A logistic regression fitted to ungrouped Bernoulli observations has residual deviance $D=-2\ell (\hat{\beta })$. Comparing it automatically with ${\chi }_{n-p}^2$ is generally invalid: the saturated dimension grows with $n$, with one observation per fitted probability. If the true model has only an intercept and success probability $1/2$, then $D/n\to 2\log 2$, whereas ${\chi }_{n-1}^2/n\to 1$. Fixed-dimensional nested-model likelihood-ratio tests can still use Wilks theorem under regularity. For absolute fit, use meaningful grouping, residual diagnostics, or a fitted-model parametric bootstrap.

With the axiom of choice, enumerate both players' binary-move strategies in an infinite game in length $\mathfrak{c}$. Each strategy admits $\mathfrak{c}$ compatible plays. At stage $\alpha$, reserve a fresh play following I's strategy as a loss for I, and a fresh play following II's strategy as a win for I. Avoid all previously reserved plays, which number less than $\mathfrak{c}$. The collection of the second reservations is a payoff set against which neither player has a winning strategy in an infinite game. This uses no regularity assumption on $\mathfrak{c}$.

Every infinite game of perfect information on a set-sized move alphabet with a Borel payoff in its discrete product topology has a winning strategy in an infinite game for one player. A proof strengthens the induction hypothesis to existence of arbitrary-depth-preserving game coverings unravelling the payoff. The closed-set announcement construction starts the induction. For countable unions, successive coverings stabilize each finite level and their inverse limit makes every constituent payoff clopen; their union is open and can be unravelled once more.

If a Sylow p-subgroup $P$ is abelian, a hypothetical central subgroup $Q\le G^{\prime }$ of order $p$ has a nontrivial linear character extending to $P$. Inducing this extension to $G$ gives a character of degree prime to $p$, hence an irreducible constituent of such degree. On $Q$ its determinant character is a nontrivial power of the chosen linear character, but every determinant character is trivial on $G^{\prime }$. This contradiction proves the displayed restriction.

The derivative of the determinant at an invertible matrix is $D{\det }_A(K)=\det A\mathrm{tr}(A^{-1}K)$. At $A=I+H$, use $\det (I+H)=1+\mathrm{tr}H+O(\Vert H{\Vert }^2)$ and $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$. Differentiating the first derivative gives the displayed symmetric bilinear map. More generally $D^2{\det }_A(H,K)=\det A[\mathrm{tr}(A^{-1}H)\mathrm{tr}(A^{-1}K)-\mathrm{tr}(A^{-1}H{A}^{-1}K)]$.

For an arbitrary square matrix $A$, the derivative of the determinant in direction $H$ is the displayed expression. Differentiating the determinant's multilinear expansion gives the sum of entries of $H$ weighted by their cofactors, equal to the displayed trace. If $A$ is invertible, $\mathrm{adj}(A)=(\det A)A^{-1}$ gives $(\det A)\mathrm{tr}(A^{-1}H)$, and division by $\det A$ gives the Jacobi determinant derivative formula for the logarithmic derivative. The adjugate expression remains valid at singular matrices.

Use the unnormalized Fourier transform and $|f|\le 1$. Normalize each derivative correlation by $N$ and align its phase. Their absolute sum is at least $\alpha N$. One Cauchy-Schwarz inequality, expansion by shift/frequency differences, and a second Cauchy-Schwarz inequality bound its fourth power by $N$ times the additive energy of a frequency graph. Parseval identity on a finite group supplies the factor $N$; exact equality of both shift sums and frequency sums is retained.

An inner derivation of a Lie algebra is ${\mathrm{ad}}_x:y\mapsto [x,y]$. The Jacobi identity makes this a derivation of a Lie algebra. For every derivation of a Lie algebra $D$, $[D,{\mathrm{ad}}_x]={\mathrm{ad}}_{Dx}$, so the inner derivations form an ideal of a Lie algebra in the derivation Lie algebra. The kernel of $x\mapsto {\mathrm{ad}}_x$ is the center of a Lie algebra.

Over a field of characteristic zero, a nilpotent operator $D$ which is a derivation of a Lie algebra exponentiates to a Lie algebra automorphism. The iterated rulefollows by induction using Pascal's identity. Regrouping the finite expansion of $[\exp (D)x,\exp (D)y]$ by total degree gives $\exp (D)[x,y]$. The inverse is $\exp (-D)$. This factorial construction must not be reduced naively to small positive characteristic.

