For a nondegenerate convex minor spherical triangle, the spherical law of cosines relates two angular sides $b,c$, their included interior angle $A$, and the opposite angular side $a$:Relabelling the vertices gives the corresponding formulas for $b,B$ and $c,C$. As with the spherical law of sines, physical side lengths must first be divided by the sphere's radius. The spherical law of cosines determines $a\in (0,\pi )$ uniquely from $b,c,A$ when the data describe such a spherical triangle. It also recovers each interior angle from three sides, for exampleUsing the cosine and the range $0<A<\pi$ selects a unique angle, avoiding the supplementary-angle ambiguity of an inverse sine calculation.

To derive the spherical law of cosines, let $u,v,w$ be the vertex unit vectors at $A,B,C$, so $u\cdot v=\cos c$, $u\cdot w=\cos b$ and $v\cdot w=\cos a$. The outgoing unit tangent vectors at $u$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their dot product is both $\cos A$ and $(\cos a-\cos b\cos c)/(\sin b\sin c)$, proving the identity. This gives a geometric explanation: spherical trigonometry compares angles in the tangent plane using vectors in the surrounding Euclidean space.

For a worked example on the unit sphere, take the verticesAt $A$, the outgoing unit tangent vectors are $(1,0,0)$ and $(0,1,0)$, so the interior angle is a right angle. The vertex dot products give $b=c=\pi /3=60^{\circ }$. The spherical law of cosines then reduces to the right-angle identity $\cos a=\cos b\cos c$, givingTo find $B$, use the relabelled spherical law of cosines:The physical side lengths on a sphere of radius $R$ are ${\ell }_a\approx 1.318116R$ and ${\ell }_b={\ell }_c=\pi R/3\approx 1.047198R$. For example, on a toy sphere with $R=10$ metres they are approximately $13.18116$, $10.47198$ and $10.47198$ metres; the interior angles stay the same. Applying the planar Pythagorean theorem directly to the angular sides would instead give $\sqrt{60^2+60^2}\approx 84.852814^{\circ }$. That flat right triangle calculation does not give the great-circle distance between $B$ and $C$.

The familiar law of cosines reappears as a local approximation. With $a,b,c$ all small in radians, substitute the small-angle approximations $\sin x\approx x$ and $\cos x\approx 1-x^2/2$ into the spherical law of cosines and keep terms through second order:Multiplication by $R^2$ gives the planar law of cosines for the physical side lengths. For a fixed small surface triangle on a progressively larger sphere, the angular sides shrink and this approximation improves. Thus the curved and flat formulas fit together: the Pythagorean theorem is recovered locally when $A$ is a right angle.

For a nondegenerate convex minor spherical triangle, with angular sides $a,b,c$ opposite interior angles $A,B,C$, the spherical law of sines statesThe sines of sides refer to central angles, so a physical side length ${\ell }_a$ on a sphere of radius $R$ must enter as $\sin ({\ell }_a/R)$. All six angles lie in $(0,\pi )$, and the denominators are positive. The reciprocal form, $\sin a/\sin A=\sin b/\sin B=\sin c/\sin C$, is equivalent. This is Tatum's sine formula, equation 3.5.1, in Celestial Mechanics, section 3.5: spherical sine and cosine formulas.

A short proof explains the symmetry of the spherical law of sines. Let $u,v,w$ be the vertex unit vectors at $A,B,C$. At $u$, the outgoing unit tangent vectors towards $v,w$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their angle is $A$, so the magnitude of their cross product is $\sin A$. Since both tangent vectors are orthogonal to $u$, the scalar triple product obeysDividing by $\sin a\sin b\sin c$ proves the spherical law of sines. Its beauty is that the same three-dimensional scalar triple product can be calculated at any vertex, producing all three ratios at once.

For the spherical triangle worked out under the spherical law of cosines, $A=90^{\circ }$, $b=c=60^{\circ }$ and $\cos a=1/4$. Hence $\sin a=\sqrt{15}/4$, and the spherical law of sines givesThe inverse sine gives $B\approx 63.434949^{\circ }$, but the supplementary angle $116.565051^{\circ }$ has the same sine. The spherical law of sines alone therefore leaves two candidates for this angle. The spherical law of cosines, using the known three sides, gives $\cos B=1/\sqrt{5}>0$ and selects $63.434949^{\circ }$; likewise $C=B$. When only two sides and an opposite angle are given, two distinct valid spherical triangles can remain. A cosine relation resolves an inverse sine branch only when enough additional data are known; it does not make every ambiguous data set unique.

