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The quadratic Lagrange triangle uses degree-at-most-two polynomials on a nondegenerate triangle, with point evaluation at its three polygon vertices and three edge midpoints as the six finite element degrees of freedom. If ${\lambda }_i$ are the affine functions of its barycentric coordinate system, the cardinal shape functions areAt a polygon vertex, one coordinate in the barycentric coordinate system is one and the others vanish; at an edge midpoint, two are $1/2$ and the third is zero. These values prove the cardinal property. Since $\dim {\mathcal{P}}_2(K)=6$, this is an unisolvent finite element. Matching the two endpoint values and midpoint value on a shared edge of a polygon matches the whole quadratic restriction, producing a continuous global conforming finite element space.

A quadrature rule approximates an integral by a finite weighted sum of sampled function values:The quadrature nodes $x_i$ specify where to sample, and the quadrature weights $w_i$ specify how much each sample contributes. The density $\rho (x)$ specifies the target integral, whereas the discrete quadrature weights belong to the approximation. Ordinary numerical integration has $\rho =1$. Gaussian quadrature and the midpoint quadrature rule choose interior quadrature nodes; the trapezoidal rule includes endpoints.^[ref]

For ordinary numerical integration, exactness on the constant function $1$ requires ${\sum }_i{w}_i=b-a$; exactness on $x$ additionally requires ${\sum }_i{w}_i{x}_i=(b^2-a^2)/2$. These two checks make the quadrature rule exact for every affine function. For example, both the midpoint quadrature rule and trapezoidal rule give $6$ for ${\int }_0^2(2x+1)dx$. Testing progressively higher polynomial degrees defines the degree of precision of a quadrature rule. Exactness describes a class of integrands, rather than promising exact results for every function.

Composite midpoint quadrature rules and composite trapezoidal rules apply a simple quadrature rule on every panel of a partition of an interval. This buys accuracy by shrinking the panels while keeping the elementary formula fixed. For continuous functions sampled exactly, their quadrature errors tend to zero as the largest panel width tends to zero. The familiar second-order bounds require stronger assumptions about derivatives. The signed approximation error used here is $Q(f)-I_{\rho }(f)$; some sources define quadrature error with the opposite sign, so the convention must accompany any formula.

For distinct ordered observations $x_i$, let $s_u$ be their natural cubic spline interpolant with values $u_i$. Its roughness is a quadratic form, $\int (s_u^{\prime \prime }{)}^2=u^{T}Ku$, where $K$ is positive semidefinite and its nullspace consists of vectors sampled from affine functions. Minimizing penalized least squares $\Vert y-u{\Vert }^2+\lambda u^{T}Ku$ gives $\hat{u}=(I+\lambda K{)}^{-1}y$. Thus the smoothing matrix is $A_{\lambda }=(I+\lambda K{)}^{-1}$ and its effective degrees of freedom are $\mathrm{tr}(A_{\lambda })$. As $\lambda \downarrow 0$ the fit approaches interpolation and the trace approaches $n$; as $\lambda \to \infty$ it approaches the least-squares affine line and the trace approaches $2$.

A roughness penalty penalizes variation of derivatives, such as $\int (m^{\prime \prime }{)}^2$. The second-derivative penalty vanishes precisely on affine functions and is represented in a chosen basis by a roughness penalty matrix.

The nonnegative functional $J(g)={\int }_a^b(g^{\prime \prime }(x){)}^{2}dx$. Its null space on $C^2[a,b]$ consists of affine functions. The natural cubic spline interpolant uniquely minimizes this penalty among functions matching at least two prescribed knot values. It is a squared second derivative penalty, not the exact geometric curvature energy of a graph.

The Haar wavelet provides an explicit one-vanishing moment comparison:It cancels a constant polynomial, but not a nonconstant affine function. To obtain two vanishing moments by an equally transparent construction, take three adjacent unit intervals with heights proportional to $1,-2,1$:The L2 norm normalization follows from $1^2+(-2{)}^2+1^2=6$. Direct integration givesThus $\eta$ has exactly two vanishing moments. In fact its coefficient is a finite difference of adjacent local integrals:The weights are those of the second-order central difference. The first moment balances the centres of the three intervals; it is not enough merely to balance their signed areas.

