A fermionic mode with canonical anticommutation relations $\{b,b^{†}\}=1$ has the symmetrically ordered Hamiltonian operator $H=\frac{E}{2}(b^{†}b-b{b}^{†})=E(b^{†}b-1/2)$. Its empty-state zero-point contribution is $-E/2$. A quantum harmonic oscillator with bosonic canonical commutation relations contributes $+E/2$ instead. For a Dirac field, sum over both particle and antiparticle modes and over spin: there are four independent oscillators per spatial momentum in four spacetime dimensions. Comparing bosonic and fermionic vacuum energies requires a common regularization in quantum field theory, matching degrees of freedom and a stated vacuum-energy subtraction convention.

For an increasing strictly concave wealth value $v$, the wealth-variable dual is the convex functionAt a unique interior maximizer where $v$ is differentiable, $z=v^{\prime }(x)$ and $J^{\prime }(z)=-x$, whenever the maximizer gives a differentiable dual. Thus $v(x)=J(z)-z{J}^{\prime }(z)$. If, additionally, $v$ is twice continuously differentiable and $v^{\prime \prime }(x)<0$ at that maximizer, the inverse function theorem yieldsStrict concavity alone does not ensure nonzero curvature. The Inada utility with vanishing curvature is an explicit example where the dual has a first derivative but fails to have a finite second derivative at the corresponding price.

This definition is the negative of the concave Legendre dual. Under the stated smooth nonzero-curvature hypotheses, it can convert the optimized portfolio term of a Hamilton-Jacobi-Bellman equation into a linear second derivative. Boundary optimizers require the appropriate one-sided or subgradient formulation instead of the interior derivative identities.

For concave increasing utility, the concave supporting-tangent inequality gives $e^{-bt}[U({\hat{c}}_t)-U(c_t)]\le Y_t({\hat{c}}_t-c_t)$. If the proposed consumption exhausts the state-price budget and every competitor has no larger budget, integration proves optimality. The objective must be well defined; a positive discount rate and finite utility bounds are a convenient sufficient condition.

The upper unitriangular group $U{T}_3({\mathbb{F}}_3)$ consists of $M(a,b,c)=\left(\begin{array}{ccc}1& a& c\\ 0& 1& b\\ 0& 0& 1\end{array}\right)$ for $a,b,c\in {\mathbb{F}}_3$. Its multiplication is $M(a,b,c)M(a^{\prime },b^{\prime },c^{\prime })=M(a+a^{\prime },b+b^{\prime },c+c^{\prime }+a{b}^{\prime })$. The elements $M(1,0,0)$ and $M(0,1,0)$ do not commute. Yet every element is $I+N$ with $N^3=0$, so characteristic three gives $(I+N{)}^3=I$. It is a nonabelian finite group of order $27$ whose nonidentity elements all have order three.

For a three-dimensional strictly upper triangular matrix $A$, the binomial theorem stops after $A^2$. Over a finite field of odd prime characteristic $p$, both coefficients $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ vanish, so every nonidentity matrix in $U{T}_3({\mathbb{F}}_p)$ has order $p$. In characteristic two, a matrix with nonzero $A^2$ instead has order four.

For a fixed outer pressure and slip velocity, retain terms linear in a disturbance of a base flow $(U,V)$. The tangential equation has the displayed form and continuity is ${\tilde{u}}_x+{\tilde{v}}_y=0$. Disturbances vanish at the wall and the tangential disturbance vanishes in the far field. A local parallel-profile approximation may freeze the coefficients even when the profile is not an exact global steady solution.

For a Lorentzian metric and timelike $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2$. The Euler-Lagrange equations of $\mathcal{L}$ give ${\nabla }_{v}v=0$, while those of $L$ give ${\nabla }_{v}v=(\dot{L}/L)v$. Taking $ds/du=L$ removes the latter tangential acceleration and gives a unit timelike tangent. Thus the two actions have the same geodesic images; the quadratic action singles out an affine parameter. They do not have the same solutions for an arbitrary fixed parameter: $t=u^2$, $\mathbf{x}=0$, $u>0$, in Minkowski spacetime solves the length equation but not the energy equation. Null curves are excluded since $L=0$.

