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Average power divides the work over a nonzero interval by the elapsed time. For the same $2\mathrm{kg}$ point mass driven from rest through $4\mathrm{m}$ by a net $3\mathrm{N}$ force, the displacement is $x(t)=a{t}^2/2$. ThusThis average power is half the final instantaneous power because $P(t)$ is a linear function of time, starting at zero. Its graph of a function bounds a triangle of area $P(t_f)t_f/2=12\mathrm{J}$; the horizontal line at average power encloses a rectangle with the same area. In general, an endpoint value of instantaneous power does not determine average power. ^[ref]

For the Cameron-Martin space of Wiener measure path $h(s)=\mu s$, the Cameron-Martin theorem says that translation of Brownian motion by this linear function changes its law through time $t$ by the density

The capacitor energy is the electrostatic energy stored when electric charge is separated. For a linear capacitor of fixed positive capacitance $C$, plate electric charge magnitude $Q$, and terminal electric potential difference $V=Q/C$, it isThe zero is the uncharged state. Thus the capacitor energy is positive for nonzero $Q$, even though the total electric charge of the two plates is zero. ^[ref].

For the parallel-plate capacitor example, capacitance $C=8.854\times 10^{-11}\mathrm{F}$ and terminal electric potential difference $V=10\mathrm{V}$ giveThe plate electric charge magnitude is $Q=0.8854\mathrm{nC}$. Doubling terminal electric potential difference with capacitance unchanged doubles the plate electric charge magnitude and quadruples the capacitor energy.

The electric charge is a linear function of terminal electric potential difference, while capacitor energy is a quadratic function. The second curve in each plot uses twice the capacitance: at a given terminal electric potential difference, both plate electric charge magnitude and capacitor energy double. These relations let us calculate how much energy an electrical circuit can receive when a capacitor discharges.

A brief charging argument explains the factor $1/2$. If the plate electric charge magnitude has reached $q$, the terminal electric potential difference is $q/C$. Transferring a further amount $dq$ of electric charge quasistatically requires work $dW=(q/C)dq$. With the geometry and permittivity fixed, the integral isUsing $Q=CV$ gives the other two forms. This integral accounts for the gradually rising terminal electric potential difference; it is the work stored in the capacitor's electric field, rather than a claim that all energy drawn from a fixed-voltage battery during charging ends up stored.

The field picture gives the same result. In the vacuum gap of a parallel-plate capacitor, the electric field has energy density $u={ϵ}_0{E}^2/2$. Multiplying by the gap volume $Ad$ gives the electrostatic energyFor the worked example, $u=4.427\times 10^{-4}J/{\mathrm{m}}^3$ and $Ad=10^{-5}{\mathrm{m}}^3$ reproduce $U=4.427\mathrm{nJ}$. The electrostatic energy resides in the electric field throughout the gap, despite overall charge neutrality.

When capacitance changes, specify the constraint. At fixed plate electric charge magnitude $Q$, increasing capacitance decreases capacitor energy $Q^2/(2C)$; doubling capacitance halves it. At fixed terminal electric potential difference $V$, increasing capacitance increases capacitor energy $C{V}^2/2$; doubling capacitance doubles it. In the latter case a connected battery supplies additional electric charge and energy, and there may also be mechanical work. Use the fixed-voltage capacitor energy balance to account for all transfers before inferring a force from changing capacitor energy.

Optimization of a linear function subject to affine constraints and membership in a closed convex cone. Choosing the positive semidefinite cone gives semidefinite programming; other choices include copositive optimization and completely positive optimization. Dual cones produce scalar-product bounds on feasible objectives.

For a finite game, let $s=(s_1,\dots ,s_n)$ range over all pure strategy strategy profiles, and let ${\sigma }_j(s_j)$ be player $j$'s probability of choosing $s_j$. Under independent mixed strategies, the probability of $s$ is ${\prod }_j{\sigma }_j(s_j)$. The displayed sum is the expected value of player $i$'s payoff. Holding everyone else's mixed strategy fixed makes $U_i$ a linear function of ${\sigma }_i$; this is why checking pure strategy deviations suffices to check all mixed strategy deviations.

Two related properties of an extensive quantity must be distinguished. Additivity says that a quantity $X$ for independent subsystems A and B obeysThe subsystems may be in different states. Extensivity expressed through a homogeneous function says that the equilibrium fundamental relation obeyswith composition held fixed. This is degree-one homogeneity along a replication ray; it does not say that internal energy is a linear function of arbitrary independent changes of entropy, volume and amount of substance.

