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In the pion rest frame, conservation of momentum makes the muon and massless neutrino momenta equal and opposite, of magnitude $p$. The neutrino energy is $pc$, while the muon satisfies $E^2=p^2{c}^2+m^2{c}^4$. Energy conservation gives $E+pc=M{c}^2$. Squaring after substituting $E=M{c}^2-pc$ yieldsThe relativistic two-body decay speed is thusChoose $S^{\prime }$ moving with velocity $+v$ along $x$ relative to $S$, so the pion has velocity $-v$ in $S^{\prime }$ and speed $v$. With ${\Gamma }_v=(1-v^2/c^2{)}^{-1/2}$, the four-dimensional Lorentz transformation isThe muon's four-momentum is $(E/c,0,p,0)$ before the boost. The same matrix givesVelocity and momentum have the same spatial direction. The unsigned deflection angle from the positive $y$ axis therefore satisfiesChoosing the opposite boost makes the pion move along positive $x$ and reverses $p_x^{\prime }$, leaving this speed-based angle formula unchanged.

Irrotational flow means $\nabla \times \mathbf{u}=0$; incompressible flow means $\nabla \cdot \mathbf{u}=0$. The given streamfunction automatically satisfies the second condition, since ${\partial }_x{\psi }_y-{\partial }_y{\psi }_x=0$. For an oriented path with unit tangent $(dx/ds,dy/ds)$, choose its right-hand unit normal $\mathbf{n}=(dy/ds,-dx/ds)$. ThenThe opposite normal reverses the sign; the source's unspecified normal orientation must be chosen consistently. The volume flux between two streamlines is thus their streamfunction difference. A closed contour has zero net flux, expressing mass conservation for constant-density flow without a source inside.

For the channel, a rigid boundary has zero normal velocity and hence constant streamfunction. Choose zero on the lower wall. Uniform inflow at infinity must have horizontal velocity $-m$, so there $\psi =-my$. The upper wall then has constant $-m$. The unbroken left wall connects to it and also has constant $-m$. The jump by $m$ between the walls at the corner represents the withdrawn flux. Finally the vorticity is $v_x-u_y=-\Delta \psi$, so irrotationality gives $\Delta \psi =0$.

Write $\psi =-my+w$. Then $w$ vanishes at $y=0,1$ and at infinity, while $w(0,y)=-m(1-y)$ for $0<y<1$. Separation of variables gives the harmonic sine mode in a half-strip $e^{-n\pi x}\sin n\pi y$. The Fourier sine series coefficients of the left boundary areTherefore the corner sink in a semi-infinite channel isFor $x>0$ the series and its differentiated series converge on compact subsets, making the sum harmonic. At $x=0$, ${\sum }_{n\ge 1}\sin (n\pi y)/n=\pi (1-y)/2$ for $0<y<1$, so the left-wall value is exactly $-m$. The other wall values and far-field limit follow directly.

The plus sign printed in the final series is inconsistent with the printed boundary conditions. It would give $m-2my$ on the left wall, not $-m$. The negative sign above is required. As a local flux check, near the corner the wall values give $\psi \sim -2m\theta /\pi$; the radial velocity is $-2m/(\pi r)$ and its outward flux across a quarter-circle is $-m$, precisely withdrawal.

Introduce nonnegative slack variables $s_1,s_3$, a nonnegative surplus variable $s_2$, and an artificial variable $a$. Phase I of the two-phase simplex method minimizes $w=a$. The initial simplex dictionary isWith the nonbasic variables zero this is feasible for the artificial problem. Enter $x_1$; the minimum ratios are $9/4,12/6,3/3$, so $s_3$ leaves. Solving its row for $x_1$ and substituting givesNow enter $x_3$. The ratios from decreasing basic variables are $5/(4/3)=15/4$ and $6/3=2$, so $a$ leaves. The resulting dictionary isSetting the nonbasic variables zero gives $a=0$, so Phase I has reached its minimum and found feasibility. Remove $a$ and reinstate the true objective $z$. Substitution of the basic variables givesEnter $x_2$. The decreasing rows give ratios $(5/3)/(2/3)=5/2$ for $x_1$ and $(7/3)/(7/3)=1$ for $s_1$, so $s_1$ leaves. Phase II then hasEvery nonbasic variable is nonnegative and every reduced cost is positive. Thus the simplex optimality criterion provesAll three original constraints are active. Positivity of all three reduced costs also proves uniqueness: equality in the objective lower bound forces $s_1=s_2=s_3=0$, which fixes the displayed solution.