The universal consequences $T_{\forall }$ of a first-order theory $T$ are all universal first-order sentences entailed by $T$. Their model class is exactly the class of substructures of models of $T$, up to structure embedding. This characterization follows from the diagram embedding criterion for universal theories.

The Dickey–Fuller test detects a unit-root autoregressive process against a stationary causal alternative. Regress $\Delta X_t$ on $X_{t-1}$, with deterministic terms appropriate to the model. Under the null, the usual regression statistic has a nonnormal Brownian motion functional limit, so ordinary normal critical values are inappropriate.

If ${\sum }_i{g}_i{f}_i=1$, expand its power $r(N-1)+1$. By the pigeonhole principle, each monomial contains some $f_i^N$. Collecting terms expresses $1$ as a linear combination of these powers. This lets one use a common denominator exponent in localization arguments.

For $F⊣G$ with adjunction unit $\eta$ and adjunction counit $\epsilon$, the identities are ${\epsilon }_{FC}F{\eta }_C=1_{FC}$ and $G{\epsilon }_D{\eta }_{GD}=1_{GD}$. They make the transpose maps $f\mapsto Gf\eta$ and $g\mapsto \epsilon Fg$ inverse.

For $F⊣G$, the unit is the natural transformation $\eta :1\Rightarrow GF$ corresponding to the identity of $FX$ under the adjunction at each $X$. It participates in the triangle identities for an adjunction.

For $F⊣G$, the counit is the natural transformation $\epsilon :FG\Rightarrow 1$ corresponding to the identity of $GY$ under the adjunction at each $Y$. It participates in the triangle identities for an adjunction.

On a finite-dimensional vector space over ${\mathbb{Q}}_p$, the triangle inequality gives an upper bound by a coordinate maximum norm. The coordinate unit sphere is compact. The other norm is continuous and nonzero there, so has a positive minimum. Scaling by powers of $p$ gives the lower bound. The resulting equivalence of norms in finite dimensions proves completeness of finite extensions, and applying the comparison to powers of a field element proves equality of two extending multiplicative absolute values on a field.

A square matrix $U$ is unipotent when $U-I$ is nilpotent. Over an algebraically closed field, this is equivalent to all eigenvalues being one. If $j$ belongs to a nilpotent ideal of an operator algebra, $I+j$ is a unipotent matrix. This realizes the radical factor in the Levi decomposition of a quiver automorphism group.

It is the largest connected normal unipotent subgroup of a linear algebraic group. A connected reductive algebraic group has trivial unipotent radical. The subgroup $1+J(E)$ of a finite-dimensional algebra's unit group is closed, connected and unipotent.

For ${\mathbb{Q}}_2({\zeta }_8)/{\mathbb{Q}}_2$, the shifted polynomial $(1+T{)}^4+1$ is Eisenstein and ${\zeta }_8-1$ is a uniformizer. The three nonidentity automorphisms have displacement valuations two, four and two. Thus the lower breaks are one and three, the upper breaks one and two, and the different exponent from ramification groups is eight.

For $L={\mathbb{Q}}_3({\zeta }_3,\alpha )$ with ${\alpha }^3=2$, the shifted cubic for $\alpha +1$ and quadratic for ${\zeta }_3-1$ are Eisenstein. Their compositum is totally ramified of degree six. The element $\pi =({\zeta }_3-1)/(\alpha +1)$ has valuation one. Order-three automorphisms move it with valuation two and transpositions with valuation one. The lower breaks are zero and one, the upper breaks zero and one-half, and the different exponent from ramification groups is seven.

Let $|f|\le M$ on $[0,1]$ and let $P$ have $r$ interior division points. At most $r$ cells of the equal-length partition $D_n$ have one of those points in their interior. Their total length is at most $r/n$, and their oscillation is at most $2M$. Every other cell lies in a single cell of $P$. ThereforeThis comparison of Darboux sums avoids the incorrect assumption that $D_n$ refines $P$ for all large $n$.

For a uniform-load bending of a clamped filament, total load $F=wL$ and tip displacement $\delta =w{L}^4/(8B)$ obey Hooke's law in the form $F=k_{\mathrm{eff}}\delta$, with $k_{\mathrm{eff}}=8B/L^3$. This stiffness depends on the specified distributed loading. A concentrated tip force instead gives $3B/L^3$, so equal total forces with different distributions are not interchangeable.