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For a convex scalar function $f(x,y)$, the partial subdifferential ${\partial }_{y}f(x,y)$ means the subdifferential of $v\mapsto f(x,v)$ at $v=y$. It is not differentiation of a set-valued subdifferential. If this section is differentiable, the partial subdifferential is the singleton containing ${\nabla }_{y}f(x,y)$.

For a convex function and $\tau >0$, the forward step is the set-valued explicit update $F_{\tau f}(x)=\{x-\tau p:p\in \partial f(x)\}$. It reduces to the usual gradient step when differentiable. In contrast, a backward step uses a subgradient at the new point and is the proximal operator for a proper closed convex function.

Adding the two subgradient inequalities for $p\in \partial f(x)$ and $q\in \partial f(y)$ givesThus the subdifferential is a monotone operator. This inequality proves uniqueness of its resolvent: two solutions of $z\in x+\tau \partial f(x)$ must coincide when $\tau >0$.

For a proper convex function finite at $x$, global minimality is equivalent to $0\in \partial f(x)$. This follows immediately from the defining subgradient inequality. Combined with a subdifferential sum rule, it converts a convex minimization problem into an inclusion, for example the equation defining a proximal operator.

If $\mathbb{E}{e}^{tX}\le e^{{\sigma }^2{t}^2/2}$, then $X^2-\mathbb{E}{X}^2$ is a sub-exponential random variable with scale at most a universal constant times ${\sigma }^2$. The Chernoff bound gives $\mathbb{P}(|X|>x)\le 2e^{-x^2/(2{\sigma }^2)}$; integrating this tail bounds $\mathbb{E}|X{|}^{2q}$ by $2(2{\sigma }^2{)}^{q}q!$. Centering preserves a factorial moment bound, so the Bernstein bound for independent sub-exponential variables apply to independent averages of these squares.

For two independent normal simple regressions with centered predictors, common unknown error variance and $m>2$ observations in each, let $S_u=\sum u_i^2$, $S_w=\sum w_i^2$. The estimated slope difference is normal with mean $b-d$ and variance ${\sigma }^2(S_u^{-1}+S_w^{-1})$. Pool the two residual sums of squares and use $s^2=RSS/(2m-4)$. Independence from the estimates gives a Student t-distribution pivot, yielding the displayed two-sided interval of confidence $1-\alpha$.

Scattering with one meson and one baryon in the initial and final states. The strong interaction conserves electric charge, baryon number, net strangeness, four-momentum, total angular momentum, and parity. Approximate isospin symmetry imposes additional relations among charge-channel amplitudes.

The intrinsic Lorentzian metric in the Polyakov action is independent of the string embedding map. Its equation relates it conformally to the induced worldsheet metric. A Weyl transformation changes the intrinsic metric without changing the classical action.

The characteristic string length is ${\ell }_s=\sqrt{{\alpha }^{\prime }}$, with $T=1/(2\pi {\alpha }^{\prime })$. Excited string states have masses of order $1/{\ell }_s$ and higher-derivative corrections become important at that scale.

Tree-level scattering of four ordered open string tachyons gives the Veneziano amplitude. Its poles factorize through intermediate open-string states.

An oscillator vacuum is annihilated by all positive-frequency oscillator annihilation operators. For a Ramond sector string this leaves a degenerate space on which the fermionic zero modes act through the Ramond zero-mode Clifford algebra. Vacuum annihilation therefore does not select a unique spinor state.

A constant dilaton contributes ${\Phi }_0\chi (\Sigma )$ to the Euclidean worldsheet action by the Gauss-Bonnet theorem. With string coupling $g_s=e^{{\Phi }_0}$ and Euler characteristic $\chi =2-2g$, the path-integral weight is $g_s^{2g-2}$. Canonically normalized external closed-string vertices add one power of $g_s$ each.

Connected oriented closed string worldsheets of genus $g$ give the spacetime effective-action term $g_s^{2g-2}{S}_g$. Each $S_g$ has a separate low-energy derivative expansion in ${\alpha }^{\prime }$. The string coupling orders quantum string loops, while ${\alpha }^{\prime }$ orders effects of finite string size.

The deviatoric part of a three-dimensional stress tensor is ${\mathbf{\sigma }}_{\mathrm{dev}}=\mathbf{\sigma }-\mathrm{tr}(\mathbf{\sigma })\mathbf{I}/3$. It is traceless. Adding an isotropic pressure does not change a normal-stress difference.

The stress tensor maps a surface normal to the traction exerted across that surface. Its divergence is the force density in local momentum balance. For an incompressible Newtonian fluid it combines pressure with twice the viscosity times the symmetric strain-rate tensor.

The incompressible linear flow $(u,v)=(ϵx-\gamma y,\gamma x-ϵy)$ has streamfunction $\psi =ϵxy-\gamma (x^2+y^2)/2$. Its velocity matrix squares to $({ϵ}^2-{\gamma }^2)I$. Thus strain-dominated flow has saddle trajectories, rotation-dominated flow has closed elliptical trajectories, and equal nonzero strain and rotation gives simple shear flow along parallel lines with a stationary line. Pure rotation is counterclockwise for positive $\gamma$.

If the output of a stoquastic circuit is $|0⟩|r_0⟩+|1⟩|r_1⟩$ with nonnegative amplitudes, its plus-outcome probability is $1/2+\mathrm{Re}⟨r_0|r_1⟩\ge 1/2$. Basis quantum witnesses with zero and plus ancilla qubits always have this property. A soundness threshold below one half cannot hold for every such input.

With the convention used here, a Stokes line is a ray where competing exponential actions have the same imaginary part. A subdominant exponential can acquire a rapidly changing multiplier there. For $e^{\pm x^2/\epsilon }$, these rays have $\arg x=m\pi /2$. The anti-Stokes lines instead have equal exponential magnitudes.

With the equal-magnitude convention, an anti-Stokes line has $\mathrm{Re}(S_1-S_2)=0$ for competing exponentials. For $e^{\pm x^2/\epsilon }$, its rays are $\arg x=\pi /4+m\pi /2$. Crossing one interchanges which exponential grows. Some literature interchanges the names, so the action-based convention should always be specified.

For $z=2n+i\mu \sqrt{n}+O(1)$, the factor $t^{2n}{e}^{z(1-t)}$ has saddle scale $t=1+u/\sqrt{n}$ and becomes $e^{-u^2-i\mu u}$. If the amplitude has a simple pole there, its leading local integral contains $e^{-u^2-i\mu u}/u$. Differentiating its Cauchy principal value with respect to $\mu$ reduces it to a Gaussian integral, producing an error function. The nonsingular amplitude $(1+t{)}^{-1}$ gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$.