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Let $f$ be bounded and holomorphic on a half-plane and continuous on its closure. A bound $|f|\le M$ on the boundary line propagates throughout the half-plane. Apply the bounded-domain principle to $f(z)z^{-1/n}$ on expanding truncated half-discs, then let $n\to \infty$.

Writing $a={\Sigma }_{11}$, $c={\Sigma }_{12}$, and $b={\Sigma }_{22}$, the vector $({\hat{\Sigma }}_{11},{\hat{\Sigma }}_{12},{\hat{\Sigma }}_{22})$ has asymptotic covariance

Local class field theory describes the abelian extensions of a local field $K$ through its multiplicative group $K^{\times }$.

The local Kronecker-Weber theorem states that every finite abelian extension of ${\mathbb{Q}}_p$ is contained in an extension obtained by adjoining roots of unity.

The cyclotomic extension ${\mathbb{Q}}_p({\zeta }_{p^n})/{\mathbb{Q}}_p$ is generated by a primitive $p^n$th root of unity. It is totally ramified and has Galois group $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$.

For a non-Archimedean local field $K$, the Weil group is the inverse image of the infinite cyclic subgroup generated by Frobenius under $\mathrm{Gal}(\overline{K}/K)\to \mathrm{Gal}(\overline{k}/k)\cong \hat{\mathbb{Z}}$. It is dense in the absolute Galois group with its profinite topology.

If one edge leaving the source has capacity $x$, every cut capacity is affine in $x$ with coefficient zero or one. In the 2024 Cambridge Part IB example, cuts of capacities $x+5$ and $14$ are both sharp, giving

If a symmetric invertible payoff matrix satisfies $A^{-1}e\ge 0$, both players may useand the game value is $(e^T{A}^{-1}e{)}^{-1}$.

If the column player has an optimal strategy at value $v$ and a row probability vector $p$ satisfies $p^{T}A\ge v{e}^T$, then $p$ is optimal: it guarantees $v$, while the optimal column strategy prevents every row strategy from exceeding $v$.

The row player solves

On its half-plane of convergence,Indeed, applying the transform to $y^{\prime }=Ay$, $y(0)=y_0$, gives $(sI-A)Y=y_0$ for every $y_0$.

Balancing outward radiative acceleration against gravity givesIt is an approximate upper luminosity for a hydrostatic star of opacity $\kappa$.

Applying Markov's inequality to the positive random variable $e^{tX}$ givesHere $M_X(t)=\mathbb{E}[e^{tX}]$ is the moment-generating function, and the useful bound requires it to be finite at the chosen $t$. Optimizing over $t>0$ gives the Chernoff bound method: an exponential function weights large values more strongly before applying the same average-budget argument.

For example, let $N$ have the Poisson distribution with parameter one. Its expected value and variance are both one, and its moment-generating function is $M_N(t)=\exp (e^t-1)$. At threshold four, choosing $t=\log 4$ givesThis $t$ minimizes the exponent $e^t-1-4t$ because its derivative is $e^t-4$, with positive second derivative. The direct Markov's inequality bound is $1/4$, and the two-sided Chebyshev's inequality bound, applied to a deviation of at least three from the mean one, is $1/9$. The exact Poisson distribution tail probability is $1-e^{-1}(1+1+1/2+1/6)\approx 0.0189882$. These compare three valid bounds obtained from progressively different information; none is an exact tail probability formula for general random variables.

For a joint density $f_{XY}$, a two-coordinate Gibbs update draws

The effective sample size discounts a correlated chain's raw length by its integrated autocorrelation time.

Random-walk Metropolis proposes a symmetric increment from the current state and accepts a proposal $y$ from $x$ with probability $\min \{1,\pi (y)/\pi (x)\}$.

Suppose $U_f$ agrees with the reversible identity-function oracle except at $x_0$, where its answer differs by a fixed one-bit string $a$. Composing $U_f$ with the known identity oracle and preparing the affected answer qubit in $|-⟩$ kicks back the phase $-1$ exactly on $|x_0⟩$, using one query to $U_f$.

For a homogeneous Poisson point process of intensity $\lambda$ in ${\mathbb{R}}^d$, mapping $x$ to $v_d|x{|}^d$, where $v_d$ is the unit-ball volume, gives a rate-$\lambda$ Poisson process on $[0,\infty )$. Mapping to $|x{|}^d$ instead gives rate $\lambda v_d$.

If $H$ is the rank-$p$ orthogonal projection onto a normal linear model's column space, thenAdding $2p{\sigma }^2$ gives $\Vert (I-H)\mu {\Vert }^2+(n+p){\sigma }^2$, so Mallows' $C_p$ is unbiased for independent-copy prediction error even when the projection model is misspecified.

If $Y,Y^{\ast }$ are independent $N(\mu ,{\sigma }^2{I}_n)$ vectors and $H$ is a rank-$p$ orthogonal projection, thenThe first term is squared approximation bias, while $p{\sigma }^2$ is fitted-model variance and $n{\sigma }^2$ is irreducible new-response noise.