For $m>n$, the $L^1$ difference is ${\int }_n^m{x}^{-1}dx=\log (m/n)$. Choosing $m=2n$ makes it $\log 2$, so the sequence is not Cauchy in $L^1$. Its squared $L^2$ difference is ${\int }_n^m{x}^{-2}dx\le 1/n$, which tends uniformly to zero for $m>n$. Hence it is Cauchy in $L^2$, with norm limit $x^{-1}$ on $(1,\infty )$.

Completeness means that every norm-Cauchy sequence has a limit belonging to the same Lp space, with $\Vert f_n-f{\Vert }_p\to 0$. To prove it, choose a subsequence $f_{n_j}$ with $\Vert f_{n_{j+1}}-f_{n_j}{\Vert }_p\le 2^{-j}$. The functionshave $p$-norm bounded by $\Vert f_{n_1}{\Vert }_p+1$, by Minkowski inequality. They increase to $H$, and Fatou's lemma gives $\int H^p\le (\Vert f_{n_1}{\Vert }_p+1{)}^p$. Thus $H$ is finite almost everywhere, so the subsequence telescopes absolutely there to a measurable limit $f$ with $|f|\le H$ and $f\in L^p$.

The same argument on the tails gives $\Vert f-f_{n_j}{\Vert }_p\le {\sum }_{k\ge j}{2}^{-k}\to 0$. Given $ϵ$, choose such a subsequence term beyond the Cauchy threshold and use $\Vert f_n-f{\Vert }_p\le \Vert f_n-f_{n_j}{\Vert }_p+\Vert f_{n_j}-f{\Vert }_p$. This proves convergence of the full sequence and completeness for every $1\le p<\infty$, on an arbitrary measure space.

For $3\pi /2<2\alpha <2\pi$, $1/2<{\lambda }_0<2/3$ and $3/2<{\lambda }_1<2$, whereas ${\lambda }_2>5/2$. Hence both of the first two harmonic modes precede the $r^2$ particular term in the corner expansion; the first still dominates as $r\to 0$. The first-mode gradient is singular. These are reentrant wedges, and the global arc condition determines the important homogeneous coefficients.

For $2\alpha <\pi /2$, ${\lambda }_0>2$ and ${\lambda }_1>6$. Every harmonic correction decays faster than $r^2$ at the corner, so the $r^{2}f(\theta )$ particular solution is the leading corner velocity. It is positive inside this acute wedge and the velocity gradient tends to zero at the vertex.

A useful description of the low-column Lyman-alpha forest at $z\simeq 3$ is a power law $f_X(N,z)=A(z)N^{-\beta }$, with $\beta$ around $1.4$–$1.5$ over the ordinary unsaturated forest range. The same column dependence holds for $f_z$ at a fixed redshift. It is not a universal unbroken power law: saturation, increased neutral fraction and changing absorber populations produce breaks. Lyman limit systems become optically thick at $N\simeq 10^{17.2}{\mathrm{cm}}^{-2}$ because $N{\sigma }_{912}\simeq 1$ with ${\sigma }_{912}\simeq 6.3\times 10^{-18}{\mathrm{cm}}^2$. Damped Lyman-alpha systems have $N\gtrsim 2\times 10^{20}{\mathrm{cm}}^{-2}$ and strong radiative damping wings. They are rare in number but contain substantial neutral mass. The highest-column distribution steepens or cuts off rather than extending indefinitely.

This sketch shows representative shapes and transition scales, not an observational fit; its ordinate is distribution per unit column, not counts per logarithmic bin. In an Einstein-de Sitter universe, $H=H_0(1+z{)}^{3/2}$, so a fixed comoving population with fixed physical cross-section above a fixed column threshold has $d\mathcal{N}/dz\propto (1+z{)}^{1/2}$. ThusThe observed redshift evolution of the Lyman-alpha forest is much stronger for commonly selected forest lines. For example, Kim, Cristiani and D'Odorico's absorption survey finds an index about $2.19$ in $d\mathcal{N}/dz\propto (1+z{)}^{\gamma }$ for $10^{13.64}<N/{\mathrm{cm}}^{-2}<10^{16}$ over $1.5<z<4$, corresponding to an increase by approximately $2^{2.19}\simeq 4.56$. This number is selection-dependent. Expansion changes the gas density, and the evolving cosmic ionizing background changes its neutral fraction through photoionization equilibrium, moving systems across a fixed neutral-column threshold. Structure also evolves. A non-evolving population with fixed physical cross-section is therefore not an adequate physical model of the observed forest.

