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For velocity $\mathbf{u}=\nabla \varphi$ and a conservative body force $-\nabla \chi$, the unsteady Bernoulli equation isIt follows by integrating the inviscid Euler equations in space; a purely time-dependent potential gauge can absorb $C(t)$.

Use a coordinate along the thin U-tube from the left surface to the right, and let the right surface rise by $\zeta$ while the left falls by $\zeta$. Incompressibility and uniform area give velocity $\dot{\zeta }$ along the whole fluid column. If its total length is $L$, the velocity-potential difference between the surfaces is $L\dot{\zeta }$; the column length stays constant as their displacements cancel. Both surfaces have atmospheric pressure and the same speed. Subtracting their Bernoulli balances, with $\chi =gy$, therefore yieldsIn the long-leg approximation underlying the requested formula, the base bend contributes negligible length and $L=2h$. ConsequentlyIf the horizontal connector or bend has nonnegligible length, it belongs in $L$ and the general frequency is $\sqrt{2g/L}$; the stated height alone would then not determine the inertia.

Use the weighted inner product $⟨u,v⟩={\int }_0^{1}u(x)v(x)xdx$. A monic quadratic orthogonal polynomial ${\pi }_2(x)=x^2+ax+b$, orthogonal to 1 and $x$, satisfiesThus $a=-6/5$, $b=3/10$, and its two zeros are $x_{-}=(6-\sqrt{6})/10$ and $x_{+}=(6+\sqrt{6})/10$. Matching the first two weighted moments givesso $w_{-}=(9-\sqrt{6})/36$ and $w_{+}=(9+\sqrt{6})/36$. The two-node Gaussian quadrature with linear weight is thereforeFor completeness, a polynomial $P$ of degree at most three can be divided as $P=q{\pi }_2+r$, where both $q$ and $r$ have degree at most one. Orthogonality makes the weighted integral of $q{\pi }_2$ zero, and this term vanishes at both nodes. The two moment equations integrate $r$ exactly. Thus the Gaussian quadrature is exact for every cubic polynomial. No two-node rule can integrate every quartic: the square of its nodal polynomial has quadrature value zero and strictly positive weighted integral.

The two given energy eigenstates are orthonormal. Resolve the initial state as $\psi (0)=\cos \theta {\psi }_0-\sin \theta {\psi }_1$. The Schrödinger equation therefore givesThe second component, which is the probability amplitude for the requested final state, is $\sin \theta \cos \theta (e^{-i{E}_{0}t/\hslash }-e^{-i{E}_{1}t/\hslash })$. By the Born rule,This includes $P(t)=0$ when the energy eigenvalues coincide or when the initial state is itself an energy eigenstate.

In vacuum magnetostatics, the two relevant Maxwell equations areOrient the unit normal $\mathbf{n}$ from the minus side to the plus side. Integrating Gauss's law for magnetism over a pillbox straddling the current sheet and shrinking its thickness gives $\mathbf{n}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})=0$.

To obtain the tangential condition, choose an arbitrary tangent unit vector $\mathbf{t}$ and an infinitesimal rectangle with its long sides of length $\ell$ parallel to $\mathbf{t}$ on the two sides of the sheet. Orient its surface normal as $\mathbf{n}\times \mathbf{t}$, so that its plus-side long edge is traversed along $\mathbf{t}$. The line integral in Ampère's law tends to $\ell \mathbf{t}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})$. The enclosed electric current tends to $\ell \mathbf{s}\cdot (\mathbf{n}\times \mathbf{t})$. HenceBecause this holds for every tangent direction and the normal jump is zero,Here $\mathbf{s}\cdot \mathbf{n}=0$, as required for a surface current density.

The force on a current sheet from the mean magnetic field isThe mean removes the sheet's equal and opposite local self-fields, leaving the field that acts on its current through the Lorentz force. Using a one-sided limiting magnetic field would incorrectly include the sheet's own force.

In polar coordinates, the Laplace equation isFor a single-valued, $2\pi$-periodic harmonic function on an annulus, separation of variables and its angular Fourier series giveIndeed a nonzero angular mode has radial equation $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, with solutions $r^{\pm n}$; the zero mode has solutions 1 and $\log r$. This expression describes the periodic annular problem here; angular terms not periodic, such as $\theta$, are excluded.

