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Put $S={\sum }_i{X}_i$, and let $z_{.95}$ be the exact standard-normal 95th percentile. Under parameter $\theta$, $S=n\theta +\sqrt{n}Z$ with $Z$ standard normal. This distribution can be justified directly: the joint density of the centered observations is proportional to $\exp [-{\sum }_i(X_i-\theta {)}^2/2]$. An orthogonal change of coordinates with first row $(1,\dots ,1)/\sqrt{n}$ has unit absolute Jacobian and preserves the sum of squares. Its density factors into $n$ one-dimensional standard-normal densities, making the first coordinate $Z=(S-n\theta )/\sqrt{n}$ standard normal.

Consider the test ${\varphi }_{\ast }=1_{\{S>\sqrt{n}{z}_{.95}\}}$. Its rejection probability isincreasing in $\theta$. Therefore its maximum rejection probability over the null $\theta \le 0$ is achieved at zero and equals $\alpha =1/20$. This proves it has the required test size.

To prove the uniformly most powerful test property, rather than quote the Neyman-Pearson lemma, fix any ${\theta }_1>0$. The likelihood ratio of the full sample against $\theta =0$ isstrictly increasing in $S$. Set $K=\exp ({\theta }_1\sqrt{n}{z}_{.95}-n{\theta }_1^2/2)$. For any competing possibly randomized test $0\le \varphi \le 1$ with composite-null size at most $\alpha$, we have ${\mathbb{E}}_0\varphi \le \alpha ={\mathbb{E}}_0{\varphi }_{\ast }$. Pointwise,because both factors are nonnegative above the threshold and both are nonpositive below it. Integrating givesThis comparison proves the likelihood-ratio optimality result needed here in full. Since the same critical region works for every ${\theta }_1>0$, it proves the uniformly most powerful test property over the entire alternative.

For $n=9$, the exact critical region and its quoted-table approximation areUsing the supplied rounded percentile $1.65$ gives $\sum X_i>4.95$. The exact cutoff uses $z_{.95}\approx 1.644854$ and hence $\overline{X}>0.548285$; the rounded cutoff is the intended numerical approximation, not a literally exact size-$0.05$ threshold. The Gaussian distribution is continuous, so equality at the cutoff has probability zero and no randomization there is needed.

For steady electromagnetic fields, Maxwell's equations give $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ and $\nabla \cdot \mathbf{B}=0$. Introduce a magnetic vector potential with $\mathbf{B}=\nabla \times \mathbf{A}$ and the permitted Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the double curl then gives $-\Delta \mathbf{A}={\mu }_0\mathbf{j}$. For localized currents and the decaying solution,Here $-\Delta (1/(4\pi |\mathbf{r}|))=\delta$ follows from harmonicity away from zero and the inward flux of the radial derivative through a small sphere, whose magnitude is one. It verifies the vector Poisson solution componentwise. Moreover steady electric charge conservation gives ${\nabla }^{\prime }\cdot \mathbf{j}=0$; integration by parts confirms the stated potential is divergence-free. Taking its curl, using $\nabla |\mathbf{r}-{\mathbf{r}}^{\prime }{|}^{-1}=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$, proves the Biot-Savart law:The zero-field condition at infinity selects the magnetic field produced by these currents, excluding an externally imposed homogeneous magnetic field.

At the origin write the source point as $r{\mathbf{e}}_r+z{\mathbf{e}}_z$. Its current is $kr{\mathbf{e}}_{\varphi }$, so the numerator of the Biot-Savart law is $k{r}^2{\mathbf{e}}_z-krz{\mathbf{e}}_r$. The radial term cancels after angular integration. Using $d{V}^{\prime }=rdrd\varphi dz$ givesThe inner integral is $2h/[r^2\sqrt{h^2+r^2}]$, as is verified by differentiating $z/[r^2\sqrt{r^2+z^2}]$. HenceFor $b,h>0$, its magnitude is ${\mu }_0|k|h(\sqrt{h^2+b^2}-h)$; it points along $+z$ for positive azimuthal current $k>0$, and along $-z$ for $k<0$. The limit $h\to \infty$ is ${\mu }_{0}k{b}^2/2$, agreeing with direct integration of the infinitely long cylindrical current distribution.

