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For a finite path in a Böhm tree, choose $q$ at least the head arity at every preceding node. Replace relevant variables by $U_q=\lambda z_1\cdots z_{q}w.w{z}_1\cdots z_q$. After exposing each node, supply its initial bound variables with $U_q$, pad its $m$ arguments to $q$, and apply the projection $\lambda z_1\cdots z_q.z_i$ selecting the next child. Each step returns that child with accumulated substitutions, so iteration reaches a substitution instance of the chosen subtree. Tupling preserves all children until selection, even when the same variable occurs at several nodes.

For metric $g=\Omega (d{x}^2+d{y}^2)$, set $F_{12}={\partial }_x{A}_y-{\partial }_y{A}_x$ and use the physical magnetic field $B=F_{12}/\Omega$. At critical coupling, define $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$. The identity $|D_x\Phi {|}^2+|D_y\Phi {|}^2=|(D_x+i{D}_y)\Phi {|}^2+F_{12}|\Phi {|}^2+{\partial }_x{j}_y-{\partial }_y{j}_x$ completes the energy into nonnegative squares plus half the magnetic flux and half the current boundary integral. For unit-magnitude vacuum boundary data, the combined boundary contribution is $\frac{1}{2}\oint (A+j)=\pi N$, since $j=d\arg \Phi -A$ there. In the more restrictive covariantly decaying sector the current integral separately vanishes and the flux is $2\pi N$. Equality for positive $N$ requires $(D_x+i{D}_y)\Phi =0$ and $B=(1-|\Phi {|}^2)/2$. The factor $\Omega$ appears instead if the symbol $B$ denotes the coordinate component $F_{12}$.

The holomorphic map $[Z_0:Z_1:Z_2]\mapsto [Z_1:Z_2]$ on ${\mathbb{CP}}^2\setminus \{[1:0:0]\}$ records the line through the missing point and the input point. Its graph closure is the incidence surface $Z_{1}b=Z_{2}a$ in ${\mathbb{CP}}^2\times {\mathbb{CP}}^1$, the blowup of a complex manifold at a point. Projection to $[a:b]$ extends the pencil holomorphically. On the exceptional divisor it is the identity map on the projective tangent directions.

The centralizer condition defining block induction refers to a defect group of a block of the actual source block. Containment of the centralizer of an unrelated p-subgroup does not suffice. For example, in characteristic $3$ take $G=A_5$, $H=A_4$ and $D=C_3$. Then $D{C}_G(D)\le H$, but the unique defect-zero block of $k{A}_4$ occurs on bimodule restriction of both three-dimensional defect-zero blocks of $k{A}_5$. The two ordinary degree-three characters have identical restriction to $A_4$, namely its degree-three character, and defect-zero reduction gives the same simple projective module. Thus the unique-block characterization fails if $D$ is not the defect group of the source block.

The Logistic cell-division process is a density-limited pure birth model on $\{0,1,\dots ,N\}$, where $N$ is a positive integer carrying capacity. With $k$ cells, each divides at rate $1-k/N$, so the total rate for $k\to k+1$ is $k(1-k/N)$. Zero and $N$ are absorbing; a positive initial population can only increase until it reaches $N$.

Let $Z_t^N=K_t^N/N$ and assume $Z_0^N\to p\in (0,1)$ in probability, for example $K_0^N=\lfloor Np\rfloor$. On every fixed finite interval $[0,T]$, $Z^N$ converges uniformly in probability toAn explicit martingale argument verifies this instance of the Kurtz fluid limit theorem. Compensating the jumps gives $Z_t^N=Z_0^N+{\int }_0^t{Z}_s^N(1-Z_s^N)ds+M_t^N$, with predictable quadratic variation at most $t/(4N)$. Doob's L2 maximal inequality gives $\mathbb{E}{\sup }_{s\le T}|M_s^N{|}^2\le T/N$. Since $z\mapsto z(1-z)$ is Lipschitz on $[0,1]$, Gronwall's inequality bounds the trajectory error by $e^T(|Z_0^N-p|+{\sup }_{s\le T}|M_s^N|)$, which tends to zero in probability.

The finite-population mean need not itself solve the logistic equation: $\mathbb{E}[Z(1-Z)]=m(1-m)-\mathrm{Var}(Z)$. The fluid limit follows because the fluctuations vanish on the specified finite horizons.

