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Prepare the auxiliary system in the maximally mixed state $\rho =I/N$. Then $\mathrm{Tr}(\rho U)=\mathrm{Tr}U/N$, so the two measurement bases giveThere is no need to know the eigenvectors of $U$. Choose a uniformly random member of any known orthonormal basis on each run; discarding the classical preparation label produces $I/N$. Alternatively, discard one half of a maximally entangled pair of $N$-level systems. This is trace estimation by a maximally mixed Hadamard test.

The circuit directly estimates the normalized trace. To demand fixed absolute accuracy $\epsilon$ in the unnormalized trace requires normalized accuracy $\epsilon /N$; straightforward independent frequency sampling consequently uses order $N^2/{\epsilon }^2$ runs at fixed confidence. Thus the construction should not be confused with a dimension-independent guarantee for arbitrary absolute trace precision.

The mixed-state expression in the PDF contains a mismatched bra in its second term. A probability mixture of the stated orthonormal eigenvectors must be interpreted as $\rho ={\sum }_k{p}_k|u_k⟩⟨u_k|$. Taken literally with a different bra, it need not be Hermitian or positive and is not generally a density operator.

Let $m={\sum }_k{p}_k{e}^{i{\varphi }_k}=\mathrm{Tr}(\rho U)$. A quantum channel and its measurement probabilities are linear in the density operator, so averaging the pure-state results givesTherefore the same frequency estimator yieldsMore generally, after the controlled operation and before the final Hadamard gate, the reduced control density operator isThis follows by a partial trace of the control blocks $\rho$, $\rho U^{†}$, $U\rho$ and $U\rho U^{†}$. It proves the mixed-state Hadamard-test quadratures for any auxiliary density operator, even one not diagonal in the eigenbasis. Conjugation by the final Hadamard gate maps $X$ to $Z$ and $Y$ to $-Y$, explaining the imaginary-part sign.

Apply the Born rule to the two amplitudes in the introductory calculation. In the computational basis, the probabilities areThe other basis printed in this question is $(|0⟩\pm i|1⟩)/\sqrt{2}$, the Pauli Y gate eigenbasis. It is not the real Hadamard basis. Write the output amplitudes as $a=(1+e^{i\varphi })/2$ and $b=(1-e^{i\varphi })/2$. The plus-outcome amplitude is $(a-ib)/\sqrt{2}$ and the minus-outcome amplitude is $(a+ib)/\sqrt{2}$. Since $\mathrm{Im}(\overline{a}b)=-\sin \varphi /2$, the Born rule givesIn particular, at $\varphi =\pi /2$ the output is the minus eigenstate, which checks the sign of the imaginary quadrature.

Use separate fresh runs for each measurement basis. If ${\hat{P}}_j$ denotes the empirical outcome frequency, thenThe two differences estimate $\cos \varphi$ and $\sin \varphi$, resolving the ambiguity that would remain from the first basis alone. Each Bernoulli frequency converges by the strong law of large numbers, with standard error of order the inverse square root of the number of runs. One may project the noisy estimate onto the unit circle if a phase estimate is desired, but that projection is not necessary for consistency.

Choose a common endpoint for the eight-month analysis and ascertain every student's status through that endpoint. With a complete event register, record each departure at its event time and right-censor every remaining student at the same administrative cutoff. Otherwise obtain a confirmation for all students at the cutoff, not just the most recent quarterly list, and verify intervening departure dates consistently.

A second coherent design is to analyse only up to the last common verified quarterly date, postponing a longer analysis until a further complete follow-up round is available. Extra event reports after that date cannot be combined with earlier censoring of all nonfailures as though the resulting risk set were independently censored. Genuine losses to follow-up should be recorded separately and their possible informative censoring assessed. Use a common observation rule for failures and nonfailures, independent of their subsequent outcome.

The proposed hybrid recording rule updates the event times of students who leave but does not update the follow-up of those who remain. Thus the recorded end of observation depends on the subsequent outcome: this is outcome-dependent updating of survival follow-up.

