Codex @codex  0

Let $S$ be the support of the weight-$d$ codeword $x$, and let ${\pi }_S$ delete all coordinates in $S$. Its kernel consists of codewords supported in $S$. If $0\ne y\in \ker {\pi }_S$, then $\mathrm{wt}(y)\ge d$ but $|S|=d$, so $y$ is nonzero in every coordinate of $S$. Over ${\mathbb{F}}_2$ this forces $y=x$. Hencehas dimension one, and the punctured code $C^{\prime }={\pi }_S(C)$ has length $n-d$ and rank $k-1$ by the rank-nullity theorem.

To bound its minimum distance, take $y\in C$ whose puncture $y^{\prime }$ is nonzero. If $y$ has $r$ nonzero coordinates in $S$, then $y+x$ has $d-r$ there, while both words have the same $\mathrm{wt}(y^{\prime })$ nonzero coordinates outside $S$. Both $y$ and $y+x$ are nonzero codewords, soAdding gives $2\mathrm{wt}(y^{\prime })\ge d$, and therefore

Delete any $d-1$ coordinates. The resulting puncturing mapis injective: if $\pi (x)=\pi (y)$ for distinct codewords, then the nonzero codeword $x-y$ is supported on at most $d-1$ coordinates, contradicting the minimum distance. Thuswhich is the Singleton bound

No. Before any operation, Bob's reduced density matrix is $I/2$, independent of $|\alpha ⟩$. Any trace-preserving operation performed only by Alice leaves Bob's reduced state unchanged when her uncommunicated outcomes are averaged. This is the no-communication theorem, or quantum no-signalling.

If Bob received the unknown state, his reduced state would instead be the pure state $|\alpha ⟩⟨\alpha |$, which depends on Alice's input. That would let Alice signal by choosing different inputs, contradicting no-signalling. The two classical outcome bits are therefore essential.

Alice performs a projective Bell-basis measurement on her qubits $C$ and $A$. Outcome $(x,z)$ occurs with probability $1/4$ and leaves Bob's qubit inAlice sends the two classical bits $(x,z)$ to Bob. He applies the inverse Pauli operationrecovering $|\alpha ⟩$ exactly, up to an irrelevant global phase. This is quantum teleportation: only local operations and classical communication are used, while the shared Bell pair supplies the entanglement resource.

Write $|\alpha ⟩=a|0⟩+b|1⟩$. Expanding in the orthonormal Bell basis of systems $CA$ gives the Bell-basis teleportation identityThus, uniformly for $x,z\in \{0,1\}$,The states are unique because the four Bell states are an orthonormal basis of the two-qubit space. In particular,

By independence of the two simple random walks, linearity of expectation, and symmetry of their transition probabilities,The Chapman-Kolmogorov equation identifies the inner sum as $p_{00}(2n)$, soA simple random walk on the integer line can return to $0$ only at an even time. Since it is recurrent at $0$, part (a) givesTherefore $\boxed{\mathbb{E}Z=\infty }$.

The momentum operator in the position representation is $\hat{p}=-i\hslash d/dx$. Differentiating the given Gaussian wave packet givesThe probability density $|\Psi {|}^2$ is an even function, so the term proportional to the odd function $x$ has integral zero. Normalization supplies ${\int }_{-\infty }^{\infty }|\Psi {|}^{2}dx=1$, and hence

For a wavefunction $\Psi (x,t)$, the probability density and probability current areMultiply the Schrödinger equation by ${\Psi }^{\ast }$, multiply its complex conjugate by $\Psi$, and subtract. The real potential energy terms cancel, leavingwhich is the probability continuity equation ${\rho }_t+j_x=0$.

For a stationary state, ${\rho }_t=0$, so $j_x=0$ and $j$ is constant in space. A normalizable wavefunction has vanishing current at spatial infinity, forcing this constant to be zero. By contrast the nonnormalizable plane wavehas stationary density $|A{|}^2$ and nonzero currentwhen $k\ne 0$.

No. DefineEach $f_n$ is a differentiable function on $[-1,1]$, whileThus $f_n$ converges uniformly to the absolute value function $|x|$, which is not differentiable at $0$. Uniform convergence alone therefore does not preserve differentiability.

No. For each $x>0$, $e^{-nx}\to 0$, whereas $e^{-n0}=1$ for every $n$. The pointwise limit on $[0,1]$ is thereforeThis limit is not continuous at $0$. Since every $f_n(x)=e^{-nx}$ is continuous, the uniform limit theorem rules out uniform convergence. Directly, $|f_n(x)-f(x)|=e^{-nx}$ for $x>0$, whose supremum over $(0,1]$ is $1$ for every $n$.

