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For $\dot{u}=u(\mu +u-v)$, $\dot{v}=\lambda u^2$ in $u>0$, the time change $d\tau /dt=u$ gives the affine linear system $u_{\tau }=\mu +u-v$, $v_{\tau }=\lambda u$. Shifting $w=v-\mu$ removes the forcing. Its eigenvalues are $(1\pm \sqrt{1-4\lambda })/2$. For $\lambda >1/4$, putting $\omega =\sqrt{4\lambda -1}/2$ givesFor fixed $\mu >0$ and initial $h,v_0$ tending to zero, the first return to small positive $u$ occurs after approximately $\pi /\omega =2\pi /\sqrt{4\lambda -1}$. This is a half-turn, not a full turn. At $\lambda =1/4$, $u=e^{\tau /2}[h+(\mu +h/2-v_0)\tau ]$ does not return and grows without bound. The loss of return diagnoses escape to infinity rather than a finite-amplitude homoclinic orbit. Physical time obeys $dt=d\tau /u$ and need not share the divergent desingularized travel time.

For the diffusionless quadratic activator-inhibitor model, choose the Dulac function $u^{-2}$. Then $u^{-2}f=1/v-b/u$ and $u^{-2}g=1-v/u^2$, giving divergence $(b-1)/u^2$. For $0<b<1$ it is strictly negative throughout the simply connected positive quadrant. Green's theorem rules out a periodic orbit: its boundary flux would be zero because the rescaled vector field is tangent to the orbit, while its interior divergence integral would be negative.

A mode with squared wavenumber $q$ in the quadratic activator-inhibitor model has negative trace $b-1-(1+d)q$ when $0<b<1$. Its determinant is $D(q)=d{q}^2+(1-db)q+b$. A positive minimum location and a negative minimum require $db>1$ and $(db-1{)}^2>4db$, equivalent to $db>3+2\sqrt{2}$. At onset, the double determinant root is $q_c=(1+\sqrt{2})/d$. In a bounded domain an admissible nonzero spatial mode must additionally lie in the open negative-determinant band.

If a real $m$-by-$n$ matrix $A$ has full column rank, with $m\ge n$, then $A^{T}A$ is positive-definite. Its upper-triangular Cholesky decomposition $A^{T}A=R^{T}R$, with positive diagonal, is unique. Defining $Q=A{R}^{-1}$ gives $Q^{T}Q=I$ and $A=QR$. Any other such factorization yields another Cholesky factor of $A^{T}A$, so its $R$ and then its $Q$ must agree. This proves existence and uniqueness; it does not assert that forming $A^{T}A$ is the most numerically stable algorithm.

Let $K$ be block diagonal with $2\times 2$ blocks $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. If a real invertible matrix $A$ commutes with $K$, its QR decomposition $A=BC$, with $B$ orthogonal and $C$ upper triangular with positive diagonal, has both $BK=KB$ and $CK=KC$. Indeed $CK{C}^{-1}=B^{-1}KB$, so $E=K^{-1}CK{C}^{-1}$ is orthogonal and upper triangular in $2\times 2$ blocks. Orthogonality forces its off-diagonal blocks to vanish. Its diagonal blocks are $J^{-1}{C}_{i}J{C}_i^{-1}$. For $C_i=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $a,d>0$, this block is $\left(\begin{array}{cc}d/a& -b/a\\ -b/a& (a^2+b^2)/(ad)\end{array}\right)$. Orthogonality of its columns forces $b=0,d=a$, and hence each block is the identity. Thus $E=I$, proving the claim.

For the normalized Black-Scholes call function, completing the square gives $F(v,m)=\Phi (d_1)-m\Phi (d_2)$, with $d_1=(-\log m+v/2)/\sqrt{v}$ and $d_2=d_1-\sqrt{v}$. The identities $d_1(v,1/m)=-d_2(v,m)$ and $\Phi (-x)=1-\Phi (x)$ implyFor a stock with risk-neutral drift $r$, maturity difference $\tau$, and positive strike $K$, put-call parity then givesThe positive sign of $r\tau$ in the reciprocal argument is forced by taking the reciprocal of $K{e}^{-r\tau }/S_t$. The identity includes $v=0$ by continuity and requires $m>0$.

The difference of the terminal power-call and power-put payoffs is $S_T^{\beta }-K^{\beta }$. Risk-neutral pricing and the geometric Brownian motion moment formula therefore give the displayed parity, with ${\lambda }_{\beta }=\beta (r-q)+{\sigma }^2\beta (\beta -1)/2$. At $\beta =1$ this reduces to dividend-adjusted put-call parity.

Suppose disjoint separating spheres cut a 3-manifold into pieces none of which is a punctured three-sphere. Compress one sphere along a disk disjoint from the others. The old sphere and the two new spheres enclose a three-times punctured sphere. At least one of the two pieces on the compressed side is not a punctured sphere; retain that new sphere and discard the other. The piece on its opposite side cannot become a punctured sphere, since capping its boundaries and applying the Three-dimensional Schoenflies theorem would make the original opposite piece a punctured sphere as well. Hence the number of spheres and the nontriviality condition can be preserved.

