The inclusionalways holds: each generator coming from $I\cap J$ lies in both extended ideals.

The reverse inclusion can fail. Letfor a field $k$. Under $A\cong k[x]$,whereas $I\cap J=(xy)$ in $R$, soThus statement (2) is true in general and statement (1) is false in general.

There is an exact sequence of $R$-modulesIf $A$ is a flat module over $R$, tensoring this sequence with $A$ preserves its left exactness. The image of each tensor product inside $A$ is the corresponding extension of an ideal, soBoth displayed inclusions therefore hold. This is the flat extension preserves finite ideal intersections property.

Conversely, assume condition (b), and let $M,N$ be nonzero. The stated property givesThese are nonzero vector spaces over the residue field $k=R/\mathfrak{m}$, so their tensor product over $k$ is nonzero. Associativity and base change giveTherefore $M{\otimes }_{R}N\ne 0$. This proves the reverse implication and the local tensor nonvanishing criterion.

Assume tensor products of two nonzero modules never vanish. If $\mathfrak{m}\ne \mathfrak{n}$ were distinct maximal ideals, then $\mathfrak{m}+\mathfrak{n}=R$ andalthough both residue fields are nonzero. Hence $R$ has one maximal ideal $\mathfrak{m}$ and is a local ring.

For every module $M$,If $\mathfrak{m}M=M$, this tensor product vanishes. Since $R/\mathfrak{m}\ne 0$, the assumed property forces $M=0$. Thus condition (a) implies condition (b).

For $Q=p$ and $P=\lambda q$,The map is therefore canonical only forThen $Q=p$ and $P=-q$. The type-one generating function for a canonical transformationgivesas required.

For $Q=q$ and $P=\lambda p$,Thus the map is canonical only forAt that value it is the identity and is generated by

For $Q=\lambda q$ and $P=\lambda p$,The one-dimensional criterion for a canonical transformation therefore givesFor either value, the type-two generating function for a canonical transformationworks, because $p=S_q=\lambda P$ and $Q=S_P=\lambda q$; when ${\lambda }^2=1$, the first equation is equivalent to $P=\lambda p$. For every other $\lambda$, the Poisson bracket is not one and the map is not canonical.

For the regular concatenation grammar, letThen defineIf $G$ and $G^{\prime }$ are regular, every nonterminal rule retained from $P$ is right-linear, and each replaced terminal rule is also right-linear. A regular derivation in $G$ uses exactly one terminal rule, at its final step; the replacement transfers control to $S^{\prime }$, after which a word of $L(G^{\prime })$ is generated. Hence

Choose a new start variable $S_0\notin V\cup V^{\prime }$. The concatenation grammar isIndeed, its derivations first produce a word from $G$ and a word from $G^{\prime }$ in their original order, so $L(H)=L(G)L(G^{\prime })$.

Part (a) givesHence both operators preserve the magnetic quantum number $m$. They are also odd under parity, while $|n,\ell ,m⟩$ has parity $(-1{)}^{\ell }$. Their matrix elements between states of equal parity vanish. The position-momentum selection rules in the first excited hydrogen eigenspace therefore leave only the coupling betweenThe $m=\pm 1$ states decouple, and all diagonal matrix elements vanish.

In the ordered basis $(|s⟩,|p⟩)$, remove the common energy $E_2$ and writeLetThe rotating-frame amplitudes obey a constant Schrödinger equationPutSince $H_{\mathrm{eff}}^2=q^{2}I$,For the initial amplitudes $(A_0,A_1)$, the exact coefficients are thereforeandRestoring the common unperturbed phase, the solution isThis is the rotating-frame solution for a harmonically driven two-level system.

The states $|n,\ell ,m⟩$ diagonalize $L^2$:The perturbation therefore introduces no mixing. The orbital-angular-momentum-squared perturbation of hydrogen leavesat energywhileremain triply degenerate atThese statements are exact because the unperturbed Hamiltonian commutes with $L^2$.

Take $n=2^k$. Every proper divisor of $n$ divides $n/2=2^{k-1}$, so the points fixed by $F^n$ that do not have exact period $n$ are precisely those fixed by $F^{n/2}$. Part (ii) therefore givesBecause different starting points are counted separately, no division by the period is made. This is the exact power-of-two periods of the doubling map.

By part (i), $F^n(x)=x$ exactly whenThus the periodic points of the doubling map fixed by $F^n$ areand their number isFor $n=2$ this gives three points: the fixed point and the two points in the two-cycle. For $n=4$ it gives fifteen: those three points and the twelve points in the three four-cycles found above.

The result follows by induction. It is true for $n=1$. IfthenTherefore

Points whose period divides two solveso, after removing the fixed point zero, the unique two-cycle is

Points whose period divides four are $j/15$ for $j=0,\dots ,14$. Removing the three points fixed by $F^2$, namely $0,5/15,10/15$, leaves twelve points of exact period four. They form the three orbitsandStarting at another point on one displayed orbit gives the corresponding cyclic rotation.

A fixed point satisfiesso $x$ is an integer. In $[0,1)$ the only possibility isEquivalently, a binary sequence fixed by the left shift is constant; $0.000\dots {}_2=0$, while $0.111\dots {}_2=1$ represents the excluded endpoint, which is the same circle point.

Concatenate every finite binary word, ordered first by length and then lexicographically:Every finite binary block occurs in this expansion. Since iterating $F$ shifts digits to the left, its orbit enters every binary cylinder and is therefore dense. The digit string is not eventually periodic, so it gives a concrete nonperiodic chaotic orbit.

Put $T_k=R_k^3$. Part (i) identifies $T_k$ as the $k$th arrival time of a ratePoisson process. Hence $T_k$ has the gamma densityThe change of variables $t=r^3$, with $dt=3r^{2}dr$, gives the kth-nearest-neighbour distance in a homogeneous Poisson point process:

Write $r=(x^2+y^2+z^2{)}^{1/2}$, so $f(x,y,z)=r^3$. For $t\ge 0$,whose volume isThus the pushforward intensity of $[0,t]$ isThe pushforward is locally finite and diffuse, so the radial volume transform of a homogeneous Poisson point process and the mapping theorem show that

Choose $\varphi \in C_c^{\infty }(\mathbb{R})$ with $\Vert \varphi {\Vert }_2=1$, and for $R>0$ setA change of variables givesThe latter tends to zero as $R\to \infty$. Since the energy is nonnegative,This is vanishing Dirichlet energy by dilation on the line.

The infimum is not attained. If an admissible $u$ had zero energy, then its weak derivative would vanish almost everywhere, so $u$ would be constant almost everywhere. The only constant in $L^2(\mathbb{R})$ is zero, contradicting $\Vert u{\Vert }_2=1$.