Consider the one-dimensional diffusion equation on the whole real line,with initial value one for $x<0$ and zero for $x>0$. This is an idealized Heaviside step function profile, not a measured material or an apparatus specification. The heat kernel gives a convolution with the initial profile:The change of variables $s=(x-y)/(2\sqrt{Dt})$ turns the convolution into a Gaussian integral tail. The resulting similarity variable $\eta =x/(2\sqrt{Dt})$ shows that the transition width grows like $\sqrt{Dt}$: multiplying elapsed time by four doubles every fixed-profile distance from the original interface.

At every positive time, $u(0,t)=1/2$, while $u(x,t)\to 1$ as $x\to -\infty$ and $u(x,t)\to 0$ as $x\to +\infty$. For every fixed $x\ne 0$, the limit as $t\to 0^{+}$ returns the specified initial value. The value assigned at the single initial discontinuity does not change this convolution. The error function therefore explains smoothing without pretending that a discontinuous initial profile remains sharp.

Direct differentiation checks the partial differential equation. Writing $\eta =x/(2\sqrt{Dt})$ givesThe model assumes constant diffusion coefficient, no advection, and no finite-domain boundary effects. For an interpretation as heat conduction, the heat equation applies to a normalized temperature difference under its own material assumptions.

Because the real error function is strictly increasing from $-1$ to $1$, its real inverse function maps $(-1,1)$ onto the real line. For $0<q<1$, the standard normal quantile is thereforeMore generally, the $q$-quantile of a normal distribution with expected value $\mu$ and positive standard deviation $\sigma$ is $\mu +\sigma \sqrt{2}{\mathrm{erf}}^{-1}(2q-1)$. This reverses the area calculation: instead of asking how much probability lies below an endpoint, a quantile asks which endpoint encloses a specified probability. The domain restriction matters; $q=0$ and $q=1$ correspond to infinite endpoints, rather than finite real values of the inverse error function.

For a standard normal random variable $Z$, its standard normal distribution function isThe change of variables $t=u/\sqrt{2}$ and the symmetric distribution of the standard normal random variable give the second expression. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable then givesIn particular, the intervals within one, two, and three standard deviations have calculated probabilities about $0.682689$, $0.954500$, and $0.997300$. These values depend on a normal distribution; they are not universal coverage guarantees for an arbitrary random variable. This is the direct connection between a Gaussian function area and the familiar central intervals.

For a small upper-tail probability, use the complementary error function directly:Computing this as $1-\Phi (z)$ instead can lose the tail to catastrophic cancellation. The Gaussian tail bound gives another useful estimate without requiring an exact error function value.

For distinct error locations and the syndrome convention $S_{\ell }={\sum }_i{E}_i{X}_i^{b+\ell }$, the displayed formula recovers the error magnitudes from the error evaluator polynomial and error locator polynomial. At $z=X_i^{-1}$, the evaluator is $E_i{X}_i^b{\prod }_{h\ne i}(1-X_h/X_i)$, whereas the locator's formal derivative is $-X_i{\prod }_{h\ne i}(1-X_h/X_i)$. Their quotient gives the formula; the denominator is nonzero because the locations are distinct.

// Target: polynomial.bigb

Scale the offered load and capacity together. Reversing the denominator of the Erlang loss formula givesFor each fixed $j$, the summand tends to $(C/\nu {)}^j$. If $\nu >C$, the dominated convergence theorem applies with the summable geometric series bound $(C/\nu {)}^j$. Otherwise each fixed finite partial sum has limiting value at least its number of terms, so the reciprocal diverges. Taking reciprocals proves the formula, including the critical case $\nu =C$.

For positive integer capacity $C$ and offered load $a>0$, let $m_C(a)$ be the carried load of an Erlang loss resource. Differentiating the upper-truncated Poisson distribution gives $m_C^{\prime }(a)=\mathrm{Var}(N)/a>0$. Differentiating its probability at $C$ gives the displayed derivative of the Erlang loss formula; it is positive because $m_C(a)<C$. Thus blocking increases continuously from zero to one, and the carried load increases from zero to $C$. These properties justify the inverse blocking coordinates in the convex potential for the Erlang fixed point.