An isolating integral is a single-valued integral of motion $I(\mathbf{x},\mathbf{v})$ whose fixed value confines an orbit to a lower-dimensional level set of phase space. Along the flow, $dI/dt=0$; for an autonomous Hamiltonian, this means $\{I,H\}=0$. Several isolating integrals are independent when their gradients are linearly independent on the regular region considered. Counting them requires a specified domain: merely local orbit labels in a flow box do not establish global single-valued integrals. The following cases concern regular globally meaningful orbit constraints.

Each independent run uses one oracle call and gives a uniformly sampled equation $z\cdot s=0$. Collect $n-1$ linearly independent equations and use Gaussian elimination over ${\mathbb{F}}_2$. Their solution space has dimension one and consists of $\{0,s\}$, so the promised nonzero solution identifies $s$ uniquely.

To quantify the expected query count for Simon sampling, write $d=n-1$. If the existing equations have rank $k<d$, their span contains $2^k$ of the $2^d$ possible samples. A fresh equation raises the rank with probability $1-2^{k-d}$. The mean waiting time is the corresponding geometric distribution mean, givingThe last sum is less than $2$: its first term is $1$ and its remaining terms are bounded by ${\sum }_{j=2}^{\infty }{2}^{1-j}=1$. Hence the expected number of quantum queries is $O(n)$. For a fixed high-confidence budget, use $m=d+c$ samples. Failure to span means that a nonzero linear functional on the $d$-dimensional sampling space annihilates every sample. Each such functional has probability $2^{-m}$, so the union bound gives failure probability at most $(2^d-1)2^{-m}<2^{-c}$. Thus $O(n+\log (1/\delta ))$ queries suffice for failure probability at most $\delta$. For $n=1$, the nonzero string is known already and no samples are needed.

Classically, a collision between distinct queried arguments immediately yields $s=x+y$. Random queries produce this collision after order $2^{n/2}$ calls, by the birthday problem. Conversely, choose the nonzero period uniformly and label the fibres randomly. Before a collision, $q$ queries can exclude at most $\left(\genfrac{}{}{0ex}{}{q}{2}\right)$ periods, their pairwise argument differences; every remaining period has the same no-collision transcript distribution. When $q=o(2^{n/2})$, those exclusions occupy a vanishing fraction of the candidate periods, and neither collision discovery nor guessing a remaining period has a constant success probability. More quantitatively, conditioned on no previous collision, the next query tests at most $q$ of at least $2^n-1-\left(\genfrac{}{}{0ex}{}{q}{2}\right)$ remaining periods; summing these conditional probabilities proves the same square-root lower bound. Therefore the optimal bounded-error classical query count is $\Theta (2^{n/2})$, exponentially larger in $n$ than the quantum count.

This comparison presumes a hidden nonzero period. Under a literal interpretation asking merely for some $s$ satisfying the printed identity, returning $s=0$ requires no queries at all, quantum or classical. Nonzero $s$ is the necessary intended promise, not a consequence of two-to-one behavior alone.

Apply the Hadamard transform to the conditional coset state. The amplitude of a first-register output $z$ isWhen $s\cdot z=1$ over ${\mathbb{F}}_2$, the two terms cancel exactly. When $s\cdot z=0$, their squared modulus is $2^{-(n-1)}$. ThusThere are $2^{n-1}$ orthogonal outputs, so these probabilities sum to one. The distribution is independent of the particular observed fibre. It is the Single-sample distribution in Simon's algorithm: each run gives a uniform random equation restricting the hidden period to the orthogonal complement over the binary field.

Only strings $k$ in the image of $f$ can be observed. Choose one preimage $x_k$ of an observed string. Under the nonzero-period promise its two preimages are $x_k$ and $x_k+s$. The Born rule gives probability $2/2^n$ to this second-register outcome, and conditional normalization leaves the first-register coset stateThe relative plus sign is retained: measuring the output register reveals which fibre was sampled but does not reveal which member of the fibre was present. Measuring its bits successively gives the same projection onto the observed full string as measuring the whole register at once.

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.