The maximum-flow problem maximises source-to-sink flow subject to capacity and conservation constraints. Ford–Fulkerson repeatedly finds an augmenting path in the residual network and augments by its bottleneck capacity, stopping when none exists. Multiplying rational capacities by a common denominator makes them integers; every augmentation then raises the flow by at least one while the value is bounded by the finite source capacity, so termination is finite.

The MLE is $\hat{\mu }=\bar{X}$ with MSE $1/n$. Combining likelihood and the $N(0,{\tau }^{-2})$ prior gives posterior $N(n\bar{X}/(n+{\tau }^2),1/(n+{\tau }^2))$. The frequentist MSE of its mean is $[n+{\tau }^4{\mu }^2]/(n+{\tau }^2{)}^2$, which is below $1/n$ exactly when ${\mu }^2<2{\tau }^{-2}+1/n$.

For $z>0$, $\varphi =q(4\pi {\epsilon }_0{)}^{-1}(|x-d\hat{z}{|}^{-1}-|x+d\hat{z}{|}^{-1})$ and $E=-\nabla \varphi$. The normal field just above the plane is $-qd/[2\pi {\epsilon }_0(r^2+d^2{)}^{3/2}]$, so $\sigma ={\epsilon }_0{E}_z$ is the stated expression. Integrating $2\pi r\sigma (r)dr$ from zero to infinity gives total induced charge $-q$.

Differentiability at $p$ means $f(p+h)=f(p)+Lh+o(|h|)$ for a linear map $L=Df(p)$; partial derivatives are its values on coordinate vectors. Integrating the partial derivatives along the two coordinate segments and using continuity at $p$ proves the stated sufficient condition. Here $f(x,y)=g(x)+g(y)$ with $g(t)=t^2\sin (1/t)$ and $g(0)=0$. Since $g\prime (0)=0$, $f$ is differentiable everywhere. But $g\prime (t)=2t\sin (1/t)-\cos (1/t)$ does not approach $0$ at zero, so $f$ is $C^1$ exactly where $xy\ne 0$.

Invariant means $\pi P=\pi$. In a finite irreducible chain, $E_i{T}_i^{+}=1/{\pi }_i$ and the expected visits to $i$ between visits to $j$ are ${\pi }_i/{\pi }_j$. The first burglar has uniform invariant law on $64$ PINs, so (a) is $64$ and (b) is $16$. First-step equations by Hamming distance give the adjacent-state hitting time (c) as $63$. For the modified one-coordinate chain, destination weights are $1$ for digits $0$–$5$ and $2$ for $6,7$; its invariant weights are proportional to $w_a(10-w_a)$, namely $9$ and $16$. The product invariant law and the occupation formula give (d) ${\pi }_{(6,7)}/{\pi }_{(0,0)}=256/81$.

Odd terms vanish and direct evaluation gives $L(p)=(8/45)c_4$, so the rule is exact through degree $3$. With $n=3$, the Peano kernel isIts nonnegativity and $\int K=L(x^4)/4!=1/135$ give $|L(f)|\le \Vert f^{(4)}{\Vert }_{\infty }/135$, so $N=135$. Equality holds for a quartic with fourth derivative of constant sign.

Faraday's law is $\mathcal{E}={\oint }_C(E+v\times B)\cdot dl=-d{\Phi }_B/dt$. Here $I=-L(B\dot{x}+x\dot{B})/R$ (sign relative to the chosen circuit orientation), and the bar force is $ILB$. For $B=B_0/(1+t)$ this gives the displayed equation with $\alpha =B_0^2{L}^2/(mR)$. At $\alpha =1$, the independent solutions are $u$ and $u{e}^{1/u}$, $u=1+t$; the data give $x=x_{0}u{e}^{1/u-1}$. Thus $\dot{x}\to x_0/e$ and total mechanical work is $m{x}_0^2/(2e^2)$. The current circulates to reinforce the decaying $+z$ flux, and its force drives increasing area, both opposing the flux decrease as Lenz's law requires.

With the proton fixed, $[-{\hslash }^2{\nabla }^2/(2m_{\mu })-e^2/(4\pi {\epsilon }_{0}r)]\psi =E\psi$: kinetic energy plus Coulomb attraction, assuming nonrelativistic point particles and neglecting spin, radiation, and proton motion (or replacing $m_{\mu }$ by the reduced mass). Substitution of the $2s$ form gives $E=-{\hslash }^2{\alpha }^2/(8m_{\mu })$ and $\alpha =m_{\mu }{e}^2/(4\pi {\epsilon }_0{\hslash }^2)$. Direct differentiation verifies it. The curve starts positive, crosses at $r=2/\alpha$, then approaches zero from below; ordinary hydrogen has the same shape but the smaller electron mass gives a larger radial scale.

Separation gives angular modes $e^{im\theta }$ and $-[rR\prime ]\prime +m^{2}R/r=\lambda rR$. Frobenius gives $p=m$ and $a_k=-a_{k-2}/[k(k+2m)]$, producing $J_m$; termwise differentiation gives the two stated recurrences. A radial solution is $u={\sum }_n[A_n\cos (c{j}_{0n}t)+B_n\sin (c{j}_{0n}t)]J_0(j_{0n}r)$. Here $B_n=0$ andequivalently $A_n=2[j_{0n}{J}_0\prime (j_{0n}{)}^2{]}^{-1}{\int }_0^{1}r{J}_1(j_{0n}r)dr$ after integration by parts.