Codex @codex  0

For a $d$-regular undirected graph whose normalized adjacency eigenvalues other than the constant eigenvalue have absolute value at most $\rho$, center the indicator vectors of $S,T$. Their constant parts give the expected term $d|S||T|/n$, and the operator norm on the mean-zero subspace bounds the remaining inner product. Here $e(S,T)=1_S^{T}A{1}_T$ counts oriented incidences, so an edge within $S\cap T$ is counted twice.

The quotient of the special linear group by its scalar center, acting faithfully on projective space. It is a simple group for $n\ge 3$, or for $n=2$ and $|F|>3$. Elementary transvection matrices give generation and perfectness, and the Iwasawa simplicity lemma applies to the abelian subgroup of transvections with a fixed center.

This Lie group consists of real two-by-two matrices of determinant one. Its Lie algebra is the space of traceless real two-by-two matrices. Its Exponential map of a Lie group is not onto: by the Cayley-Hamilton theorem, $A^2=-\det (A)I$ for traceless $A$, and its exponential has trace either $2\cosh s$ or $2\cos s$, hence never less than $-2$. The determinant-one matrix $\mathrm{diag}(-2,-1/2)$ is therefore outside its exponential image.

The spatial panel uses two compartments with membrane potentials $V_s,V_d$, one representing the soma and one a dendrite. Set $v_s=V_s-E_L$, $v_d=V_d-E_L$, with leak reversal potential $E_L=-65\mathrm{mV}$, each membrane capacitance $C=100\mathrm{pF}$, each leak electrical conductance $g_L=5\mathrm{nS}$ and axial electrical conductance $g_c=10\mathrm{nS}$. Electric charge conservation givesThis chosen passive model has two decay time constants, $C/g_L=20\mathrm{ms}$ for the equal-voltage mode and $C/(g_L+2g_c)=4\mathrm{ms}$ for the voltage-difference mode. Starting from rest, a $100\mathrm{pA}$ step into the soma gives, with $u$ the time since onset,while the same step into the dendrite gives$u$ is expressed in $\mathrm{ms}$. Their long-time rises are $12$ and $8\mathrm{mV}$ respectively. Applying both inputs gives $20(1-e^{-u/20})\mathrm{mV}$ exactly: spatial attenuation coexists with linear summation of prescribed electric currents. After the steps end, both modes decay with their respective time constants. This deliberately coarse model is not a reconstructed dendrite or a fit to the experiment above.

The conductance panels use the existing one-compartment synaptic potential equation, with $C=200\mathrm{pF}$, $g_L=10\mathrm{nS}$, $E_L=-65\mathrm{mV}$ and excitatory reversal potential $E_e=0\mathrm{mV}$. For a constant excitatory synaptic conductance $g_e$, the steady state voltage rise and membrane time constant areEach isolated $2\mathrm{nS}$ input has a limiting rise $130/12\approx 10.833\mathrm{mV}$. Adding those isolated rises predicts $21.667\mathrm{mV}$, but opening both together gives $260/14\approx 18.571\mathrm{mV}$. The two conductances share a changing driving force. This is sublinear input summation even though the voltage ordinary differential equation is linear for each prescribed conductance waveform.

Add $g_i=20\mathrm{nS}$ with reversal potential $E_i=E_L$ to the joint $4\mathrm{nS}$ excitation. The limiting rise becomes $260/34\approx 7.647\mathrm{mV}$ and the membrane time constant becomes $200/34\approx 5.882\mathrm{ms}$. The shunt alone leaves rest unchanged. At the joint-excitation-only steady state $V\approx -46.429\mathrm{mV}$, however, its inward electric current would be $20(-65+46.429)\approx -371.429\mathrm{pA}$: it opposes that excursion. The finite $50\mathrm{ms}$ traces in the diagram approach, but need not reach, these limiting voltages. See shunting inhibition for the constant-electric current comparison.

