For a single claim $x$, quota share reinsurance with retained fraction $\alpha \in [0,1]$ makes the direct insurer pay $\alpha x$ and the reinsurer pay $(1-\alpha )x$. Under excess of loss reinsurance with retention $M\ge 0$, the direct insurer pays $\min (x,M)$ and the reinsurer pays the positive part $(x-M{)}_{+}$. Thus the concise payout pairs areThe cap in excess of loss reinsurance applies separately to every claim; it is not a cap on the entire annual aggregate.

For the following variance calculations take $\lambda >0$ and $\mathbb{E}[X^2]<\infty$, so the displayed variances are finite. For any per-claim payout $Y$, the law of total variance in a compound Poisson distribution givesThe final term is the raw second moment, not the single-claim variance. Both parties' totals are retained compound Poisson aggregates, with different payout functions of the same claims; they are generally dependent.

For the constructed output, its defining quadratic says $h(X)=Y$ identically. Since the preceding branch calculation proves that its law is $IG(\mu ,\lambda )$, any variable with that law has the same pushforward distribution:One can also verify this chi-squared transform of an inverse Gaussian variable directly. The density contributions from the inverse branches areThe final equality follows from $x_{-}{x}_{+}={\mu }^2$. This confirms the transformed law independently of which reciprocal root was selected.

The smooth exponential sequence isLocal logarithms make it exact as a sequence of sheaves. A line bundle has a transition class in $H^1(M,{\mathcal{C}}^{\infty }({\mathbb{C}}^{\times }))$; its image under the connecting map defines the First Chern class. For a holomorphic line bundle the holomorphic exponential sequence gives the same class by naturality.

Here is the explicit Čech-de Rham curvature descent. On a good cover, write $e_j=e_i{g}_{ij}$ and choose logarithms $g_{ij}=\exp (2\pi i{f}_{ij})$. The integer cocyclerepresents $c_1(E)$. By the proved frame formula,Put $B_i=-A_i/(2\pi i)$ and $F=i\Theta /(2\pi )=d{B}_i$. Then $\delta B=-df$ and $\delta f=c$. In the Čech-de Rham double complex, with total differential $D_{\mathrm{tot}}=\delta +(-1{)}^{p}d$ on Čech degree $p$,Thus the global closed differential form and the integer cocycle represent the same class under de Rham theorem:This proves integrality, including the sign and normalization. It identifies the image of the integral class; vector-bundle curvature alone cannot recover torsion classes lost in passage to complex coefficients.

Two connections on the same line bundle differ by a global scalar one-form $a$. Their vector-bundle curvatures satisfy ${\Theta }_1-\Theta =da$, so the normalized representatives differ by an exact differential form. Hence the class is independent of the connection.

Use the integral normalization of the Fubini-Study form. On the chart $Z_j\ne 0$, define the integrally normalized Fubini-Study form byThe potentials on overlaps differ by the logarithm of the squared modulus of a nowhere-zero holomorphic function, whose $\partial \bar{\partial }$ is zero. Thus the forms patch and are closed. The dual of the tautological metric on $\mathcal{O}(1)$ has local squared norm $h=(1+\sum |z_k{|}^2{)}^{-1}$. Its Chern connection has $A=\partial \log h$ and vector-bundle curvature $\Theta =\bar{\partial }\partial \log h=\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$. Therefore $[{\omega }_{\mathrm{int}}]=c_1(\mathcal{O}(1))$, proving the requested integrality. With the unnormalized convention ${\omega }_{FS}=i\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$, this is ${\omega }_{\mathrm{int}}={\omega }_{FS}/(2\pi )$.

Take $\mathbb{N}=\{1,2,\dots \}$. We first prove the Ellis–Numakura lemma in the one-sided continuity convention needed here. Let $S$ be a nonempty compact Hausdorff space with an associative multiplication for which each map $x\mapsto xa$ is continuous. Among its nonempty closed subsemigroups, there is an inclusion-minimal one, $M$: a decreasing chain has nonempty intersection by compactness, and that intersection is a closed subsemigroup, so Zorn's lemma applies.