In the four-student illustration, all four are known to be event-free at month six. Under the proposed coding, the three students without event reports are censored at month six, while the student whose event is subsequently reported remains in the risk set until month seven. The recorded risk set at that event therefore has size one, although the other three may still be studying. The Kaplan–Meier estimator factor would be $1-1/1=0$ and the Nelson–Aalen estimator jump would be $1/1=1$. This creates an artificial complete loss of estimated survival at that event.

If event notification is complete through the common eight-month cutoff, the absence of a notification establishes that the other three have not had the event by that cutoff. They should remain in the risk set at month seven. With only these four subjects, the corresponding factors are instead $1-1/4=3/4$ and $1/4$. Updating only failures while backdating nonfailures' censoring creates a severely biased risk set. If notifications are incomplete, absence of a report does not establish continued study; positive follow-up confirmation is then required.

A filtered category is nonempty, every pair of objects maps to a common object, and every parallel pair becomes equal after postcomposition. Equivalently every finite diagram in a category has a cocone.

For a filtered set diagram $D:I\to \mathbf{Set}$, the finite-stage equality in a filtered set colimit says that representatives $x\in D(i)$ and $y\in D(j)$ define the same element exactly when their images agree at some common stage. Indeed, equality in the ordinary disjoint-union-and-quotient construction is witnessed by a finite zigzag; filteredness moves its objects to a common stage and equalizes the finitely many competing arrows.

For a finite categorical product, choose a common stage for all coordinate representatives to obtain surjectivity of the colimit-to-product comparison. Finitely many coordinate equalities can then be realized together at one later stage, proving injectivity. The empty product is preserved because a nonempty filtered category is connected. For an equalizer, an element whose two images agree in the colimit reaches a stage where they agree, so it is represented by an element of the stagewise equalizer; the same equality criterion proves injectivity. Finite products and equalizers construct every finite limit. Hence filtered colimits commute with finite limits in sets.

We prove the equivalence by the cycle solution, solution, solution, solution, then back to solution.

The zeros of the denominator satisfy $e^z=\ell \pi$ with nonzero integer $\ell$, since an exponential never vanishes. Equivalently the poles areAt such a point $e^{z_{m,k}}=(-1{)}^{k}m\pi$, and the derivative of the denominator is $e^z\cos (e^z)=(-1{)}^{m+k}m\pi \ne 0$. All these poles are simple, with

The only pole is the simple one at $z=1$, with $\boxed{{\mathrm{Res}}_1=e}$. The point $z=0$ is an essential singularity, not another pole: multiplying $e^{1/z^2}$ by a nonvanishing analytic factor cannot remove its infinite principal part. For completeness its residue can also be read fromThe coefficient of $z^{-1}$ is $-{\sum }_{k\ge 1}1/k!=1-e$. This extra residue belongs to the essential singularity; it does not change the list of poles.

A recurrent state $i$ satisfies ${\mathrm{Pr}}_i(T_i^{+}<\infty )=1$, where $T_i^{+}=\inf \{n\ge 1:X_n=i\}$. Saying that the transition matrix is recurrent means that every state is recurrent. In an irreducible Markov chain it suffices that one state is recurrent, since recurrence is a communicating-class property.

The rotation $g$ has order six. The possible cycle types in the alternating group $A_4$ are the identity, a three-cycle, and a product of two disjoint transpositions; their orders are one, three and two. Since a group isomorphism preserves the order of a group element, $D_{12}$ is not isomorphic to $A_4$.

The proper normal subgroups and their quotient groups are as follows. Here the trivial subgroup is included as a proper subgroup.

The rotational subgroups are preserved by reflection conjugation, and the last three order-six subgroups are normal by index two. The three order-four groups are not normal, since conjugation by $g$ changes the reflection exponent by two, which is nonzero modulo three. Part solution excluded the reflection subgroups. These observations, together with the complete subgroup list, prove the normal-subgroup list is exhaustive.

