For a nonnegative integer with decimal expansion $a={\sum }_i{a}_{i}10^i$, the congruence $10\equiv 1(\mathrm{mod}d)$ gives $a\equiv {\sum }_i{a}_i(\mathrm{mod}d)$ whenever $d\mid 9$. Thus divisibility by three or nine is equivalent to divisibility of the sum of its decimal digits by the same number. More generally a base-$b$ expansion has the analogous rule for every divisor of $b-1$. This follows by repeated addition and multiplication in modular arithmetic; it does not require an empirical rule about decimal notation.

Let $y$ be piecewise twice continuously differentiable, with finite one-sided values and derivatives at $c$, and write $[y{]}_c=y(c+)-y(c-)$. Its distributional derivative is $Dy=y_{\mathrm{reg}}^{\prime }+[y{]}_c{\delta }_c$, and differentiating again givesHere ${\delta }_c$ is the Dirac delta at $c$, and the subscript reg means the ordinary derivative on the two open intervals, interpreted as a regular distribution. To prove the first formula, split $-\int y{\phi }^{\prime }$ at $c$ and use integration by parts on each piece for a test function $\phi$. The two boundary contributions add to $[y{]}_c\phi (c)$. Applying the same argument to the regular first derivative proves the second formula. Thus a derivative jump produces a delta source, whereas a value jump produces the derivative of a delta source.

For a function smooth on either side of a joining point $a$, integration by parts on the two intervals adds a Dirac mass whose coefficient is the right trace minus the left trace. If the one-sided derivatives are locally integrable, a first weak derivative can be represented by a locally integrable function across the join only when the jump vanishes. Derivative jumps themselves do not obstruct first-order Sobolev membership: a continuous piecewise Lipschitz function still has a bounded weak first derivative.

The dislocation core is the small region near a dislocation line where the smooth linear-elastic continuum description fails. For a screw dislocation, the far-field displacement gradient scales as the Burgers-vector magnitude divided by radial distance, so the predicted elastic stress diverges on the axis. Atomic structure, nonlinear response or local plastic deformation replaces that singular prediction in the core.

Serve requests in arrival order. This preserves queue order but can cause repeated long seeks when successive requests lie far apart.

Serve the pending request whose track is nearest to the current head position, with an explicit tie rule when necessary. This minimizes the next seek distance, rather than the total distance of every future request; a continuing stream of nearby requests can starve a distant request.

Sweep in one direction, servicing requests encountered in track order, continue to the physical endpoint, and then reverse. A calculation must state the initial direction and whether movement after the final requested service is included. LOOK disk scheduling instead turns around at the last pending request in that direction.

Sweep in one direction through pending requests, reversing at the last pending track rather than travelling to the disk's physical endpoint. It shares the directional sweep of SCAN disk scheduling but can avoid movement into an empty end region.

Total absolute track displacement along a head route $t_0,t_1,\dots$. Include physical turnaround points in the route if the scheduling policy visits them, even if they do not service a request.

A hydrograph records discharge against time at a fixed location. A nonlinear kinematic wave can steepen an upstream increase into a sharp front; longitudinal diffusion or additional storage rounds the resulting rise.

absolute convergence of a Dirichlet series depends only on the real part of its variable. The set of real parts giving convergence is upward closed by the comparison test. If that set is nonempty and not all of the real line, its finite infimum is the abscissa: every larger real part gives convergence, and every smaller real part gives divergence. Boundary convergence is a separate question.

Let $S(x)={\sum }_{n\le x}{a}_n$ satisfy $|S(x)|\le C{x}^{\alpha }$ for $x\ge 1$. The Abel summation formula gives $A(s)=s{\int }_1^{\infty }S(t)t^{-s-1}dt$ when $\mathrm{Re}s>\alpha$, since its finite-sum boundary term tends to zero. Split this integral into blocks $I_m(s)={\int }_m^{m+1}S(t)t^{-s-1}dt$. For every $\delta >0$, the estimate $|I_m(s)|\le C{m}^{-1-\delta }$ holds uniformly on $\mathrm{Re}s>\alpha +\delta$. Each block is an entire function, by differentiation under the finite-range integral on compact sets. The Weierstrass M-test makes their sum locally uniformly convergent and therefore holomorphic; multiplication by $s$ preserves holomorphy.

This proof distinguishes uniform convergence of the block sum on a whole half-plane from local uniform convergence after multiplication by $s$. The latter is all that holomorphy needs. In particular, a bound ${\sum }_{n\le x}\mu (n)=O(x^{\alpha })$ gives a holomorphic continuation of the reciprocal Riemann zeta function to this half-plane and excludes zeta zeros there, with the pole at one treated by cancellation.

Let $V$ be a closed linear subspace of a real Hilbert space, and let $Q(v)=a(v,v)$ for a symmetric bounded bilinear form that is nonnegative on the ambient space and coercive on $V$. Then $Q$ has a unique minimizer on every closed affine subspace $g+V$. If $Q(v_j)\to m={\inf }_{g+V}Q$, the midpoint belongs to $g+V$ and the parallelogram law gives $Q(v_j-v_k)\le 2Q(v_j)+2Q(v_k)-4m\to 0$. Coercivity on $V$ makes the minimizing sequence Cauchy; completeness and continuity give a minimizer. The vanishing first variation is $a(u,v)=0$ for all $v\in V$. Two minimizers differ by $w\in V$ with $Q(w)=0$, so coercivity gives uniqueness. This proves the symmetric existence theorem directly, without invoking the Lax-Milgram theorem.

