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The transition operator of an additive random walk preserves convex functions: apply their convexity inequality to $x+\xi$ and $y+\xi$, then integrate. The pointwise maximum of convex functions preserves convexity too, so induction proves convexity of the Snell envelope Bellman functions. Expectations are allowed to be extended-valued until finiteness is justified. Integrability of rewards on a finite horizon implies finiteness of these functions at every state for $t\ge 1$, by integrability of translated convex random-walk rewards and the bound $W_t(x)\le {\sum }_{j=0}^{T-t}\mathbb{E}[f(x+Z_j{)}^{+}]$. At time zero only $X_0=0$ is observed; if $W_0$ is infinite elsewhere, its finite value at zero has a constant convex extension representing the same Snell envelope. Full statewise Bellman recursion at time zero requires an additional translated-integrability hypothesis.

In this temporal sense, the optical coherence length is $L_c=c{\tau }_c$, where $c$ is the vacuum speed of light and ${\tau }_c$ is an explicitly chosen optical coherence time. It measures a delay as a vacuum-equivalent path difference; it does not measure transverse spatial optical coherence. ^[ref].

For the stipulated $1\mathrm{GHz}$ profiles, the $1/e$ convention gives $L_c\approx 9.54269\mathrm{cm}$ for the Lorentzian and $15.8896\mathrm{cm}$ for the Gaussian. In a vacuum Michelson interferometer, changing one mirror position by $x$ changes the round-trip delay by $2x/c$, so the mirror travel to either of these delays is $L_c/2$. In a medium, convert physical travel using the actual group delay, which can depend on frequency. For a narrow line centered at vacuum wavelength ${\lambda }_0$, differentiating $\nu =c/\lambda$ gives $\Delta \nu \approx c\Delta \lambda /{\lambda }_0^2$; this small-width approximation must use consistently defined spectral widths.

Allow terminal positions of an infinite game of perfect information to carry a losing-player label. A clopen set of infinite winning plays still gives a determined infinite game. At a position where all infinite continuations have one winner, the other player can win only by forcing a terminal loss for that winner. The usual winning-position attractor in an infinite game decides this reachability game. The tree of remaining undecided positions is well-founded: every infinite branch would eventually have its clopen payoff decided by a prefix. Assign the already decided winners upwards by well-founded recursion. Terminal labels are essential; a terminal node created by a covering need not be lost by the player whose turn would have come next.

For $0<S<1$ and $V=1-S$, the translational branch of one-sided kinetic solidification dispersion has $\sigma =(S/V-\Gamma ){\alpha }^2+O({\alpha }^4)$. Expand $p=V+(\sigma +{\alpha }^2)/V+O({\alpha }^4)$ and substitute to prove this limit. At zero growth, $p_0=[V+\sqrt{V^2+4{\alpha }^2}]/2$, and the destabilizing right side has sign ${\alpha }^2(SV/p_0^2-\Gamma )$. For nonnegative real $\sigma$, the residual has derivative at least one, so there is one growing root precisely when $\Gamma <SV/p_0^2$. Thus instability exists exactly when $\Gamma <S/V$, with unstable band ${\alpha }^2<p_c(p_c-V)$, $p_c=\sqrt{SV/\Gamma }$ for $\Gamma >0$. At threshold the next term is $\sigma =-2S{\alpha }^4/V^3+O({\alpha }^6)$. Surface energy removes short-wave growth even when long waves remain unstable.

The cubic follows by eliminating $\sigma =p^2-Vp-{\alpha }^2$ from one-sided kinetic solidification dispersion, with $V=1-S$. For $\Gamma =0$, the interface branch tends to $\sigma =SV-SV/\alpha +O({\alpha }^{-2})$. For fixed $0<\Gamma <1$, substituting $p=\sqrt{1-\Gamma }\alpha +d$ gives $2d-V=S\Gamma /(1-\Gamma )$, hence $\sigma =-\Gamma {\alpha }^2+[S\Gamma /\sqrt{1-\Gamma }]\alpha +O(1)$. At $\Gamma =1$ the balance is $p^3\sim S{\alpha }^2$, giving $\sigma =-{\alpha }^2+S^{2/3}{\alpha }^{4/3}+o({\alpha }^{4/3})$. For $1<\Gamma <1+2S/V$, $p\to S/(\Gamma -1)>V/2$ and $\sigma =-{\alpha }^2+p^2-Vp+o(1)$. Stronger capillarity can remove the real isolated interface mode from the chosen thermal branch; it does not create growth. These distinctions explain why replacing $p$ by $\alpha$ is not a controlled fixed-capillarity limit.