For a bound state take $b>0$ and set $\psi (r)=e^{-br}v(r)$. Substitution into the radial equation and cancellation of the common exponential giveSeek a power series $v(r)={\sum }_{j\ge 0}{c}_j{r}^j$ with $c_0\ne 0$. Equating coefficients of $r^j$ yields the polynomial construction of hydrogen S states recurrenceFor any integer $N\ge 1$, choose $b=a/(2N)$. The numerator vanishes at $j=N-1$, while it is nonzero for $j<N-1$, so $v$ is a nonzero polynomial of degree $N-1$. Therefore $\psi$ is continuous at $r=0$ and decays exponentially as $r\to \infty$. The three-dimensional normalization integral isIt is finite near zero because $\psi$ is bounded, and finite at infinity because an exponential dominates every polynomial. Multiplying by its positive reciprocal square root gives a normalizable, nonzero wavefunction. The recurrence ensures the differential equation for $r>0$; at the singular endpoint the regular continuous extension has $2{\psi }^{\prime }(0)+a\psi (0)=0$, so the potentially singular coefficient cancels. No nonexistent value of $1/r$ at zero is substituted.

Since $E=-{\hslash }^2{b}^2/(2m)$, the choice of $b$ givesFor $N=1$, the normalized example is $\psi (r)=\sqrt{b^3/\pi }{e}^{-br}$. For $N=2$, $v(r)$ is proportional to $1-br$. These illustrate the polynomial-times-exponential Coulomb bound states produced for every required energy. The radial derivative at the origin is generally nonzero, a Coulomb cusp consistent with the requested continuity.

For $y=e^{-x}$, substitution in the unnormalized equation gives $[(x+2)-(x+1)-1]e^{-x}=0$. For $y=ax+b$, the remaining expression is $(x+1)a-(ax+b)=a-b$. Thus an independent second solution is $x+1$. Their Wronskian is $-(x+2)e^{-x}$, nonzero on $x\ge 0$.

The left boundary solution is $u(x)=x+1$, since $u^{\prime }(0)=u(0)=1$; the decaying right solution is $v(x)=e^{-x}$. A Robin half-line Green function from left and right solutions must be continuous at $x=\xi$, while its derivative jumps by one because the leading coefficient of $L$ is one. These conditions giveThereforeThe two values agree at $x=\xi$, and $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=(\xi +1)/(\xi +2)+1/(\xi +2)=1$. Each branch solves the homogeneous equation and its relevant boundary condition, so these checks prove the distributional identity $LG=\delta$ rather than merely postulate the kernel.

Integrate this Green function against the forcing $-(\xi +2)e^{-\xi }$. Splitting at $\xi =x$ givesIt has $y(0)=y^{\prime }(0)=1$ and decays at infinity. Direct substitution gives $L[y]=-(x+2)e^{-x}$. Uniqueness follows because a decaying homogeneous solution must be a multiple of $e^{-x}$, and its left boundary condition would require $-c=c$, forcing $c=0$.

The intended open region lies inside the larger circle and outside the smaller one. The two circles are tangent at zero, a boundary point, so $w=1/z$ is holomorphic throughout the region. For a circle $|z-ia|=a$, its equation is $|z{|}^2=2a\mathrm{Im}z$, which becomes $\mathrm{Im}w=-1/(2a)$. The side inside that circle becomes $\mathrm{Im}w<-1/(2a)$. Hence the tangent-circle crescent conformal map begins withThe affine coordinate $\zeta =4w+2i$ sends this strip bijectively to $0<\mathrm{Im}\zeta <1$. Exponentiation $\eta =e^{\pi \zeta }$ is then a conformal bijection to the upper half-plane: its argument ranges strictly from zero to $\pi$, and the inverse logarithm has that unique argument. Finally $(\eta -i)/(\eta +i)$ is a Cayley transform between the half-plane and disk onto the unit disc, since for $\mathrm{Im}\eta >0$ its modulus is less than one and the inverse is $\eta =i(1+u)/(1-u)$. As $e^{\pi \zeta }=e^{4\pi /z}$, a complete answer isEvery stage has nonzero derivative and is one-to-one onto its indicated range. The denominator cannot vanish in the upper half-plane. Unlike a region between disjoint nested circles, this tangent crescent has no enclosed hole, so there is no annulus obstruction to mapping it onto a disc.