For rational offered load, every step of the stable recursion for the Erlang loss formula is rational. Comparing the final numerator and denominator with rational thresholds therefore certifies signs without floating-point uncertainty. Opposite certified signs of a continuous fixed-point residual imply a root by the intermediate value theorem; several alternating intervals can prove multiple Erlang fixed points under alternative routing.

For irrational $\alpha$, the circle rotation $T(x)=x+\alpha (\mathrm{mod}1)$ preserves Lebesgue measure and is an ergodic transformation.

For nonzero coefficients of $A^{\ast }V$ in $\dot{B}$ and $AB$ in $\dot{V}$, a relative phase rotation makes $c_1$ real positive. Positive rescalings of $A,V,B$ then normalize both magnitudes to one. The phase of $c_1{c}_3$ is invariant under these rescalings and remains as the phase of normalized $c_3$. Zero couplings cannot be normalized to one; they are degenerate parameter cases.

For real $c_2,d_4$ and the simplified magnetic amplitude equations, the pure-velocity branch is $V=s{e}^{i{\omega }_{3}t}$ with $A=B=0$. In its rotating frame the magnetic linearization on $(B,\mathrm{Re}A,\mathrm{Im}A)$ has rows $({\mu }_1-e{s}^2,2s,0)$, $(c_{2}s,{\mu }_2,-\Delta )$ and $(0,\Delta ,{\mu }_2)$, where $\Delta ={\omega }_2-{\omega }_3$. Its zero determinant gives the displayed onset relation. A nonzero magnetic branch needs a kernel involving both magnetic amplitudes, a transverse parameter crossing and nondegenerate nonlinear saturation; determinant zero alone is insufficient in degenerate cases.

For complex velocity amplitudes $v_x=v_y=V$ and magnetic amplitudes $b_x=-b_y=B$ in Alfvén-speed units, phase averaging gives the displayed stress. For fixed horizontal velocity energy, equal radial and azimuthal components maximize their positive Reynolds stress. Opposite equal magnetic components maximize the outward tension contribution of the Maxwell stress tensor, since $-B_x{B}_y$ is then positive. These conclusions follow from $2|ab|\le |a{|}^2+|b{|}^2$. The vertical channel transports angular momentum outward and extracts energy from decreasing angular velocity.

For $f(x)=x^2$ at any real $a$, polynomial factorization givesIf $|x-a|<1$, the triangle inequality yields $|x+a|\le |x-a|+2|a|<2|a|+1$. Therefore the displayed choice of $\delta$ makes $|x-a|<\delta$ imply $|f(x)-f(a)|<\epsilon$. This proves continuity and gives an explicit local error bound, rather than an inference from a drawing.

At $a=1$ and $\epsilon =0.6$, this sufficient choice is $\delta =0.2$. The permitted function inputs satisfy $0.8<x<1.2$; their actual function outputs lie in $(0.64,1.44)$, inside the requested band $(0.4,1.6)$. A sufficient input window need not be the largest possible one.

Fix the second moment $v>0$. Put $B=\sqrt{5v}$ and $E(u)=3(B^2-u^2{)}_{+}/(4B^3)$. For every nonnegative unit-integral kernel for density estimation $K$ with the same second moment and finite $R(K)$, the mass and second-moment constraints give $\int (B^2-u^2)(K-E)=0$. Outside $[-B,B]$, $E=0$ and $(B^2-u^2)K\le 0$, so $\int E(K-E)\ge 0$. Expanding the square gives $R(K)-R(E)=\int (K-E{)}^2+2\int E(K-E)\ge 0$, with equality only when $K=E$ almost everywhere. Since $R(K{)}^2{\mu }_2(K)$ is scale invariant, rescaled Epanechnikov kernels minimize the optimized asymptotic mean integrated squared error over all such second moments. This conclusion concerns nonnegative second-order kernels; signed higher-order kernels obey a different bias expansion.