Additivity gives integer replication by repeated addition. Subdividing a uniform bulk sample into $b$ equivalent parts gives rational scaling $a/b$; continuity then extends it to positive real factors. This argument requires negligible boundary contributions and a consistently prepared state. Conversely, a degree-one homogeneous function need not be additive between different rays: $f(x,y)=\sqrt{x^2+y^2}$ has $f(\lambda x,\lambda y)=\lambda f(x,y)$, but $f(1,0)+f(0,1)=2\ne \sqrt{2}=f(1,1)$. Thus a scaling identity alone cannot justify summing arbitrary states.

For finite samples, the surface correction to thermodynamic free energy can matter. Under geometric enlargement with volume multiplied by $\lambda$, surface area grows as ${\lambda }^{2/3}$, so a surface energy term is not a bulk extensive quantity. Replicating separated droplets retains every original surface, whereas merging them changes the total surface area. Likewise, appreciable mutual gravitational potential energy makes interacting subsystems fail the independent-copy premise. The thermodynamic limit explains when bulk scaling becomes accurate; it does not remove these qualifications for every small or strongly interacting system.

A frequency-dependent payoff is an expected payoff depending on the population composition $x$. Under random pairwise encounters with a fixed payoff matrix it is the linear function $f_i(x)=(Ax{)}_i$. The relative advantage $f_i(x)-f_j(x)$ can change sign as opponents become more or less common. The replicator equation converts this feedback into changes of frequency; nonlinear frequency-dependent payoffs allow interactions not represented by a fixed pairwise table.

Kinetic energy is the energy associated with motion. For a nonrelativistic point mass of mass $m>0$, measured in a specified inertial frame,Here $\mathbf{v}$ is velocity and $v$ is speed; $T$ is another common symbol for kinetic energy. The dot product makes $K$ nonnegative regardless of the direction of velocity. At fixed mass, doubling speed quadruples kinetic energy: a $2\mathrm{kg}$ point mass at $2m/s$ has $4\mathrm{J}$, whereas at $4m/s$ it has $16\mathrm{J}$. The SI units are $\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$. Different reference frames can assign different velocities and therefore different kinetic energies. ^[ref]

A short outline explains why kinetic energy has this formula. In a nonrelativistic inertial frame, constant mass and Newton's second law give, using the chain rule,Integrating over time proves the work-energy theorem, $W_{\mathrm{net}}=K_f-K_i$. The net force is the sum of all forces, and $W_{\mathrm{net}}$ is their total work, not merely the work of a chosen driving force. This route to final speed is useful when the forces as functions of position are known but the full motion as a function of time would be harder to find. ^[ref]

In the running example, a $2\mathrm{kg}$ point mass starts at rest and receives a constant net $3\mathrm{N}$ force through $4\mathrm{m}$. Its net work is $12\mathrm{J}$, soAt any intermediate distance $x\ge 0$, $K(x)=(3\mathrm{N})x$ and $v(x)=\sqrt{2Fx/m}$: kinetic energy is a linear function of distance, while speed is proportional to the square root of distance. If a net opposing $3\mathrm{N}$ braking force subsequently acts for $4\mathrm{m}$, it does $-12\mathrm{J}$ of work and brings this point mass to rest. Negative net work decreases kinetic energy; it does not make kinetic energy negative. Beyond the stopping point the motion must be reconsidered, since the point mass cannot keep moving forward with a negative $K$.

Kinetic energy also makes the exchange with potential energy intuitive: a falling body's loss of gravitational potential energy can appear as increased kinetic energy. The formula above applies when speed is small compared with the speed of light; at higher speeds, special relativity changes the expression. ^[ref]

A scalar linear differential equation of order $n\ge 1$ has the formHere the coefficients $a_j$ and the forcing term $g$ are prescribed functions of the independent variable $t$. A classical solution of a differential equation is an $n$-times differentiable function satisfying the identity at every point of its interval. The unknown function $y$ and its derivatives enter to the first power, without products between them or coefficients depending on $y$. Equivalently, the differential operator $L$ is a linear map: for constant scalars $\alpha ,\beta$,The forcing term may be zero, giving a homogeneous linear differential equation, or nonzero, giving an inhomogeneous linear differential equation. Linearity of $L$ does not mean that every function graph of a solution of a differential equation is a straight line.