If $F=b\circ a$ with $a:{\mathbb{R}}^2\to \mathbb{R}$ and $b:\mathbb{R}\to {\mathbb{R}}^2$, the chain rule gives $DF=DbDa$, the product of a two-by-one and a one-by-two matrix. Its rank is at most one, so its Jacobian determinant is identically zero.

Conversely, continuity and $f_y(0,0)\ne 0$ give a neighborhood $U$ on which $f_y$ stays nonzero. The identity $f_x{g}_y-f_y{g}_x=0$ then implies $g_x=f_x{g}_y/f_y$. Therefore the smooth function $e=g_y/f_y$ satisfiesFor a graph curve $c(t)=(t,\alpha (t))$ with $F\circ c$ constant, the first component must have derivative zero. Thus the necessary equation isIt is also sufficient: it makes $d(f\circ c)/dt=0$, and then $d(g\circ c)/dt=ed(f\circ c)/dt=0$.

To justify the existence through every nearby point, choose a closed rectangle about the origin inside $U$. The right-hand side $h(t,y)=-f_x/f_y$ is smooth there, hence bounded and locally Lipschitz in $y$. The Picard-Lindelöf theorem supplies a unique local solution through every initial point $(t_0,y_0)$ in a smaller open rectangle $V$. The boundedness of $h$ and a positive distance from the smaller rectangle to the outer boundary permit a common sufficiently short time interval about each $t_0$ on which the solution stays in $U$. Each point of $V$ therefore lies on one of these graph curves, and the preceding chain-rule calculation proves that $F$ is constant along it. This proves rank-one level curves from a characteristic differential equation using local ODE existence. The curves are local in their time parameter; no unjustified global existence for all real $t$ is required.

The second fundamental form records the normal component of second-order bending of the surface. For a surface curve $\gamma (t)=\sigma (u(t),v(t))$, the terms containing $u^{\prime \prime }$ and $v^{\prime \prime }$ are tangent and thereforeThus on a unit tangent direction it gives the normal curvature; for a general coordinate direction the normal curvature is the ratio of the second to the first fundamental form. Equivalently the quadratic part of displacement out of the tangent plane is one half of this form. Reversing the chosen unit normal changes its sign.

Now suppose $L=M=N=0$. Differentiate $\mathbf{N}\cdot {\sigma }_u=\mathbf{N}\cdot {\sigma }_v=0$. This givesAlso differentiation of $\mathbf{N}\cdot \mathbf{N}=1$ gives ${\mathbf{N}}_u\cdot \mathbf{N}={\mathbf{N}}_v\cdot \mathbf{N}=0$. The parametrization is regular, so ${\sigma }_u,{\sigma }_v,\mathbf{N}$ form a basis of three-dimensional space. Both normal derivatives are orthogonal to this basis and hence vanish:The parameter ball is connected; integrating these derivatives along a straight segment between parameter points shows that $\mathbf{N}={\mathbf{N}}_0$ is constant. Finally both derivatives of $\sigma \cdot {\mathbf{N}}_0$ vanish, because ${\sigma }_u,{\sigma }_v$ are tangent. Thus the zero second fundamental form implies planar image statement isThe PDF has ${\mathbf{N}}_v$ in the second derivative equality; the apparent ${\mathbf{N}}_t$ in the TeX is not a distinct coordinate.