The initial cost of the one-period quadratic hedge is $\mathbb{E}[{\rho }^{\ast }\xi ]$. This pricing weight satisfies $\mathbb{E}[{\rho }^{\ast }R]=1$ and $\mathbb{E}[{\rho }^{\ast }{S}_1]=S_0$. Among square-integrable weights satisfying these equations it has the smallest squared norm:For another admissible weight $\rho$, its difference from ${\rho }^{\ast }$ is orthogonal to $1$ and all coordinates of $S_1$, hence to ${\rho }^{\ast }$. The Pythagorean theorem in an inner-product space gives $\mathbb{E}[{\rho }^2]=\mathbb{E}[({\rho }^{\ast }{)}^2]+\mathbb{E}[(\rho -{\rho }^{\ast }{)}^2]$. This weight can be signed; positivity, and hence interpretation as a state-price density, requires additional assumptions.

If $S_1$ has a multivariate normal distribution with mean $\mu$ and invertible covariance $V$, and $g$ has bounded gradient, Gaussian integration by parts gives $\mathrm{Cov}(S_1,g(S_1))=V\mathbb{E}[\nabla g(S_1)]$. Consequently the risky holdings of its one-period quadratic hedge equal the expected payoff gradient. The initial cost is $R^{-1}\mathbb{E}[g(S_1)]+(S_0-R^{-1}\mu )\cdot \mathbb{E}[\nabla g(S_1)]$. Bounded gradient ensures linear growth and square integrability, even if the payoff is not bounded.

A finite signed measure with total mass one under which discounted asset gains have zero integrals. Its density can be negative, so it need not be a probability measure or give arbitrage-free prices to every nonnegative contingent claim. A square-integrable signed density is useful in one-period least-squares hedging, even when it is not an equivalent martingale measure.

Let $m=\mathbb{E}Y$ and $C=\mathrm{Cov}(Y)$ be an invertible covariance matrix of discounted gains. The displayed signed martingale measure leaves unchanged the mean-zero random variables orthogonal to the martingale part $Y-m$: if $\mathbb{E}L=0$ and $\mathbb{E}[(Y-m)L]=0$, then $\mathbb{E}[Z_{\ast }L]=0$. Conversely, preservation of all these orthogonal directions forces the density to lie in the span of $1,Y-m$, and its pricing constraints give the displayed formula. If $Z_{\ast }>0$ it is an equivalent martingale measure; if $Z_{\ast }\ge 0$ it is a dominated martingale measure. Positivity is an extra condition, not a consequence of absence of arbitrage alone. In one period this density also minimizes $\mathbb{E}{Z}^2$ among square-integrable signed pricing densities, because every other such density differs by a vector orthogonal to $1,Y-m$. The initial capital in the unrestricted one-period quadratic hedge of a discounted payoff $H$ is $\mathbb{E}[Z_{\ast }H]$.

For a discounted contingent claim $H$ and gain vector $Y$, fix initial capital $v$ and minimize $\mathbb{E}(H-v-{\theta }^{T}Y{)}^2$. Differentiation gives the least-squares normal equations $G\theta =\mathbb{E}[Y(H-v)]$, where $G=\mathbb{E}[Y{Y}^T]$ is assumed invertible. This gives the displayed hedge. Optimizing capital as well instead centers the gains and uses their covariance matrix: ${\theta }^{\ast }=\mathrm{Cov}(Y{)}^{-1}\mathrm{Cov}(Y,H)$ and $v^{\ast }=\mathbb{E}H-({\theta }^{\ast }{)}^T\mathbb{E}Y$. Fixing the initial capital is therefore a different quadratic problem.

For positive ${\beta }_0$, the one-loop running coupling tends to zero at high energy and grows as energy decreases. Near the strong-coupling scale, omitted perturbative terms cease to be small; the apparent infrared pole is not a controlled prediction of a physical infinity. Continuing the algebraic formula below $\Lambda$ gives a negative coupling and is not a valid physical continuation. Confinement and hadron formation are nonperturbative phenomena, not a theorem proved by this one-loop pole. The integration constant obeys $\Lambda ={\mu }_0\exp [-1/(2{\beta }_0{\alpha }_0)]$, so an order-one coefficient alone cannot determine an absolute mass scale; a reference coupling or empirical input is also necessary.