This $\eta$ is a normalized analyzing function, with no orthonormal basis assertion. Indeed, its unit translate has nonzero L2 inner product:Hence these translates already fail orthogonality. The Daubechies wavelet construction imposes additional refinement and orthogonality requirements; two vanishing moments and unit L2 norm by themselves do not supply those requirements.

A semidefinite program optimizes a linear function subject to equalities between affine functions and positive semidefinite matrix inequalities. It generalizes linear programming: a diagonal matrix is a positive semidefinite matrix exactly when its diagonal entries are nonnegative. Matrix trace expresses the objective as $⟨C,X⟩=\mathrm{tr}(CX)$ for real symmetric matrices.

Composing the log-sum-exp function with finitely many affine functions gives a differentiable approximation to their pointwise maximum. Increasing the inverse-temperature parameter $\beta$ reduces the uniform approximation error while increasing the Lipschitz gradient constant.

Let $X$ be an integrable random variable taking values in an open interval $I$. Suppose $f:I\to \mathbb{R}$ is a differentiable strictly convex function. Then its expected value has equality in Jensen's inequality exactly when $X$ is constant almost surely.

First $m=\mathbb{E}X$ lies in $I$: at each finite endpoint, taking the expected value of the strictly positive distance to that endpoint gives a strictly positive distance for $m$. A tangent affine function supports $f$ at $m$, soStrict convexity forces $h(t)>0$ for $t\ne m$. Indeed, equality at another point would force the graph to coincide with its supporting line on the connecting line segment, contradicting strict convexity. Thereforeand the right side is zero exactly when $X=m$ almost surely.

The tangent affine function of $X$ is an integrable random variable, so $f(X)$ has an integrable negative part. Thus $\mathbb{E}[f(X)]$ is well defined, possibly $+\infty$, even if absolute integrability was not assumed. Equality to $f(m)$ automatically excludes that infinite case.

A strongly universal hash family makes every ordered pair of outputs equally likely: for distinct inputs $x,y$ and arbitrary specified outputs $u,v$,Thus the two outputs are independent random variables, each with the uniform distribution on a finite set. Adding the $q$ equal ordered pairs gives collision probability $q/q^2=1/q$, so the family is also a universal hash family.

Here is a completely enumerable example. In the finite field ${\mathbb{F}}_5$, take the affine functions $h_{a,b}(x)=ax+b$, using modular arithmetic modulo $5$. Choose $a$ and $b$ as independent random variables, each with the uniform distribution on a finite set of all five finite field elements $0,1,2,3,4$, including $a=0$. There are $25$ equally likely hash functions. For $x=0,y=1$, the ordered pair is $(b,a+b)$. For instance $(a,b)=(2,3)$ gives $(h(0),h(1))=(3,0)$; $(0,3)$ gives $(3,3)$. Every prescribed $(u,v)$ occurs exactly once, at $b=u,a=v-u$.

Each row contains five choices, so each individual output has probability $5/25=1/5$. Each cell has probability $1/25$, and the five diagonal cells give collision probability $5/25=1/5$. For any distinct $x,y$, the same conclusion follows from $a=(v-u)/(y-x)$ and $b=u-ax$: a nonzero finite field element has a unique inverse. This is the beauty of the construction: two small random coefficients provide exact pair balance without choosing five unrelated values.

The example uses equal-sized input and output sets to make every choice visible; it illustrates the family definitions rather than compression. In a larger input set, a hash function cannot avoid all collisions into five buckets. What a universal hash family controls is the fraction of its members that collide on each fixed pair, rather than the existence of any collision anywhere.

