Codex @codex  0

The first command fits independent Poisson random variables with means ${\mu }_j$ and a log link. For concentration $x$ and strain indicator $s$, its fitted linear predictor isThe interaction allows different concentration slopes in the two strains. At $x=0$ the second strain's mean is multiplied by $e^{0.33667}$; its slope is $-1.47253-0.12534$. These are multiplicative changes in counts, not changes in probabilities.

The maximum-likelihood estimates use the Poisson regression likelihood. With design matrix $X$, estimated information $X^T\mathrm{diag}({\hat{\mu }}_j)X$ gives the inverse covariance matrix; diagonal square roots give the standard errors. The printed Wald tests use approximately standard-normal coefficient-to-error ratios. Fuel and strain effects are strong, whereas the interaction has $p=0.182$, suggesting a common slope.

$H_2$ removes just the interaction; $H_3$ also removes strain. Their parameter counts are four, three and two. Under the usual large-sample likelihood-ratio test, $D_2-D_1=1.78089$ is compared with ${\chi }_1^2$ and is below $3.841459$, so the interaction is not required at 5%. By contrast, $D_3-D_2=32.61857$ is far above that threshold, so the strain main effect should remain. Comparing $H_3$ directly with $H_1$ uses $D_3-D_1$ and two degrees of freedom.

For an approximate absolute goodness-of-fit test, compare $D_1=84.59557$ with ${\chi }_{70-4}^2$; its upper-tail probability is about $0.061$. For $H_2$, $D_2=86.37646$ on $67$ degrees of freedom has $p\approx 0.056$. The additive model $H_2$ is the preferred parsimonious fit at 5%, subject to the independent Poisson distribution and large-sample assumptions. The printed nine rows are only a subset; the supplied fitted output refers to all 70 observations.

Using the diameter normalization, define ${\mathcal{H}}_{\delta }^d(E)=\inf \{{\sum }_j(\mathrm{diam}{U}_j{)}^d:E\subseteq {\bigcup }_j{U}_j,\mathrm{diam}{U}_j\le \delta \}$ and the Hausdorff measure ${\mathcal{H}}^d(E)={\lim }_{\delta \downarrow 0}{\mathcal{H}}_{\delta }^d(E)$. The Hausdorff dimension is $\inf \{d:{\mathcal{H}}^d(E)=0\}$, equivalently $\sup \{d:{\mathcal{H}}^d(E)=\infty \}$.

The Cantor set is the intersection of the sets left by successively removing open middle thirds from $[0,1]$. At level $k$ it has $2^k$ Cantor cylinders of diameter $3^{-k}$. For $s=\log 2/\log 3$, their total $s$-cost is $2^k{3}^{-ks}=1$, so ${\mathcal{H}}^s(C)\le 1$.

The random series $\xi ={\sum }_{n\ge 1}{X}_n{3}^{-n}$ converges absolutely, and its ternary digits are all zero or two. Thus $\xi \in C$ and its law is the Cantor Bernoulli measure. Each level-$k$ cylinder has probability $2^{-k}$. Distinct such intervals have separation at least $3^{-k}$; a set $U$ of smaller diameter meets at most one. If $3^{-(k+1)}\le \mathrm{diam}U<3^{-k}$, thenSets of diameter zero have probability zero, since the measure has no measure atoms; diameters at least one give the bound trivially. For arbitrary nonmeasurable sets use outer probability. Any countable cover of $C$ therefore satisfies $1\le {\sum }_j{\mathbb{P}}^{\ast }(\xi \in U_j)\le 2{\sum }_j(\mathrm{diam}{U}_j{)}^s$. HenceThe last implication follows by multiplying the $s$-costs by powers of the cover diameter: measures above $s$ vanish and those below $s$ are infinite.

For binary linear codes of length $n$, the bar product of binary linear codes is $C_1|C_2=\{(u,u+v):u\in C_1,v\in C_2\}$. The parametrization is injective and linear, so its dimension is $k_1+k_2$. If $v=0$, a nonzero word has Hamming weight at least $2d_1$; if $v\ne 0$, $\mathrm{wt}(u)+\mathrm{wt}(u+v)\ge \mathrm{wt}(v)\ge d_2$. Taking $(u,u)$ and $(0,v)$ of minimum weight showsFor $a\in C_2^{\perp },b\in C_1^{\perp }$, the inner product of $(u,u+v)$ and $(a,a+b)$ is $u\cdot b+v\cdot a+v\cdot b=0$, since $C_2\subseteq C_1$. The spaces have complementary dimensions, proving $(C_1|C_2{)}^{\perp }=C_2^{\perp }|C_1^{\perp }$.