The Rao-Blackwell theorem says that, for a sufficient statistic $T$ and a finite-second-moment estimator $\delta (X)$, the conditional estimator ${\delta }_T(T)=E_{\theta }[\delta (X)\mid T]$ can be chosen independently of the unknown parameter and has no greater mean squared error for estimating any specified function $g(\theta )$. Sufficiency makes the conditional distribution of the data given $T$ parameter-independent. Conditional expectation preserves the expectation, hence preserves unbiasedness. Writing $\delta -g=(\delta -{\delta }_T)+({\delta }_T-g)$, the cross term has expectation zero, because $E[\delta -{\delta }_T\mid T]=0$. ThereforeEquality holds exactly when $\delta ={\delta }_T$ almost surely. This proves the squared-loss form, including the usual unbiased-estimator variance statement.

Here $\theta >0$ is necessary for the stated uniform interval. Put $L=X_{(1)}={\min }_i{X}_i$ and $R=X_{(n)}={\max }_i{X}_i$. The joint density factors asThe factorization theorem for sufficient statistics shows that $T=(L,R)$ is sufficient. Also $E[X_1]=2\theta$, so $X_1/2$ is unbiased and has variance ${\theta }^2/12$.

For $n\ge 2$, conditional on distinct extrema $l,r$, exchangeability gives probability $1/n$ that $X_1$ is each endpoint; with probability $(n-2)/n$ it is one of the remaining observations, uniform on $(l,r)$. Its conditional mean is consequentlyThus the Rao-Blackwell estimator for a uniform scale interval iswhich is unbiased and has no greater mean squared error. For $n=1$, $L=R=X_1$ and the same formula reduces to the original estimator. As a quantitative check, the covariance of two uniform order statistics givesFor instance, after writing $X_i=\theta +2\theta U_i$, both extreme uniform variances are $n/[(n+1{)}^2(n+2)]$ and their covariance is $1/[(n+1{)}^2(n+2)]$. These formulas follow by integrating the joint extreme density $n(n-1)(r-l{)}^{n-2}$ on $0<l<r<1$; they also display strict improvement for $n>1$.

Fix $z_0\in E$ and $ϵ>0$. Choose $N$ so that ${\sup }_E|f_N-f|<ϵ/3$, and then choose $\delta >0$ so that $|f_N(z)-f_N(z_0)|<ϵ/3$ for $z\in E$ with $|z-z_0|<\delta$. The triangle inequality gives $|f(z)-f(z_0)|<ϵ$. This proves the uniform limit theorem on the relative topology of $E$, whether or not $E$ is open.

The Weierstrass M-test states that if $|u_n(z)|\le M_n$ for every $z\in E$ and $\sum M_n<\infty$, then $\sum u_n$ converges uniformly and absolutely on $E$. Indeed, the supremum of a tail is at most the corresponding tail of $\sum M_n$, proving the uniform Cauchy criterion.

For the given secant function terms, a denominator can vanish only at $z=n(2k+1)$, an integer. Thus every individual term is continuous off the integers. Fix $z_0\notin \mathbb{Z}$, and choose a closed disk $K$ about $z_0$ containing no integer. Let $R={\sup }_K|\mathrm{Re}z|$ and take $N>2R$. For $n\ge N$, $|\pi \mathrm{Re}z/(2n)|<\pi /4$; the permitted cosine inequality givesThe Weierstrass M-test makes the tail uniformly convergent on $K$. The finitely many earlier terms are continuous there, so the uniform limit theorem makes their sum continuous at $z_0$. Therefore $f$ is continuous on $\mathbb{C}\setminus \mathbb{Z}$. The argument is local; a global M-test on the whole punctured plane would fail near its poles.