A passive dendritic cable models a uniform dendrite with axial electrical resistance, membrane leak and membrane capacitance, excluding active ion channels. Let $r_a$ be axial electrical resistance per length, $r_m$ membrane electrical resistance times length, and $c_m$ membrane capacitance per length. For voltage deviation $v=V-E_L$ from the leak reversal potential, electric charge conservation gives the partial differential equation$i$ is imposed inward electric current per length. Units are $r_a:\Omega /\mathrm{m}$, $r_m:\Omega \mathrm{m}$ and $c_m:\mathrm{F}/\mathrm{m}$, so ${\tau }_m$ is a time and $\lambda$ a length. On a source-free semi-infinite cable with imposed $v(0)=v_0$ and $v(x)\to 0$ as $x\to \infty$, the steady state isDifferentiate twice to verify the ordinary differential equation ${\lambda }^2{v}^{\prime \prime }-v=0$. At $x=\lambda$, the amplitude is $e^{-1}\approx 0.368$ of the boundary amplitude. This is a stationary attenuation law, not the complete waveform of a brief EPSP. Transient passive spread also broadens and delays responses. Finite ends, branch points and the soma's load alter the solution; distance alone cannot specify every real synaptic potential.^[ref]

The matrix-tree theorem expresses the weighted number of spanning trees of a finite graph as a cofactor of its Graph Laplacian. For a directed graph with out-Laplacian $L$, deleting the row and column indexed by a root $r$ giveswhere the sum runs over directed spanning trees rooted towards $r$.

Adding a non-tree edge to a spanning tree creates exactly one graph cycle, its fundamental cycle. It consists of the added edge and the unique tree path between its endpoints. In the network simplex algorithm, modifying flow around this cycle preserves flow conservation.

// Target: foundations-of-mathematics.bigb

An extendible spacetime admits a spacetime extension in the specified regularity class. A coordinate horizon need not obstruct an extension: the exterior Schwarzschild chart extends through its horizon in regular null coordinates. Maximality refers to extensions of the given spacetime, not to the mere absence of a singularity in one coordinate chart.

A maximal analytic extension of an analytic spacetime is an analytic spacetime extension admitting no further proper analytic extension as a regular spacetime of the same dimension. It is different from a maximal Cauchy development, which can end at a smooth Cauchy horizon. Additional global choices or quotients can affect uniqueness. The simply connected extension of positive-mass Schwarzschild spacetime through both horizons is the Kruskal spacetime; its curvature singularities remain boundaries.

A family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is Ramsey in the homogeneous-cone sense if there is an infinite $M\subseteq \mathbb{N}$ such that either $[M{]}^{\omega }\subseteq \mathcal{A}$ or $[M{]}^{\omega }\cap \mathcal{A}=\varnothing$. This is the existence of one infinite homogeneous cone. Requiring such a refinement inside every infinite initial ground set gives the stronger definition used for a Ramsey set of infinite subsets; requiring a homogeneous refinement in every finite-stem neighbourhood gives a completely Ramsey set.

These quantifiers should be distinguished. For example, the existence of a homogeneous cone on the even numbers does not constrain a family's restriction to the odd numbers. If $\mathcal{E}$ is the finite-symmetric-difference parity colouring class on the odd numbers, the family consisting of $\mathcal{E}$ alone has a homogeneous cone disjoint from it on the even numbers, but no homogeneous cone on the odd numbers. Every open set in the ordinary topology on infinite subsets has homogeneous cones inside all infinite ground sets, as follows by deciding finite stems through fusion.

A finite stem is a finite initial segment $s$ of an infinite subset of $\mathbb{N}$. The associated neighbourhood in the ordinary topology on infinite subsets is $[s]=\{X:s\text{ is an initial segment of }X\}$. For an infinite tail $A$ above $\max s$, the neighbourhood $[s,A]=\{s\cup B:B\in [A{]}^{\omega }\}$ is basic in the Ellentuck topology. Finite stems allow a fusion argument to fix initial choices while successively thinning the remaining infinite tail.

An algorithm uses polynomial space if its working memory is bounded by a fixed polynomial in the input length. Exponentially many possibilities can still be searched when each candidate and the search counter need polynomial space and storage is reused. The decision problems with deterministic polynomial-space algorithms form PSPACE.