The monomials $1,x,\dots ,x^n$ are linearly independent and span, so $\boxed{{\dim }_{\mathbb{C}}V=n+1}$. Use the factorial monomial basis for polynomial differentiation $p_j(x)=x^j/j!$, $0\le j\le n$. Direct differentiation gives $e_i(p_j)={\delta }_{ij}$. HenceIndeed every $\varphi \in V$ has the Taylor expansion $\varphi ={\sum }_{j=0}^n{e}_j(\varphi )p_j$; evaluation on the $p_j$ proves independence of the functionals. Derivative orders include zero. If one uses a convention excluding zero from $\mathbb{N}$, $e_0$ must explicitly be included, since all positive-order derivatives annihilate constants and cannot span the dual space.

Differentiation satisfies $D{p}_0=0$ and $D{p}_j=p_{j-1}$ for $j\ge 1$. With column vectors of coefficients, its matrix $A$ has entries $A_{ij}=1$ if $j=i+1$ and zero otherwise, for indices $0\le i,j\le n$:The dual map is $D^{\ast }(\ell )=\ell \circ D$, not a conjugate-linear adjoint. Thus $D^{\ast }{e}_i=e_{i+1}$ for $i<n$ and $D^{\ast }{e}_n=0$. Therefore $\boxed{[D^{\ast }]=[D{]}^T}$ in the corresponding dual basis. For $n=0$ both matrices are the one-by-one zero matrix.

For a linear map $\theta :U\to V$, its rank of a linear map is $r(\theta )=\dim \mathrm{im}\theta$, and its nullity is $n(\theta )=\dim \ker \theta$. Choose a basis $e_1,\dots ,e_s$ of its kernel and extend it to a basis $e_1,\dots ,e_d$ of $U$. Then $\theta e_{s+1},\dots ,\theta e_d$ span its image. They are linearly independent: a linear combination mapped to zero would belong to the kernel, contradicting independence of the extended basis unless all its coefficients vanish. Hence $r(\theta )=d-s$, provingFor endomorphisms of $U$, define $(\theta +\varphi )(u)=\theta (u)+\varphi (u)$ and $(\theta \varphi )(u)=\theta (\varphi (u))$. The image of the sum lies in the sum of the two images. ThusTo prove the product inequality, restrict $\varphi$ to $K=\ker (\theta \varphi )$. This map has kernel $\ker \varphi$ and image $\mathrm{im}\varphi \cap \ker \theta$: every vector in the intersection has a preimage in $K$. Applying the rank-nullity theorem to the restricted map givesNow put $d=\dim U$ and $s=r(\theta )+r(\varphi )$. If both equalities hold, $r(\theta +\varphi )=s\le d$, while $n(\theta \varphi )=2d-s\le d$ forces $s\ge d$. Therefore $s=d$, the sum has full rank, and the product has nullity $d$. The sum is an isomorphism and the product is zero.

Conversely, if $\theta \varphi =0$, then $\mathrm{im}\varphi \subseteq \ker \theta$, giving $s\le d$. If the sum is also an isomorphism, its rank is $d\le s$. Thus $s=d$, and the two inequalities become equalities. This proves the equality in the rank-sum and nullity-product inequalities in both directions.

For maximization of $f(x)$ subject to $g_j(x)\le 0$ and $h_k(x)=0$, the Lagrangian sufficiency theorem says: if a feasible $x^{\ast }$ globally maximizes $L(x)=f(x)-{\sum }_j{\lambda }_j{g}_j(x)+{\sum }_k{\nu }_k{h}_k(x)$, where ${\lambda }_j\ge 0$, and ${\lambda }_j{g}_j(x^{\ast })=0$, then $x^{\ast }$ globally maximizes $f$ over the feasible set. Indeed, at every feasible $x$,This requires a global maximum of the optimization Lagrangian; a stationary point alone is not sufficient without additional concavity or another justification.