To use the usual order-$n$ initial value problem theory on an interval $I$, assume that $a_n(t)\ne 0$ throughout $I$ and that all $a_j$ and $g$ are continuous functions there. Dividing by $a_n$ normalizes the highest derivative; specifying $y(t_0),y^{\prime }(t_0),\dots ,y^{(n-1)}(t_0)$ at $t_0\in I$ then selects one solution of a differential equation throughout $I$. A zero of $a_n$ needs separate analysis: one cannot divide through it and retain these conclusions. The first-order normalization appears in OpenStax's definition and standard form of linear equations; the corresponding system theorem is in MIT's existence and uniqueness notes for linear systems.

The coefficients need not be linear functions of $t$. For example, $t^2{y}^{\prime }+y=0$ is a linear differential equation on $t>0$, although $t^2$ is quadratic. Its solution of a differential equation with initial condition $y(1)=1$ is $y(t)=e^{1/t-1}$: its derivative is $y^{\prime }=-y/t^2$, so substitution gives $t^2{y}^{\prime }+y=0$, and $y(2)=e^{-1/2}\approx 0.60653$. The zero of the highest coefficient at $t=0$ explains the domain restriction. In contrast, $y^{\prime }+y^2=0$, $y{y}^{\prime }+y=0$ and $y^{\prime }+\sin y=0$ are nonlinear ordinary differential equations because of their dependence on the unknown function. Variable coefficients and nonlinear ordinary differential equations are different issues; see MIT's definitions and examples of first-order linear equations.

The practical benefit of linearity is that it separates the effect of the initial condition from the effect of the forcing term. Once one particular solution $y_p$ satisfies $L[y_p]=g$, every solution of a differential equation isHere $\ker L$ is the kernel of a linear map, consisting of solutions of a differential equation to the associated homogeneous linear differential equation. Indeed, $L[y-y_p]=g-g=0$, and conversely $L[y_p+h]=g+0=g$. Thus the family for a fixed forcing term is an affine space obtained by translating $\ker L$. If $g$ is not identically zero, the zero function is absent, so this family is not a vector space under the usual operations on functions. This short argument explains why a general solution is a particular solution plus a homogeneous solution.

Linear programming optimizes a linear function subject to finitely many linear equalities and inequalities.

A Nash equilibrium is a strategy profile in which every player's strategy is a best response to the others. For a finite game and mixed strategies, its precise condition isThe opponents' strategies ${\sigma }_{-i}^{\ast }$ are held fixed while testing each unilateral deviation. Equality is allowed: a player may have other equally good best responses. Because expected payoff is a linear function of the player's own mixture, it suffices to test every pure strategy. Equivalently, every choice in the support of a mixed strategy must attain the largest pure-choice expected payoff, and no unused choice may exceed it.

The two worked examples illustrate different uses of this condition. Matching pennies has no pure strategy Nash equilibrium, but its unique mixed strategy Nash equilibrium removes each player's vulnerability to predictable play. The asymmetric coordination example has three Nash equilibria, and its mixed one fails Pareto efficiency. A Nash equilibrium tests individual incentives, whereas Pareto efficiency compares feasible outcomes that may require several players to change together. In the prisoner's dilemma, for example, mutual defection is a Nash equilibrium even though mutual cooperation gives both players a larger payoff because $R>P$.

Existence in a finite game is guaranteed by Nash's theorem. The theorem does not select one Nash equilibrium, promise Pareto efficiency, or assert convergence of a learning process. Nash's ^[ref] used the Kakutani fixed-point theorem; his ^[ref] gave a proof using the Brouwer fixed-point theorem. The connection to a fixed point captures the idea of choices that remain mutually optimal when considered together.

Let $G=(L\cup R,E)$ be the given bipartite graph. A matching in a graph selects edges with no shared endpoint. For each edge $e$, introduce an indicator $x_e$ for selection. The integer program for a maximum cardinality matching isIts linear programming relaxation iswith the same objective function. Explicit upper bounds $x_e\le 1$ may be included, but they already follow from a constraint at either endpoint.

We prove that every extreme point of $P$ is integral by showing that any feasible point with a fractional coordinate admits a two-sided perturbation. Given such an $x$, form the fractional support graph $H$ with exactly those edges for which $0<x_e<1$. This is a nonempty bipartite graph. A vertex incident to one of these edges cannot also meet an edge with value $1$, because its incident sum would exceed $1$.