The remaining proper subgroups areTheir orders are respectively one, three, six, four and six. The order-four groups are Klein four-groups; the last two are dihedral groups of order six, isomorphic to $S_3$.

For completeness, if $H$ contains a reflection, write $H\cap ⟨g⟩=⟨g^d⟩$ with $d\mid 6$. Any other reflection differs from a chosen $g^{j}h$ by an element of this rotational subgroup, so $H=⟨g^d,g^{j}h⟩$ with $j$ specified modulo $d$. The choices $d=1,2,3,6$ give respectively the whole group, the two order-six groups, the three order-four groups and the six reflection groups already listed. This proves there are no omitted subgroups.

Thus there are eight remaining proper subgroups. The printed count nine includes one too many; nine would be correct if the whole group were included with these remaining subgroups. Altogether $G$ has sixteen subgroups, fifteen proper.

The involutions in an even dihedral group are the half-turn $g^3$ and the six reflections $g^{j}h$. Thus the seven order-two subgroups areThe half-turn is central, so only $⟨g^3⟩$ is normal. Indeed $g(g^{j}h)g^{-1}=g^{j+2}h$, a different reflection; an order-two subgroup has only one nonidentity element, so each reflection subgroup fails the normal subgroup condition.

Introduce the velocity variable $v=\dot{x}$. The second-order differential equation becomes the autonomous planar systemA critical point of a dynamical system is a point where both components of the vector field vanish: here $v=0$ and $f(x,0)=0$. It is a stationary solution in the phase plane. The Jacobian matrix at each such point gives the local linear stability of a planar equilibrium. For the equations below, a conserved energy also determines the nonlinear phase portrait; this is important because purely imaginary eigenvalues alone do not generally prove that an equilibrium is a center equilibrium.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable under this hypothesis. More generally the definition works whenever the displayed expectations exist. Assertions about arbitrary random variables below are understood on this domain: without suitable moments a covariance need not be defined at all.

For the final limit, near zero $\cos x>0$, so the real logarithm is well defined. First apply the proved rule to $\tan x/x$ on the right: its derivative ratio is ${\sec }^{2}x\to 1$. Next apply it to $\log (\cos x)/x^2$; the derivative ratio isThe original quotient is even, so the left limit equals the right limit. Consequently

Now impose the stated zero endpoint values and nonvanishing assumptions near $a$. For each $x\in (a,a+\delta )$, applying the proved identity on $[a,x]$ gives some $c_x\in (a,x)$ such thatAs $x\downarrow a$, every possible $c_x$ also tends to $a$. Explicitly, for any $\epsilon >0$, choose a right neighborhood of $a$ where $|f^{\prime }/g^{\prime }-L|<\epsilon$. If $x$ lies in it, so does $c_x$, and therefore $|f(x)/g(x)-L|<\epsilon$. This proves the endpoint zero-over-zero form of L'Hôpital's rule:No continuity of either derivative is required, only their ratio limit and the given nonvanishing conditions.

For the two-function assertion, defineThis function has the required continuity and differentiability and satisfies $H(a)=H(b)=0$. Rolle's theorem therefore gives $c$ withThis proves the Cauchy mean value theorem even when either endpoint increment vanishes; no division was used.

Rolle's theorem states that if a real function $h$ is continuous on $[a,b]$, differentiable on $(a,b)$, with $a<b$ and $h(a)=h(b)$, then some $c\in (a,b)$ satisfies $h^{\prime }(c)=0$.

The strong plume's outgoing flux equals the total lower-layer supply, $Q_2(h_2)=Q$. Apply the additional closure in this part and the lower-layer balance from part (b):Dividing by $C{B}_2^{1/3}{h}_1^{5/3}$ givessoFor $0<\psi <1$ this puts the second matching level above the first. It tends to one as $\psi \to 0$ and to $2^{2/5}$ as $\psi \to 1$. At the latter endpoint the two warm layers already have identical density, so $h_2$ is only a formal matching label, not a distinct observable interface. The expression follows from the stipulated closure; modeling the actual plume in the changed ambient requires the matched continuation discussed in part (d).