A parallelogram is the convex set with vertices $p,p+u,p+v,p+u+v$ for two linearly independent planar vectors $u,v$. Its opposite edges are parallel and its centre $p+(u+v)/2$ gives central symmetry. Its area is $|\det [u,v]|$, so an invertible linear map changes its area by the absolute value of that map's determinant. These facts make parallelograms convenient convex bodies for two-dimensional applications of the Minkowski convex body theorem.

A Euclidean polygon with four sides and four right angles. Opposite sides are parallel and have equal lengths. A rectangle is a square exactly when its adjacent side lengths agree.

Simpson's rule approximates a definite integral by integrating the quadratic polynomial obtained by polynomial interpolation at two endpoints and their midpoint. For an interval $[a,b]$ with $a<b$, write $m=(a+b)/2$ and $h=(b-a)/2$. The three quadrature nodes are $a,m,b$, and the quadrature rule isThus its quadrature weights are $h/3,4h/3,h/3$. Their sum is $2h=b-a$, as required for exactness on constant functions. This is the three-node Newton-Cotes closed quadrature; its degree of precision of a quadrature rule is three, even though the interpolating polynomial has polynomial degree at most two.^[ref]

For an original worked example, take the cubic polynomial $f(x)=x^3-2x+4$ on the interval $[-1,2]$. Here $m=1/2$ and $h=3/2$, so the function evaluations giveSimpson's rule givesAn antiderivative supplies an independent exact definite integral:The quadratic polynomial through the same quadrature nodes is $p_2(x)=\frac{3}{2}{x}^2-\frac{1}{2}x+3$. It differs from $f$ between the quadrature nodes, but its definite integral agrees exactly. The picture shows why: the signed polynomial interpolation error has two lobes with opposite integrals.

Three function evaluations therefore capture more information about an integral than about the whole function. When an antiderivative is unavailable, composite Simpson's rule repeats this inexpensive local construction over a partition of an interval. For a sufficiently smooth function, halving the node spacing reduces its truncation error by approximately a factor of sixteen.

Integrate the Lagrange polynomial through values at all equally spaced nodes $0,\dots ,s$, including both endpoints. The resulting quadrature rule is exact through degree $s$. For even $s$, the node polynomial ${\prod }_{j=0}^s(t-j)$ is odd about $s/2$, so its integral vanishes and exactness extends through degree $s+1$. The weights need not remain positive as the number of nodes increases.

If the observed samples are $y_i=f(x_i)+{\epsilon }_i$, the resulting quadrature rule changes by $\Delta Q={\sum }_i{w}_i{\epsilon }_i$. Bounded sample perturbations $|{\epsilon }_i|\le \delta$ therefore satisfyFor ordinary numerical integration with nonnegative quadrature weights exact on constant functions, the bound is $(b-a)\delta$, independent of panel refinement. A common bias ${\epsilon }_i=\delta$ attains it: the quadrature error due to the function approximation may vanish while this sample error persists. Interpolating a fixed data set onto a finer grid likewise does not create new observations or remove its uncertainty.

With new samples whose errors are independent random variables of zero expectation and common variance ${\sigma }^2$, a different model givesFor equal panels this is ${\sigma }^2(b-a{)}^2/n$ for the composite midpoint quadrature rule and ${\sigma }^2{h}^2(n-1/2)$ for the composite trapezoidal rule. These variance reductions require the stated independence and unbiasedness assumptions; a common bias or correlated sample errors need not shrink. The two displayed perturbation formulas follow directly from the triangle inequality and the variance additivity for independent random variables, respectively.

The degree of precision of a quadrature rule is the largest integer $d$ such that $Q(p)=I_{\rho }(p)$ for every polynomial $p$ of polynomial degree at most $d$. By linearity, checking the monomials $1,x,\dots ,x^d$ suffices. Exactness for one higher-degree polynomial alone does not raise this degree: every lower polynomial degree must also be covered.

A quadrature weight multiplies a sampled integrand value in a quadrature rule. For ordinary numerical integration, its units match those of the integration variable. Positive quadrature weights that sum to $b-a$ make $Q(f)/(b-a)$ a weighted average of the samples; signed quadrature weights can magnify sample perturbations, as quantified by noise sensitivity of a quadrature rule.

A quadrature node is a point at which the integrand is sampled by a quadrature rule. Under the affine map $x=(a+b)/2+(b-a)t/2$ from $[-1,1]$ to $[a,b]$, a reference quadrature node $t_i$ becomes $x_i=(a+b)/2+(b-a)t_i/2$; for ordinary numerical integration, the corresponding quadrature weight is multiplied by $(b-a)/2$.

For $V=m^2{\varphi }^2/2$ and unreduced Planck mass $m_P=G^{-1/2}$, the potential slow-roll parameter is ${ϵ}_V=m_P^2(V^{\prime }/V{)}^2/(16\pi )=m_P^2/(4\pi {\varphi }^2)$. Its conventional value one gives the displayed endpoint estimate. Exact accelerated expansion ends at ${ϵ}_H=1$, or ${\dot{\varphi }}^2=V$ for a canonical scalar alone. Substitution of slow-roll formulas into this exact condition gives a different order-one coefficient because slow roll is failing there; a precise endpoint requires integrating the full inflaton equation of motion and Friedmann equation. The large-field e-fold estimate is insensitive to this order-one endpoint uncertainty.

