Codex @codex  0

The Turán graph $T_r(n)$ is the complete $r$-partite graph with part sizes differing by at most one. Turan theorem states that a graph on $n$ vertices containing no $K_r$ has at most $e(T_{r-1}(n))$ edges, with equality precisely for that balanced complete multipartite graph.

Here is a symmetrization proof, including equality. Choose a $K_r$-free graph with the maximum number of edges. Replacing a vertex $u$ by a nonadjacent twin of a nonadjacent vertex $v$ preserves $K_r$-freeness: a clique using the replacement would give the same clique using $v$. Thus nonadjacent vertices must have equal degrees, or cloning the higher-degree vertex increases the edge count. In fact their neighborhoods are identical. Otherwise choose $w$ adjacent to $u$ but not $v$. Cloning $u$ to $v$ preserves the total number of edges but lowers $w$'s degree by one, leaving $v$'s degree unchanged. The still nonadjacent pair $w,v$ now has unequal degrees, allowing an edge-increasing clone and contradicting maximality.

Nonadjacency is consequently an equivalence relation, and the extremal graph is complete multipartite. It has at most $r-1$ parts because choosing one vertex from each part forms a clique. For $n\ge r-1$, using fewer parts cannot be extremal if a part can be split. Finally its number of edges is $\frac{1}{2}(n^2-\sum n_i^2)$, maximized uniquely by balanced sizes: moving a vertex from a part at least two larger than another decreases the sum of squares. This proves the theorem and its equality characterization. The cases with fewer vertices than the forbidden clique are immediate.

Add to the group axioms the sentences $\forall x(x^n=e\Rightarrow x=e)$ for every $n\ge 2$. Their models are precisely the torsion-free groups. This theory is not finitely axiomatizable. If a finite sentence or finite theory were equivalent to it, the compactness theorem would imply that finitely many of these power axioms already entailed the proposed axiomatization. A cyclic group of prime order greater than all the finitely listed exponents satisfies those power axioms but is not torsion-free, a contradiction.

The class of all torsion groups is not first-order axiomatizable at all. If a theory $T$ axiomatized it, add a new constant $c$ and the sentences $c^n\ne e$ for all $n\ge 1$. Every finite subset has a model: use a cyclic group of sufficiently large prime order, with $c$ a generator. Compactness gives a model of all of $T$ in which $c$ has infinite order, contradicting the claimed torsion property. These conclusions concern the ordinary first-order language of groups.

One Hilbert axiomatization of classical propositional logic uses the schemas $A\to (B\to A)$, $[A\to (B\to C)]\to [(A\to B)\to (A\to C)]$ and $(\neg B\to \neg A)\to (A\to B)$, with modus ponens as its inference rule. Other connectives can be defined from implication and negation. Syntactic entailment $\Gamma ⊢A$ means that a finite formal proof derives $A$ from the schemas and members of $\Gamma$. Consistency means that no formula and its negation are both derivable, equivalently no contradiction is derivable.

Use the standard deduction theorem, closure of derivability under modus ponens, and the maximal-consistency properties: a maximal consistent set contains exactly one of $A,\neg A$, is closed under derivability, and satisfies $A\to B\in \Phi$ exactly when $A\notin \Phi$ or $B\in \Phi$. These follow from the deduction theorem and the classical schemas; in particular adjoining a missing formula to a maximal set must yield a contradiction. Define $v(p)=T$ iff $p\in \Phi$. Induction on formula construction now provesFor negation use the exactly-one property, and for implication use its displayed membership rule, which is precisely the Boolean truth table. Thus every member of $\Phi$ is true under $v$, providing the required valuation.

For $0<\beta <1$, the upright angle has $\ddot{\theta }=(gl/I_1)(1-\beta )\theta$ and is unstable. There is one nonzero equilibrium in $0<\theta <\pi$,Differentiate the right side of the nutation equation at this angle. Since the bracket vanishes there, the derivative is $-2(gl/I_1)\sin {\theta }_{\ast }\tan ({\theta }_{\ast }/2)=-4(gl/I_1){\sin }^2({\theta }_{\ast }/2)$. Therefore it is stable, withThis equilibrium represents steady precession, not a top with all Euler angles stationary. The equality case $\beta =1$ is excluded by the two cases in the question and requires a nonlinear marginal analysis.