If each bond has closed probability at most $q$ in one-independent bond percolation on the square lattice, a set of $\ell$ bonds contains a matching in a graph of size at least $\ell /4$: divide horizontal bonds by the parity of their left endpoint and vertical bonds by the parity of their lower endpoint. Each of the four classes is a matching. Mutual independence in the largest class proves the circuit bound. There are at most $4\ell 3^{\ell }$ dual circuits of length $\ell$ surrounding the origin. For sufficiently small $q$, the sum of these circuit probabilities is less than one, so the origin has a positive chance of belonging to an infinite percolation cluster.

Place side-$2n$ square lattice boxes around the points $3nv$, $v\in {\mathbb{Z}}^2$. The width-$2n$ rectangle joining adjacent boxes has length $5n$. Declare a coarse bond open when its long rectangle is crossed and each endpoint box is crossed in the transverse direction. Nonincident coarse bonds use disjoint fine-bond sets, while incident good bonds give connected open paths inside their common endpoint box. When fixed-aspect crossing probabilities tend to one, these coarse bond probabilities tend to one. The Peierls circuit bound for one-independent percolation then yields an infinite fine-lattice open cluster.

For pricing drift ${\mu }_Q$, a unit zero-coupon bond price $P(t,r;T)$ satisfies $P_t+{\mu }_Q{P}_r+{\eta }^2{P}_{rr}/2-rP=0$, with $P(T,r;T)=1$. Itô formula makes its bank-account-discounted value a local martingale; suitable integrability gives $P(t,r;T)={\mathbb{E}}^Q[\exp (-{\int }_t^T{r}_{s}ds)\mid r_t=r]$. The rate kills the diffusion semigroup, as in the Feynman-Kac formula.

If all traded zero-coupon bonds depend on one scalar rate driven by one Brownian motion, their diffusion exposures form one vector $v_i=\eta P_{i,r}$. Their instantaneous covariance matrix is $v{v}^T$, of rank at most one. Nondegenerate maturity changes are therefore instantaneously perfectly correlated, with correlation $+1$ or $-1$, though finite-horizon bond-return correlations need not be one.

For smooth derivative prices depending on a Markov short rate, Itô formula gives physical drift $m=V_t+b{V}_r+{\sigma }^2{V}_{rr}/2$ and exposure $s=\sigma V_r$. The common short-rate market price of risk gives $m-rV=\theta s$, yielding this equation. Under the risk-neutral measure, the rate drift is $b-\sigma \theta$ and the bank-discounted price is a martingale under suitable integrability. A unit zero-coupon bond has terminal value one and price ${\mathbb{E}}^Q[\exp (-{\int }_t^T{r}_{s}ds)\mid r_t=r]$.

If $p$ divides the primitive level $M$ of a trivial-character newform and $N=M{p}^r$, put $e_j=p^{jk/2}f(p^{j}z)$. Then $U_p{e}_0=a_p{e}_0$ and $U_p{e}_j=p^{k/2}{e}_{j-1}$ for $j\ge 1$. The normalized Fricke involution acts by $e_j\mapsto \epsilon e_{r-j}$, where $W_{M}f=\epsilon f$. These two operators act irreducibly: a nonzero invariant subspace contains an eigenvector of $U_p$; its eigenvalue is $a_p$ or zero. Either it contains $e_0$ immediately or applying the reversal and a power of $U_p$ produces $e_0$, using $a_p^2\ne p^k$. Reversal then produces $e_r$, and its successive images produce every basis vector.

If $M\mid N$ and $D\mid N/M$, conjugating ${\Gamma }_0(N)$ by $\mathrm{diag}(D,1)$ gives matrices $\left(\begin{array}{cc}a& Db\\ c/D& d\end{array}\right)$ in ${\Gamma }_0(M)$. This proves the weight transformation of $f(D\tau )$. At any rational modular cusp, factor the rational change of variable into an integral cusp representative followed by an upper-triangular rational matrix of positive slope. The original cusp expansion then stays holomorphic and has zero constant term. Thus this map takes cusp forms of level $M$ to those of level $N$.

Author of the 1827 treatise developing the electrical transport relation now called Ohm's law. ^[ref].

For a finite-state homogeneous continuous-time Markov chain with generator matrix $Q$, integrate the indicator of occupying state $j$ over $[0,T]$. Interchanging the bounded time integral and expected value gives the displayed integral of transition probabilities. It counts all visits, not only the first holding time. Summing over states gives $T$. A zero eigenvalue of $Q$ does not cause a problem for this integral, although it prevents casually writing it with an inverse matrix $Q^{-1}$.