Decay at the origin leaves only positive powers in the interior, and decay at infinity leaves only negative powers in the exterior. Continuity on the unit circle matches the corresponding Fourier coefficients. If the common cosine coefficient is $K_n$, the interior and exterior terms are $K_n{r}^n\cos n\theta$ and $K_n{r}^{-n}\cos n\theta$. Their exterior-minus-interior radial derivative at the circle is $-2n{K}_n\cos n\theta$. Thus harmonic matching across a circle with a derivative jump gives $K_2=-1/4$, $K_4=-1/8$, and zero for every other sine or cosine mode. ConsequentlyEach branch satisfies the Laplace equation and its respective decay condition. Both give $-\cos 2\theta /4-\cos 4\theta /8$ at $r=1$, and direct differentiation gives the required jump $\cos 2\theta +\cos 4\theta$. The same mode calculation leaves no nonzero homogeneous solution in this class.

The quotient $h=f/g$ is a holomorphic function because the denominator has no zeros, and $|h|=1$. Write $h=u+iv$. The Cauchy-Riemann equations say $u_y=-v_x$ and $v_y=u_x$. Differentiating $u^2+v^2=1$ in both real directions givesThe coefficient determinant is $-(u^2+v^2)=-1$, so $u_x=v_x=0$, and the Cauchy-Riemann equations also give $u_y=v_y=0$. Thus $h$ is locally constant. Since a domain is connected, it is constant throughout $\Omega$. Its constant value has modulus one and can be written $e^{i\alpha }$ with $\alpha \in \mathbb{R}$. Therefore $f=e^{i\alpha }g$ throughout $\Omega$.

This is the constant-modulus holomorphic function principle. Connectedness is essential to a single global phase: on two disjoint discs, taking $g=1$ and choosing $f=1$ on one disc and $f=-1$ on the other gives equal moduli without one common phase.

Put ${\epsilon }_n={\sup }_{[-1,1]}|F_n-F|$. Uniform convergence means ${\epsilon }_n\to 0$, and continuity ensures that the functions have Riemann integrals. The integral triangle inequality givesThis proves that uniform convergence preserves integrals on a compact interval.

For the derivative assertion, take $F_n(x)=\sin (n^{2}x)/n$ and $F(x)=0$. These functions are differentiable, with $\sup |F_n|\le 1/n$, so their convergence is uniform. However, $F_n^{\prime }(0)=n$ whereas $F^{\prime }(0)=0$. The derivatives therefore need not converge even at one interior point.

We prove Simplicity of the alternating group A5 by finding all its conjugacy classes. The alternating group has group order $|A_5|=5!/2=60$. Its even permutations have cycle types $1$, $(abc)$, $(ab)(cd)$ and $(abcde)$. There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles, $5\cdot 3=15$ products of disjoint transpositions, and $4!=24$ five-cycles.

For a three-cycle, a commuting permutation must rotate its three-element support and may interchange the two fixed points. Its centralizer in $S_5$ has order six, and only the three rotations are even. Thus its conjugacy class in $A_5$ has size $60/3=20$ and contains every three-cycle. The centralizer of $(12)(34)$ in $S_5$ has order eight: one may interchange the two pairs and independently interchange the elements within each pair. It contains the odd element $(12)$, so exactly half of these commuting elements are even. The conjugacy class in $A_5$ consequently has size $60/4=15$. A permutation commuting with a five-cycle is determined by the image of one point and must be a power of that cycle. All five powers are even, so each such conjugacy class has size $60/5=12$. The 24 five-cycles form exactly two classes.

A normal subgroup contains the identity and is a union of whole conjugacy classes. Its possible orders therefore lie inBy Lagrange's theorem, its order must divide 60. Only 1 and 60 in this list do so. Hence $A_5$ has no proper nontrivial normal subgroup and is a simple group.

