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Let $X,Y$ be independent fair bits, and let $Z=X\oplus Y$ be their exclusive or: $Z=1$ precisely when $X\ne Y$. The four possible triples are $000,011,101,110$, each with probability $1/4$. Each coordinate is a fair bit. Writing out all eight patterns givesEvery pair has all four cells $00,01,10,11$ with probability $1/4$:For $(X,Z)$, the four allowed triples project to $00,01,11,10$; for $(Y,Z)$ they project to $00,11,01,10$. Each projection lists each pair exactly once, proving pairwise independence. But $\mathbb{P}(X=0,Y=0,Z=0)=1/4$ differs from the product $1/8$, so the three are not independent random variables. Indeed, any two determine the third. A single coordinate reveals nothing about another, while two coordinates together reveal it completely.

The exclusive or example is the complementary-bit version of the two-toss exercise in Grinstead and Snell, Introduction to Probability, section 4.1, exercise 7, printed page 151. It is a compact reason to check the full product rule whenever a calculation uses three-way or higher events.

A Y-piece is a compact curvature-$-1$ hyperbolic surface homeomorphic to a pair of pants, with three geodesic boundary curves. Its labelled boundary lengths ${\ell }_1,{\ell }_2,{\ell }_3>0$ determine it up to isometry. Cutting along the three mutually perpendicular seams produces two congruent right-angled hyperbolic hexagons with alternating sides ${\ell }_i/2$. Conversely, doubling the hexagon along its other three sides constructs the Y-piece for arbitrary positive lengths.

For a steady accretion disk with arbitrary rotation law and a zero-torque inner boundary condition at $R=3R_S$, the outward viscous torque in an accretion disk is $\mathcal{G}=\dot{M}(j-j_{\mathrm{in}})$. The dissipation per unit planar area summed over both disk faces is $D=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)=\nu \Sigma (R{\Omega }^{\prime }{)}^2$. Integrating $2\pi RD$ gives $L=\dot{M}GM/(8R_S)=\dot{M}{c}^2/16$ in the Paczyński-Wiita potential. The radiative flux from one face is $D/2$, preventing a factor-of-two error in the luminosity.

The limiting inward departure from the marginal Paczyński-Wiita circular orbit at $r=3a$ conserves the displayed specific angular momentum and specific orbital energy. In the Paczyński-Wiita potential, its inward speed satisfies $v_r^2=GM(3a-r{)}^3/[4a{r}^2(r-a)]$. The exactly marginal circular orbit does not depart without a perturbation; the formula describes the separatrix limit of inward releases.

For a stable isotropic solid in $y<0$, use $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$. An incident P wave with frequency $\omega$, tangential wavenumber $\kappa$ and upward normal wavenumber $a$ reflects into a P potential $A{e}^{i(\kappa x-ay-\omega t)}$ and an SV potential $B{e}^{i(\kappa x-by-\omega t)}$, where $b^2={\omega }^2/c_s^2-{\kappa }^2$. Vanishing of both displacements at the rigid plane gives $\kappa (1+A)-bB=0$, $a(1-A)-\kappa B=0$. Hence $A=(ab-{\kappa }^2)/(ab+{\kappa }^2)$ and $B=2a\kappa /(ab+{\kappa }^2)$. Because $c_p>c_s$ and ${\kappa }^2\le {\omega }^2/c_p^2$, $b^2>0$ and the converted S wave propagates rather than being evanescent.

Let $t=T(\mathbf{x})$ be a smooth wavefront and suppose the first jump in time derivatives of displacement field is $[{\partial }_t^k\mathbf{u}]=a^{[k]}\mathbf{n}$, with $k\ge 2$. Substitution of $\mathbf{u}=a^{[k]}\mathbf{n}(t-T{)}_{+}^k/k!+\cdots$ into the elastic equation first gives the eikonal equation $|\nabla T|=1/\alpha$ and longitudinal polarization $\mathbf{n}=\alpha \nabla T$. Projecting the next-order compatibility condition onto $\mathbf{n}$ gives $2\rho \alpha \mathbf{n}\cdot \nabla a^{[k]}+a^{[k]}\nabla \cdot (\rho \alpha \mathbf{n})=0$, hence the displayed conservation law. Coefficients and wavefront must be sufficiently differentiable for this classical transport equation; its local form fails at a caustic.

