For a complex unital Banach algebra and a circle enclosing its element's Banach algebra spectrum, the displayed identity holds for $n\ge 0$. On a sufficiently large circle, expand the Neumann series and integrate it term by term. At any resolvent point the local series $R({\lambda }_0+h)={\sum }_{j\ge 0}(-h{)}^{j}R({\lambda }_0{)}^{j+1}$ proves norm analyticity. Apply continuous complex linear functionals and the scalar Cauchy theorem to deform the large circle to any radius greater than the maximum spectral modulus; the Hahn-Banach theorem separates points and restores the vector integral identity. Its norm bound $\Vert x^n\Vert \le R^{n+1}{\max }_{|\lambda |=R}\Vert R(\lambda )\Vert$ gives the reverse inequality in the spectral radius formula.

For a closed subalgebra of a Banach algebra $B\subseteq A$ with common identity, the algebra resolvent set ${\rho }_B(x)$ is clopen in ${\rho }_A(x)$. A point in ${\rho }_A(x)$ has a small connected disc on which membership of ${\rho }_B(x)$ is constant, so it cannot lie in $\partial {\sigma }_B(x)$. Also ${\sigma }_A(x)\subseteq {\sigma }_B(x)$; a point in the interior of the former is in the interior of the latter. Together these observations prove the boundary inclusion, without assuming a spectral-boundary theorem.

For solenoidal velocity and magnetic field, with constant magnetic diffusivity $\eta$, the resistive induction equation impliesThe product rule gives $D_{t}P=\mathbf{x}\cdot D_t\mathbf{B}+\mathbf{u}\cdot \mathbf{B}$ and $\Delta P=\mathbf{x}\cdot \Delta \mathbf{B}+2\nabla \cdot \mathbf{B}$. Consequently the radial velocity, through $Q$, is the source term for the radial magnetic field.

// Target: fluid-mechanics.bigb

For a planar magnetic field written as $\mathbf{B}=\nabla \times (A\hat{\mathbf{z}})$ and a circular velocity $\mathbf{u}=s\Omega (s)\hat{\mathbf{\varphi }}$, the resistive induction equation reduces, after choosing the additive gauge transformation, to ${\partial }_{t}A+\Omega {\partial }_{\varphi }A=\eta {\nabla }^{2}A$. A Fourier mode $A=a(s,t)e^{im[\varphi -\Omega (s)t]}$ therefore obeys the equation in the title. Indeed, each radial derivative acting on the exponential replaces ${\partial }_s$ by ${\partial }_s-imt{\Omega }^{\prime }$, while the azimuthal Laplacian contributes $-m^2/s^2$. At zero magnetic diffusivity, $a$ is constant in time, but the physical magnetic field can still be amplified algebraically by differential rotation.

A probe's sheath and leads carry heat between the sensing region and the surroundings through thermal conduction. In a stable, sufficiently uniform heat bath, compare the indicated temperature as insertion depth increases, allowing settling each time. A plateau supports adequate immersion for that apparatus and required uncertainty; it does not prove that every depth or medium works. Keep the sensing element and reference thermometer at comparable depths, and check bath gradients independently, so moving the probe does not merely sample a different temperature. The documented platinum resistance thermometer comparison calibration includes these controls.

Measuring electric current creates Joule heating, raising the sensing temperature. At fixed bath temperature, compare settled readings at $I$ and $\sqrt{2}I$ and extrapolate the approximately linear $I^2$ dependence toward zero current.^[ref]

Under the local model $R(I)=R_{\mathrm{zero}}+k{I}^2$,For synthetic readings $109.629\Omega$ at $1\mathrm{mA}$ and $109.633\Omega$ at $\sqrt{2}\mathrm{mA}$, the extrapolated value is $109.625\Omega$. The lower-current excess $0.004\Omega$ corresponds to $0.01039\dots {}^{\circ }\mathrm{C}$ in the Pt100 local linear model. At $109.625\Omega$, $1\mathrm{mA}$ dissipates $109.625\mu \mathrm{W}$; power alone does not specify a temperature rise without a thermal coupling model. These stipulated values are not measurements.

Repeat the comparison after settling and test the $I^2$ approximation if necessary. Changing from liquid to air changes heat transfer; a self-heating correction need not transfer unchanged. The saturation-pressure equilibrium measurement provides an existing example of controlling this disturbance in resistance thermometry.

A resistance thermometer using platinum as its sensing material. It provides the temperature readout in the saturation-pressure equilibrium measurement.

Transduction turns electrical resistance into a temperature indication. A uniform wire has $R={\rho }_{\mathrm{el}}\ell /A$, with electrical resistivity ${\rho }_{\mathrm{el}}$, length $\ell$, and cross-sectional area $A$. Stronger atomic vibrations at higher temperature scatter electrons more, increasing platinum's electrical resistivity; impurities and crystallographic defects also contribute.^[ref]

A small known electric current $I$ and measured electric potential difference $V$ give $R=V/I$ through Ohm's law; a resistance bridge can instead compare the element with a reference resistor. Inverting the calibrated response then gives the sensing element's temperature. This makes electrical readout convenient, but the platinum resistance thermometer lead resistance error, resistance thermometer self-heating, and resistance thermometer immersion test explain why a stable display alone is insufficient. The gas thermometer supplies a contrasting route through an equation of state; the saturation-pressure equilibrium measurement shows why a trustworthy independent temperature readout matters.

