Put $g=\lambda /(4{\omega }^2)$. Normalization through order $g^2$ givesThusTracing over $B$ removes the cross terms. The reduced density operator of a weakly coupled oscillator pair is thereforewhose trace is one to the required order.

The unperturbed ground state is $|0,0⟩$ with energy $\hslash \omega$. Sincethe first-order energy correction vanishes and first-order nondegenerate perturbation theory gives

The region $D$ is compact and positively invariant and contains the unique fixed point. If that fixed point is unstable, choose an initial point in $D$ which is not on its stable set. Its nonempty compact $\omega$-limit set cannot be the fixed point. The Poincaré-Bendixson theorem therefore supplies a periodic orbit. Hence a sufficient parameter condition, and the one selected by the preceding analysis, is

On $x=0$, $\dot{x}=ay\ge 0$; on $y=0$, $\dot{y}=b>0$. On $x=x^{\ast }$, the identity $\dot{x}+\dot{y}=b-x$ gives a negative outward component because $x^{\ast }>b$. On $y=b/a$, one has $\dot{y}=-x^{2}b/a\le 0$. Finally, the sloping edge has outward normal proportional to $(1,1)$, and thereThus the vector field points inward or is tangent on every edge, so trajectories cannot leave the closed polygon $D$.

Adding the two equations gives $\dot{x}+\dot{y}=b-x$, so a fixed point hasAt this point the Jacobian has determinant $a+b^2>0$ and traceThe fixed point is asymptotically stable exactly when $T<0$, orThe equality curve exists only for $a\le 1/8$ and isFor $0<a<1/8$ the fixed point is unstable between these two curves and stable outside them; for $a>1/8$ it is stable everywhere.

Along a periodic orbit the vector field is tangent and makes one full turn, so its index around the orbit is $+1$. If the enclosed region contained no fixed point, the normalized vector field would extend continuously across the disc and its boundary map would have degree zero. This contradicts the index $+1$. Thus every planar periodic orbit encloses a fixed point.

The Poincaré-Bendixson theorem says that a nonempty compact $\omega$-limit set of a planar $C^1$ flow which contains only finitely many fixed points is either a fixed point, a periodic orbit, or a union of fixed points and connecting trajectories. In particular, if it contains no fixed point, it is a periodic orbit.

For any $\epsilon >0$,ThereforeThe last term tends to zero. Taking the limsup and then $\epsilon \downarrow 0$ proves the claim.

By the union bound,For an $L^2$ random variable, $t^2\mathbb{P}(|X_1|>t)\to 0$. Taking $t=\epsilon \sqrt{n}$ makes the right side tend to zero, proving $M_n/\sqrt{n}\to 0$ in probability.

Again the weights tend to zero, so finite-rank truncations prove compactness. Products of $m$ consecutive weights are bounded by $1/(m+1)!$, making the spectral radius zero. Therefore

The weighted shift is compact because truncating after the first $N$ weights gives finite-rank operators converging in norm, as $1/n\to 0$. Moreoverso its spectral radius is zero. Since the spectrum is nonempty,

This is the unilateral left shift. It is not compact because $S{e}_{n+1}=e_n$. For every $|\lambda |<1$,is an eigenvector with eigenvalue $\lambda$. The norm bound and closedness then give

This is the unilateral right shift. It is not compact because the orthonormal sequence $(T{e}_n)=(e_{n+1})$ has no convergent subsequence. Its norm is one. For $|\lambda |>1$ the Neumann series gives a resolvent; for $|\lambda |<1$, $\bar{\lambda }$ is an eigenvalue of the adjoint left shift, so $\lambda$ lies in the spectrum of $T$. Closedness supplies the boundary. Hence

True. At the vertices the side conditions forceOn each side, $g$ remains in that same side, so the boundary restriction is homotopic within the boundary to the identity and has degree one. If it extended continuously over the filled triangle with image in the boundary circle, that degree-one boundary map would be null-homotopic, which is impossible. Equivalently, a sufficiently fine simplicial approximation would contradict the parity form of Sperner's lemma proved in part (b).

False. The constant map to $E$ satisfies it: $E\in I$ and also $E\in J\subset J\cup K$.

Not necessarily, once the subdivision has at least two segments per side. Colour $D,E,F$ red, green, blue respectively and colour every other grid vertex red. No small triangle contains both exceptional vertices $E$ and $F$, and every small triangle therefore uses at most two colours. For the unsubdivided triangle the assertion is of course true.

Count red-green segments modulo two. On $DE$, part (a) says their number is odd; the other two outer sides contain no red-green segment. Every interior segment belongs to two small triangles and therefore contributes zero modulo two, while each boundary segment belongs to one. Thus the sum, over all small triangles, of their numbers of red-green edges is odd.

A triangle with all three colours has exactly one red-green edge. A triangle using only red and green has zero or two, and any other nontrichromatic triangle has zero red-green edges. Hence the parity sum counts precisely the trichromatic triangles modulo two, proving that their number is odd.

Alice performs the operation and measurement from part (iii) and sends the one-bit outcome to Bob. If the bit is $0$, Bob does nothing. If it is $1$, Bob applies $ZX$, which maps $|{\psi }_2⟩$ to $|{\psi }_1⟩$ by part (i). In either case Bob finishes with the known target state $|{\psi }_1⟩$, using one shared Bell pair and one classical bit.

Using the decomposition from part (ii), $U_{\theta }^{-1}$ maps Alice's $|{\psi }_1⟩$ to $|0⟩$ and $|{\psi }_2⟩$ to $|1⟩$. Alice obtains $0$ or $1$, each with probability $1/2$. Conditional on outcome $0$, Bob has $|{\psi }_1⟩$; conditional on outcome $1$, Bob has $|{\psi }_2⟩$.

The matrix $U_{\theta }$ is real orthogonal. Part (a) givesExpanding the left side in the computational basis yields