A p-subgroup is p-radical if it equals the p-core of its normalizer. Every defect group of a block is p-radical by defect groups are centralizer-conjugate Sylow intersections. Being an intersection of arbitrary Sylow subgroups alone is weaker than the centralizer-conjugate statement used in that argument.

The integer $h=|G:G^{\prime }{|}_p$ divides $s={\sum }_{\chi \in {\mathrm{Irr}}_{p^{\prime }}(G)}\chi (1{)}^2$. The group is p-nilpotent exactly when $p\nmid s/h$. To see this, let $N=O^p(G)$. Invariant p-prime-degree characters of $N$ extend, and each has exactly $h$ extensions of p-prime degree. Conjugation by the p-group $G/N$ then gives $s/h\equiv |N|(\mathrm{mod}p)$. Thus the quotient is prime to $p$ precisely when $N$ is a normal p-complement.

If a prime $p$ divides the degree of every nonlinear irreducible character of a finite group, the group has a normal p-complement. Indeed, the p-prime-degree characters are exactly the linear characters. Their squared-degree sum is $|G:G^{\prime }|$, whose p-part is $|G:G^{\prime }{|}_p$; Isaacs' character-degree criterion for p-nilpotency applies.

Put $\overline{G}=G/O_{p^{\prime }}(G)$ using the p-prime core of a finite group. A finite group is p-constrained if $C_{\overline{G}}(O_p(\overline{G}))\le O_p(\overline{G})$. If $O_{p^{\prime }}(G)=1$, the p-core of a finite group has its centralizer contained in itself. Then block idempotents centralize a normal p-subgroup places every block idempotent in $k{O}_p(G)$, a local ring, so $kG$ has only the principal block.

On a profinite group, convolution is first defined for functions on a finite quotient by summing products of the two measures' coset coefficients, then extended by uniform approximation. Finite-quotient associativity proves associativity. For an abelian group it is commutative; the Dirac measure at the identity is the identity element. Under the measure realization of an Iwasawa algebra, it becomes group-algebra multiplication.

The coefficient of $T^n$ is $\int \left(\genfrac{}{}{0ex}{}{x}{n}\right)d\mu (x)$. The Mahler expansion of continuous p-adic functions identifies $I$-valued p-adic measures on ${\mathbb{Z}}_p$ with $I[[T]]$. If $D=(1+T)d/dT$, then $D^k{\mathcal{A}}_{\mu }(0)=\int x^{k}d\mu (x)$. A measure supported on ${\mathbb{Z}}_p^{\times }$ has ${\sum }_{{\zeta }^p=1}{\mathcal{A}}_{\mu }(\zeta (1+T)-1)=0$, because the sum of ${\zeta }^x$ is zero for $p\nmid x$.

Let $I$ be a complete discrete valuation ring with a continuous ${\mathbb{Z}}_p$-algebra structure. Every continuous function on ${\mathbb{Z}}_p$ with values in $I$ has a unique uniformly convergent expansion in the binomial polynomials, with $a_n={\Delta }^{n}f(0)$. This is the coefficient-ring version of the Mahler theorem. To see that $a_n\to 0$, approximate the function modulo any power of the maximal ideal by a function of period $p^k$. In the expansion of $(E-1{)}^{p^k}$, all intermediate binomial coefficients are divisible by $p$, while $(E^{p^k}-1)$ kills the periodic function. Iterating makes its differences vanish modulo the prescribed ideal power. Finite differences do not increase the uniform norm, so the same conclusion holds for the original function. Binomial inversion gives agreement at the nonnegative integers, and their density in the p-adic integers gives the expansion everywhere. This gives the dual description of p-adic measures by arbitrary bounded sequences of binomial moments, and hence the Amice transform.