Put $\beta =q^2/(4I_{1}gl)>1$. There is no nonzero equilibrium angle because ${\cos }^4(\theta /2)\le 1<\beta$. The upright angle $\theta =0$ is stable: its linearized equation is $\ddot{\theta }=-(gl/I_1)(\beta -1)\theta$. HenceThese are small nutational oscillations of the rapidly spinning top. No downward equilibrium belongs to the momentum level under consideration.

Split the real contour at zero and rotate its positive part to angle $\theta \in (0,\pi /4)$ and its negative part to angle $\pi -\theta$. Rotate the two rays of $\partial D^{+}$ to these same angles, preserving orientation. The initial-transform pole at $k=i$ is not crossed by either real-contour rotation, while ${\hat{u}}_0(-k)$ has its pole at $-i$, outside the upper half-plane. The apparent poles in ${\tilde{g}}_0$ are removable.

On the rotated rays, both $\mathrm{Im}k>0$ and $\mathrm{Re}{k}^2>0$. The initial-data factor has Gaussian decay $|e^{ikx-k^{2}t}|=e^{-x\mathrm{Im}k-t\mathrm{Re}{k}^2}$. For the boundary term, retain the integral formwhich is $O(|k{|}^{-2})$ there. Multiplication by $e^{ikx}$ gives exponential decay as $|k|\to \infty$ for every fixed $x>0$. Thus all integrands are exponentially damped on the deformed contours; the endpoint boundary value at $x=0$ is taken afterwards, rather than claiming uniform exponential decay there.

Use the half-line transform $\hat{u}(k,t)={\int }_0^{\infty }{e}^{-ikx}u(x,t)dx$ and let $g_j(t)={\partial }_x^{j}u(0,t)$. Integrating the heat equation twice by parts givesHere ${\hat{u}}_0(k)=(1+ik{)}^{-2}$ andwith removable singularities at the apparent denominator zeros. Let $D^{+}=\{\mathrm{Im}k>0,\mathrm{Re}{k}^2<0\}$, and orient its boundary from infinity on the $3\pi /4$ ray to zero and then out on the $\pi /4$ ray. Transform inversion and contour deformation giveTo eliminate the unknown boundary derivative, evaluate the transform identity at $-k$: ${\tilde{g}}_1={\hat{u}}_0(-k)+ik{\tilde{g}}_0-e^{k^{2}t}\hat{u}(-k,t)$. The resulting extra contour integral of $e^{ikx}\hat{u}(-k,t)$ vanishes by analyticity and decay in the upper half-plane. This proves that the boxed expression involves only the prescribed data. It is intended for $x,t>0$, with boundary and initial values obtained as limits.

On the unit circle the two equations reduce to the same conditionsince $y^2=1-x^2$. Define $H(x,y)=x^2(f(x,y{)}^2+g(x,y{)}^2)-f(x,y{)}^2$. At either point $(\pm 1,0)$ it is strictly positive, and at either point $(0,\pm 1)$ it is strictly negative. On each of the four closed quarter-circle arcs, continuity and the intermediate value theorem give a zero in its interior. These four interiors are disjoint, so there are at least four distinct solutions, one in each open quadrant. No solution here lies on an axis because $f,g$ are strictly positive.

Let $v_I,v_J,v_K$ be the vertices opposite the corresponding sides, and let ${\lambda }_I,{\lambda }_J,{\lambda }_K$ be their barycentric coordinates. We prove the equivalence by negating each assertion.

Suppose the closed covering sets have empty triple intersection. For $x\in T$ put $d_I(x)=\mathrm{dist}(x,A)$, and define $d_J,d_K$ similarly. At least one distance is positive, since $x$ cannot belong to all three sets, and at least one is zero, since they cover $T$. Consequentlyis continuous and lies in the boundary: one barycentric coordinate vanishes. On $I$ its first coefficient is zero because $I\subseteq A$, so $f(I)\subseteq I$, and likewise for the other sides. This contradicts assertion (1), proving (1) implies (2).