For a scalar conservation law law $u_t+f(u{)}_x=0$, the Kruzhkov entropy inequalities requirein distributions for every real constant $c$, together with the initial trace. This means the integral against every nonnegative compactly supported test function is nonpositive after applying the distributional derivatives. For a smooth solution the entropy balance is an equality away from $u=c$. For a viscous approximation $u_t+f(u{)}_x=\epsilon u_{xx}$, smooth convex approximations $\eta$ of $|u-c|$ give ${\partial }_t\eta (u)+{\partial }_{x}q(u)=\epsilon {\partial }_{xx}\eta (u)-\epsilon {\eta }^{\prime \prime }(u)u_x^2\le \epsilon {\partial }_{xx}\eta (u)$, with $q^{\prime }={\eta }^{\prime }{f}^{\prime }$. Letting the entropy smoothing vanish and then taking a bounded, strongly locally convergent zero-viscosity limit proves the inequality. The discrete entropy inequality of a monotone conservative scheme is its numerical analogue.

For a nonempty subset of a hypercube graph, split it into sections of sizes $a,b$. The proof of the entropy edge bound uses $H_2(t)\ge 2t$, $t=\min (a,b)/(a+b)$. Strict concavity makes equality possible only at $t=0$ or $t=1/2$. In the first case one section is empty; in the second the sections coincide, since all possible crossing edges must be present. Induction forces a coordinate subcube. Conversely every coordinate subcube has the displayed edge count. At size $2^{n-1}$ these are exactly the two halves determined by fixing one coordinate.

For independent positive random variables $Y_1,Y_2$ with densities, finite differential entropies and finite ${\mu }_i=\mathbb{E}{\log }_2{Y}_i$, applying the entropy power inequality to $Z_i=\ln Y_i$ givesIndeed $h(Z_i)=h(Y_i)-{\mu }_i$, while $h(Y_1{Y}_2)=h(Z_1+Z_2)+{\mu }_1+{\mu }_2$, provided the sum's differential entropy is defined. Positivity alone does not ensure these expressions exist: $Y=e^T$ with $T$ having Cauchy distribution has undefined $\mathbb{E}{\log }_{2}Y$.

// Target: algebra.bigb

For a discrete memoryless source of entropy $H$, every uniquely decodable binary code has expected length at least $H$. Shannon code lengths satisfy $H\le \overline{\ell }<H+1$, and block coding can make the expected length per source symbol arbitrarily close to $H$.

For a flux that is a differentiable convex function with unique minimum at $s_{\ast }$, define $f^{+}(u)=f(u)-f(s_{\ast })$ for $u>s_{\ast }$ and zero otherwise, and $f^{-}(u)=f(u)-f(s_{\ast })$ for $u<s_{\ast }$ and zero otherwise. The Engquist-Osher flux is $F(a,b)=f(s_{\ast })+f^{+}(a)+f^{-}(b)$. Thus transsonic rarefaction wave states $a\le s_{\ast }\le b$ give $F=f(s_{\ast })$, while reversed transsonic states give $F=f(a)+f(b)-f(s_{\ast })$. This follows by integrating the positive and negative parts of $f^{\prime }$.

The endomorphism bundle connection is ${\tilde{D}}_X\alpha =D_X\circ \alpha -\alpha \circ D_X$ on sections. Expanding twice makes the two middle terms cancel, proving the displayed commutator identity. Subtracting the derivative along the Lie bracket of vector fields gives $\tilde{R}(X,Y)\alpha =[R(X,Y),\alpha ]$. The operators $D_X$ are differential operators, rather than fibrewise endomorphisms; the corrected curvature operator is fibrewise linear.

For fixed $x_1,\dots ,x_d$, the vector $({\alpha }_n(x_j){)}_{j=1}^d$ converges in distribution to $(B(G(x_j)){)}_{j=1}^d$, where $B$ is a standard Brownian bridge. To prove it, write the vector as $n^{-1/2}$ times the sum of independent centred bounded indicator vectors. Their covariance matrix is $G(\min (x_i,x_j))-G(x_i)G(x_j)$. The multivariate central limit theorem gives a centred multivariate normal distribution with this covariance, exactly the bridge vector. Singular covariance matrices, including endpoint times or repeated values of $G$, cause no problem. This is a finite-dimensional assertion; a full function-space limit additionally requires tightness.