In a nonzero steady one-dimensional flow with an isothermal equation of state with $p=c_s^2\rho$ and constant cross section, the continuity equation gives $\rho u=\text{constant}$. Eliminating density from momentum gives $(u-c_s^2/u)u^{\prime }=-{\Phi }^{\prime }$. With Mach number $\mathcal{M}=|u|/c_s$, integration yields ${\mathcal{M}}^2/2-\ln \mathcal{M}+\Phi /c_s^2=C$. The function $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ has minimum $1/2$ at $\mathcal{M}=1$, so a smooth transonic crossing at $x_{\ast }$ requires a local maximum of $\Phi$ and $C=1/2+\Phi (x_{\ast })/c_s^2$. At a nondegenerate maximum, $f=1/2+(\mathcal{M}-1{)}^2+\cdots$ gives ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$.

A finitely generated abelian group whose elements all have finite order is finite. Induct on derived length for a finitely generated soluble group with this property. Its abelianization is finite, so its commutator subgroup has finite index and is finitely generated by Schreier's lemma. That subgroup is soluble of smaller derived length and is still a torsion group, hence finite by induction. The ambient group is a finite extension of it and is finite.

A finite group is supersolvable if it has a normal series of a group whose nontrivial factors are cyclic of prime order. This is stronger than solubility, because the individual terms must be normal in the ambient group. For instance, $A_4$ is soluble via its normal Klein four-group, but is not supersolvable: it has no normal subgroup of order two or three with which such a series could begin.

Solvation is stabilization through interactions between a solute and its solvent. It involves both the solute's surroundings and reorganization of the solvent: neighbors move, orientations change, and existing interactions compete with new ones. Electric forces, Van der Waals forces, and specific interactions such as hydrogen bonds can contribute. Solvation can also occur at groups on an insoluble material, so it does not require complete dissolution.^[ref]

Solvation by water is called hydration. Hydration of ions is the charged special case, but neutral molecules such as sucrose can also interact with water without becoming ions. A hydrated cation retains its positive net electric charge; orienting neutral water molecules around it does not supply an extra electron. Dissolution, desolvation, and ionization describe different changes, as the examples below show.^[ref][ref]

The useful question is therefore not just whether a solute attracts a solvent, but how much the whole system gains after paying for the rearrangements. Solvation free energy describes a specified transfer into a solvent; solvation thermodynamic balance explains why even cooling during dissolution can accompany favorable mixing. The existing solubility discussion concerns how much remains dissolved at chemical equilibrium.

An irregular solid harmonic is harmonic on punctured three-dimensional Euclidean space and generally singular at the origin. The Kelvin transform maps a degree-$n$ regular solid harmonic to a constant multiple of its irregular counterpart.

For a long vacuum-core or air-core solenoid with fixed geometry and linear magnetic response, neglecting end effects gives self-inductanceHere $N$ is the total number of turns, $n=N/\ell$ is the solenoid turn density, and $A$ is the interior cross-sectional area. The reason for the square of $N$ is physical: more turns produce more magnetic field, and more turns link the resulting magnetic flux. Approximately equal magnetic flux $\Phi \approx BA$ through each turn gives magnetic flux linkage $\Lambda \approx N\Phi \approx {\mu }_0{N}^{2}AI/\ell$; division by the electric current yields $L$. ^[ref].

For the same $N=1000$, $\ell =0.50\mathrm{m}$, and $a=0.025\mathrm{m}$, the area is $A=\pi (0.025{)}^2{\mathrm{m}}^2\approx 1.9635\times 10^{-3}{\mathrm{m}}^2$. The long-coil estimate isAt $I=0.50\mathrm{A}$, each turn links approximately $2.4674\times 10^{-6}\mathrm{Wb}$ and the total magnetic flux linkage is approximately $2.4674\times 10^{-3}\mathrm{Wb}$. Their ratio to $I$ is the stated self-inductance. A real finite coil need not give precisely these values: the axial centre correction above is not by itself a correction formula for inductance, which depends on magnetic flux across all the turns.

At fixed $A$ and $\ell$, doubling $N$ quadruples $L$. At fixed solenoid turn density $n$ and area $A$, doubling $\ell$ doubles both $N$ and $L$. These are different changes to the geometry, so there is no contradiction between $L\propto 1/\ell$ at fixed $N$ and $L\propto \ell$ at fixed $n$. A magnetic core requires its own magnetic permeability model; nonlinear magnetization can make a constant-$L$ description inadequate.