Keep the circular portion of a Hankel contour at fixed nonzero radius. Its numerator $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$ is an entire function of $z$, because the contour avoids zero and its tails decay exponentially, uniformly on compact parameter sets. At a positive integer $n$, the integrand is single-valued and entire in $t$, so the bank integrals cancel and the circular integral is zero. Thus $I(n)=0$, canceling the simple zero of $2i\sin (\pi z)$ in the Hankel representation of the gamma function. At a nonpositive integer $-n$, the circular integral instead gives $2\pi i/n!$, yielding residue $(-1{)}^n/n!$.

At fixed degree $n$ in positive characteristic $p$, two Specht modules belong to the same block of a group algebra exactly when their labels have the same core of a partition at modulus $p$. The simple labels in that block are precisely the regular partitions with that core. The theorem follows by combining the residue-content criterion for symmetric-group blocks with the Jucys–Murphy description of the center of a symmetric-group algebra.

For residual fraction $0<s<1$, differentiation of $V=Q\tau (e^{-st/\tau }-e^{-t/\tau })/(1-s)$ gives its unique positive maximum at the displayed time. The maximum volume is $Q\tau s^{s/(1-s)}$. When $s=0$ the mobile volume approaches $Q\tau$ monotonically, with no finite maximum. The limiting $s\to 1$ expression is $V=Qt{e}^{-t/\tau }$ with peak at $t=\tau$.

A residual synchronous time displacement ${\xi }^0=C(\mathbf{x})/a$ changes a fluid density contrast by $-{\bar{\rho }}^{\prime }{\xi }^0/\bar{\rho }$. For constant equation of state this gives the displayed mode: it scales as ${\tau }^{-2}$ during radiation domination and as ${\tau }^{-3}$ during matter domination. Such density changes alone do not indicate physical structure growth. Choosing the cold-matter rest frame fixes the residual time displacement, while a spatial relabeling fixes the remaining integration convention for the metric trace.

For an undirected capacitated network, write the net edge flow as $y_{ij}=-y_{ji}$, and the directed residual capacity as $c_{ij}-y_{ij}$. On every source–sink cut, residual capacity is the original cut capacity minus the current net flow value $f$. Taking the minimum over cuts and applying the max-flow min-cut theorem to each network gives the displayed remaining-flow identity.

A normed representation acts by individually bounded operators on a normed vector space. This condition alone does not assume that $a\mapsto \pi (a)$ is continuous in the operator norm, nor that the representation space is complete. A continuous representation has a common bound $\Vert \pi (a)\Vert \le C\Vert a\Vert$.

Algebraic irreducibility excludes every proper nonzero invariant vector subspace, including those that are not closed. It is stronger than topological irreducibility, which excludes only closed invariant subspaces. For a nonzero irreducible action, every nonzero vector is cyclic in the algebraic sense: $A\xi =X$.

A convex terminal payoff has a convex pricing extension in the binomial market, because risk-neutral pricing is a positive weighted sum of rescaled copies of the payoff. Let the stock factors be $u,d$, the riskless gross factor be $R$, and $q=(R-d)/(u-d)$. The replicating portfolio in a binomial market satisfiesThe weights sum to one. Convexity orders the secant slopes on the adjacent intervals $[d^{2}s,uds]$ and $[uds,u^{2}s]$, proving the displayed inequalities. The inequalities need not be strict: an affine payoff has constant stock holding.

For a finite commuting family in a unital Banach algebra, divide each $x_j$ by $r(x_j)+\epsilon /2$. The spectral radius formula makes each normalized element's powers bounded. Commutativity lets every product be reordered into one power of each generator, so the entire generated semigroup is bounded. Renorming a bounded multiplicative semigroup gives the displayed bounds simultaneously. Applying this to two commuting elements proves submultiplicativity and subadditivity of their spectral radius.

For $1<s\le \sqrt{2}$, set $h_s(x)=-(s-1)x$ and $J_s=[-(s-1),s-1]$. The tent map sends $J_s$ to $[1+s-s^2,1]$, with disjoint interiors, and $T_s^2(J_s)\subseteq J_s$. Directly,Successive squaring eventually brings every $s>1$ into $(\sqrt{2},2]$. A horseshoe for an interval map of an iterate in the renormalized coordinate lifts to a horseshoe of an iterate of the original map.

// Target: dynamical-systems.bigb

A standard thermal Cosmic neutrino background with one populated neutrino and one antineutrino helicity per species has $n_{\nu }+n_{\bar{\nu }}=3n_{\gamma }/11$ after electron-positron annihilation in cosmology. If this species is nonrelativistic today and dominates a spatially flat matter universe, equate $m_{\nu }(n_{\nu }+n_{\bar{\nu }})$ to the critical density $3H_0^2{M}_P^2/(8\pi )$, using the unreduced Planck mass. The photon number density gives the displayed conditional mass. Extra populated states or several massive species change the abundance count; vacuum-dominated cosmology changes the density available for matter.

Suppose $0<p_L\le p\le p_U<1$ and $0<a_L\le a\le a_U<1$. The formula for relative risk from group compositions increases with $a$ and decreases with $p$, since $a/(1-a)$ increases and $(1-p)/p$ decreases. This monotonicity proves the displayed bounds. If percentages are rounded to the nearest whole percentage point, each corresponding proportion lies within $0.005$ of its displayed value. These bounds describe rounding uncertainty; they are not a confidence interval for sampling uncertainty.

Compare expected utility of $W(1+\varphi )$, where $\mathbb{E}\varphi =0$, with the certainty equivalent $W(1-\psi )$. A second-order Taylor expansion gives $U(W)+W^2{U}^{\prime \prime }(W)\mathbb{E}{\varphi }^2/2$ for the risky side and $U(W)-W{U}^{\prime }(W)\psi$ for the sure side. Equality gives the displayed proportional risk premium. This requires a genuinely small-shock family with controlled remainders; small variance alone does not control rare large wealth shocks.