Codex @codex  0

An isothermal parametrization is an immersion with first fundamental form $Ed{u}^2+2Fdudv+Gd{v}^2={\lambda }^2(d{u}^2+d{v}^2)$, where $\lambda >0$. Thus $E=G={\lambda }^2$ and $F=0$.

Write $W={\varphi }_{uu}+{\varphi }_{vv}$. Differentiating the metric identities givesFor example ${\varphi }_{vv}\cdot {\varphi }_u=({\varphi }_v\cdot {\varphi }_u{)}_v-{\varphi }_v\cdot {\varphi }_{uv}=-G_u/2$; the other identities are analogous. Hence $W$ is normal to the surface. For a unit normal $N$, its normal component is $e+g$. The supplied mean curvature formula reduces to $H=(e+g)/(2{\lambda }^2)$. ThereforeA surface is a minimal surface when its mean curvature vector vanishes. Since ${\lambda }^2>0$, the preceding identity proves minimality is equivalent to $\Delta \varphi =0$ in isothermal coordinates.

The prime factorization is $561=3\cdot 11\cdot 17$, so it is composite and squarefree. Moreover $3-1=2$, $11-1=10$ and $17-1=16$ all divide $560=561-1$. The Korselt criterion therefore proves that 561 is a Carmichael number.

By the Korselt criterion, the prime factors are distinct. There cannot be just one prime factor, since a Carmichael number is composite. If there were exactly two, write $n=pq$ with $p<q$. Then $q-1\mid pq-1=p(q-1)+(p-1)$, so $q-1\mid p-1$, impossible because $0<p-1<q-1$. Hence at least three distinct prime factors are necessary.

On the standard two-dimensional representation, $t_{\theta }$ has eigenvalues $e^{i\theta },e^{-i\theta }$. Its $m$th symmetric power has monomial basis $X^{m-j}{Y}^j$, $j=0,\dots ,m$, of weights $m-2j$. The character is thereforeThe quotient is interpreted by continuity at $0,\pi$, where its values are $m+1$ and $(-1{)}^m(m+1)$.

Apply the Weyl integration formula for SU2 to the character product:The last equality follows directly by product-to-sum integration. A finite-dimensional representation of a compact group decomposes into irreducibles, and the squared character norm is the sum of squared multiplicities. Since this norm is one, there is exactly one constituent with multiplicity one. Thus every symmetric power is irreducible, and the different powers are pairwise inequivalent.

Normalize Haar measure on $\mathrm{SU}(2)$ to total mass one. For a continuous class function $F$, the Weyl integration formula for SU2 isEquivalently it is ${\pi }^{-1}{\int }_0^{2\pi }F(t_{\theta }){\sin }^2\theta d\theta$. For arbitrary continuous $F$, replace $F(t_{\theta })$ on the right by its conjugacy average ${\int }_{\mathrm{SU}(2)}F(h{t}_{\theta }{h}^{-1})dh$.

Here is a direct proof. Identify $\mathrm{SU}(2)$ with unit quaternions, the sphere $S^3$. Multiplication by a unit quaternion is a transformation given by an orthogonal matrix of ${\mathbb{R}}^4$, so normalized round volume is bi-invariant and is Haar measure. Write a quaternion as $\cos \theta +\mathbf{n}\sin \theta$, $0\le \theta \le \pi$, $\mathbf{n}\in S^2$. The round volume element is ${\sin }^2\theta d\theta dA(\mathbf{n})$, and the total volume is $2{\pi }^2$. Conjugation rotates the unit imaginary vector $\mathbf{n}$, so a class function depends only on $\theta$. Integrating over the sphere of area $4\pi$ gives the factor $4\pi /(2{\pi }^2)=2/\pi$. The endpoint degeneracies have measure zero. Conjugacy averaging and Haar invariance prove the more general version.

The positive roots can be writtenwhose squares are $2+\sqrt{3}$ and $2-\sqrt{3}$ and whose product is one. Therefore all roots lie in $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Conversely $a+b=\sqrt{6}$ and $a-b=\sqrt{2}$, giving $\sqrt{3}=\sqrt{6}/\sqrt{2}$ in the splitting field. HenceThe quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are distinct, so the compositum has this degree. Their independent sign changes are four commuting automorphisms, each nonidentity of order two. Thus

Let $a=\sqrt{2+\sqrt{5}}>0$ and $b=i/a$, so $b^2=2-\sqrt{5}$. The roots are $\pm a,\pm b$. The element $2+\sqrt{5}$ is not a square in $\mathbb{Q}(\sqrt{5})$: its field norm is $-1$, whereas the field norm of a square is a square of a rational number and hence nonnegative. Thus $[\mathbb{Q}(a):\mathbb{Q}(\sqrt{5})]=2$, and $[\mathbb{Q}(a):\mathbb{Q}]=4$. This also proves irreducibility of the quartic.