The discriminant of a binary quadratic form is $d=b^2-4ac$; it is unchanged by a unimodular matrix change of integral variables and satisfies $d\equiv 0$ or $1(\mathrm{mod}4)$. For such a $d$, the discriminant criterion for prime representation by a binary quadratic form says that some integral binary quadratic form of discriminant of a binary quadratic form $d$ represents an odd prime $p$ exactly when $d$ is a square modulo $p$, including zero. Equivalently,These versions agree: a square root modulo $p$ can be chosen even when $d\equiv 0(\mathrm{mod}4)$ and odd when $d\equiv 1(\mathrm{mod}4)$, by adding $p$ if needed, and then the Chinese remainder theorem applies.

For completeness, if $f(x,y)=p$, then $(x,y)$ is primitive, since any common divisor would have its square divide $p$. Complete $(x,y)$ to a basis of ${\mathbb{Z}}^2$ with determinant one using Bézout's identity. The transformed binary quadratic form has leading coefficient $p$ and discriminant of a binary quadratic form $B^2-4pC=d$. Conversely, a suitable $B$ makes $[p,B,(B^2-d)/(4p)]$ an integral binary quadratic form representing $p$ at $(1,0)$.

For $x^2+3y^2$, the discriminant of a binary quadratic form is $-12$. The prime $3$ is represented. If $p\ne 3$ is represented, reducing modulo $3$ gives $p\equiv x^2\equiv 1(\mathrm{mod}3)$, which also excludes $p=2$. To prove sufficiency for $p\equiv 1(\mathrm{mod}3)$, quadratic reciprocity givesThe criterion constructs a positive definite binary quadratic form $[p,B,C]$ of discriminant of a binary quadratic form $-12$. It is primitive: a common divisor of its coefficients would divide $p$, and if that divisor were $p$, then $p^2$ would divide $12$, impossible for $p\ne 2,3$.

We must still establish that this form is equivalent to $x^2+3y^2$, rather than merely some form of the same discriminant of a binary quadratic form. In a positive definite integral binary quadratic form, replace $x$ by $x+ky$ to arrange $|b|\le a$. If $c<a$, apply $(x,y)\mapsto (-y,x)$ and repeat. Each such exchange strictly decreases the positive integer leading coefficient, so this reduction algorithm for a positive definite binary quadratic form terminates with $|b|\le a\le c$. Then $12=4ac-b^2\ge 3a^2$, so $a\le 2$. With $a=1$, parity forces $b=0$, giving $c=3$. With $a=2$, the only integral choices are $b=\pm 2,c=2$, and these are not primitive. Thus every primitive positive definite binary quadratic form of discriminant of a binary quadratic form $-12$ is equivalent to $[1,0,3]$, and preserves its represented integers. Therefore

With polar unit vectors ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, the second term is $\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero divergence and zero vorticity away from the axis. Its singular axis carries a line vortex; adding the constant velocity field gives a uniform flow plus that line vortex.

The counterclockwise circulation around a positively oriented loop is $\oint \mathbf{u}\cdot d\mathbf{r}$. For a radius-$R$ circle, the uniform flow contributes zero and the line vortex contributesAny loop winding once around the axis has the same circulation, by Stokes theorem in the nonsingular intervening region.

On $C_R$, $\mathbf{n}={\mathbf{e}}_r$, $\mathbf{u}\cdot \mathbf{n}=U\cos \theta$, and $dl=Rd\theta$. ThusIn fact this equality holds for every $R>0$ for the stated field, since the remaining vector integral is $(0,\pi ,0)$.

This is the advective part of the momentum flux. To obtain the force, include pressure. For steady irrotational flow, the Bernoulli equation givesThe constant terms integrate to zero against $\mathbf{n}$; the remaining pressure contribution is $\oint p\mathbf{n}dl=\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The total outward momentum flux plus pressure term is therefore $\rho \mathbf{\Gamma }\times \mathbf{U}$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, isThis is the Kutta–Joukowski theorem, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional aerofoil, faster-decaying far-field terms do not change this limit. The displayed advective integral supplies half the lift; pressure supplies the other half.