Fix $a\in M$. The set $Ma$ is nonempty and compact, hence closed in the Hausdorff space, and is a subsemigroup becauseMinimality gives $Ma=M$. Thus the set $T=\{x\in M:xa=a\}$ is nonempty. It is closed by the stated continuity, and it is a subsemigroup, since $(xy)a=x(ya)=xa=a$ for $x,y\in T$. Minimality again gives $T=M$, so $a\in T$ and $\boxed{a^2=a}$, giving a semigroup idempotent. If one instead uses the opposite one-sided continuity convention, the same proof uses $aM$ and $\{x:ax=a\}$; no joint continuity is required.

Apply this lemma to the Stone-Čech compactification of the natural numbers with its given addition. In the addition on the Stone-Čech compactification of the natural numbers convention,The continuous variable is $p$ when $q$ is fixed. The granted compactness, Hausdorff property and associativity therefore imply there exists an idempotent ultrafilter $p$ with $p+p=p$.

On the positive integers such a $p$ is nonprincipal: a principal ultrafilter at $n$ adds to itself to give the one at $2n$, which is different. Hence every cofinite set belongs to $p$. If zero is included in one's convention for $\mathbb{N}$, use the same compact semigroup argument on the closed space of nonprincipal ultrafilters: it is nonempty by compactness of the infinite discrete set's compactification, and the displayed addition sends two free ultrafilters to a free ultrafilter. This avoids the trivial principal idempotent at zero.

To deduce Hindman's theorem, let $A$ be the unique colour class belonging to $p$. DefineIdempotence gives $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma also gives $A^{\ast }-n\in p$ whenever $n\in A^{\ast }$. Here is its proof: $A-n\in p$, and applying idempotence to $A-n$ givesIntersecting this set with $A-n$ produces exactly $A^{\ast }-n$.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_r$ have been chosen with every nonempty finite sum in $A^{\ast }$. The finite intersectionbelongs to $p$, and is therefore nonempty. Choose $x_{r+1}$ from it. The new sums are $x_{r+1}$ and $s+x_{r+1}$, and they all lie in $A^{\ast }$. Induction yieldsThis proves every finite colouring of the positive integers admits a monochromatic finite-sums set generated by a strictly increasing infinite sequence, the required Hindman theorem.

Clear a common denominator first, so that all coefficients are nonzero integers. This changes neither the solutions nor partition regularity. We prove the one-equation criterion directly; no form of Rado's theorem is used as an assumption.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Colour a positive integer by its first nonzero base-$p$ digit: writing $x=p^{v_p(x)}u$ with $p\nmid u$, use the colour $u(\mathrm{mod}p)\in \{1,\dots ,p-1\}$. Suppose the equation has a monochromatic solution. Let $v$ be the smallest p-adic valuation of its entries and put $I=\{i:v_p(x_i)=v\}$, a nonempty set. Divide the equation by $p^v$ and reduce modulo $p$. All entries indexed by $I$ contribute the same nonzero digit $u$, and all other terms vanish. HenceIt follows that $p$ divides ${\sum }_{i\in I}{a}_i$. The absolute value of this sum is less than $p$, so $\boxed{\sum _{i\in I}{a}_i=0}$.

For sufficiency, take a nonempty zero-sum index set $I$ and put $S={\sum }_{i\notin I}{a}_i$. If $S=0$, the full coefficient sum is zero and the constant vector $(1,\dots ,1)$ is already a monochromatic solution in every colouring. Otherwise choose $j\in I$, put $t=|a_j|$, and set $b_j=-\mathrm{sgn}(a_j)S$, $b_i=0$ for $i\in I\setminus \{j\}$. ThenAdd the same sufficiently large nonnegative integer $H$ to all the $b_i$, obtaining $c_i=b_i+H\ge 0$. The sum is unchanged because ${\sum }_{i\in I}{a}_i=0$. Let $K={\max }_{i\in I}{c}_i$.

Apply the scaled Brauer progression theorem, proved above from the permitted Van der Waerden theorem, to obtain one-colour numbers $td$ and $a,a+d,\dots ,a+Kd$. DefineAll entries are positive and monochromatic, andTherefore the row is partition regular exactly when some nonempty coefficient subset sums to zero. Repetition of entries is allowed by the definition of partition regularity.

The final property concerns a pointwise periodic self-map. Such a map is a bijection: each point has a predecessor in its finite cycle, giving surjectivity; if $f(x)=f(y)$, choose a common multiple $N$ of periods of $x,y$ and apply $f^{N-1}$ to get $x=y$. Its finite cycles are therefore disjoint.

