For a plate occupying $x>0$, the boundary part of the first variation of its isotropic thin-plate bending energy is, up to an overall nonzero sign,At a free edge the variation and its normal derivative can be prescribed independently. Their coefficients must therefore vanish, giving the two displayed natural boundary conditions: zero bending moment and zero effective transverse shear. The mixed-derivative coefficient includes the derivative along the edge of the twisting moment; keeping only the normal derivative of $w_{xx}+\nu w_{yy}$ would miss that contribution.

Suppose the rational canonical form of $T$ is $C(f_1)\oplus \cdots \oplus C(f_r)$, with monic nonconstant invariant factors of a linear operator satisfying $f_1\mid \cdots \mid f_r$. Each block has minimal polynomial and characteristic polynomial $f_j$, because a cyclic operator has equal minimal and characteristic polynomials. The determinant of a block diagonal matrix is the product of block determinants, giving $c_T={\prod }_j{f}_j$. A polynomial kills the direct sum exactly when it kills each block, equivalently when every $f_j$ divides it. Thus $m_T$ is their least common multiple, which the divisibility chain reduces to $f_r$. On the zero vector space both polynomials are conventionally one.

A Newton-Cotes quadrature is an interpolatory quadrature rule whose quadrature nodes are equally spaced. Integrating the Lagrange interpolation polynomial gives quadrature weights $W_j={\int }_a^b{\ell }_j(x)dx$. A Newton-Cotes closed quadrature includes both endpoints; an open rule uses interior quadrature nodes. Simpson's rule is the closed rule with three quadrature nodes. Its quadrature weights are $(b-a)(1,4,1)/6$.

A compatible transfinite family of bijections can enumerate the square of an ordinal by adjoining a new row and then the remaining new column. Starting with $g_0=\varnothing$, extend at a successor by $g_{\alpha +1}(\alpha ,\beta )={\gamma }_{\alpha }+\beta$ for $\beta \le \alpha$ and $g_{\alpha +1}(\beta ,\alpha )={\gamma }_{\alpha }+(\alpha +1)+\beta$ for $\beta <\alpha$. Thus ${\gamma }_{\alpha +1}={\gamma }_{\alpha }+(\alpha +1)+\alpha$. At limits take compatible unions. The two new domain pieces are disjoint; the column is empty at the initial step $\alpha =0$. Each step enumerates consecutive ordinal intervals, proving bijectivity by transfinite induction. Finite stages have ${\gamma }_n=n^2$. Successor and limit cardinality arguments then give the square equality for well-orderable infinite cardinals without using that equality to justify the construction.

For two equal-mass particles with orthonormal coordinates $y_{\pm }=(x_1\pm x_2)/\sqrt{2}$ confined to $|y_{\pm }|<a$, the Hamiltonian is the sum of two width-$2a$ infinite square well Hamiltonians. Exchange preserves $y_{+}$ and reverses $y_{-}$. The relative-coordinate box mode is even for odd index $s$ and odd for even index $s$. Pure spatial bosonic exchange symmetry therefore allows odd $s$, while pure spatial fermionic exchange symmetry allows even $s$.

The bosonic ground mode is $(r,s)=(1,1)$, while its first excited mode $(2,1)$ is degenerate in energy with the fermionic ground mode $(1,2)$. Angular spin degrees of freedom, if present, must be included in the full exchange symmetry rather than assigned the spatial rule automatically. The mean relative distance is $⟨|y_{-}|⟩=a(1/2-2/{\pi }^2)$ in the lowest even mode and $a/2$ in the lowest odd mode, obtained by integrating the normalized squared cosine and sine. Multiplication by $\sqrt{2}$ gives the particle separation.

Let $G$ be finite and let $H$ have index $p$, the smallest prime divisor of $|G|$. The group action on left cosets gives $\Phi :G\to S_p$, with $\ker \Phi \subseteq H$. The image order divides both $|G|$ and $p!$. By the choice of $p$, the only possible nontrivial prime factor common to these orders is $p$, and $p!$ contains only one factor of $p$. Thus the image order is $1$ or $p$. The action is transitive on $p>1$ cosets, so its image is nontrivial and has order $p$. The first isomorphism theorem gives $|\ker \Phi |=|G|/p=|H|$, hence $\ker \Phi =H$. A kernel of a group homomorphism is a normal subgroup, proving the claim. For example, every subgroup of order $25$ in a group of order $75$ is normal.