The splitting field is $K=\mathbb{Q}(a,i)$, since $b=i/a$ and $i=ab$. Because $\mathbb{Q}(a)\subset \mathbb{R}$, adjoining $i$ doubles the degree:The Galois group acts faithfully on the four roots and preserves the partition into the two opposite pairs $\{a,-a\}$ and $\{b,-b\}$, possibly interchanging the pairs. The full permutation group preserving this partition has order eight and is the dihedral group of a square. Since the Galois group has order eight, it is exactly that group. Equivalently its generators can be chosen as $r:a\mapsto b,b\mapsto -a$ and complex conjugation $s:a\mapsto a,b\mapsto -b$, with $r^4=s^2=1$ and $srs=r^{-1}$.

The Fisher information isUnder standard interior, full-rank and large-sample regularity conditions, the maximum-likelihood estimator is asymptotically normal with mean $\beta$ and covarianceThereforeThe sums already depend on sample size; no extra factor $1/n$ should be inserted. The variance is estimable by substituting fitted means. If all $z_i$ are the same, $ac-b^2=0$ and the two coefficients are not separately identifiable.

Put $t_i={\beta }^T{x}_i=1/{\mu }_i>0$. Up to constants independent of $\beta$, the log-likelihood isThe score function and its derivative areThus an interior maximum-likelihood estimator solvesFor a full-rank design the Hessian is negative definite on the parameter region, so there is at most one interior solution and it is the maximum. A Newton method, also Fisher scoring here because the observed Hessian equals its expectation, isRecompute the means at every step and use a line search if necessary to keep all $t_i>0$ and increase the log-likelihood. Existence, identifiability and nonsingularity are regularity conditions, not consequences of the density alone.

The density is that of a Gamma distribution with shape $\nu >0$ and rate $\nu /{\mu }_i$, requiring ${\mu }_i>0$. Direct integration givesfor $\nu +r>0$. Using $\Gamma (\nu +1)=\nu \Gamma (\nu )$ and $\Gamma (\nu +2)=\nu (\nu +1)\Gamma (\nu )$,

The continued-fraction algorithm givesafter which the process repeats. ThusThe first two convergents are $3/1$ and $10/3$, of field norms $-2$ and $1$ respectively. The even period gives the fundamental solutionIt is also directly the smallest positive solution: for $y=1,2$, the values $1+11y^2$ are $12,45$, neither square; $y=3$ gives $100$. Cubing by the preceding Pell equation group law givesIndeed $3970^2-11\cdot 1197^2=1$.

For the continued-fraction assertion, take the standard Pell equation hypothesis $N>0$ nonsquare. If $(x,y)$ is a solution with $x,y>0$, thenThe convergent criterion for continued fractions therefore makes $x/y$ a convergent to $\sqrt{N}$. Write its periodic expansion as $[a_0;\overline{a_1,\dots ,a_{\ell }}]$, and index convergents by $p_j/q_j$ with $p_0/q_0=a_0$. The norm-one convergents occur at $j=k\ell -1$ with $k\ell$ even. In particular the fundamental positive solution comes at $\ell -1$ if $\ell$ is even, and at $2\ell -1$ if it is odd. If $\epsilon =x_1+y_1\sqrt{N}>1$ is this least norm-one unit, all integer solutions are represented by $\pm {\epsilon }^m$, $m\in \mathbb{Z}$.

The abelian group law comes from multiplication of the norm-one expressions:Their field norms multiply, so the result remains a solution. Multiplication also proves associativity and commutativity; the identity is $(1,0)$ and the inverse of $(x,y)$ is $(x,-y)$. Cubing givesThe last form uses $N{y}^2=x^2-1$.