// Target: analysis.bigb

For a pure birth process with fixed initial state and finite statewise rates, the jump-count compensator gives $dM=(e^{\theta }-1)M_{-}[dN-\lambda (X_{-})dt]$, so $M$ is a local martingale up to explosion. If $\lambda \le C$, a rate-$C$ Poisson candidate clock with accepted jumps of probability $\lambda (X_{-})/C$ prevents explosion and dominates the jump count. Its finite exponential moments dominate every stopped $M$ on each finite horizon, upgrading the local martingale to a true martingale. A random initial state requires $\mathbb{E}{e}^{\theta X_0}<\infty$; uniformly bounded rates do not supply this missing initial integrability.

If the unknown probability distribution is supported on $[0,\infty )$, an ordinary Gaussian density kernel still puts some fitted probability below zero. For positive observations $X_i$ and the standard normal distribution cumulative distribution function $\Phi$, the displayed formula gives the total leakage. For the statistical sample $(0.1,0.4,0.9)$ and smoothing bandwidth $h=0.5$, it isThus about $22.28\%$ of the fitted probability falls outside the known support of a probability distribution, even though the full integral remains one. The leakage panel in smoothing bandwidth shades this area. Simply deleting the negative part leaves area $0.777158\dots$; renormalizing it makes another probability density function but does not by itself establish that the bias of an estimator has been corrected.

The issue also appears in the expectation of the kernel density estimator. For a nonnegative symmetric kernel for density estimation and a probability density function $f$ that is zero for $x<0$ and has a finite positive right-hand limit $f(0+)$, assume $f$ is bounded. Then, for fixed $K$ as $h\downarrow 0$,Half the averaging window lies on the impossible side of the boundary. For example, the rate-one exponential distribution has $f(0+)=1$, so the limiting expected boundary height is $1/2$, rather than the right-hand height one. This refers to estimating the right-hand limit; changing a probability density function at a single point does not change its probability distribution. Decreasing the smoothing bandwidth narrows the affected region but does not remove this boundary-height error. This is why the interior second-order pointwise bias bound for kernel density estimation cannot be applied across a jump at the boundary.

An adjoint-invariant symmetric bilinear form $B$ on a Lie algebra makes the displayed tensor alternating. If $B$ is invariant under the full group Adjoint representation, its left-translated three-form is also right-invariant. On left-invariant fields, the exterior derivative is $d\eta (W,X,Y,Z)=-2B(W,[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]])=0$ by the Jacobi identity. The Killing form supplies such a $B$ on every semisimple Lie group. An overall sign change in the lowered structure-constant convention negates the form but not its invariance or closedness.

Replace two p-values $P_i,P_j$ by zeros in the Benjamini-Hochberg procedure at level $\alpha$, keeping the original denominator $m$, and let $R^{(ij)}$ be the resulting rejection count. Equivalently, remove the two values and apply the step-up critical values $\alpha (k+2)/m$ to the remaining ordered p-values; their rejection count is $R_{-ij}=R^{(ij)}-2$. For $r\ge 2$,If both original values are rejected, lowering them changes no ordered rank greater than $r$, so the rejection count stays $r$. Conversely, if the modified count is $r$ and both original values meet the rank-$r$ threshold, reinserting them leaves at least $r$ qualifying values. Monotonicity under lowering gives at most $r$ original rejections. This identity is deterministic and does not require independence or a distributional assumption.

For a subset of the unbiased Boolean hypercube of density $\alpha$, use normalized Fourier-Walsh transform coefficients. Applying Beckner's inequality with $\rho =1/\sqrt{3}$ gives $\Vert T_{\rho }{1}_A{\Vert }_2^2\le \Vert 1_A{\Vert }_{4/3}^2={\alpha }^{3/2}$. On levels at most two the weights ${\rho }^{2|S|}$ are at least $1/9$, giving the displayed bound. Parseval's identity gives total Fourier mass $\alpha$, so when $9\sqrt{\alpha }<1/2$ the higher levels contain strictly more than the lower levels.

The coefficient $b$ is the leading relative null-mean discrepancy of a regular likelihood-ratio test statistic. It is distinct from the finite-sample divisor $1+b/n$ used in a Bartlett correction. Different conventions call either of these a correction factor, so the normalization should always be stated explicitly.