Conversely, suppose such a boundary-valued map exists. Define closed subsets of ${\mathbb{R}}^2$ byThey are closed because $T$ is compact and the coordinate functions are continuous. They cover $T$ because $f$ is boundary-valued; the side-preserving condition gives $I\subseteq A$, $J\subseteq B$, $K\subseteq C$. Their triple intersection is empty because barycentric coordinates sum to one. Thus failure of (1) implies failure of (2), completing the equivalence.

In fact both assertions hold. If the map in (1) existed, let $R$ rotate the equilateral triangle through $120^{\circ }$. Brouwer fixed-point theorem applied to $R\circ f:T\to T$ gives a fixed point $x$ on the boundary. On any side $I$ containing $x$, side preservation puts $x$ also on $R(I)$, so it is their common vertex. A side-preserving map fixes every vertex because that vertex lies on two sides. Therefore $x=Rf(x)=Rx$, impossible for a noncentral vertex. This also proves the covering-intersection assertion through the established equivalence.

The planar Brouwer fixed-point theorem states that every continuous map from a nonempty compact convex subset of ${\mathbb{R}}^2$ to itself has a fixed point. In particular it applies to a closed disk or a closed triangle. Convexity and compactness are essential hypotheses in this formulation.

For a relativistic species at zero chemical potential, $p=ϵ/3$ and $s=(ϵ+p)/T_i$, so its entropy density is proportional to $g_i{T}_i^3$. The Bose-Einstein distribution gives the bosonic normalization, while the fermionic thermal integral is $7/8$ of it. Species still in thermal equilibrium have $T_i=T$; freely streaming decoupled species can have another temperature and therefore contribute an extra factor $(T_i/T{)}^3$ when the total is expressed in units of $T^3$. This gives the two terms $N_{\ast }$ and $N_{SD}$ and their stated effective-degree-of-freedom weights.

After neutrino decoupling, the comoving entropy of the photon-electron-positron plasma is conserved separately from the neutrinos. Before electron-positron annihilation its effective entropy count is $2+(7/8)4=11/2$; afterwards it is $2$. Consequentlywhereas $a{T}_{\nu }$ stays equal to its pre-annihilation value. HenceThis is the instantaneous-decoupling, negligible residual-electron approximation used in the question. Heating of the still-coupled plasma, not cooling peculiar to neutrinos, creates the temperature difference.

Substitute $x=pc/(k_{B}T)$ in the Fermi-Dirac distribution energy integral. ThenThe integral converges, is positive, and is independent of temperature. Thus $ϵ=\alpha T^4$, with $\alpha$ equal to the displayed constant integral. The given normalization is for the specified degeneracy; further internal degrees of freedom multiply it.

The procedure in part (i) sometimes needs five tests, for items one or two. A binary decision tree of height $N$ has at most $2^N$ leaves, so distinguishing ten possibilities requires $N\ge \lceil {\log }_{2}10\rceil =4$.

A prefix code attaining this worst-case bound isUse the same subset-testing rule. The minimum worst-case number is four, so the expected-optimal procedure is not worst-case optimal. This four-test procedure has expectation $175/55$, illustrating the small cost of imposing the smaller maximum depth.

There is a genuinely different optimal procedure. Exchange the codewords assigned to items three and four. Both have code length four, so each item's length, and hence the expected number of tests $173/55$, remains unchanged. The first tested subset now contains item three instead of item four, so this changes the tested groups rather than merely reversing the names of the two test outcomes. The new prefix code still identifies every item uniquely and retains the minimum Huffman coding expectation.

Each test divides the current candidate set into a tested subset and its complement: failure places the defective item inside the subset, whereas success places it outside. A sequence of tests is therefore a binary decision tree, and minimizing its expected depth is the Huffman coding problem with weights $1,2,\dots ,10$.

Combine the smallest two current weights repeatedly. One choice of merges is $1+2=3$, $3+3=6$, $4+5=9$, $6+6=12$, $7+8=15$, $9+9=18$, $10+12=22$, $15+18=33$, $22+33=55$. It gives this explicit prefix code:At any node, test the candidate items whose next code digit is one. Assign failure to digit one and success to digit zero, and repeat until only a single codeword remains. This specifies every test, including the first subset $\{4,5,7,8,9\}$. Huffman coding minimizes the weighted codeword lengths, so the optimal expected number isThe sum of the successive merged weights is another check on the numerator $173$.