Put $q=p/(p-1)$ and $C=({\sum }_i|a_i{|}^q{)}^{1/q}$. If every $a_i=0$, the maximum is zero and every feasible vector is optimal. Otherwise setSince $p(q-1)=q$, ${\sum }_i|x_i^{\ast }{|}^p=1$. For $g(x)={\sum }_i|x_i{|}^p-1$, the optimization Lagrangian is $L(x)={\sum }_i{a}_i{x}_i-\lambda g(x)$. Its gradient vanishes at $x^{\ast }$ becauseThe function ${\sum }_i|x_i{|}^p$ is a strictly convex function for $p>1$, so $L$ is a strictly concave function. Its supporting-hyperplane inequality proves that this stationary point is its global maximum. Complementary slackness and feasibility hold, so the sufficiency theorem applies. Substitution givesThe maximizing vector is unique when $a\ne 0$. It is the norming vector for Holder inequality, and the value is the corresponding dual norm.

For the independent normal distribution observations, the log likelihood isIts derivative is ${\sum }_i(X_i-\theta )$ and its second derivative is $-n<0$, so the unique maximum-likelihood estimate isFor the prior $\theta \sim N(\mu ,{\tau }^{-1})$ with $\tau >0$, multiplying prior density by likelihood and completing the square givesThus normal-normal conjugacy gives the posterior distributionFor quadratic loss, the conditional expected loss of action $a$ is the posterior variance plus $(a-\mathbb{E}[\theta \mid X]{)}^2$. Its unique minimizer is the posterior mean, so the Bayes estimator isThis weights the sample and prior means by their respective precisions.

The displayed evolution law uses the usual assumptions of incompressible flow, constant density and conservative body force. With $\mathbf{\omega }=\nabla \times \mathbf{u}$, the Euler equation isUse $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ and take a curl. Gradients have zero curl, andHere $\nabla \cdot \mathbf{\omega }=0$ because it is a curl, and $\nabla \cdot \mathbf{u}=0$ by incompressibility. The vorticity equation is thereforeThe material derivative combines local change with advection by fluid particles. The right-hand side is vortex stretching and tilting: spatial variation of velocity along a vortex line changes its strength and direction. Along each particle path this is a homogeneous linear equation for vorticity. For a smooth velocity field its solution with zero initial vorticity remains zero by uniqueness. Thus an initially irrotational flow remains irrotational while these assumptions and smoothness hold.

For a plane flow $\mathbf{u}=(u(x,y,t),v(x,y,t),0)$, the vorticity is $\mathbf{\omega }=(0,0,\zeta )$ and $(\mathbf{\omega }\cdot \nabla )\mathbf{u}=\zeta {\partial }_z\mathbf{u}=0$. Hence $D\zeta /Dt=0$. Every particle retains its initial value, so an initially uniform value givesInviscid motion alone is insufficient for the printed simplified law: compressible variable-density motion generally adds $-\mathbf{\omega }\nabla \cdot \mathbf{u}+\nabla \rho \times \nabla p/{\rho }^2$. Those terms vanish under the assumptions used above.

Interpret $m$ as the rest mass of each initial ball and assume that no energy or momentum escapes the collision. Let $\gamma =(1-v^2/c^2{)}^{-1/2}$ be its Lorentz factor. The total initial energy and momentum areTheir conservation determines the final four-momentum. For a single lump, $P={\gamma }^{\prime }{m}^{\prime }{v}^{\prime }$ and $E={\gamma }^{\prime }{m}^{\prime }{c}^2$, henceThe final invariant mass follows from $m^{\prime 2}{c}^4=E^2-c^2{P}^2$:This is the rest mass of a coalescing relativistic collision. For $v>0$, $m^{\prime }>2m$: some kinetic energy becomes internal energy in the final rest frame. Setting $m^{\prime }=2m$ would discard that energy and contradict conservation. At small speeds $v^{\prime }\sim v/2$, while $v=0$ gives $m^{\prime }=2m$.