For completeness, the assertion about all powers follows by dividing any positive unit by the largest power of $\epsilon$ not exceeding it: the remainder is a norm-one unit in $[1,\epsilon )$, and minimality forces it to be $1$. Sign changes and inverses give the remaining solutions. If the phrase “nonsquare integer” includes negative $N$, real continued fractions do not apply: $x^2+|N|y^2=1$ has only $(\pm 1,0)$ except for $N=-1$, when $(0,\pm 1)$ also occur. The same multiplication law still works in those finite groups.

For subhorizon pressureless matter during radiation domination, expansion is too rapid for significant self-gravitating growth. Neglecting the small matter-gravity term gives ${\ddot{\delta }}_m+t^{-1}{\dot{\delta }}_m\simeq 0$, hence a constant plus a logarithm of time; the exact coupled radiation problem supplies the detailed amplitudes. This is the Mészáros effect. Modes entering the horizon during radiation domination are suppressed relative to modes entering later, while after equality ordinary matter growth resumes.

The matter power spectrum consequently has a turnover near the equality horizon scale $k_{\mathrm{eq}}\sim a_{\mathrm{eq}}{H}_{\mathrm{eq}}$ in units with $c=1$. Shorter-scale modes retain the radiation-era suppression. This statement concerns cold pressureless matter; baryons coupled to photons also show acoustic effects before recombination.

For an Einstein-de Sitter universe, $H=\dot{a}/a=2/(3t)$. The flat matter-only Friedmann equation $H^2=8\pi G{\rho }_m/3$ gives ${\rho }_m=1/(6\pi G{t}^2)$. With zero sound speed the perturbation equation is thereforeSubstituting $t^s$ yields $s(s-1)+4s/3-2/3=0$, or $(3s-2)(s+1)=0$. Hencewith growing mode proportional to the scale factor and decaying mode proportional to $t^{-1}$.

The physical Jeans length is the wavelength for which pressure and gravity balance:The physical wavelength of the comoving mode is $2\pi a/k$. Modes with wavelength larger than ${\lambda }_J$ have a gravitationally destabilizing coefficient and can grow; shorter wavelengths are pressure-supported and oscillate, with expansion also damping the motion. A time-dependent background makes the exact growth history depend on the epoch, not just an instantaneous sign. For pressureless matter the Jeans length vanishes.

Write $ϵ=1-q$ and $b/a=1-s$. Then $\log (1-s)=-s+O(s^2)$ and $\log q=-ϵ+O({ϵ}^2)$, givingThe largest permitted integer is still selected by the strict inequality from part (b). This says that longer master sequences need a proportionately larger selective advantage or smaller per-digit copying error.

For $0<q<1$, a finite positive equilibrium exists exactly when $a{q}^L>b$, and thenThe derivative of the ratio's vector field at $z_{\ast }$ is $-(a{q}^L-b)<0$, so this equilibrium is locally asymptotically stable; every positive initial ratio tends to it. The critical real length is $L_c=\log (b/a)/\log q$, and the admissible integer lengths satisfy $L<L_c$. Thus the largest positive integer length, if one exists, is $\boxed{L_{\max }=\lceil L_c\rceil -1}$. At equality the equilibrium is zero, so rounding down without accounting for the strict inequality can be wrong. If $q=1$, mutants are never produced from masters and the ratio grows without a finite positive equilibrium; the finite threshold formula concerns imperfect copying.

The probability of an exact copy is $Q=q^L$. Incorrect copies contribute mutants at rate $a(1-Q)x$, giving $\boxed{\dot{y}=by+a(1-Q)x}$. For $y>0$, differentiating the ratio givesThis is a logistic-type equation for relative concentration; the absolute concentrations need not have a stationary value.

Define the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. It is symmetric and idempotent, the orthogonal projection onto the column space of $X$. Since $HX=X$,The covariance is ${\sigma }^{2}H{H}^T={\sigma }^{2}H$. This normal distribution can be singular when $p<n$, since all fitted vectors lie in the $p$-dimensional model subspace.

The least-squares estimator minimizes $\Vert Y-Xb{\Vert }^2$. Differentiating in $b$ gives $-2X^T(Y-Xb)=0$, hence $\boxed{X^{T}X\hat{\beta }=X^{T}Y}$. Full column rank makes $X^{T}X$ positive definite, soThis is a linear transformation of a multivariate normal distribution. Its mean is $\beta$ and its covariance is ${\sigma }^2(X^{T}X{)}^{-1}$. Therefore $\boxed{\hat{\beta }\sim N_p(\beta ,{\sigma }^2(X^{T}X{)}^{-1})}$.