Reflections in the sides of a spherical triangle of angles $\pi /p,\pi /q,\pi /r$ generate a finite group when its images tessellate the sphere. The triangle area is $\pi (1/p+1/q+1/r-1)$ by the spherical excess formula. Label its three vertex types and continue those labels across each reflected side; they are consistent around every vertex and, because the sphere is simply connected, globally. The generated group preserves these labels, so a chamber stabilizer fixes all three vertices and is trivial. Dividing sphere area by chamber area gives its order. Up to permutation the possibilities are $(2,2,n)$ for $n\ge 2$, and $(2,3,3),(2,3,4),(2,3,5)$, of orders $4n,24,48,120$.

For well depth $U$ and radius $a$, let $u=r\psi$ for a regular spherically symmetric wavefunction. At energy $-U<E<0$, the interior is $u=C\sin (kr)$ and the exterior is proportional to $e^{-\kappa r}$, with $k^2=2m(U+E)/{\hslash }^2$ and ${\kappa }^2=-2mE/{\hslash }^2$. Matching gives $k\cot (ka)=-\kappa$. The first negative-energy solution exists strictly above the displayed threshold. At equality the exterior zero-energy wave behaves as $1/r$ and is not square-integrable, so the threshold is not itself a bound state.

These combinations of Spherical Bessel functions solve the radial Helmholtz equation. For time dependence $e^{-i\omega t}$, $h_n^{(1)}(kr)$ is outgoing and $h_n^{(2)}(kr)$ incoming. At fixed order and large argument, $h_n^{(1)}(t)\sim (-i{)}^{n+1}{e}^{it}/t$. Their recurrence is $h_{n+1}=(2n+1)h_n/t-h_{n-1}$.

For integer $\ell \ge 1$, put $I_{\ell }={\int }_0^{\infty }{j}_{\ell }(x{)}^{2}dx/x$ and $J_{\ell }={\int }_0^{\infty }{j}_{\ell }(x)j_{\ell -1}(x)dx$. Integrating $j_{\ell }^{\prime }=j_{\ell -1}-(\ell +1)j_{\ell }/x$ gives $J_{\ell }=(\ell +1)I_{\ell }$. For $\ell >1$, integrating $j_{\ell -1}^{\prime }=(\ell -1)j_{\ell -1}/x-j_{\ell }$ also gives $J_{\ell }=(\ell -1)I_{\ell -1}$. For $\ell =1$, the boundary values $j_0(0)=1$ and $j_0(\infty )=0$ give $J_1=1/2$, hence $I_1=1/4$. The recurrence proves the displayed identity. The integral without the square is different; for example ${\int }_0^{\infty }{j}_1(x)dx/x=\pi /4$.

A spherical shell is the region between two concentric spheres. A thin mass shell at radius $r$, thickness $dr$ and mass density $\rho (r)$ has mass $4\pi r^2\rho (r)dr$ to first order. Decomposing a spherical body into such shells makes the spherical shell theorem applicable to its gravitational potential energy.

For a statistically homogeneous, isotropic, reflection-symmetric solenoidal vector random field in three dimensions, the nonzero-wavevector spectral tensor is a scalar multiple of the transverse projector of a vector field. Isotropy gives $A(k){\delta }_{ij}+D(k)k_i{k}_j$, while zero divergence gives $A+D{k}^2=0$. Its trace is $2A$, giving the displayed normalization. Without reflection symmetry an additional antisymmetric, helical term can occur.

If an entire Banach algebra-valued function has bounded spectral radius, that spectral radius is constant. The spectral-radius three-circle inequality on circles about an arbitrary point, with the outer radius tending to infinity, bounds every interior value by the maximum on any smaller circle. Upper semicontinuity of the spectral radius then lets the smaller radius tend to zero. Every value is bounded by the value at the arbitrary center; exchanging centers gives equality. The algebra-valued function itself can be nonconstant, for example $zN$ for a fixed nonzero square-zero matrix $N$.

In a complex unital Banach algebra, the spectral radius formula makes the root norms of $(x/t{)}^n$ tend to $r(x)/t$. If $t>r(x)$, those norms are eventually bounded above by a geometric sequence and the powers vanish. Conversely, vanishing powers eventually have norm less than one; the infimum version of the radius formula then forces $r(x)/t<1$. Consequently $r(x)=\inf \{t>0:(x/t{)}^n\to 0\}$, including the case $r(x)=0$.

A spectral method has spectral accuracy for a class of smooth data when its error decreases faster than any fixed inverse power of the truncation size. For periodic Fourier series, this follows from arbitrarily many integrations by parts when every derivative matches across the periodic boundary. An analytic periodic extension to a complex strip gives exponential coefficient decay. Analyticity inside one period and matching only endpoint values do not suffice: a jump in a derivative of the periodic extension yields algebraic decay instead.

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.