The usual associative, identity, two-sided inverse and commutativity axioms form an independent axiom set. A three-element commutative unital operation with self-inverses can fail associativity; a constant-zero operation can satisfy the other three axioms while failing the identity condition; natural-number addition fails the inverse condition; and a noncommutative group fails commutativity. Each example must be checked against all three retained axioms.

An improper integral whose absolute-value integral is finite. The triangle inequality for an integral bounds each truncated tail by the corresponding tail of $|f|$, so the Cauchy criterion for an improper integral gives ordinary convergence. Absolute convergence is stronger than a Cauchy principal value, which may exist only because two divergent sides cancel. The continuous sine-product integrand $\sin (ax)\sin (bx)/x^2$ has a removable value $ab$ at zero and an integrable $1/x^2$ tail; its auxiliary complex contour integrand can nevertheless require a principal value.

For a locally integrable real or complex function, the displayed tail condition is equivalent to ordinary convergence of its improper integral at infinity. Apply the Cauchy criterion for a function limit to its truncated integrals, using completeness of the real or complex numbers. The necessity follows by subtracting two nearby-to-limit truncations; sufficiency follows by first taking a convergent sequence of truncations and then bounding all other late truncations against it. The analogous criterion applies at a finite singular endpoint. Bounding the modulus of every tail by its absolute-value integral proves that an absolutely convergent improper integral converges ordinarily.

For $p>0$, the function $x\mapsto \sin x/x^p$ is Lebesgue integrable on $(0,\infty )$ exactly when $1<p<2$. Near zero, $\sin x/x\to 1$, so absolute integrability is equivalent to that of $x^{1-p}$, which requires $p<2$. At infinity, $p>1$ gives the upper bound $x^{-p}$. For $0<p\le 1$, on each interval $[k\pi ,(k+1)\pi ]$,and the corresponding P-series diverges.

The signed improper integral nevertheless converges for every $0<p<2$: near zero the same local comparison applies, and at infinity the Dirichlet test for improper integrals uses the bounded antiderivative of sine and the decreasing factor $x^{-p}\to 0$. Thus $0<p\le 1$ gives conditional convergence, while $1<p<2$ gives absolute integrability. In particular, substituting $t=1/x$ turns ${\int }_0^1|x^{-1}\sin (1/x)|dx$ into the divergent tail ${\int }_1^{\infty }|\sin t|/tdt$.

Let $h$ be measurable, integrable over a positive period $L$, and have mean zero there. Its positive and negative parts have equal integrals $H$. On the $k$-th period the signed integral of $h(x)/x$ has absolute value at mostThis follows by bounding the positive and negative weighted integrals between $H/((k+1)L)$ and $H/(kL)$. The bounds are summable, so the integrals at period endpoints converge. A leftover partial period has absolute integral at most $\Vert h{\Vert }_{L^1(0,L)}/(kL)$, tending to zero, which proves convergence for arbitrary upper endpoints too.

In contrast, each full-period integral of $|h(x)|/x$ is at least $2H/((k+1)L)$. It diverges in total unless $H=0$, equivalently $h=0$ almost everywhere. Thus a nonzero mean-zero periodic function gives a convergent improper integral here but no Lebesgue integrability.

In the half-plane $y>0$ with an impermeable wall, place a point vortex of circulation $\Gamma$ in the background velocity $(\kappa x,-\kappa y)$, where $\kappa >0$. The image vortex at a plane wall has circulation $-\Gamma$ at the reflected location and contributes horizontal velocity $\Gamma /(4\pi y)$. Excluding the vortex's own singular velocity gives the displayed motion equations.

Put $c=\Gamma /(4\pi )$. Their first integral is $\kappa xy+c\log y$, since its time derivative is $\kappa [(\kappa x+c/y)y-\kappa xy]+c(-\kappa )=0$. The explicit trajectory is $y=y_0{e}^{-\kappa t}$ and $x=e^{\kappa t}(x_0+ct/y_0)$. These formulas show how image-induced drift changes the hyperbolic trajectories of the background flow while preserving a modified stream-function-type invariant.