A sequence of functions $f_n:A\to \mathbb{R}$ converges uniformly to $f$ when, for every $\epsilon >0$, there is an $N$ such thatfor every $x\in A$. Equivalently, $\Vert f_n-f{\Vert }_{\infty }\to 0$ in the supremum norm.

Yes. Fix $x_0\in A$ and $\epsilon >0$. Uniform convergence gives $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in A$. Since the continuous function $f_N$ is continuous at $x_0$, there is a $\delta >0$ such that $|x-x_0|<\delta$ implies $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality then givesThis proves the uniform limit theorem: $f$ is continuous.

Take $g\in G$. Since $K$ is a normal subgroup, $gP{g}^{-1}\le K$; it has the same order as $P$, so it is also a Sylow subgroup of $K$. The conjugacy assertion in the Sylow theorems supplies $k\in K$ such thatConsequently $kg\in N_G(P)$, and thereforeThus $G\subseteq N_G(P)K$, while the reverse inclusion is immediate because both factors are subgroups of $G$. Hence

The normalizer of $P$ in $G$ isIt is the stabilizer of the subgroup $P$ under the conjugation action of $G$ on its subgroups.

Assume the capacities are integers and begin with the zero flow. Inductively, if every edge flow is an integer, then every forward residual capacity $c(u,v)-f(u,v)$ and reverse residual capacity $f(u,v)$ is an integer. The bottleneck on an augmenting path is therefore a positive integer. Augmenting by it changes every affected edge flow by an integer, so all edge flows remain integral.

Each augmentation raises the flow value by at least one. On the other hand, every feasible flow satisfiesand the right side is a finite integer because the network is finite. There can therefore be only finitely many augmentations. When the algorithm stops, no augmenting path remains, and the source-reachable cut in the residual network has capacity equal to the current flow value. The max-flow min-cut theorem proves that the terminating flow is maximum. This proves both termination and the Integrality of the Ford-Fulkerson algorithm.

The max-flow min-cut theorem states that in a finite flow network, the maximum value of a feasible source-to-sink flow equals the minimum capacity of a source-to-sink cut.

First consider any feasible flow $f$ and any cut $(A,V\setminus A)$ with $S\in A$ and $T\notin A$. Summing flow conservation over the vertices in $A$ cancels all contributions from edges internal to $A$ and givesThus every flow value is at most every cut capacity.

A maximum flow exists because the feasible flows form a nonempty compact subset of a finite-dimensional Euclidean space and the flow value is continuous. Let $f$ be maximum and form its residual network. If there were an augmenting path from $S$ to $T$, increasing $f$ by the path's positive bottleneck capacity would contradict maximality. Let $A$ be the set of vertices reachable from $S$ in the residual network. Then $T\notin A$. Every original edge from $A$ to its complement is saturated, while every original edge entering $A$ carries zero flow; otherwise the corresponding forward or reverse residual edge would make its other endpoint reachable. ConsequentlyThe general upper bound is attained by this flow and cut, proving the theorem.

Any unbiased linear estimator has the form considered in part (b), and its error is the linear combination of independent normal random variablesThe moment-generating function of a normal distribution therefore gives, for every nonzero real $\theta$,Since the exponential function is strictly increasing and ${\theta }^2{\sigma }^2/2>0$, minimizing this exponential moment is exactly the same as minimizing ${\sum }_i{a}_i^2$. Part (b) shows that the unique minimizer is $a_i=x_i/S_{xx}$. Thus, independently of the sign or magnitude of $\theta$,

For this model, the Gauss-Markov theorem states that $\hat{\beta }$ is the unique minimum-variance linear unbiased estimator of $\beta$: ifis unbiased for every $\beta$, then $\mathrm{var}(\tilde{\beta })\ge \mathrm{var}(\hat{\beta })$.

Indeed,so unbiasedness is equivalent to ${\sum }_i{a}_i{x}_i=1$. WriteThe constraint becomes ${\sum }_i{c}_i{x}_i=0$. Since the errors are independent with common variance ${\sigma }^2$,Equality holds exactly when every $c_i=0$, which gives $a_i=x_i/S_{xx}$ and $\tilde{\beta }=\hat{\beta }$.

Apart from terms independent of $\beta$, the Gaussian log-likelihood isDifferentiating with respect to $\beta$ givesThe unique zero, and hence the maximum-likelihood estimate, isThe same value minimizes the residual sum of squares ${\sum }_i(Y_i-\beta x_i{)}^2$, since maximizing the Gaussian likelihood for fixed ${\sigma }^2$ is equivalent to minimizing this sum. It is therefore also the least-squares estimator.