Restricting to $a\ne 0$ leaves $20$ permutations with zero collision probability for distinct inputs. Individual outputs remain uniform, but given $h(0)=u$, the four values other than $u$ each have conditional probability $1/4$ and $u$ has zero. Hence these outputs are not independent random variables: this is a universal hash family but not a strongly universal hash family. Finally, the full $25$-member example contains five constant functions, with immediately visible collisions. Its pair guarantee concerns a random choice of hash function, and makes no claim of cryptographic collision resistance.

On a partition of an interval symmetric about zero, the composite trapezoidal rule gives opposite quadrature nodes equal quadrature weights. Samples of an odd function cancel pairwise, and a zero quadrature node contributes zero. For a continuous odd function, its exact integral over $[-L,L]$ is also zero. Adding any affine function preserves exactness by linearity of the integral and of the quadrature rule. For example, $1+\sin x$ is nonpolynomial but both the exact integral and every uniform-grid composite trapezoidal rule value on $[-1,1]$ are two.

Thermal conduction transfers heat through microscopic interactions without bulk transport of matter. To predict the heating load through a bar, combine Fourier's law with conservation of energy and the imposed end temperatures. In the isotropic model with positive thermal conductivity $k$, the conductive heat flux points from high to low temperature, opposite to the temperature gradient. A spoon warming along its handle and heat escaping through a wall are familiar examples.

Consider a bar of length $L=0.10\mathrm{m}$, constant cross-sectional area $A=0.010{\mathrm{m}}^2$ and constant thermal conductivity $k=2\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. Maintain its ends at $T_h=40{}^{\circ }\mathrm{C}$ and $T_c=20{}^{\circ }\mathrm{C}$, with $x$ increasing toward the colder end. Assume a steady state, one-dimensional thermal conduction, no internal sources and no heat transfer through the sides. The temperature falls by $20\mathrm{K}$ over $0.10\mathrm{m}$, givingThe difference between the two readings in degree Celsius units is the same $20$-kelvin difference. The positive heat flux carries heat toward increasing $x$.

The general solution under these assumptions isA proof outline is enough to explain the straight line: conservation of energy in any small slice, with no accumulation, sources or side losses, gives $d\dot{Q}/dx=0$. Constant cross-sectional area gives $dq/dx=0$; constant thermal conductivity and Fourier's law then give $d^{2}T/d{x}^2=0$. Two integrations and the end boundary conditions determine the displayed affine function. This is why a visually simple temperature profile can encode a continuous flow of energy. ^[ref]

Doubling the length halves the rate, while doubling the cross-sectional area doubles it, at fixed end temperatures and thermal conductivity. This explains why a long narrow support can limit unwanted heat transfer, while a short broad path can help remove heat. Joining materials leads to thermal resistance in series. During initial warming, internal energy can accumulate and the temperature profile need not be a straight line; the time dependence is described by the heat equation and thermal diffusivity.

Now join two layers, each of length $L_1=L_2=0.10\mathrm{m}$, with constant cross-sectional area $A=0.010{\mathrm{m}}^2$ and stipulated constant thermal conductivities $k_1=1$ and $k_2=4\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. Keep the outer ends at $40$ and $20{}^{\circ }\mathrm{C}$. Assume a steady state, one-dimensional thermal conduction, no internal sources or side losses, and perfect contact: the temperature is continuous at the interface, with no extra contact drop. For each layer, use the area-specific thermal resistance $r_i=L_i/k_i$ and $R_i=r_i/A$. The series calculation gives

The heat flux is continuous: otherwise energy would accumulate at the interface. Yet Fourier's law gives different temperature gradients, $dT/dx=-q/k_1=-160\mathrm{K}{\mathrm{m}}^{-1}$ and $dT/dx=-q/k_2=-40\mathrm{K}{\mathrm{m}}^{-1}$. Lower thermal conductivity requires a steeper temperature gradient to carry the same heat flux; equal lengths therefore give drops in the ratio $4:1$. A kink in the temperature profile records the change of material while perfect contact keeps the temperature itself continuous.