For the Euler-type operator $-x^2{\partial }_x^2-x{\partial }_x+{\partial }_y^2$, set $z=\ln |x|$ on either half-plane $x\ne 0$. The chain rule gives $x{\partial }_x={\partial }_z$ and $x^2{\partial }_x^2+x{\partial }_x={\partial }_z^2$. The operator becomes ${\partial }_y^2-{\partial }_z^2$. Its characteristic coordinates $\xi =y+z$, $\eta =y-z$ reduce it further to $4{\partial }_{\xi }{\partial }_{\eta }$. Thus the constant-forcing equation $-x^2{u}_{xx}-x{u}_x+u_{yy}=c$ has solutions $u=(c/4)\xi \eta +g(\xi )+h(\eta )$, with arbitrary sufficiently differentiable functions $g,h$. The logarithmic change of variables cannot by itself describe smooth passage across $x=0$.

On $x>0$, the linear partial differential equation $u_{xx}-x^2{u}_{yy}-u_x/x=0$ has characteristic curves $y\pm x^2/2=\text{constant}$. The change of variables $\xi =x^2-2y$, $\eta =x^2+2y$ has Jacobian determinant $8x\ne 0$. The chain rule givesThus its canonical form of a second-order partial differential equation is $U_{\xi \eta }=0$, and its general twice differentiable solution is $u=f(x^2-2y)+g(x^2+2y)$. The first derivative term exactly cancels the derivatives of the coordinate coefficients.

More generally, for positive differentiable $a(x)$, choose $s(x)$ with $s^{\prime }(x)=a(x)$. Then the same cancellation givesThe characteristic coordinates are $s\pm y$. The quadratic example is $a(x)=x$, $s=x^2/2$, with both coordinates multiplied by two.

This partial differential equation has discriminant $y$, so it is a hyperbolic partial differential equation above $y=0$ and an elliptic partial differential equation below it. In its hyperbolic region the displayed characteristic coordinates have Jacobian determinant $18\sqrt{y}$. The chain rule givesThe first derivative terms arise from differentiating the coordinate coefficients and cannot be discarded. The coordinate change degenerates at the line where the two characteristic directions merge.

A first-order structure has finite-tuple homogeneity when any two finite tuples of the same parameter-free complete type are related by an automorphism. For countably infinite first-order structures this agrees with extension of finite partial elementary maps to automorphisms. Countable atomic models have this property by a back-and-forth method using isolating logical formulas. Uncountable atomic models can fail it.

Let $n$ be even and let $X_1,\dots ,X_n$ be independent Bernoulli variables with parameter $1/2$, arranged on a circle. The probability that some consecutive half-circle $I$ has $|{\sum }_{i\in I}{X}_i-{\sum }_{i\notin I}{X}_i|\ge w$ is at most $2n\exp (-2w^2/n)$. For a fixed $I$, the variable $Z_I={\sum }_{i\in I}{X}_i+{\sum }_{i\notin I}(1-X_i)$ has the binomial distribution $\mathrm{Bin}(n,1/2)$, and the signed imbalance is $Z_I-n/2$. Apply Hoeffding inequality to this single binomial variable, then take a union bound over the at most $n$ intervals. Consequently $w=c\sqrt{n\log n}$ suffices for uniform balance with probability tending to one whenever $c>1/\sqrt{2}$.

For an inviscid irrotational unit-speed layer turning through a right angle, the complex potential of a two-dimensional potential flow occupies the strip $0<\mathrm{Im}w<1$ and its complex velocity $q$ occupies a fourth-quadrant quarter disk. The maps $r=q^2$, $s=(r-1)/(r+1)$ and $s^2$ take this disk to the upper half-plane, sending upstream velocity one, corner stagnation zero, and downstream velocity $-i$ to $0,1,\infty$. The map $e^{\pi w}$ has the same normalization, proving the displayed relation.

For a planar reservoir draining below a vertical lip of height one into a unit-speed stream, the potential strip $0<\mathrm{Im}w<a$ and first-quadrant unit velocity disk satisfyOn the fluid free boundary put $q=e^{-i\theta }$, $-\pi /2\le \theta <0$. Then $dz/d\theta =-2a{e}^{i\theta }\cot \theta /\pi$, so the downstream height is $1-2a/\pi$. Equating it to $a$ gives $a=\pi /(\pi +2)$.