Since $H=\dot{a}/a$, differentiation gives the Hubble parameter identitySubtracting the given Friedmann equation from the Raychaudhuri equation therefore yieldsOn the other hand, differentiating the Friedmann equation givesSubstitute $\dot{H}=-4\pi G{\dot{\varphi }}^2$ and divide by $(8\pi G/3)\dot{\varphi }$, which is permitted by the assumption $\dot{\varphi }\ne 0$. The result is the homogeneous inflaton equation of motionThe middle term is Hubble friction.

In the hypergeometric symbol, select the exponent $1-c$ at zero and factor out its local power by writingThe Exponential shifting of a Papperitz symbol makes the exponents of $v$ equal to $0,c-1$ at zero, $0,c-a-b$ at one, and $a-c+1,b-c+1$ at infinity. These are the hypergeometric parametersbecause $1-C=c-1$ and $C-A-B=c-a-b$. The solution for $v$ that is analytic at zero and equals one there is therefore the Gauss hypergeometric function $F(A,B,C;z)$. Thus the second local hypergeometric solution isFor noninteger $c$ this is linearly independent from $F(a,b,c;z)$; at resonant values, the usual limiting or logarithmic modification may be required.

A Möbius transformation of a Papperitz symbol changes the independent variable by a Möbius transformation. In particular,maps $a,b,c$ to $0,1,\infty$, respectively. Because a Möbius map is locally biholomorphic away from its pole, it carries each exponent pair with its singular point and givesNow define a dependent-variable rescaling of a Papperitz symbol by writing the old dependent variable asso that $v=u^{-\alpha }(1-u{)}^{-\beta }y$. This dependent-variable rescaling of a Papperitz symbol subtracts $\alpha$ from both exponents at zero, subtracts $\beta$ from both exponents at one, and adds $\alpha +\beta$ to both exponents at infinity. Hence $v$ has symbolDefineThe Fuchs relation gives ${\alpha }^{\prime }-\alpha =1-C$ and ${\beta }^{\prime }-\beta =C-A-B$. After renaming $u$ as $z$, the symbol is thereforewhich is the Gauss hypergeometric equation. Renaming $A,B,C$ as $a,b,c$ gives exactly the displayed symbol in the question.

Let $\omega$ solve the equation represented by the original symbol and define a dependent-variable rescaling of a Papperitz symbolwith a consistent local choice of branch. Near $a$, the multiplier is a nonzero analytic factor times $(z-a{)}^{\delta }$, so both exponents at $a$ increase by $\delta$. Near $b$, it is a nonzero analytic factor times $(z-b{)}^{-\delta }$, so both exponents at $b$ decrease by $\delta$. It is analytic and nonzero near $c$, so the exponents there are unchanged. Therefore the transformed equation has symbolThe opposite shifts also preserve the Fuchs relation.

The Papperitz symbolrepresents the two-dimensional solution space of a homogeneous linear differential equation for a second-order Fuchsian differential equation on the Riemann sphere. Its only singularities are the three regular singular points $a,b,c$. The two entries beneath each point are its characteristic exponents at a regular singular point: for example, in the nonresonant case a local basis near $a$ has leading behaviours $(z-a{)}^{\alpha }$ and $(z-a{)}^{{\alpha }^{\prime }}$. If one of the points is infinity, exponent $\rho$ conventionally means behaviour $z^{-\rho }$ there.

The six exponents obey the Fuchs relationSubject to this relation, the symbol determines the corresponding Riemann differential equation; resonant differences between exponents may produce Logarithmic Frobenius solutions without changing the exponent data recorded by the symbol.

Both variables are ordinal categorical variables, and the proposed alternative is directional: higher scores should tend to occur at higher prices. Assign increasing numerical scores to low, medium, high price and to poor, good, excellent quality, and use a one-sided linear-by-linear association test. Its statistic is based on the sample covariance, or equivalently the correlation, of those scores under the fixed margins.

This test uses one degree of freedom and directs its rejection region toward positive association. The four-degree-of-freedom likelihood-ratio test of independence in a contingency table spends power on every possible departure from independence, including unordered and negative patterns, so the ordered one-sided test is more powerful when the stated alternative is credible.

The null hypothesis is that price and score are independent, equivalently that the independence log-linear model for a two-way contingency table is correct. Against the saturated alternative, the reported Poisson deviance is the likelihood-ratio test statisticUnder the null and the usual large-sample regularity assumptions, Wilks theorem givesThe observations must be independent, the cell probabilities must not lie on the boundary of the parameter space, and the expected counts must be large enough for the chi-squared asymptotic approximation. The fitted counts are either $31/3$ or $28/3$, so the customary expected-count check is comfortably satisfied. Since(equivalently, the p-value is about $0.329$), we do not reject the null at the $1\%$ significance level. The data provide no significant lack of fit for independence.

Write $y_{i+}={\sum }_j{y}_{ij}$ and ${\mu }_{i+}={\sum }_j{\mu }_{ij}$. Under the model in part (a),and $p_j$ is independent of the price category $i$. The Poisson-multinomial conditioning identity givesMore explicitly, the likelihood function for row $i$ factors asThe first factor is the likelihood of the row total and the second is the conditional multinomial likelihood. The free price effect ${\alpha }_i$ makes the first factor maximal at ${\hat{\mu }}_{i+}=y_{i+}$, while maximizing the remaining factors givesConsequently both fits have the same maximum-likelihood fitted valueHere every row total is $30$, the column totals are $31,28,31$, and $y_{++}=90$, so every fitted row isThis factorization is the Poisson trick.

Let $Y_{ij}$ be the number of ice creams in price category $i\in \{\mathrm{high},\mathrm{low},\mathrm{medium}\}$ and score category $j\in \{\mathrm{excellent},\mathrm{good},\mathrm{poor}\}$. The fitted Poisson regression assumes that the nine counts are independent random variables withwhere the logarithmic link function is used and excellent is the reference level in a regression factor, so ${\beta }_{\mathrm{excellent}}=0$. Equivalently,This is the independence log-linear model for a two-way contingency table: price and score have main effects but no interaction.

Apply part (b) repeatedly. After the first puncturing, the code has rank $k-1$, length $n-d$, and minimum distance at least $\lceil d/2\rceil$. Repeating with a minimum-weight codeword at each stage givesbecause nested ceilings obeyThuswhich is the binary Griesmer bound.

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 Schrodinger 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 $n$.