With the usual Dirac adjoint, Hermitian ${\gamma }^5$ and $\{{\gamma }^0,{\gamma }^5\}=0$, taking the adjoint gives ${\psi }^{†}{\gamma }^5{\gamma }^0\psi =-{\psi }^{†}{\gamma }^0{\gamma }^5\psi$. Consequently a real pseudoscalar with coupling $-g\varphi \bar{\psi }{\gamma }^5\psi$ has a Hermitian interaction only if $g^{\ast }=-g$. Writing $g=i{g}_P$ with real $g_P$ gives the familiar interaction $-i{g}_P\varphi \bar{\psi }{\gamma }^5\psi$. Formal Euler-Lagrange field equations obtained with independent spinor and adjoint variables can still be computed for the original symbolic coupling; it must not silently be identified with a real physical coupling.

Zero distance in a pseudometric defines an equivalence relation: the triangle inequality proves transitivity. If $x$ and $x^{\prime }$ have zero distance, as do $y$ and $y^{\prime }$, the triangle inequality in both directions gives $d(x,y)=d(x^{\prime },y^{\prime })$. Thus distance descends to the quotient and becomes a metric, since zero quotient distance means the classes coincide.

This first-order operator vanishes exactly on J-holomorphic curves. Linearization at a map $u$ gives a Fredholm operator between suitable Sobolev spaces of sections. At a constant map it is the Dolbeault operator on the trivial pulled-back tangent bundle.

A J-holomorphic sphere is a J-holomorphic curve with domain the complex projective line.

A J-holomorphic curve is simple if it does not factor through a holomorphic branched covering of degree greater than one. For a nonconstant curve this is equivalent to having a point where its derivative is nonzero and its image has exactly one preimage.

A J-holomorphic curve on a punctured disc with finite energy and image contained in a compact target extends smoothly across the puncture when the target has a smooth tame almost complex structure and compatible bounded geometry. Local monotonicity controls escape from small target balls, and the elliptic small-energy estimate followed by elliptic regularity gives the extension. This is the step turning a finite-energy plane obtained by rescaling into a J-holomorphic sphere.

For $n\ge 2$ and positive integers $d_1,\dots ,d_n$ with ${\sum }_v{d}_v=2(n-1)$, the number of labelled trees having vertex degree $d_v$ at label $v$ isBy the Prüfer degree formula, this is the multinomial coefficient counting sequences with $d_v-1$ occurrences of each label. If a degree is zero or the sum is wrong, there are no such trees. For the example degree sequence $(2,1,3,1,2,1)$, there are $4!/(1!0!2!0!1!0!)=12$ labelled trees, one for each ordering of the labels $1,3,3,5$. Knowing the vertex degrees thus constrains, but usually does not determine, the tree.

Proline is a cyclic alpha-amino acid whose amine nitrogen is joined to two carbon positions within a five-membered ring. This explains the exception to ${\mathrm{H}}_2\mathrm{N}-\mathrm{CH}(R)-\mathrm{COOH}$ in amino acid. ^[ref]

If $A⊢B$, eliminate the letters private to $A$ byEvery assignment satisfying $A$ satisfies one disjunct of $C$. If an assignment satisfies $C$, change only the eliminated letters to its witnessing assignment. The modified assignment satisfies $A$ and hence $B$, whose letters were unchanged. Thus $A\models C\models B$, and the propositional completeness theorem gives the two derivations. The interpolant uses only common letters; an empty common vocabulary is handled by logical truth constants.

For $\nu =1/2$ and initially pristine gas, $M=M_0(1-Z/(2y))$ and the formed stellar mass is $S=2(M_0-M)=M_{0}Z/y$. If a fixed initial mass function supplies $\kappa$ surviving tracers per unit formed mass, then $N(<Z)=\kappa M_{0}Z/y$ and $dN/dZ=\kappa M_0/y$ until the current gas abundance. Complete exhaustion gives a uniform normalized metallicity distribution function on $0<Z<2y$. This is uniform in linear metallicity, not in its logarithm.

A three-uniform hypergraph is bipartite when its vertex set has a partition into two parts such that every hyperedge meets both parts. All such triples on parts of sizes $a,b$ number $a\left(\genfrac{}{}{0ex}{}{b}{2}\right)+b\left(\genfrac{}{}{0ex}{}{a}{2}\right)=ab(a+b-2)/2$. For fixed total order this is maximized by balanced parts.

In a gauge-preserving regularization and consistent renormalization convention, the proper-vertex Ward-Takahashi identity in QED equates the vertex and electron wave-function renormalization constants. With ${\psi }_0=Z_2^{1/2}\psi$ and $A_0=Z_3^{1/2}A$, the vertex coefficient gives $e_0={\mu }_R^{ϵ}{Z}_{1}e/(Z_2\sqrt{Z_3})$, proving the displayed charge relation in $d=4-2ϵ$. Gauge invariance restricts counterterms but does not forbid charge running.