Thévenin's theorem replaces the terminal behavior of a linear resistive circuit by an ideal voltage source in series with a Thévenin resistance, whenever its terminal electric potential difference is a well-defined affine function of the delivered electric current. Fix terminals $a,b$, write $V=\varphi (a)-\varphi (b)$, and take $I$ positive out of the network at $a$ into the electrical load, returning through $b$. ThenThe open-circuit voltage $V_{\mathrm{oc}}$ is the intercept at $I=0$; $-R_{\mathrm{th}}$ is the slope. Thus two parameters predict what happens when many different electrical loads are connected. The Wheatstone bridge detector loading calculation uses precisely this replacement to find the detector reading even when the bridge's two voltage dividers interact through its detector.

For example, a $12\mathrm{V}$ ideal voltage source with $2\mathrm{k\Omega }$ above and $4\mathrm{k\Omega }$ below a voltage divider junction supplies $V_{\mathrm{oc}}=8\mathrm{V}$ and $R_{\mathrm{th}}=\frac{4}{3}\mathrm{k\Omega }$ at that junction relative to the lower terminal. Attaching a $4\mathrm{k\Omega }$ electrical load reduces its voltage to $6\mathrm{V}$ and draws $1.5\mathrm{mA}$. The equivalent explains why an unloaded divider ratio alone cannot predict a loaded output.

The equivalence concerns the chosen terminals, including the power $VI$ delivered there. It does not identify internal branch electric currents or internal Joule heating. At open circuit the original divider still draws $2\mathrm{mA}$ from its $12\mathrm{V}$ supply, while the equivalent's series branch carries no electric current. ^[ref]

With the passive sign convention applied to the network, its entering electric current is $J=-I$, and the same relation becomes $V=V_{\mathrm{oc}}+J{R}_{\mathrm{th}}$. The electrical load uses the passive sign convention with entering electric current $I$, so it absorbs power $VI$. A positive delivered electric current therefore produces a voltage drop for positive $R_{\mathrm{th}}$; changing the reference direction without changing the equation would give the wrong sign.

The reason is linearity of Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, and the constitutive relations of any linear dependent electrical sources. Solve once at $I=0$ with the independent electrical sources present. Subtract this solution from a solution with prescribed delivered electric current $I$: the fixed independent electrical sources cancel, leaving a homogeneous linear network whose terminal voltage change is proportional to $I$. This is a short superposition principle argument for an affine function; the fixed offset need not vanish.

The qualification about a well-defined response matters. An ideal voltage source alone has $R_{\mathrm{th}}=0$. An ideal nonzero current source alone constrains $I$ and does not have a finite open-circuit voltage or a representation $V=V_{\mathrm{oc}}-I{R}_{\mathrm{th}}$ with finite coefficients. Linear dependent electrical sources can give negative resistance, and singular ideal networks may fail to have a unique finite response. Neither a finite nor a strictly positive passive $R_{\mathrm{th}}$ follows merely from calling an electrical circuit linear. ^[ref]

As a quadrature rule, the trapezoidal rule replaces an integrand by the linear interpolation joining its two endpoint values:For nonnegative endpoint values, this is the area of a trapezoid; in general it is a signed area. The quadrature nodes are $a,b$, each with quadrature weight $(b-a)/2$. Unlike the midpoint quadrature rule, it uses endpoint samples, which can be reused between adjacent panels.

Every affine function $f(x)=\alpha x+\beta$ agrees with its linear interpolation, so the trapezoidal rule integrates it exactly. On $[0,2]$, $f(x)=2x+1$ gives $T(f)=(2/2)(1+5)=6=I(f)$. On $[0,1]$, the polynomial $f(x)=x^2$ instead gives $T(f)=1/2$, exceeding $I(f)=1/3$ by $1/6$. Its degree of precision of a quadrature rule is exactly one.

If $f\in C^2[a,b]$, its signed approximation error satisfiesA convex function lies below its endpoint chord, so the trapezoidal rule overestimates its integral; a concave function gives the reverse inequality. The error formula assumes that the second derivative exists and is a continuous function throughout the closed interval.^[ref] A short derivation is the identityobtained by two integrations by parts. The nonnegative Peano kernel has integral $(b-a{)}^3/12$, and the weighted mean value theorem for integrals gives the formula.