If ${\mathbb{E}}_{0}W=d(1+b/n+O(n^{-2}))$, the Bartlett-corrected statistic has mean $d+O(n^{-2})$. It is commonly calibrated against a chi-squared distribution with $d$ degrees of freedom. Equivalently, to the displayed order, one may multiply $W$ by $1-b/n$.

Every finite coloring of a barrier in better-quasi-order theory has an infinite restriction on which all barrier members have the same color. It extends finite-dimensional Ramsey's theorem from fixed-length subsets to variable-length initial segments. A proof proceeds by well-founded induction on the initial-segment tree and a fusion construction selecting successively nested infinite tails. It implies that every finite set with the equality order is a better-quasi-ordering. The barrier/front partition proof is given in Section 3.2 of arxiv.org/pdf/1604.05866 .

For a balanced incomplete block design with $r>\lambda$, $N{N}^T$ has eigenvalues $r-\lambda >0$ on the zero-sum subspace and $r+(t-1)\lambda >0$ on the constant vector. Thus its rank is $t$. As $N$ is $t\times b$, $t=\mathrm{rank}(N{N}^T)\le \mathrm{rank}(N)\le b$.

A balanced incomplete block design is symmetric when its number of blocks equals its number of treatments. The identity $bk=tr$ then gives $r=k$. Its square incidence matrix of a set system $N$ obeys $N{N}^T=(r-\lambda )I+\lambda J$, with eigenvalues $r-\lambda$ of multiplicity $t-1$ and $r^2$ of multiplicity one.

For squared loss $E={\sum }_k(t_k-z_k{)}^2/2$, let $e_k=(t_k-z_k)g_k^{\prime }(x_k)$ and $e_j=g_j^{\prime }(x_j){\sum }_k{w}_{jk}{e}_k$. The chain rule gives $\partial E/\partial w_{jk}=-e_k{z}_j$ and $\partial E/\partial w_{ij}=-e_j{z}_i$. A gradient descent step of size $\eta$ uses $\Delta w_{jk}=\eta e_k{z}_j$ and $\Delta w_{ij}=\eta e_j{z}_i$, with every derivative evaluated at the same pre-update weights.

For ${\mathcal{L}}_{\mathrm{int}}=-g\varphi \bar{\psi }{\gamma }^5\psi$, the variational Dirac equation and adjoint Dirac equation are $(i/\partial -m-g\varphi {\gamma }^5)\psi =0$ and $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }+g\varphi \bar{\psi }{\gamma }^5=0$. Substitute these into the derivative of $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. Anticommuting the chirality matrix past ${\gamma }^{\mu }$ gives two identical mass terms and two identical interaction terms, proving the displayed identity. These are classical field equations; the formula does not assert an unrenormalized quantum composite-operator identity. With Hermitian coupling $g=i{g}_P$, the second term is $-2g_P\varphi \bar{\psi }\psi$.

For $\mathcal{L}=\bar{\psi }(i/\partial -S+iP{\gamma }^5)\psi$, with real scalar backgrounds $S,P$, the Dirac equation is $(i/\partial -S+iP{\gamma }^5)\psi =0$ and the adjoint Dirac equation is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+S\bar{\psi }-iP\bar{\psi }{\gamma }^5=0$. Use $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ and $({\gamma }^5{)}^2=1$ in the derivative of $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ to obtain the displayed classical identity. Taking $S=m+g\varphi$ and $P=G\Phi$ includes both scalar and Hermitian pseudoscalar Yukawa couplings.

For an ordinary scalar field, its derivative is a polar covector, while the axial current is a pseudovector. Their contraction is invariant under the proper orthochronous Lorentz group but changes sign under parity. If the field is a pseudoscalar, the extra transformation sign makes this coupling parity even. In four spacetime dimensions its mass dimension is five, so Lorentz invariance alone does not imply power-counting renormalizability.

An axial charge is the spatial integral of the time component of an axial current. For a Dirac field and an internal generator $t$, it is $\int {\psi }^{†}{\gamma }_{5}t\psi d^{3}x$. Its canonical action rotates left- and right-handed fields oppositely. It is conserved only when the current divergence and flux at infinity vanish; Dirac masses and a chiral anomaly can obstruct conservation. The nonsinglet quark generators have no color anomaly and obey the nonsinglet axial charge algebra of quark densities.