Read the arrows in the original diagram as follows: $0$ moves to $1^{\prime }$ with probability one; from an unprimed $n\ge 1$, probability $q$ leads to $n-1$ and probability $p$ leads to $(n+1{)}^{\prime }$; from $n^{\prime }$ the corresponding destinations are $n$ and $(n+1{)}^{\prime }$. This is the two-lane Markov chain with a deterministic restart. It is irreducible: every state can reach zero by a finite path of $q$ transitions, and zero can reach every primed state by outward $p$ transitions and then every unprimed state by a $q$ transition. Thus all states have the same recurrence classification.

Let $a_n$ and $b_n$ be the hitting probabilities of zero starting from $n$ and $n^{\prime }$, with $a_0=1$. First-step conditioning givesSubtracting yields $b_n=(1+q)a_n-q{a}_{n-1}$. Eliminating $b_{n+1}$ givesIts characteristic roots are $1$ and $r=q/[p(1+q)]=q/(1-q^2)$. Put $q_c=(\sqrt{5}-1)/2$, so $r=1$ precisely when $q=q_c$.

For $q>q_c$, the general solution is $a_n=A+B{r}^n$. The bound $0\le a_n\le 1$ and $r>1$ force $B=0$, and $a_0=1$ then gives $a_n=b_n=1$. For $q=q_c$, the general solution is $a_n=A+Bn$; boundedness again gives $a_n=b_n=1$. In both cases, the return probability from zero is $b_1=1$, so the Markov chain is recurrent.

For $q<q_c$, $0<r<1$, and the pairis a nonnegative bounded solution of the first-step equations. To identify it as the actual hitting probability, use that hitting probability is the minimal nonnegative harmonic extension. This fact follows directly by iteration: a nonnegative solution with boundary value one at zero dominates the probability of hitting zero within $N$ steps, for every $N$. The actual bounded solution has form $a_n=A+(1-A)r^n$, where nonnegativity as $n\to \infty$ requires $A\ge 0$. Minimality bounds it above by $r^n$, while this form bounds it below by $r^n$. Hence it is exactly the displayed solution. Starting at zero, the first return time has finite-return probability $b_1=q^2/p<1$. The Markov chain is therefore transient.

It remains to distinguish positive and null recurrence. Write $u_n={\pi }_n$ and $v_n={\pi }_{n^{\prime }}$ for invariant probabilities. Balance at zero and at $1^{\prime }$ gives ${\pi }_0=q{u}_1$ and $v_1={\pi }_0$. For the finite cut consisting of zero and both lanes up to level $n$, invariance equates its outward and inward probability fluxes:Balance at the unprimed state $n$ also gives $u_n=q(u_{n+1}+v_n)$. Combining these identities yieldsTherefore every invariant probability distribution would necessarily have $u_n=({\pi }_0/q)t^{n-1}$ and $v_n={\pi }_0{t}^{n-1}$. Conversely, these weights satisfy all individual balance equations, including $v_{n+1}=p(u_n+v_n)$, so their summability is the only remaining condition.

When $q>q_c$, $t<1$, and normalization givesAn irreducible Markov chain has an invariant probability distribution exactly when it is positive recurrent; equivalently its mean return time is finite and ${\pi }_0$ is its reciprocal. Thus this case is positive recurrent. At $q=q_c$, $t=1$ and the nonzero invariant weights cannot be summed to one. Irreducibility would force ${\pi }_0>0$ in any invariant probability distribution. Since recurrence was already proved but no such distribution exists, this case is null recurrent.

The complete classification is therefore transient for $0<q<q_c$, null recurrent at $q=q_c$, and positive recurrent for $q_c<q<1$. A drift argument using only the unprimed arrows would miss the primed lane and give the wrong threshold.

A simple hypothesis specifies one probability distribution. Let the null hypothesis and alternative have probability density functions $f_0,f_1$ with respect to a common measure. A possibly randomized statistical test is a measurable function $\phi (x)\in [0,1]$, its conditional probability of rejecting the null hypothesis. Its size of a statistical test is ${\mathbb{E}}_0\phi$, and its statistical power at the alternative is ${\mathbb{E}}_1\phi$. A most powerful test at level $\alpha$ maximizes power among tests with size at most $\alpha$.