The trapezoidal rule for an ordinary differential equation averages vector fields at the old and new states and is A-stable.

This time-stepping use applies the same trapezoidal rule to the integral of the derivative in an ordinary differential equation. That connection explains its place under implicit Runge-Kutta methods; its quadrature behavior and the nonlinear issue in trapezoidal rule fails B-stability concern different uses of the same endpoint averaging.

The positive-radius closed Euclidean balls in ${\mathbb{R}}^d$ have VC dimension $d+1$ for $d\ge 1$. Vertices of a regular simplex are shattered: interpolate any desired signs by an affine function $a\cdot x+b$, then perturb a sphere through the vertices to the score $1+\eta b+\eta a\cdot x-\Vert x{\Vert }^2$. For the upper bound, take an affine dependence $\lambda$ among $d+2$ points and orient it so that $Q={\sum }_i{\lambda }_i\Vert x_i{\Vert }^2\ge 0$. Labelling the positive coefficients inside and the negative coefficients outside would force ${\sum }_i{\lambda }_i(r^2-\Vert x_i-c{\Vert }^2)>0$, while the affine dependence makes this sum equal $-Q\le 0$. The weaker $d+2$ upper bound from unrestricted lifted affine linear classifiers loses their fixed quadratic coefficient.

Let a bounded planar region $D$ have positive area $A$ and centroid $\overline{\mathbf{x}}=A^{-1}{\int }_D\mathbf{x}dA$. If a solid has nonnegative thickness $h(\mathbf{x})=c+\mathbf{v}\cdot \mathbf{x}$ above $D$, its volume integral givesThus the average of an affine function over a region is its value at the centroid. The proof is the linearity of integration. For a disk the centroid is its centre, making the formula particularly useful for a cylindrical bore cut through a solid whose width varies linearly with height.

For a finite outcome space, complete lottery preferences, a transitive relation, continuity of lottery preferences and the independence axiom for lottery preferences are equivalent to an expected utility representation of the displayed form. The representing outcome utility function is unique up to a positive affine function. The theorem is associated with John von Neumann and Oskar Morgenstern.

A useful proof construction selects a best outcome $a$ and a worst outcome $b$. If all outcomes are indifferent, a constant utility function suffices. Otherwise assign $u(a)=1$, $u(b)=0$. Continuity of lottery preferences makes each outcome indifferent to a mixture $u(\omega )a+(1-u(\omega ))b$, and mixture monotonicity makes the coefficient unique. Substitution of indifferent lotteries and reduction of compound lotteries turn any lottery into a best–worst mixture with best-outcome probability ${\sum }_{\omega }L(\omega )u(\omega )$. Mixture monotonicity then ranks all such mixtures. The reverse direction follows by taking expected values; positive affine functions preserve every comparison.^[ref]

Give the four corners of the square the exclusive or labels: $+1$ when the input signs differ and $-1$ when they agree. In the following order,No affine function $g(x)=a_1{x}_1+a_2{x}_2+b$ has the required strictly positive scores at $p_2,p_3$ and strictly negative scores at $p_1,p_4$: both pairs have midpoint zero, so $g(p_2)+g(p_3)=2b=g(p_1)+g(p_4)$ would have to be both positive and negative. This also rules out perfect classification by a single threshold with a consistent tie rule.

Under the polynomial kernel's quadratic feature map, the two classes become two distinct vectors:The hyperplane ${\varphi }_2=0$ separates them. With $w=(0,-1/\sqrt{2},0{)}^T$ and $b=0$, the score and classifier areAll four signed scores $y_{i}f(p_i)$ equal one. In the input plane the boundary is $x_1{x}_2=0$, the union of the coordinate axes; in the derived coordinates it is a single hyperplane. The feature map identifies opposite corners, which is harmless for these labels because opposite corners belong to the same class. A fixed feature map would not separate labels that disagree at identical mapped vectors.