If $X$ is finite, choose the least common multiple of the finitely many cycle lengths. This positive integer $N$ gives $f^N={\mathrm{id}}_X$. The empty set satisfies the property with $N=1$.

Conversely, in the usual set theory with choice, every infinite $X$ contains a countably infinite set. Divide such a subset into disjoint finite blocks $B_j$ of sizes $j+1$, $j\ge 1$, make $f$ a cyclic permutation on each block, and fix every other point. Every point returns to itself, but an identity iterate would have to be divisible by every block length $j+1$. No positive integer has that property: take a block longer than that integer. Thus the uniform period criterion for pointwise periodic maps givesThe axiom of choice assumption is stated here because extracting a countably infinite subset of an arbitrary infinite set is not an unrestricted theorem in set theory without choice.

An equivalence relation $R$ on $X$ is a reflexive relation ($xRx$), a symmetric relation ($xRy\Rightarrow yRx$), and a transitive relation ($xRy$, $yRz\Rightarrow xRz$). Its equivalence class at $x$ is $[x{]}_R=\{y\in X:xRy\}$. Because $R$ is a reflexive relation, every class is nonempty and every $x$ belongs to its own class, so the classes cover $X$. If $[x{]}_R$ and $[z{]}_R$ share an element $y$, symmetry and transitivity give $xRz$. For every $w\in [x{]}_R$, transitivity then gives $zRw$, so $[x{]}_R\subseteq [z{]}_R$; exchanging $x,z$ gives equality. Thus distinct classes are disjoint. The equivalence classes form a set partition of $X$. On the empty set, the empty family is the corresponding partition.

A local odd mass density slope $\Sigma (x)-\Sigma (-x)=2{\Sigma }_0\beta x/r_0$ changes the cancelling one-sided torques into this net torque. A local outer cutoff $L$ adds the factor $1-(H/L{)}^2$. Positive slope transfers net angular momentum from satellite to disk. An infinite linear mass density law is only a formal local extension, since it cannot remain nonnegative on both sides.

In a homogeneous Ohmic conductor, the charge continuity equation and Gauss's law imply exponential decay of the bulk charge density, $\rho (t)=\rho (0)e^{-\sigma t/ϵ}$. For an isolated finite body, bulk charge moves to its boundary; the accumulating surface charge density preserves total electric charge. The formula uses a constant permittivity $ϵ$ and the corresponding free-charge convention, or ${ϵ}_0$ for total charge in the vacuum form of Gauss's law.

If $r^3\gg {\beta }^2\nu$, the broad western Stommel boundary layer has thickness $r/\beta$, but small viscosity supplies no-slip skins of thickness $(\nu /r{)}^{1/2}$ at both east and west walls. The western solution therefore contains two decaying exponentials, while the east has the rapidly decaying skin. Setting viscosity identically to zero removes the modes needed to satisfy all four lateral no-slip conditions.

For equal finite-dimensional closed subspaces of a Hilbert space $T,S$, positivity of $\cos {\theta }_{T,S}$ makes $P_S{|}_T$ invertible. The displayed reconstruction is the unique element of $T$ whose orthogonal projection onto $S$ matches that of $f$. It is the oblique projection onto $T$ along $S^{\perp }$. Its operator norm is the value of the secant function at the directed subspace angle, and its error is at most that secant function value times the best orthogonal projection error. The lower error bound follows from the Pythagorean identity. Zero-dimensional cases are handled directly.

For a cyclic extension of p-adic fields, the Herbrand quotient of the local multiplicative group together with Hilbert theorem 90 gives this index. If the ramification index and residue degree are $e,f$, the norm valuation formula $v_K(Nx)=f{v}_L(x)$ then gives the unit norm index $e$. In particular all units are norms in an unramified extension.

Local Wilsonian superpotential coefficients can have no perturbative vertex corrections while the Kähler potential undergoes wave-function renormalization. Canonical normalization then converts a quadratic coefficient $m$ and cubic coefficient $g$ to the displayed running coefficients. Non-renormalization of a holomorphic F-term is therefore compatible with running physical masses and interactions. The distinction requires keeping a local Wilsonian effective action separate from infrared-singular one-particle-irreducible effects and from eliminating entire massive fields.