Write ${\mu }_r=\left(\genfrac{}{}{0ex}{}{n}{r}\right)p^{\left(\genfrac{}{}{0ex}{}{r}{2}\right)}$ and $T_s=\left(\genfrac{}{}{0ex}{}{r}{s}\right)\left(\genfrac{}{}{0ex}{}{n-r}{r-s}\right)p^{-\left(\genfrac{}{}{0ex}{}{s}{2}\right)}/\left(\genfrac{}{}{0ex}{}{n}{r}\right)$. The first mean condition implies $p^{-1}\le (en/r{)}^{2/(r-1)}$ eventually. For $2\le s\le r/2$, this yields $T_s\le (C/s{)}^s$, and each fixed-$s$ term tends to zero when $n\ge r^3$. The second mean condition implies $p\le ((r+1)/n{)}^{2/r}$ eventually. Writing $s=r-j$ gives $T_{r-j}={\mu }_r^{-1}\left(\genfrac{}{}{0ex}{}{r}{j}\right)\left(\genfrac{}{}{0ex}{}{n-r}{j}\right)p^{j(2r-j-1)/2}\le {\mu }_r^{-1}(C/j{)}^j$ for $1\le j\le r/2$, and $T_r={\mu }_r^{-1}$. Both bounding series are summable. The overlap formula for the variance of a clique count therefore gives relative variance tending to zero; the second moment method proves an $r$-clique exists with high probability.

Underfitting occurs when a prediction rule fails to capture useful structure in the target relation. A straight-line polynomial regression fitted to a genuinely quadratic noiseless target illustrates a representational limitation. Poor prediction alone does not identify the cause: compare candidate rules using cross-validation or a validation set, while keeping the final test set reserved for assessment.

The structure containing the angiosperm female gametophyte; a fertilized ovule can develop into a seed.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value at most one. Some prime $p$ must have value less than one, or the value would be trivial on every rational number. Two distinct such primes are impossible by a Bezout identity. Prime factorization then gives the displayed formula. Distinct primes give inequivalent topologies, since powers of $p$ tend to zero only for the $p$-adic absolute value among these values.

Take two reservoirs separated by a semipermeable membrane that passes water and retains one neutral solute. Stipulate $T=300\mathrm{K}$, $c_A=10\mathrm{mol}{\mathrm{m}}^{-3}$ and $c_B=30\mathrm{mol}{\mathrm{m}}^{-3}$, equivalent to $10$ and $30\mathrm{mmol}{\mathrm{L}}^{-1}$. Treat both chemical solutions as dilute and ideal. These are invented model conditions, not measurements or recommended biological conditions.

For pressures measured at the same height beside the membrane, define $\Delta P=P_B-P_A$. The model predictsAt $\Delta P=0$, net water transfer favors A to B. At $\Delta P\simeq 49.89\mathrm{kPa}$, the opposing pressure balances the osmotic pressure difference, giving zero net transfer with unequal concentrations. At $\Delta P=75\mathrm{kPa}$, the sign reverses and net water transfer favors B to A, toward the more dilute reservoir. Raising pressure changes the total water potential, rather than changing the membrane into a one-way gate.

Fixed concentrations require large maintained reservoirs or an instantaneous comparison. In a closed membrane osmometer, entering water causes dilution of the receiving chemical solution and changes its volume; the final balance must use final concentrations. A supported cell wall provides another way to develop the opposing pressure, as explained under turgor pressure.

A circular distant barycentric Kepler orbit around a circular star–planet binary need not have constant star-centred osculating orbital elements. Let $\mu$ be the planet's fraction of the binary mass, $\alpha =a_{\mathrm{pl}}/a\ll 1$, and $\epsilon =\mu /\sqrt{\alpha }\ll 1$. Subtracting the stellar reflex velocity produces radial and tangential perturbations $(-v_0\epsilon \sin \Delta ,v_0\epsilon \cos \Delta )$, where $\Delta$ is the planet's longitude minus the particle's longitude. The vis-viva equation gives the displayed semi-major axis variation. The eccentricity vector components follow from $e_r=r{v}_{\theta }^2/(G{M}_{\star })-1$ and $e_{\theta }=-r{v}_r{v}_{\theta }/(G{M}_{\star })$. The errors are $O(\mu +{\epsilon }^2)$; the spatial translation itself is only $O(\mu \alpha )$. These are changes of instantaneous orbital description, even when the assumed barycentric trajectory is fixed.

Tensor products of the assigned vectors represent strong graph powers, and tensor products of the handle have squared inner products equal to products of the original ones. The vectors assigned to an independent set are orthonormal. Bessel inequality then bounds its size by the largest inverse squared handle inner product. Taking roots and minimizing proves $c(G)\le \vartheta (G)$.