The Neyman-Pearson lemma says to choose $k\ge 0$ and a test ${\phi }_{\ast }$ of size $\alpha$ with ${\phi }_{\ast }=1$ where $f_1>k{f}_0$ and ${\phi }_{\ast }=0$ where $f_1<k{f}_0$, choosing values on the equality set to reach the specified size. Equivalently, threshold the likelihood ratio, treating positive alternative density with zero null density as an infinite ratio. Randomization on a threshold atom ensures exact size when needed. If $k=0$, all remaining null-only mass can be used to reach size without changing power.

For any test $\phi$ of size at most $\alpha$, the signs of the two factors givepointwise. Integrating provesThis is the required optimality proof. If $k>0$ and another test is equally powerful, both inequalities must be equalities. It must have size $\alpha$ and agree with ${\phi }_{\ast }$ wherever $f_1\ne k{f}_0$, apart from null sets for the relevant densities. Thus freedom is confined to the equality set and sets null under both hypotheses; at $k=0$, additional size need not improve power. This also explains the nonuniqueness below.

For the unit-scale Laplace distribution, put $\delta ={\theta }_1-{\theta }_0>0$. Its likelihood ratio isIt is nondecreasing, so an upper-tail critical region has the Neyman-Pearson lemma form, even if the cutoff lies inside a constant-ratio tail. Solving ${\mathbb{P}}_{{\theta }_0}(X>c_{\alpha })=\alpha$ gives the most powerful test for a Laplace location shift,Continuity of the distribution makes this nonrandomized test exactly size $\alpha$. It need not be the unique most powerful test: within a flat extreme likelihood ratio region, other selections having the same null probability are equally good.

Its statistical power isHere $\beta$ denotes power, not the probability of a type-II error. For $\alpha =0.05$, $c_{\alpha }={\theta }_0+\log 10$. Thus the power is $0.05e^{\delta }$ for $\delta \le \log 10$, and $1-5e^{-\delta }$ otherwise. The first branch is at most $1/2$. On the second branch, power at least $0.95$ is equivalent to $5e^{-\delta }\le 0.05$. Therefore the required pairs are exactly ${\theta }_1-{\theta }_0\ge \log 100$, with no restriction on the common location.

Let $r=Ax-b$ and consider an arbitrary variation $x+th$. The objective for the linear least-squares problem isAt a minimizer the derivative at $t=0$ must vanish for every $h$, so $A^{T}r=0$. Conversely, when this holds, the same expansion at $t=1$ gives $\Vert A(x+h)-b{\Vert }^2=\Vert r{\Vert }^2+\Vert Ah{\Vert }^2\ge \Vert r{\Vert }^2$ for every $h$. Thus $x$ minimizes the residual norm if and only if $A^T(Ax-b)=0$. These least-squares normal equations do not require $A$ to have full rank; without full rank the minimizing vector need not be unique.

The original PDF has $A_{22}=1$, not the 2 in the converted TeX. Using the PDF's matrix givesThe least-squares normal equations consequently have the unique solutionAs a direct check, $Ax-b=(-279,-95,238,-61{)}^T/123$, whose inner product with either column of $A$ is zero. The positive determinant verifies full column rank for this particular problem.

Let $L_{ik}=1$ when $k\le i$ and $L_{ik}=0$ otherwise. The matrix product has entriesso $A_n=L{L}^T$. Both $L$ and its matrix transpose are triangular matrices with diagonal entries one. Multiplicativity of the determinant therefore gives $\det A_n=(\det L{)}^2=1$ for every $n\ge 1$. This determinant of the minimum-index matrix also follows by subtracting each preceding row from the next, working from the bottom upwards: the resulting matrix is upper triangular with unit diagonal.

Take the momentum operator $p=-i\hslash d/dx$ and sufficiently regular real $f$. Acting on a test wave function, the commutator is $[p,f]=-i\hslash f^{\prime }$. Expanding the factorized quantum Hamiltonian givesso $V(x)=(f(x{)}^2-\hslash f^{\prime }(x))/(2m)$, which is real.