For the point interaction in a second-order differential equation $L_{\alpha }={\partial }_x^2+\alpha {\delta }_c$ on $(0,\ell )$, use homogeneous Dirichlet boundary conditions. The unperturbed Green function iswith a positive unit derivative jump at $x=\xi$. Whenever $1+\alpha G_0(c,c)\ne 0$,To derive this rank-one operator correction, move the delta coefficient to the right-hand side and apply the unperturbed inverse: $G_{\alpha }(x,\xi )=G_0(x,\xi )-\alpha G_0(x,c)G_{\alpha }(c,\xi )$. Setting $x=c$ determines $G_{\alpha }(c,\xi )$ and gives the formula. Its denominator vanishes exactly at Dirichlet resonance for a point interaction.

The kernel is symmetric. It is continuous at $x=c$ and $x=\xi$, satisfies $[{\partial }_x{G}_{\alpha }{]}_{\xi }=1$ when $\xi \ne c$, and satisfies $[{\partial }_x{G}_{\alpha }{]}_c=-\alpha G_{\alpha }(c,\xi )$ away from the source. When $\xi =c$, the single combined condition is $[{\partial }_x{G}_{\alpha }{]}_c=1-\alpha G_{\alpha }(c,c)$. These are the transmission conditions for a delta coefficient, including a source at the interaction.

For $0<x<1$, consider the self-adjoint differential operator $L=(pd/dx{)}^{\prime }+q$ with $p=e^{-x}/(1-x)$ and $q=-e^{-x}/(1-x{)}^2$. The homogeneous solutions $u=x$ and $v=e^x-ex$ satisfy the left and right zero Dirichlet boundary conditions. Their Wronskian is $(x-1)e^x$, and $pW=-1$. Thus the displayed Dirichlet Green function is continuous, zero at both endpoints, and has unit flux jump $p(s)[G_x(s^{+},s)-G_x(s^{-},s)]=1$.

The right solution has a double zero, $v(s)\sim e(1-s{)}^2/2$. Consequently $G(x,s)F(s)$ is integrable at $s=1$ for fixed interior $x$ when $F(s)=-e^{-s}/(1-s{)}^2$, even though $F$ itself is not integrable there. The endpoint zero of the Wronskian causes no loss of this construction because the weighted Wronskian $pW$ stays nonzero. This illustrates the need to retain the integrating factor when constructing a Green function at a singular endpoint.

For $L=-d^2/d{x}^2+1$ on $[-1,1]$ with zero endpoint values, the displayed Green kernel is positive and symmetric. Each side solves the homogeneous equation; the kernel is continuous at $x=s$ and its first derivative has jump $-1$. Hence $LG(\cdot ,s)={\delta }_s$. Integrating against a forcing function solves $Ly=f$. For the indicator of the right half-interval, the value at zero is $(\cosh 1-1)/(2\cosh 1)$, obtained by integrating $\sinh 1\sinh (1-s)/\sinh 2$ over $0\le s\le 1$. The derivative-jump sign changes if the operator is written with the opposite overall sign.

If a group has carrier $A$ and a disjoint set $B$ is in bijection with $A$, identify $A\cup B$ with $A\times C_2$, sending the original carrier to $A\times \{0\}$. Transporting the direct-product multiplication gives a larger group operation that agrees exactly with the old one on $A$. A maximal partial group on a fixed carrier universe therefore has complement of cardinality strictly below its own, using cardinal comparability.

For a scalar fixed-endpoint problem on a bounded interval, suppose the continuous integrand $f(x,y,p)$ is convex in $p$ and satisfies $f(x,y,p)\ge \Theta (|p|)-C$, where $\Theta$ is nonnegative convex and $\Theta (r)/r\to \infty$. If there is one admissible absolutely continuous function of finite energy, a global minimizer exists in $W^{1,1}$.

The lower bound gives uniform integrability of derivatives in a minimizing sequence. Fixed endpoint values then give boundedness and equicontinuity of the functions. Extract uniform convergence of the functions and weak $L^1$ convergence of their derivatives; the limit retains the endpoint values. Convexity in the derivative gives weak lower semicontinuity of convex gradient energies, so the limit attains the infimum. Strict convexity of the full integrand on a convex admissible class additionally gives uniqueness.

For $I_{\epsilon }(u)={\int }_0^1[(u^4-x{)}^2+\epsilon |u^{\prime }{|}^{4/3}]dx$ with endpoint values zero and one, the zero-penalty minimizer $x^{1/4}$ has infinite derivative penalty. Use $u=x^{1/4+\delta }$ instead. The mismatch is at most $(32/27){\delta }^2$, while the derivative penalty is $(3\epsilon /(4\delta ))(1/4+\delta {)}^{4/3}$. Balancing ${\delta }^2$ and $\epsilon /\delta$ by $\delta ={\epsilon }^{1/3}$ gives an upper energy bound of order ${\epsilon }^{2/3}$. Nonnegativity then proves convergence of minimum energies to zero. This is a recovery construction; no claim of a matching lower asymptotic constant is needed.

A Lavrentiev gap is a strict change in an infimum when the admissible regularity class is restricted, despite retaining the same endpoint values. For example $I(u)={\int }_0^1(u^3-x{)}^2{u}^{\prime 6}dx$ with $u(0)=0,u(1)=1$ has value zero at the absolutely continuous function $u=x^{1/3}$. Every Lipschitz competitor has a uniform positive lower energy bound. Indeed its ratio to $x^{1/3}$ starts at zero, so let $x_0$ be its first crossing of $u(x)=x^{1/3}/2$. Before that point, $|u^3-x|\ge 7x/8$. Holder inequality givesThus $I(u)\ge (49/4096)(3/5{)}^5/x_0\ge (49/4096)(3/5{)}^5>0$. This proves a gap directly. The example's integrand lacks a uniform superlinear lower bound along its zero set; it illustrates the admissible-class issue rather than by itself furnishing a counterexample under stronger coercivity assumptions.