First suppose $H$ contains a graph cycle. Choose a simple one, $e_1,\dots ,e_q$. Its length $q$ is even since $H$ is a bipartite graph. Define a nonzero vector $h$ by alternating $+1,-1$ along the graph cycle and putting $0$ on other edges. At each vertex on the graph cycle, the two changed coordinates have opposite signs, so its incident sum is unchanged. ChooseThen $x^{+}=x+\epsilon h$ and $x^{-}=x-\epsilon h$ are distinct feasible points: no coordinate leaves $[0,1]$ and all incident sums are unchanged. Since $x=(x^{+}+x^{-})/2$, it is not an extreme point.

Otherwise $H$ is a forest. Choose a connected component of a graph containing an edge, and a simple graph path in it between two tree leaves. Alternate $+1,-1$ along the graph path to define $h$. Every internal vertex of the graph path still has unchanged incident sum, including any additional edges branching from it. At either endpoint, exactly one incident edge has fractional value; all other incident coordinates are $0$, since a coordinate $1$ is impossible there. Thus each endpoint has strictly positive slack $1-{\sum }_{e\ni v}{x}_e$ in its degree constraint. Choose $\epsilon >0$ smaller than these two slacks and smaller than every $x_e$ and $1-x_e$ along the graph path. Both $x+\epsilon h$ and $x-\epsilon h$ remain feasible: the endpoint sums can increase or decrease by $\epsilon$, and the internal sums do not change. Again $x$ is the midpoint of distinct feasible points, contradicting extremality.

Both cases exclude fractional coordinates at an extreme point. ThereforeThe nonempty bounded linear polyhedron $P$ has an optimal extreme point for the stated linear function, so the relaxation yields a maximum cardinality matching. This is stronger than the existence of one integral optimum: it proves integrality of every extreme point. It does not claim that every optimal feasible point is integral; convex mixtures of distinct optimal matchings in a graph can be fractional.

An extreme point $p\in P$ cannot be written as a nontrivial convex combination of two distinct points of $P$: if $p=\lambda u+(1-\lambda )v$ with $u,v\in P$ and $0<\lambda <1$, then $u=v=p$.

A vertex of a linear polyhedron is a feasible point at which the active constraints include $n$ linearly independent row normals. Equivalently, some $n$ independent bounding hyperplanes intersect uniquely at that point. This definition does not require $P$ to be bounded or full-dimensional. For a linear polyhedron, it is also equivalent to the point uniquely maximizing a suitable linear function: sum the selected active inequalities to construct one such linear function.

The ordinary least squares fit estimates the conditional mean of the continuous safety response as a linear function of the income coding:The regression coefficient for income is positive. Since larger response values mean greater safety, one additional coded income unit corresponds to an estimated increase of $0.2369$ safety-scale units in the fitted mean. Following the question's stated units, a unit is a £10,000 income step. The observed coding also uses income bands with open end categories, so this is a linear trend per coded band step; it is not evidence of a known constant effect for every extra pound, especially within the open top band. The output's abbreviated variable label does not override the stated units.

The intercept is the fitted mean at income code zero. It is estimated as $6.5933$, but zero is outside the observed income codes, so that value is an extrapolated model baseline, not an observed estimate for a particular income category. Its small p-value tests a zero intercept; it is not the main test of the income association.

For the income coefficient, the standard error is $0.0604$. The reported test statistic and two-sided test giveproviding strong evidence against the null hypothesis of zero linear slope under the stated model assumptions. A $95\%$ Student t confidence interval, using $1727$ residual statistical degrees of freedom, isThis is an interval for the population linear coefficient under the model, not an interval containing $95\%$ of individual safety scores. The large-sample estimate is precise enough to detect a positive association, but that does not establish substantial individual predictive power.

The coefficient of determination is $R^2=0.008828$: the fitted line accounts for about $0.883\%$ of the in-sample variation about the response sample mean. The adjusted coefficient of determination is about $0.825\%$. Thus income alone explains very little of the observed variation, despite the small slope p-value. The residual standard error of $3.997$ response units describes substantial scatter about the fitted line. This is an estimated residual standard deviation, rather than the standard error of the slope or a complete individual prediction interval.

The overall F-test has $F=15.38$ on $1,1727$ statistical degrees of freedom and $p=9.129\times 10^{-5}$. With one predictor and an intercept, it tests the same zero-slope hypothesis as the coefficient Student's t-test; $F=t^2$, up to rounding. It is not an additional independent confirmation. In this simple linear regression, the Pearson correlation coefficient has the slope's sign and magnitude $\sqrt{R^2}\simeq 0.0940$, another description of the weak linear association.

