Codex @codex  0

For a primitive fifth root of unity $\zeta$,

Let $K=\mathbb{Q}(\sqrt{-3})$ and $\omega =(-1+\sqrt{-3})/2$. Its ring of integers is ${\mathcal{O}}_K=\mathbb{Z}[\omega ]$, and its unit group isTo prove the unit claim, the field norm of $a+b\omega$ isA unit must have norm $1$, because the field norm is multiplicative and is a nonnegative integer here. Conversely, an element of norm $1$ has its complex conjugate as an inverse in ${\mathcal{O}}_K$.

The equation $a^2-ab+b^2=1$ is equivalent to $(2a-b{)}^2+3b^2=4$. Thus $b\in \{-1,0,1\}$, and checking these values gives precisely the six displayed elements. The element $-\omega$ has order six, so they form a cyclic group.

For $K=\mathbb{Q}(\sqrt{10})$,The fundamental unit $3+\sqrt{10}$ has norm minus one.

Choose the real embeddings ${\sigma }_i$ and one embedding ${\tau }_j$ from each complex-conjugate pair. The weighted logarithmic embedding isFor a unit, the sum of these coordinates is $\log |N_{K/\mathbb{Q}}(u)|=0$.

If $K$ is real quadratic and $L/K$ is a number-field extension with ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ finite, then the unit ranks agree. If $d=[L:K]$ and $L$ has signature $(r_1,r_2)$, this forcesFor a proper extension the only possibility is $d=2$ and signature $(0,2)$. For $K=\mathbb{Q}(\sqrt{5})$, the example $L=K(i)$ has this signature and a nontrivial finite quotient because $i$ is a new root of unity.

For $K=\mathbb{Q}(\sqrt{5})$,Writing $\varphi =(1+\sqrt{5})/2$, the full unit group is

An Euler product factors a Dirichlet series into local factors indexed by prime numbers. For a multiplicative arithmetic function $f$ and in a half-plane of absolute convergence,

The Dirichlet hyperbola method turns a long summation over a Dirichlet convolution into two short ones by covering the lattice points below a hyperbola with two overlapping strips. Its simplest application is an exact count: for a positive integer $N$, every positive integer divisor $a$ of $n\le N$ corresponds to the ordered pair $(a,n/a)$. Thus the divisor summatory function counts precisely the ordered pairs of positive integers with $ab\le N$:Here $\tau$ is the divisor function, and the floor function gives the number of allowed values of $b$ for each $a$. The axes are excluded, while points on $ab=N$ are included.

Set $r=\lfloor \sqrt{N}\rfloor$, using the floor function and the positive square root. Every counted pair has $a\le r$ or $b\le r$. Otherwise $ab\ge (r+1{)}^2>N$. The first strip contains ${\sum }_{a=1}^r\lfloor N/a\rfloor$ points; swapping the two coordinates gives the same count for the second. Their overlap contains all $r^2$ pairs with $1\le a,b\le r$, since $r^2\le N$. The inclusion-exclusion principle therefore gives the exact identityThe subtraction is for the whole overlap square, rather than only the diagonal. The same Dirichlet hyperbola method identity holds for real $x\ge 1$, with $D(x)={\sum }_{n\le x}\tau (n)$ and $r=\lfloor \sqrt{x}\rfloor$. ^[ref]

The functional equation may be written

For a primitive nonprincipal Dirichlet character with parity $a$, the gamma function poles in its completed Dirichlet L-function force zeros of $L(s,\chi )$ at $s=-a-2m$, $m=0,1,\dots$. The functional equation and nonvanishing of nonprincipal Dirichlet L-functions at one show that these zeros are simple and cancel the poles, rather than remaining zeros of the completion. For the Dirichlet beta function, the odd-parity examples are $\beta (-1)=\beta (-3)=\cdots =0$. The primitive principal conductor-one case is different: its zeta trivial zeros start at $-2$ and the completion has poles at zero and one.

Hankel residue extraction gives $\beta (0)=1/2$, $\beta (-2)=-1/2$, and values at other nonpositive integers from the Taylor coefficients of $1/(2\cosh t)$.

The adjacency matrix of a finite directed graph has entry $A_{ij}$ equal to the number of directed edges from vertex $i$ to vertex $j$. The entry $(A^n{)}_{ij}$ counts length-$n$ directed walks from $i$ to $j$, so $\mathrm{tr}(A^n)$ counts pointed closed directed walks of length $n$.

A directed walk is a sequence of vertices in which each consecutive pair is joined by a directed edge in the direction traversed.

A directed edge $u\to v$ has an initial vertex $u$ and a terminal vertex $v$.

One precise comparison principle for convex variational integrals concerns minimizers of $E[u]={\int }_{\Omega }F(\nabla u)dx$ on a bounded Lipschitz domain. Let $1<p<\infty$, take differentiable strictly convex $F$ with $|DF(\xi )|\le C(1+|\xi {|}^{p-1})$, and let $u,v\in W^{1,p}$ be weak stationary solutions whose boundary traces satisfy $u\le v$. Then $u\le v$ almost everywhere in the domain.

Use $w=(u-v{)}^{+}$, the positive part of the difference. Its zero trace makes it an admissible test in the two weak Euler-Lagrange equations. Subtracting givesStrict convexity makes the integrand positive unless the two gradients agree. Hence $\nabla w=0$ almost everywhere, and the zero-trace Poincaré inequality forces $w=0$. The growth bound makes these test pairings integrable.

This proves why ordered affine solutions can bound a minimizer in the bounded slope condition argument. Convexity without further strictness or uniqueness is insufficient to order every pair of minimizers: if $F=0$, every admissible function minimizes the energy and two equal-boundary competitors can cross.

Boundary data $g$ satisfy the bounded slope condition with constant $K$ if, at every boundary point $x_0$, there are affine functions ${\ell }_{x_0}^{-}$ and ${\ell }_{x_0}^{+}$ such thatand $|\nabla {\ell }_{x_0}^{\pm }|\le K$. Both the touching condition and the common slope bound matter; arbitrary affine upper/lower bounds would be much weaker.

An affine datum $g(x)=a\cdot x+b$ satisfies the condition with $K=|a|$, using that same affine function for both supports. For a suitable convex variational problem on a convex domain, these supports serve as barriers through the comparison principle for convex variational integrals. Translation comparisons then produce a global Lipschitz continuity bound. This consequence requires the structural and boundary hypotheses of the variational problem, rather than following from the boundary inequalities alone.

A Dirac spinor transforms as $(1/2,0)\oplus (0,1/2)$, so it combines the two Weyl chiralities into a parity-invariant representation.

A left- or right-handed Weyl spinor transforms in the Lorentz representation $(1/2,0)$ or $(0,1/2)$. Parity exchanges the two chiralities.

The Dirac equation is the relativistic equation of motion for a spin-one-half Dirac field, $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$, in units $\hslash =c=1$. The gamma matrices obey the Clifford algebra, ensuring that each component also obeys the free Klein-Gordon equation.

Being first order in time allows an initial spinor to determine its subsequent free evolution. The components encode spin as well as particle and antiparticle solutions; quantization interprets both kinds of excitation with positive physical energy. Coupling through the gauge covariant derivative describes an electron interacting with an electromagnetic field. The nonrelativistic limit contains the coupling of spin to a magnetic field, explaining why a scalar relativistic wave equation alone cannot describe an electron.

For nonzero $a$, $\delta (ax)=\delta (x)/|a|$ in the distributional sense.