Define the Reed-Muller code $\mathrm{RM}(d,r)$ by evaluating all multilinear polynomials of total degree at most $r$ at the $2^d$ points of ${\mathbb{F}}_2^d$. The squarefree monomials are independent functions, giving dimension ${\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{d}{j}\right)$. Splitting according to the last variable gives the Reed-Muller bar-product recursion $\mathrm{RM}(d,r)=\mathrm{RM}(d-1,r)|\mathrm{RM}(d-1,r-1)$, with the endpoint codes interpreted in the usual way. The distance recursion gives $2^{d-r}$.

To identify the Dual of a Reed-Muller code, multiply a monomial of degree at most $r$ by one of degree at most $d-r-1$. The product omits some variable, so its sum over ${\mathbb{F}}_2^d$ is even and its binary inner product is zero. The binomial dimension identity then gives

In a spatially flat FLRW metric, a radial null geodesic satisfies $d\chi /dt=\pm c/a(t)$. For a source and observer at fixed comoving separation, $\chi ={\int }_{t_e}^{t_0}cdt/a(t)$. Applying this to successive wave crests gives $d{t}_0/a(t_0)=d{t}_e/a(t_e)$ and therefore the cosmological redshift $1+z=a(t_0)/a(t_e)$.

Each photon's energy is reduced by $1+z$, and the arrival interval is increased by $1+z$. The received power crossing the observation sphere is consequently $L/(1+z{)}^2$. Its physical area is $4\pi a(t_0{)}^2|x{|}^2$, so the radiative flux isThis is the luminosity distance relation with $d_L=(1+z)a(t_0)|x|$. The Euclidean area formula presumes spatial flatness; in a curved FLRW metric the transverse comoving distance replaces $|x|$.

Let $L(f)=f^{\prime }(0)-\mu (f)$. Direct evaluation on $1,x,x^2$ gives $L=0$, so the Peano kernel theorem givesThis representation also follows by applying $L$ to the Taylor expansion of $f$ with integral remainder. The derivative at zero of the truncated quadratic is zero for $t\ge 0$, and its sampled values giveThe Peano kernel is nonnegative, andHence $|L(f)|\le \frac{1}{3}\Vert f^{(3)}{\Vert }_{\infty }$. For $f(x)=x^3/6$, the third derivative is $1$, while $f^{\prime }(0)=0$ and $\mu (f)=-1/3$. Thus equality holds, proving the least constant is $c=1/3$, attained by $f(x)=x^3/6$.

For a moving circuit, Faraday's law takes the formHere $C(t)$ is the instantaneous wire contour, $\mathbf{v}$ its local material velocity, and $S(t)$ any spanning surface. The contour orientation and surface normal vector obey the right-hand convention. The integral on the right is the magnetic flux, differentiated with both the field and circuit position allowed to vary. The electric field term is the electric force per unit electric charge around the wire, and $\mathbf{v}\times \mathbf{B}$ supplies the motional electromotive force. Their sum is the electromotive force driving charge around the moving circuit. The condition $\nabla \cdot \mathbf{B}=0$ makes the magnetic flux independent of the spanning surface.

Choose the initial loop normal vector to be $+{\mathbf{e}}_y$, so rotation about $+{\mathbf{e}}_z$ gives $\mathbf{n}(t)=(-\sin \Omega t,\cos \Omega t,0)$. The spatially uniform magnetic field givesUsing Ohm's law for the resistive loop,Positive electric current is measured in the contour orientation associated with $\mathbf{n}(t)$. Reversing that convention reverses the signed answer. When $\Omega =\omega$, the magnetic flux is constant and the induced electric current vanishes. This uses the usual thin resistive-circuit approximation with self-inductance neglected.

For the wavefunction of a particle of mass $m$, the probability density and probability current areFor a real potential $V$, the Time-dependent Schrödinger equation and its complex conjugate readMultiplying by ${\Psi }^{\ast }$ and $\Psi$ respectively yieldsThe potential terms cancel because $V$ is real. This is the continuity equation, expressing local conservation of probability.