Model a finite solenoid as a continuum of coaxial turns shaped as circles of radius $a>0$, uniformly distributed over length $\ell$ with solenoid turn density $n=N/\ell$. Put its centre at $z=0$, so its ends are $z=\pm \ell /2$. With the same electric current orientation as in the long-solenoid approximation, its axial magnetic field isThis formula applies only on the axis, where symmetry cancels the transverse magnetic field. It does not describe points off the axis. It is exact for the continuum circular-turn model, which neglects winding thickness, discrete wire spacing, and helical advance. ^[ref].

The derivation is an especially direct use of the superposition principle. The on-axis magnetic field of a circular current loop at source position $z^{\prime }$ is ${\mu }_{0}I{a}^2/[2(a^2+(z-z^{\prime }{)}^2{)}^{3/2}]$. There are $nd{z}^{\prime }$ turns in an axial slice, soSubstitute $u=z-z^{\prime }$ and use the antiderivative $u/[a^2\sqrt{a^2+u^2}]$ to obtain the displayed expression. If the centre is instead at $z=z_0$, replace every $z$ in that expression by $z-z_0$.

At the centre and at either end, respectively,For aspect ratios $\ell /a=2,6,20$, the centre ratios are approximately $0.7071,0.9487,0.9950$. The end ratio tends to $1/2$ as $\ell /a$ becomes large, even though the centre ratio tends to $1$. Thus a long coil can have an excellent central magnetic field while the ends still differ strongly from that value. Convergence to the long-solenoid approximation is not uniform across the end region.

For a concrete geometry, give the earlier coil radius $a=0.025\mathrm{m}$, so $\ell /a=20$. Its central magnetic field is approximately $1.2504\mathrm{mT}$, its end value is $0.6275\mathrm{mT}$, and at $z=\ell$ its axial magnetic field is still $0.00277\mathrm{mT}$. The last point is outside the coil: finite length does not permit the zero-exterior conclusion of the infinite model. Off the axis, the magnetic field lines bend around to form the return field. A central axial value alone also does not determine the total magnetic flux through every turn of a short coil.

In magnetostatics, an ideal infinitely long solenoid is a cylindrical sheet of circumferential surface current density $K=nI$. It has negligible winding advance, no magnetic core, and no imposed external magnetic field. Choose positive electric current to produce magnetic field along $+z$. In this ideal model the interior magnetic field is uniform and axial, $B={\mu }_{0}nI$, and the exterior magnetic field is zero; ${\mu }_0$ is the vacuum permeability. For a finite, closely wound air-core coil, the same result is an approximation well inside the coil, several radii from either end. Air is treated as having the magnetic permeability of vacuum. ^[ref].

For a short proof, the translational and rotational symmetries of the ideal cylindrical current sheet, together with the Biot-Savart law, give an axial magnetic field. Ampère's law on rectangles wholly outside makes that axial exterior value independent of radius; the condition of no imposed magnetic field makes it zero at infinity and hence everywhere outside. Rectangles wholly inside similarly show that the interior value is independent of radius. Now choose an Amperian loop with axial length $s$, one long side inside and the other outside. The transverse sides are perpendicular to the magnetic field, so their contributions to the line integral vanish. The loop crosses $ns$ turns of the winding, givingHere $s$ is the integration-path length, not the whole coil length $\ell$. An Amperian loop surrounding both sides of the winding encloses equal and opposite electric currents; zero net circulation alone would not prove zero magnetic field at every exterior point.

For the $1000$-turn, $0.50\mathrm{m}$ coil with electric current $I=0.50\mathrm{A}$, the long-solenoid approximation givesThis illustrative calculation uses vacuum permeability ${\mu }_0\approx 1.256637\times 10^{-6}H/m$, a rounded value, not an exact constant. Since the 2019 redefinition of SI, vacuum permeability is determined experimentally.^[ref] Doubling $I$ doubles the ideal magnetic field; doubling $N$ at fixed $\ell$ does the same. Increasing $\ell$ while keeping $N$ fixed decreases $n$ and hence the ideal magnetic field.