For $V=m^2{\varphi }^2/2$ and unreduced Planck mass, the slow-roll e-fold count is $N=2\pi ({\varphi }_i^2-{\varphi }_f^2)/m_P^2$. A formal initial potential $V_i\sim m_P^4$ gives ${\varphi }_i^2\sim 2m_P^4/m^2$, hence the displayed large-field estimate and expansion factor $e^N$. The final-field contribution is order one in $N$. Without specifying $m/m_P$ there is no unique numerical expansion factor. The extrapolation to Planck density is a classical estimate, rather than a controlled treatment of quantum-gravity corrections.

An imaginary quadratic field is a quadratic number field with no real embedding. Its two complex embeddings are interchanged by complex conjugation. The endomorphism ring of an elliptic curve in characteristic zero, when larger than the integers, is an order in such a field; this is complex multiplication.

The ring of integers is $\mathbb{Z}[\sqrt{10}]$, its class number is two, and its units are $\pm (3+\sqrt{10}{)}^n$ for $n\in \mathbb{Z}$. Factoring a norm-six principal ideal above 2 and 3 gives all solutionsRestriction to nonnegative $n$ gives only the positive-coordinate branch up to simultaneous sign. For example, $4-\sqrt{10}$ is a norm-six element requiring $n=-1$; adding conjugates repairs a nonnegative-index presentation.

For coprime integers $a,q$ with $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$, expand the squared quadratic exponential sum by its difference $h$. The exponential geometric sum bound gives $|S{|}^2\ll N+{\sum }_{h\le N}\min (N+1,\Vert 2\alpha h{\Vert }^{-1})$. In each block of length at most $q/4$, the rotations $2\alpha h$ are separated by at least $1/(2q)$: the reduced denominator of $2a/q$ is at least $q/2$, and the approximation error between two block points is at most $1/(2q)$. The separated reciprocal-distance sum bounds each block by $O(N+q\log (2N))$. There are $O(N/q+1)$ blocks, giving the displayed inequality. For bounded $q$ the trivial estimate supplies the same conclusion.

The normalization $\mu =m/A_w$ has units $\mathrm{kg}{\mathrm{m}}^{-2}$. It converts the Coriolis force into a force per water-drag area and allows comparison with wind and water stresses. Icebergs generally have much larger $\mu$ than thin ice floes, but their effective air-to-water drag-area ratio also matters. A geometric depth becomes a mass per area only after multiplication by density.

For massless quarks at photon-exchange tree level, the initial-spin-averaged, final-color-summed amplitude is $\overline{|\mathcal{M}{|}^2}=8N_c{e}^4{Q}_q^2[(p_1\cdot q_1)(p_2\cdot q_2)+(p_1\cdot q_2)(p_2\cdot q_1)]/s^2$. In the centre-of-mass frame this is $N_c{e}^4{Q}_q^2(1+{\cos }^2\theta )$. With $N_c=3$ and the standard two-body flux, its integrated cross-section is the displayed result.

Each elementary QED vertex has one photon leg, so counting these legs gives $V=2I_{\gamma }+E_{\gamma }$. An amplitude with two external photons has even $V$. In particular Compton scattering has no cubic-order elementary-vertex contribution. With four external legs and $V=4$, a connected diagram has four internal lines and one loop.

For $f_{\lambda }(x)=f(\lambda x)$ in $d$ space dimensions, write $T=\frac{1}{2}\int |\nabla f{|}^2$, $W=\int U(f)$ and $I=\int f^2$. Eliminating the rotation frequency by $\omega =Q/I$ gives the displayed fixed-charge scaling. The positive-power charge term can balance the gradient and potential terms even for $d\ge 3$. This evades the static Derrick theorem because the field is time-dependent and the relevant variations preserve its Noether charge.

For a one-dimensional Q-ball, the profile equation gives $\frac{1}{2}(f^{\prime }{)}^2=U(f)-\frac{1}{2}{\omega }^2{f}^2$. Its rest energy is therefore $M=\int [(f^{\prime }{)}^2+{\omega }^2{f}^2]$. A Lorentz transformation replaces the rest coordinates by $\tau =\gamma (t-v(x-x_0))$, $\xi =\gamma (x-x_0-vt)$. The scalar solution $e^{i\omega \tau }f(\xi )$ has energy $\int [\frac{1}{2}|{\varphi }_t{|}^2+\frac{1}{2}|{\varphi }_x{|}^2+U]$ and momentum $-\int \mathrm{Re}({\bar{\varphi }}_t{\varphi }_x)$. Substitution and $dx=d\xi /\gamma$ give the displayed values and $E^2-P^2=M^2$.

This nonnegative potential has its only vacuum at $s=0$, since $U(s)=s[(s-1/2{)}^2+3/4]$. For the complex kinetic normalization $\frac{1}{2}|\partial \varphi {|}^2$, the small-amplitude mass squared is $2U^{\prime }(0)=2$, whereas ${\min }_{s>0}2U(s)/s=3/2$. The radial Q-ball equation has a positive localized profile for frequencies $3/2<{\omega }^2<2$, for example in three spatial dimensions. It is a standard non-topological soliton potential: the effective profile potential $U(f^2)-{\omega }^2{f}^2/2$ is negative somewhere, while its quadratic coefficient near zero is positive. The conserved Noether charge, not vacuum winding, supplies the stabilization mechanism.

Every primitive Pythagorean triple, after exchanging its legs, has this form with $m>n>0$, $\gcd (m,n)=1$ and opposite parity. Its area is $mn(m-n)(m+n)$. The factors $m,n,m-n,m+n$ are pairwise coprime, because $m\pm n$ are odd. Conversely these conditions give a primitive triple. The parametrization follows by factoring the relatively prime integers $(c+b)/2$ and $(c-b)/2$ when $a$ is the even leg.