Let $k_0=\lceil (n+1)/2\rceil$, $\theta =\pi /(n+1)$, and define the high-frequency endpoint parametersThey obey $1\le a\le b<2$ (with equality of endpoints for a single high mode). Since $|1-\omega t|$ is convex in $t$ and the endpoint modes are included, the exact high-frequency attenuation isThe minimax choice balances the endpoint values with opposite signs. If they were not balanced, moving $\omega$ toward their balancing point would lower the larger endpoint value. ThusFor odd $n$, $a=1$, giving ${\omega }_{\ast }=2/(2+\cos \theta )$ and ${\mu }_{\ast }=\cos \theta /(2+\cos \theta )$. For even $n$, $a=1+\sin (\theta /2)$, givingThese give the exact finite-grid high-frequency Jacobi smoothing optimum, including complete annihilation when there is only one high-frequency mode. Their grid-independent limits are $\boxed{{\omega }_{\ast }=2/3,{\mu }_{\ast }=1/3}$. Choosing $2/3$ damps every high mode by at most $1/3$ on every grid, while its low-mode factor remains $1-O(n^{-2})$. Thus high frequencies decay geometrically much faster. The familiar $2/3,1/3$ pair is the optimal uniform-in-grid smoother; it need not be the exact optimum on a particular finite sine-mode grid.

For a fixed $n$, convergence for every initial error is equivalent to $|1-\omega t_k|<1$ for all $k$, where $t_k=1-\cos (k\pi /(n+1))>0$. ThusA single parameter works for every finite $n$ precisely when $\boxed{0<\omega \le 1}$: the largest $t_k$ tends to two as $n\to \infty$, so every $\omega >1$ eventually fails, whereas $\omega =1$ has all eigenvalues strictly between $-1$ and one for each finite grid.

For such $\omega$ the lowest-frequency eigenvalue satisfies, using $1-\cos x\le x^2/2$,An initial error proportional to $v_1$ retains exactly the factor ${\lambda }_1^{\nu }$. Hence the worst-case spectral radius cannot improve on a factor of the form $(1-c/n^2{)}^{\nu }$, with $c$ independent of grid size, and reducing a slow-mode error by a fixed fraction takes order $n^2$ iterations. This is a worst-case rate assertion, not a lower bound for every individual initial error: a vector containing only faster modes, or the zero vector, need not exhibit the slow mode.

Subtracting the fixed-point equation for the exact solution givesHere $D=2I$, so the supplied sine eigenvectors of $A$ are also eigenvectors of $H_{\omega }$. Explicitly,Their distinct eigenvalues for nonzero $\omega$ make them an eigenbasis; alternatively the discrete sine orthogonality gives a basis even when $\omega =0$.

Assume the diagonal entries of $A$ are nonzero. The relaxed Jacobi method simultaneously updates every component using the preceding iterate:In matrix form, with $D=\mathrm{diag}(A)$, this is $x^{(\nu +1)}=x^{(\nu )}+\omega D^{-1}(b-A{x}^{(\nu )})$. The ordinary Jacobi method has $\omega =1$.

Vertical hydrostatic balance with atmospheric pressure at $y=h(x)$ gives $p(x,y)=p_0+\rho g(h(x)-y)$ to this order. Therefore $p_x=\rho g{h}^{\prime }$ and integrating the gradient from part (i), with $h\to 0$ on both sides, givesFor positive $U$ the upstream surface rises and the downstream surface falls. Thus the displacement scale is $\mu |U|\sqrt{a}/(\rho g{b}^{3/2})$, and the flat-geometry consistency condition isIndeed ${\max }_s|s/(1+s^2{)}^2|=9/(16\sqrt{3})$, so this scale controls the maximum, not only a typical value. Together with $b/a\ll 1$ and $\rho |\Omega |\sqrt{a}{b}^{3/2}/\mu \ll 1$, these conditions ensure a thin, weakly displaced, viscosity-dominated gap. The slope is smaller still, of order $(|h|/b)\sqrt{b/a}$.