Inclusion-exclusion applied to the events that a permutation fixes a specified point gives the displayed formula. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ choices of $k$ fixed points and $(n-k)!$ permutations fixing them. The probability that a uniformly chosen permutation is a derangement tends to $e^{-1}$.

// Target: analysis.bigb

In a slender steady porous gravity current, the dominant balance is $(h^{\beta }{)}_X=(h^{\beta }{h}_Y{)}_Y$. If its width and height scales are $W,H$, fixed downslope volume flux gives $H^{\beta }W=\text{constant}$, and transverse spreading gives $W^2\sim HX$. Eliminating $H$ yields the displayed width exponent. The porosity exponent affects transient storage but not this steady balance.

Color the vertices of a periodic graph independently red or blue with equal probabilities, and declare a vertex of its line graph open exactly when the corresponding edge has equal-colored endpoints. The resulting site percolation has marginal $1/2$ and is a finite-range dependent random field: sets at line-graph distance greater than one use disjoint color inputs. Red and blue open clusters cannot join. If the original graph has an infinite red component and an infinite blue component, its line graph consequently has at least two infinite percolation clusters. One explicit amenable graph with that property is obtained by replacing each square-lattice vertex by a complete graph on ten vertices, joining neighboring blocks completely. Blocks containing both colors have independent probability $511/512$; a Peierls argument proves that they percolate, and their infinite cluster carries both a red and a blue component. Thus finite-range dependence, even with all site marginals nondegenerate, does not substitute for finite energy in percolation uniqueness.

A bond percolation law is one-independent if the bond states in any two sets with disjoint endpoint sets are independent as collections. In particular, states of bonds in a matching in a graph are mutually independent. The law need not be invariant under translations or positively associated.

For events $A_i$, a directed graph is an independence digraph if $A_i$ is independent of the sigma-algebra generated by the events indexed by its non-outneighbors, excluding $i$. The direction records which joint independence condition is available. A cyclically oriented triangle imposes pairwise independence; it does not impose independence of all three events together.

A Riemannian metric on a smooth manifold gives a positive section of the density bundle, locally $d{\mu }_g=\sqrt{\det (g_{ij})}|d{x}^1\cdots d{x}^n|$. The absolute Jacobian determinant makes this independent of coordinates without requiring orientability. With an orientation of a smooth manifold, it corresponds to the Riemannian volume form.

The mapping class group of a compact connected oriented topological surface, for orientation-preserving maps fixing its boundary pointwise and isotopies relative to that boundary, is generated by Dehn twists about simple closed curves. Boundary-parallel twists are allowed. For a closed surface of positive genus a finite generating collection exists. Permuting boundary components requires additional generators if that larger group is used.

A nondecreasing function bounding the area of every null word of length at most $n$ is an isoperimetric function for a finite group presentation. The Dehn function is the minimal such bound. A subrecursive bound means one dominated by a total computable function. Arbitrarily enlarged bounds are not invariants: even a group presentation with zero area function admits both linear and exponential upper bounds, which are not equivalent under the usual coarse comparison.

Choose bounded-length substitutions $T$ from the first generating set to the second, and $U$ back. Each substituted relator of the second group presentation has bounded area in the first, and every generator correction $a(UT(a){)}^{-1}$ has bounded area. Substitution into a minimal second-presentation diagram and filling the correction strips givesThe reverse construction gives the opposite inequality. Thus the optimal Dehn functions are coarsely equivalent, and the existence of a recursive isoperimetric bound, hence solvability of the word problem for a group, is independent of the finite group presentation.

For a nonconstant rational function of degree $n$, its derivative has degree between $n-1$ and $2n$. A finite pole of order $m$ becomes one of order $m+1$. If infinity is a pole of order $p-q>0$, it contributes $p-q-1$ to the derivative's pole count, where $p,q$ are the degrees of coprime numerator and denominator. This proves the lower bound; the quotient rule denominator $Q^2$ proves the upper bound. The functions $z^n$ and $1/(z^n-1)$ attain the bounds.

On the unit two-sphere this formula maps onto the torus: every torus point has arguments corresponding to $x,y\in [-1/2,1/2]$, and $x^2+y^2\le 1/2$ leaves a real $z$ on the sphere. Scaling both arguments continuously to zero is a null-homotopy, so the degree is zero. Surjectivity alone does not force nonzero degree.