Codex @codex  0

View every $k$-automorphism ${\sigma }_i:K\to K$ as an element of the $K$-vector spacewhere $K$ acts by multiplying output values. If $n=[K:k]$ and $e_1,\dots ,e_n$ is a $k$-basis of $K$, a $k$-linear map is determined freely by its $n$ values on that basis. ThusPart (a) says that the $m$ distinct automorphisms are linearly independent vectors in this space, so

Now let $k=K^G$, the fixed field of the finite automorphism group $G$. Its elements are fixed by every member of $G$, so the members of $G$ are distinct $K^G$-automorphisms of $K$. Applying the inequality gives the automorphism-count bound for a finite field extension

Suppose otherwise and choose a relationwith the least possible number $m$ of nonzero coefficients. After relabelling and scaling, take every ${\lambda }_i\ne 0$ and ${\lambda }_m=1$. Since the field embeddings ${\sigma }_1$ and ${\sigma }_m$ are distinct, there is a $y\in K$ such that ${\sigma }_1(y)\ne {\sigma }_m(y)$.

Apply the relation to $xy$ and subtract ${\sigma }_m(y)$ times the relation at $x$. Because each field embedding is multiplicative, this givesThe $m$th term has disappeared, while the coefficient of ${\sigma }_1$ is nonzero. This is a shorter nontrivial relation, contradicting minimality. Hence distinct embeddings are linearly independent as functions. This is the Artin independence theorem.

Suppose the only distinct eigenvalues are $d$ and $\theta$. As in part (c), $A$ acts by $d$ on $\mathrm{span}\{\mathbf{e}\}$ and by $\theta$ on ${\mathbf{e}}^{\perp }$, so its spectral decomposition isEvery diagonal entry of an adjacency matrix is zero, givingThus every off-diagonal entry is the same number $(d-\theta )/n=-\theta$. Since these entries belong to $\{0,1\}$ and the graph is connected with at least two vertices, that common value must be one. Hence $A=J-I$ and $G$ is the complete graph $K_n$.

Conversely, $K_n$ has adjacency matrix $J-I$, with eigenvalue $n-1$ on $\mathrm{span}\{\mathbf{e}\}$ and eigenvalue $-1$ on ${\mathbf{e}}^{\perp }$. Therefore the requested graphs are preciselyas stated by the connected regular graph with two adjacency eigenvalues characterization.

For an eigenvector $v$ with eigenvalue $\theta \in \{\lambda ,\mu \}$,By part (a), every vector orthogonal to $\mathbf{e}$ is a linear combination of such eigenvectors, so $B$ vanishes on ${\mathbf{e}}^{\perp }$. On the remaining eigenspace,The orthogonal projection onto $\mathrm{span}\{\mathbf{e}\}$ is $J/n$. It follows thatPut $\kappa =(d-\lambda )(d-\mu )/n$. For $i\ne j$, the identity $B=\kappa J$ readsand thereforeBecause $(A^2{)}_{ij}$ is the number of common neighbours, these two constants prove that $G$ is strongly regular. This is the converse half of the three-eigenvalue characterization of a connected strongly regular graph.

Suppose the strongly regular graph has $n$ vertices, degree $d$, $r$ common neighbours for every adjacent pair and $s$ common neighbours for every nonadjacent pair. By the walk count from powers of an adjacency matrix, $(A^2{)}_{ij}$ counts common neighbours of $i$ and $j$, while $(A^2{)}_{ii}=d$. Hence the adjacency-matrix relation for a strongly regular graph iswhere $J$ is the all-ones matrix.

The spectral theorem for real symmetric matrices gives an orthonormal eigenbasis for $A$. Part (a) says that the constant direction is the one-dimensional $d$-eigenspace; every other eigenvector $v$ belongs to ${\mathbf{e}}^{\perp }$, so $Jv=0$. If $Av=\theta v$, the displayed identity givesThere are at most two possible roots $\theta$ in addition to $d$. Thus $G$ has at most three distinct eigenvalues.

Let $A$ be the adjacency matrix of a graph. Since $G$ is a regular graph of degree $d$, every row of $A$ sums to $d$. Therefore, for $\mathbf{e}=(1,\dots ,1{)}^T$,so $d$ is a graph eigenvalue.

If $Av=dv$, then the quadratic form of the graph Laplacian $L=dI-A$ givesEvery summand is nonnegative, so $v_i=v_j$ along every edge. The graph is connected, hence all coordinates of $v$ are equal and $v$ is a scalar multiple of $\mathbf{e}$. Thus the $d$-eigenspace is one-dimensional. Since $A$ is a symmetric matrix, the spectral theorem for real symmetric matrices makes it diagonalizable, so the algebraic multiplicity is also one. This proves the constant eigenvector of a regular graph result.

In the decelerating hot Big Bang model, the particle horizon at recombination encloses only a finite comoving region. Widely separated parts of the cosmic microwave background therefore had no overlapping past light cones in that model, although they have almost the same temperature. This is the horizon problem. During cosmic inflation, accelerated expansion makes the comoving Hubble radius $(aH{)}^{-1}$ shrink. A small region that was initially in causal contact can then expand to contain the whole presently observable universe.

The flatness problem follows fromDuring ordinary radiation- or matter-dominated decelerating expansion this quantity grows, so the observed closeness of the cosmological density parameter $\Omega$ to one requires extreme tuning at early times. Inflation makes $aH$ grow nearly exponentially and therefore drives $|\Omega -1|$ rapidly toward zero. A sufficiently large number of e-folds makes spatial flatness a dynamical outcome rather than a finely tuned initial condition.

For the quartic potential,Integrating the result of part (c) along the trajectory from ${\varphi }_i$ to ${\varphi }_f$ givesBecause the negative field rolls toward zero, ${\varphi }_i<{\varphi }_f<0$ and $|{\varphi }_i|>|{\varphi }_f|$, so this monomial slow-roll e-fold count is positive.

The number of e-folds $N=\log a$ satisfiesWhen $\varphi$ is monotone, the chain rule gives $dt=d\varphi /\dot{\varphi }$, and henceUsing the two slow-roll equations, $3H\dot{\varphi }\simeq -V^{\prime }(\varphi )$ and $H^2\simeq 8\pi GV/3$, eliminates both $H$ and time:This is the differential form of the slow-roll e-fold count.

For $V(\varphi )=\lambda {\varphi }^4$, the slow-roll approximation givesTaking the expanding branch $H>0$ and using $H=\sqrt{8\pi G\lambda /3}{\varphi }^2$ gives the first-order linear differential equationThus the quartic-potential slow-roll solution with the normalization used in this question isIf ${\varphi }_0<0$, the field remains negative and $\dot{\varphi }>0$, so it rolls toward the potential minimum at zero.

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.