Set $Q=p-if$. Its zero-mode equation iswith solutionIt satisfies $H{\psi }_0=Q^{†}Q{\psi }_0/(2m)=0$ as a differential equation. It is an actual quantum state only when it is square-integrable. If $f(x)=c{x}^n+\cdots$, $c\ne 0$, its antiderivative has leading term $c{x}^{n+1}/(n+1)$. For even $n$, this leading term has opposite signs at the two ends of the line, so the wave function grows at one end and cannot be normalized. For odd $n$, it has the same sign at both ends. Hence normalizability requires odd $n$, and in fact holds exactly when $n$ is odd and $c>0$. A negative leading coefficient makes it grow at both ends; $f=0$ gives a nonnormalizable constant. These are the normalizable zero mode of a polynomial factorized Hamiltonian conditions.

For a normalized state in the operator domain, integration by parts, with the appropriate vanishing boundary term, givesThus every energy eigenvalue is nonnegative. Equality forces $Q\psi =0$ almost everywhere, whose first-order equation has the one-dimensional solution space above. Whenever its zero mode is normalizable, every energy eigenstate not proportional to ${\psi }_0$ has strictly positive energy. If the formal zero mode is not normalizable, there is no zero-energy eigenstate; the positivity statement still applies to every normalizable eigenstate. This argument concerns eigenvalues and does not assert a strictly positive bound on an arbitrary continuous spectrum.

For the quantum harmonic oscillator, take $f(x)=m\omega x$ with $m,\omega >0$. ThenIts zero mode is normalizable and has no competitor of lower energy. Using the Gaussian integral for its normalization gives the concise answerup to an arbitrary constant phase.

Put $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }$. The magnetic field is the curl of the magnetic vector potential. Since $\mathbf{J}({\mathbf{r}}^{\prime })$ does not depend on the differentiated variable $\mathbf{r}$,Taking the curl under the source integral consequently gives the Biot-Savart law,Outside the compact source this differentiation is ordinary differentiation of a smooth kernel. For regular volume currents, the locally integrable $R^{-2}$ derivative kernel gives the same formula at interior points.

For the loop, choose positive current circulating counterclockwise as viewed from the positive $z$-axis. Parametrize its source position and directed line element byAt $(0,0,z)$ the denominator is the constant $(a^2+z^2{)}^{3/2}$, whileThe transverse terms integrate to zero around the loop. The Biot-Savart law therefore yieldsThe magnetic field is axial, with the stated magnitude for $I>0$; reversing the electric current reverses its direction.

Use coordinates $\mathbf{r}$ measured from the center of mass $G$, so $\int \mathbf{r}dm=0$. The displacement from $P$ is $\mathbf{r}-\mathbf{X}$. Expand the defining inertia tensor:All terms linear in $\mathbf{r}$ integrate to zero. The remaining constant terms give the tensor parallel-axis theorem,For the uniform cube, let $M=8\rho a^3$. Symmetry makes the off-diagonal inertia tensor entries zero. Direct integration gives $\int x_i^{2}dm=M{a}^2/3$ for each coordinate, and therefore $I(G)=(2M{a}^2/3)I_3$. With $\mathbf{X}=a(1,1,1{)}^T$, the tensor parallel-axis theorem givesThe unit eigenvector $u=(1,1,1{)}^T/\sqrt{3}$ has eigenvalue $2M{a}^2/3$. Every vector in the plane $x+y+z=0$ is annihilated by $11^T$ and hence has eigenvalue $11M{a}^2/3$. One orthonormal basis of that eigenspace isThe principal moments are $2M{a}^2/3$ along the body diagonal and $11M{a}^2/3$ twice in its perpendicular plane. Any orthonormal pair in that plane supplies the remaining principal axes.

Define a candidate antiderivative by integration along the straight segment from zero:For $z$ in the open unit disc and sufficiently small $h$, the triangle with vertices $0,z,z+h$ lies within the disc, since the disc is convex. By the Cauchy integral theorem, its boundary contour integral vanishes. ConsequentlyFor $h\ne 0$, subtract $f(z)$ after division by $h$. The absolute value of the remainder is at most ${\sup }_{0\le t\le 1}|f(z+th)-f(z)|$, which tends to zero by continuity of the holomorphic function $f$ at $z$. Hence the complex derivative exists and $g^{\prime }(z)=f(z)$ for every point of the disc. This proves that $g$ is a holomorphic function and supplies the requested primitive directly from the Cauchy integral theorem.