal. Their expectations, by the Markov property, are $m_i{p}_i$ and $m_{i+1}{q}_{i+1}$. Hence $m_{i+1}=m_i{p}_i/q_{i+1}$ and $m_i=w_i$. Tonelli's theorem applied to the nonnegative occupation counts gives the excursion weights of a reflected birth-death chain formulaConsequently positive recurrence implies the requested finite sum. Conversely, if ${\sum }_i{w}_i<\infty$, then $w_i\to 0$. But ${\rho }_i{w}_i=p_0/p_i\ge p_0$, so ${\rho }_i\ge p_0/w_i\to \infty$ and the already proved recurrence criterion holds. The occupation calculation is therefore valid and gives a finite mean return time. ThusAdding the finite $w_0=1$ term does not change convergence.

A linear map preserves linear combinations: $\psi (av+bw)=a\psi (v)+b\psi (w)$ for all vectors and real scalars. The rank-nullity theorem gives $\dim V=\dim \ker \psi +\dim \mathrm{im}\psi$. For an endomorphism of the same finite-dimensional vector space, surjectivity means rank $\dim V$, equivalent to zero kernel and hence injectivity.

If $\psi \varphi =I$, then every $v=\psi (\varphi v)$ lies in the image, so $\psi$ is surjective and consequently injective. Now $\psi (\varphi \psi v)=\psi v$; injectivity implies $\varphi \psi v=v$ for every $v$. Thus the right inverse is also a left inverse: $\boxed{\varphi \psi =I}$. This is the one-sided inverses of finite-dimensional endomorphisms property.

Apply it to the linear maps on ${\mathbb{R}}^n$ represented by $A$ and $B$: $AB=I$ implies $\boxed{BA=I}$. Equal finite dimensions are essential to the argument.

Write $r=1-\alpha -\beta >0$. Differentiating the probability generating function at $(1,1)$ givesExpanding its denominator as a geometric series and then applying the binomial theorem givesCoefficient identification proves $\boxed{\mathrm{Pr}(X=x,Y=y)=r\left(\genfrac{}{}{0ex}{}{x+y}{x}\right){\alpha }^x{\beta }^y}$, a one-stop negative multinomial distribution.

For an independent identically distributed sample let $S_X=\sum X_i$ and $S_Y=\sum Y_i$. Ignoring parameter-independent factors, the log-likelihood is $\ell =n\ln r+S_X\ln \alpha +S_Y\ln \beta$. When both sums are positive, its stationary equations are $S_X/\alpha =n/r$ and $S_Y/\beta =n/r$. Together with $r+\alpha +\beta =1$, these giveThe Hessian is $-(n/r^2)11^T-\mathrm{diag}(S_X/{\alpha }^2,S_Y/{\beta }^2)$, negative definite in this case, so the stationary point is the unique global maximum.

There is an important boundary qualification: if $S_X=0$, the derivative with respect to $\alpha$ is $-n/r<0$, so the likelihood increases as $\alpha$ decreases to zero. The displayed formula then gives a boundary estimator, not an attained MLE in the stated open parameter space $\alpha ,\beta >0$. The analogous issue holds for $S_Y=0$. The formulas are MLEs on the extension allowing zero category probabilities, and interior MLEs when both counts are positive.

For bias, put $p=\alpha +\beta$ and $U=(S_X+S_Y)/n$. Then $\mathbb{E}[U]=p/(1-p)$ and $\hat{\alpha }+\hat{\beta }=g(U)$ with $g(u)=u/(1+u)$. Its second derivative is $-2/(1+u{)}^3<0$, and $U$ is nondegenerate. Strict Jensen's inequality givesIn fact both estimates are downward biased. Conditional on $X_i+Y_i$, the first count is binomial with success probability $\alpha /p$; hence conditional on $S=S_X+S_Y$, $S_X$ is also binomial with that probability. ThereforeThis proves the finite-sample bias in a negative multinomial probability estimator, using the boundary-extended formulas whenever necessary.

Finally, the strong law of large numbers applies because the nonnegative variables have finite means: almost surely $\overline{X}\to \alpha /r$ and $\overline{Y}\to \beta /r$. Continuity of the estimator function then givesand similarly for $\hat{\beta }$. Thus the estimates are strongly consistent despite finite-sample bias. Almost surely both sample sums eventually become positive, so the open-parameter MLE issue affects finite samples rather than this asymptotic conclusion.

For a normalized wavefunction, $⟨A{⟩}_{\psi }$ is the statistical mean of measurements of the quantum observable $A$ over identically prepared systems, not necessarily the outcome of one measurement. Assume the state is in the required operator domains and the second moments are finite. Hermiticity givesThe commutator expectation is purely imaginary, so $i⟨[A,B]⟩$ is real. A real quadratic nonnegative for every $\lambda$ has nonpositive discriminant; if its quadratic coefficient vanishes, its linear coefficient must also vanish. ConsequentlyApply the same inequality to $A-⟨A⟩I$ and $B-⟨B⟩I$. Their commutator is unchanged and their second moments are the variances, yielding the quantum uncertainty relationFor the harmonic oscillator, $p=-i\hslash {\partial }_x$ gives $[x,p]\psi =i\hslash \psi$ by the product rule. The uncentered inequality above therefore implies $⟨x^2⟩⟨p^2⟩\ge {\hslash }^2/4$. The arithmetic-geometric mean inequality now givesNo assumption that the position or momentum means vanish is needed.