The regression uses $1729$ complete observations and estimates two coefficients, giving $1729-2=1727$ residual statistical degrees of freedom. The deletion of $88$ observations is complete-case analysis; extending its conclusions to the full target population requires suitable assumptions about the missing-data mechanism. Classical reported standard errors also rely on an adequate linear conditional mean, independent errors and homoscedasticity; a normal linear model gives the finite-sample Student t-distribution and F-test calibration, while large-sample approximations can justify similar tests under weaker distributional assumptions. The printed summary alone does not establish these assumptions. Regression diagnostics and, where appropriate, heteroskedasticity-consistent standard errors address different parts of that assessment.

Finally, this unadjusted observational study estimates an association. Other covariates, such as age or experience status, may be related to both income and reported safety. Without additional design or causal identification assumptions, the positive fitted slope does not establish that increasing income would cause an increase in safety.

Use the three standard reduction assumptions: $m$ relaxes much faster than $V,n,h$ and can be replaced by $m_{\infty }(V)$; the two slower gates have the same time constant, ${\tau }_n(V)={\tau }_h(V)=\tau (V)$; and their limiting open fractions have a voltage-independent sum, $n_{\infty }(V)+h_{\infty }(V)=K$. The constant-sum assumption is essential: equality of two time constants by itself does not preserve a moving voltage-dependent sum.

For $q=n+h$, addition of the two gate ordinary differential equations givesThe specified initial condition is $q(0)=K$. Uniqueness for this first-order linear differential equation gives $q(t)=K$, hence $h=K-n$ for all time. The remaining two ordinary differential equations areThese assumptions give a planar reduction of the Hodgkin-Huxley model. The FitzHugh-Nagumo model keeps its fast activation and slow recovery variable geometry, approximating the voltage response by a cubic polynomial and the recovery response by a linear function after nondimensionalization. The polynomial equations in part (c) are a further model simplification; they are not an exact algebraic consequence of arbitrary original gate functions.

For unit resource requirements, treat the resource-blocking events as approximately independent. Let $B_j$ be the approximate blocking probability at resource $j$. Screening class-$r$ traffic by every resource other than $j$ produces the reduced-load approximationThe modeled resource is excluded from its own screening product. For general integer requirements, the standard generalized approximation is $a_j=(1-B_j{)}^{-1}{\sum }_r{A}_{jr}{\alpha }_r{\prod }_k(1-B_k{)}^{A_{kr}}$; in the unit-requirement case this reduces to the displayed formula. The approximate route acceptance is ${\prod }_j(1-B_j{)}^{A_{jr}}$. The independence used here is an approximation to the correlated exact stationary distribution, not an extra property of that law.

Here is a proof of fixed-routing existence and uniqueness. Suppose first that every retained capacity $C_j$ is a positive integer. Links of zero capacity permanently reject their routes and can be removed. If $K$ has the upper-truncated Poisson distribution with parameter $a$, its mean isThe derivative follows by differentiating the normalized weights: $dP(K=k)/da=P(K=k)(k-m_C(a))/a$. The probability at the upper endpoint similarly givesThus $E(\cdot ,C)$ increases from zero to one, while $m_C$ increases from zero to $C$. This is the monotonicity of the Erlang blocking probability.

Set $y_j=-\log (1-B_j)$. For $y\ge 0$ defineHere the inverse is with respect to offered load, and the value at zero is zero. This function is the carried load evaluated at the inverse blocking coordinate, so it is continuous and strictly increasing, with limit $C_j$ as $y\to \infty$. Consider the convex potential for the Erlang fixed pointEach exponential of a linear function is convex. Each integral is strictly convex because its derivative $u_j$ is strictly increasing. Their sum is strictly convex on the nonnegative orthant. It is also coercive: for large $y_j$, $u_j(y_j)$ is bounded below by a positive constant, so the integral grows without bound; the other terms are nonnegative. Hence $F$ attains a unique minimum.

At a positive coordinate, the first-order condition isAt $y_j=0$, the derivative equals minus the right-hand side and is nonpositive. A boundary minimum can therefore occur there only when that load is zero, in which case the derivative is zero and the same equation holds. Since $1-B_j=e^{-y_j}$, division by this positive acceptance probability makes the first-order condition precisely the reduced-load equations above. Conversely every fixed point is such a minimum. Thus fixed routing has exactly one Erlang fixed point, including zero blocking at unused resources. The proof does not rely on a potentially nonconvergent simultaneous substitution algorithm.

