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Let the Markov chain state after each toss be $(S,I)$, where $S$ is player $A_1$'s net winnings and $I$ is the next player to toss. From $(S,i)$ it moves to $(S+1,1)$ with probability $p_i$ and to $(S-1,2)$ with probability $q_i=1-p_i$. The initial state is $(0,1)$, and a return means hitting any state with $S=0$ at positive time.

From $(1,1)$, the probability of hitting zero isIndeed, in the positive-drift case the bounded harmonic solution ofuses the decaying root $q_2/p_1$ and boundary value $h_2(0)=1$, yielding $h_1(1)=q_1/p_2$; with nonpositive drift the walk hits zero surely. By the reflected argument, from $(-1,2)$ the probability isConditioning on the first toss, the required return probability isBoth formulas equal one at the boundary $p_1+p_2=1$.

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain, $\dim V=\dim \ker T+\dim \mathrm{im}T$. Apply it to $\beta {|}_{\mathrm{im}\alpha }$. Its image is $\mathrm{im}(\beta \alpha )$ and its kernel is $\mathrm{im}\alpha \cap \ker \beta$, soSince $\mathrm{im}(\alpha \gamma )\subseteq \mathrm{im}\alpha$,Using the preceding identity for both restrictions and rearranging gives

Let $P=u{u}^T/\Vert u{\Vert }^2$. This is an orthogonal projection matrix, so $P^T=P$ and $P^2=P$. Therefore the Householder transformationsatisfies $H_u^T{H}_u=(I-2P{)}^2=I$. If $\Vert a\Vert =\Vert b\Vert$ and $a\ne b$, thenand hence $H_{a-b}a=a-(a-b)=b$.

Apply a Householder transformation to the first column of $A$ to map it to a multiple of $e_1$, then apply transformations supported on the trailing coordinates to clear each later column below its diagonal. Their product is orthogonal and produces $R$; reversing the product gives $A=QR$.

For the displayed matrix, one resulting factorization isDirect multiplication gives the stated $A$, and $Q^{T}Q=I$.

For a localized steady current, defineAssume the external field varies negligibly across the current distribution. Expanding it to first order, the uniform term gives no net force because $\int \mathbf{J}dV=0$. Using the supplied integral identity and $\nabla \cdot \mathbf{B}=0$, the first-order term in $\int \mathbf{J}\times \mathbf{B}dV$ reduces toSince $\mathbf{F}=-\nabla U$, the magnetic dipole potential energy is

Hereincreases away from the $z$-axis. A dipole parallel to $\mathbf{B}$ has energy $-m|\mathbf{B}|$ and is drawn toward stronger field, hence outward from the axis. An antiparallel dipole has energy $+m|\mathbf{B}|$ and is driven toward the weak-field axis. The field therefore spatially separates the two orientations.

For an S wave, put $u(r)=r\psi (r)$. A bound state has $E=-\epsilon <0$, and the radial Schrödinger equation giveswhereContinuity of $u,u^{\prime }$ at $a$ givesThe first bound state appears at threshold $\kappa \downarrow 0$, where $ka=\pi /2$. The zero-energy threshold itself is not normalizable, so a deuteron bound state exists precisely whenThe wavefunction is $\psi =u/r$, with normalization fixing the remaining constant.

The convolution is $(f\ast g)(x)={\int }_{-\infty }^{\infty }f(y)g(x-y)dy$. Under the usual integrability hypotheses, Fubini's theorem and $u=x-y$ give

The supplied transform and the Fourier shift theorem giveSincethe shift and differentiation rules yieldThe coincident boundary values make the endpoint convention immaterial.

A compact space is one for which every open cover has a finite subcover. Let $\mathcal{U}$ cover $[0,1]$, and letThe set is nonempty. If $s=\sup A<1$, choose $U\in \mathcal{U}$ containing $s$. An interval about $s$ lies in $U$; a point of $A$ just to the left of $s$ then extends the finite cover beyond $s$, contradicting the definition of the supremum. Hence $s=1$, and the same neighbourhood argument covers $1$, proving $[0,1]$ compact.

The continuous map $t\mapsto (\cos 2\pi t,\sin 2\pi t)$ maps $[0,1]$ onto $S^1$. A continuous image of a compact space is compact: pull an open cover back, choose a finite subcover, and push the corresponding sets forward. Thus $S^1$ is compact.

In the unusual topology on $X={\mathbb{R}}^2\setminus \{0\}$, a proper closed set is a finite union of punctured lines through the origin. Take any nonempty member $U$ of an open cover. Its complement is such a finite union. Each punctured line has the indiscrete subspace topology, because every other vector subspace meets it only at the removed origin, and is therefore compact. Finitely many finite subcovers, together with $U$, cover $X$. Hence $X$ is compact.