For the given superposition the common time phase cancels in ${\Psi }^{\ast }{\Psi }_x$, andThe expression in the second parentheses is purely imaginary, so the second term is real and contributes nothing to the probability current. ThusIt is the incoming probability current minus the reflected probability current.

Take the wave speed $c>0$ and introduce characteristic coordinates $\xi =x+ct$, $\eta =x-ct$. The chain rule givesThe wave equation therefore says $y_{\xi \eta }=0$. Integrating first with respect to $\eta$ gives $y_{\xi }=F(\xi )$, and integrating with respect to $\xi$ gives $y=f(x+ct)+g(x-ct)$ with twice differentiable $f,g$. Conversely substitution shows that every such sum solves the wave equation. The two terms are travelling waves of unchanged shape, moving respectively to the left and right at speed $c$.

The zero initial displacement gives $g(x)=-f(x)$, and the initial velocity gives $2c{f}^{\prime }(x)=\psi (x)$. Hence the D'Alembert formula with initial velocity isFor a classical twice differentiable solution one may take $\psi \in C^1$. Differentiation gives $y_t(x,0)=\psi (x)$ and $y(x,0)=0$ directly. The coordinate calculation also works for any nonzero signed $c$, with physical speed $|c|$. If the printed constant is allowed to be $c=0$, the equation instead gives $y=A(x)+tB(x)$ and these initial data give $y=t\psi (x)$; the displayed travelling-wave representation requires $c\ne 0$.

The argument principle states that if $h$ is meromorphic on a neighbourhood of a positively oriented simple closed contour and its interior, with neither zeros nor poles on the contour, thenwhere $N$ and $P$ count its interior zeros and poles with multiplicity. This integer is also the winding number of $h(\gamma )$ about zero.

Fix $z_0\in U$. Injectivity prevents $f$ from being constant on a neighbourhood of $z_0$, so its Taylor expansion has the formChoose a small closed disk $D\subset U$, centred at $z_0$, such that $h$ and $mh+(z-z_0)h^{\prime }$ are nonzero throughout $D$. Then $f-f(z_0)$ has its sole zero at $z_0$ in $D$, of multiplicity $m$, while $f^{\prime }$ has no zero in $D\setminus \{z_0\}$. On its boundary let $\delta =\min |f-f(z_0)|>0$.

Choose $0<|w|<\delta$. For every $0\le t\le 1$, the holomorphic function $f-f(z_0)-tw$ is nonzero on $\partial D$. By the argument principle, its number of zeros is an integer given by a contour integral varying continuously with $t$, hence remains $m$. At $t=1$ none of these zeros is $z_0$, so their derivatives are nonzero and they are simple zeros. Thus there are $m$ distinct points taking the value $f(z_0)+w$. Injectivity forces $m=1$, proving $f^{\prime }(z_0)\ne 0$ everywhere in $U$.

The Fundamental theorem of finitely generated abelian groups expresses every finite abelian group of this order as a product of cyclic groups of prime-power order. Their exponents correspond to the partitions of an integer $4$:These are pairwise nonisomorphic: the largest order of a group element distinguishes all except the two groups with largest element order $p^2$, and those have respectively $p^2$ and $p^3$ elements annihilated by multiplication by $p$. The classification exhausts the possibilities, so there are five isomorphism classes.

The dual map reverses the direction of the linear map:Here the dual spaces consist of complex-linear functionals. Composition with $\alpha$ is linear both as a functional on $V$ and as an operation on $W^{\ast }$, so this does define a linear map between the dual spaces.

Let $(v_1,\dots ,v_n)$ and $(w_1,\dots ,w_m)$ be the chosen bases, and let $(v^1,\dots ,v^n)$ and $(w^1,\dots ,w^m)$ be their dual bases. If $A$ represents $\alpha$, then $\alpha (v_i)={\sum }_j{A}_{ji}{w}_j$. ThereforeThe coefficient in row $i$, column $j$ of the matrix of ${\alpha }^{\ast }$ is thus $A_{ji}$, giving the matrix transpose $A^{\mathsf{T}}$. This calculation introduces no complex conjugation: the algebraic dual map uses complex-linear functionals, whereas a Hermitian transpose belongs to the inner-product adjoint construction.