For the Poisson process of claims and relative safety loading $\rho$, the premium rate in the classical risk model is $c=(1+\rho )\lambda \mu$. Dividing the adjustment coefficient equation $\lambda (M(r)-1)=cr$ by $\lambda$ explains why the arrival rate cancels. Interpret $r_{\infty }$ as the upper endpoint of the finite moment-generating function domain, so $M(r)<\infty$ for $0\le r<r_{\infty }$ and its stated divergence occurs at that endpoint.

Set $F(r)=M(r)-1-(1+\rho )\mu r$. ThenDifferentiation is valid inside the finite-transform domain, and its positive neighbourhood gives finite moments of every order. Thus $F$ is strictly convex and negative just to the right of zero. If $r_{\infty }<\infty$, divergence of $M(r)$ makes $F(r)\to \infty$ at the endpoint. If $r_{\infty }=\infty$, choose $a>0$ with $\mathbb{P}(X_1\ge a)>0$; then $M(r)\ge \mathbb{P}(X_1\ge a)e^{ar}$, which eventually exceeds every linear function. Again $F(r)\to \infty$. The intermediate value theorem gives a positive zero.

There is at most one positive zero: strict convexity and $F(0)=0$ make the secant slope $F(r)/r$ strictly increasing for $r>0$. It starts at $-\rho \mu$ and crosses zero exactly once. HenceThis proves the secant-slope existence criterion for an adjustment coefficient in the present setting.

Put $m_j=\mathbb{E}{X}_1^j$. For $u>0$, the positive remainder of the exponential series gives $e^u>1+u+u^2/2$. Apply it to $u=R{X}_1$ and take expectations:Dividing by $R>0$ yieldsKeeping the cubic term similarly givesThe polynomial $g(r)=m_3{r}^2+3m_{2}r-6\rho \mu$ is strictly increasing for $r\ge 0$, is negative at zero and tends to infinity. Let $r_2$ be its unique positive zero. The last inequality says $g(R)<0$, soFinally, $3m_2{r}_1=6\rho \mu$, whence $g(r_1)=m_3{r}_1^2>0$. Monotonicity gives $r_2<r_1$. Thus the requested upper bounds can in fact be sharpened to the strict ordering $0<R<r_2<r_1$. The bounds themselves need not lie in the finite-transform domain; the argument only evaluates the actual transform at $R$ and polynomial moments elsewhere. These are polynomial moment bounds for the adjustment coefficient.

For rate-$\alpha$ exponential distribution claims,The positive root satisfies $1/(\alpha -R)=(1+\rho )/\alpha$ after dividing out $R$, soThe cubic-bound equation reduces to $r_2^2+\alpha r_2-\rho {\alpha }^2=0$, selecting the displayed positive root. In particular $R<\alpha$, as required by the transform domain, for every $\rho >0$.

For each fixed $x$ in the nonnegative orthant, the map $A\mapsto x^{T}Ax$ is a continuous linear function on the vector space of real symmetric matrices. Thusis an intersection of closed half-spaces, using the Frobenius inner product $⟨A,B{⟩}_F=\mathrm{tr}(AB)$ on this space. Arbitrary intersections of closed sets are closed, so $K$ is closed.

If $A,B\in K$ and $s,t\ge 0$, then $x^T(sA+tB)x=s{x}^{T}Ax+t{x}^{T}Bx\ge 0$ for every $x\ge 0$. Therefore $sA+tB\in K$, including the zero matrix when $s=t=0$. HenceThe copositive cone is defined in the real space of symmetric matrices; symmetry is needed later to recover every matrix entry from these quadratic forms.

The required linear program isThere are $2n$ inequality constraints and one equality constraint, all independent of $x$.

The feasible set is the capped simplex, a nonempty compact set. A linear function attains its maximum at an extreme point; the previous part identifies these as the indicators of $k$-element subsets. At such a vector, the objective is the sum of the selected coordinates. Maximizing selects the $k$ largest coordinates, including when some are negative or tied. Equivalently, $f_k$ is the support function called the sum of the largest components.

This is the Baumslag-Solitar group $G=\mathrm{BS}(1,2)$. Its defining relation gives, by mathematical induction,In the cyclic subgroup $H=⟨a⟩$ with generator $a$,whereas in $G$, with generators $a,b$,If inclusion were a $(\lambda ,\epsilon )$-quasi-isometric embedding, its lower bound would implyfor every $n$, which is impossible because an exponential function eventually dominates every linear function. Hence $H$ is not quasi-isometrically embedded; it is an exponentially distorted subgroup.