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For $a>0$, this algebraic curve can be constructed using a circle of radius $a$ centred at $(a,0)$ and its tangent line $x=2a$. On a ray from the origin, subtract the distance to its nonzero circle intersection from the distance to the line. The resulting radial distance is $r=2a{\sin }^2\theta /\cos \theta$, for $-\pi /2<\theta <\pi /2$. Thus a parametric curve for it isEliminating $t$ gives the displayed cubic equation. It has a cusp at the origin, horizontal tangent direction there, and vertical asymptote $x=2a$. Its name honours Diocles.

For a trajectory with incoming and outgoing straight asymptotes, the deflection angle is the angle between its limiting incoming and outgoing velocity directions. For an attractive hyperbolic orbit of eccentricity $e$ and impact parameter $d$, $\sin (\delta /2)=1/e$ and $\tan (\delta /2)=GM/(d{v}^2)$, where $v$ is the incoming speed at infinity. The incoming velocity points opposite to the incoming radial direction; ignoring that reversal gives the wrong angle.

For nonzero real $a$, the folium of Descartes is the algebraic curve $x^3+y^3=3axy$, with rational parametrization of an algebraic curveIt is symmetric about $y=x$ and has a node at the origin, with the coordinate axes as tangents. For $a>0$, $0\le t<\infty$ traces a loop in the first quadrant, returning to the origin as $t\to \infty$. Its enclosed area, by Green's theorem, isThe other branches approach the asymptote $x+y+a=0$ as $t\to -1$. Negative $a$ rotates the positive-$a$ picture through $\pi$.

A hyperbola is a conic section with two unbounded branches. In suitable coordinates its equation is $x^2/a^2-y^2/b^2=1$ with $a,b>0$. The unit hyperbola is parametrized by $x=\cosh t$, $y=\sinh t$ on its right branch, relating it to the hyperbolic functions.

For example, $x^2/9-y^2/16=1$ has foci of a conic at $(\pm 5,0)$, because $5^2=3^2+4^2$. Its focal-distance difference has absolute value $6$, while its eccentricity of a conic is $5/3>1$. At the point $(3,0)$, the focal distances are $8$ and $2$, verifying that difference. The asymptotes are $y=\pm 4x/3$; a distant point approaches those directions while remaining on one of the two branches. The focal-distance difference, rather than a constant sum, distinguishes this hyperbola from an ellipse.

For two distinct intersecting straight lines through the origin, with independent real linear equations $L_1=0$ and $L_2=0$, the nondegenerate centred hyperbolas having exactly those asymptotes areIndeed, the homogeneous quadratic form of a centred hyperbola vanishes in its two asymptotic directions and hence is a nonzero scalar multiple of $L_1{L}_2$. Its equation has no linear terms because the origin is its centre, and its constant term is nonzero by nondegeneracy. Conversely, the invertible linear transformation $(x,y)\mapsto (L_1,L_2)$ takes the displayed curve to $uv=c$, a hyperbola, and its ends approach $u=0$ or $v=0$. Distinct values of $c$ give disjoint level sets. In particular, the rectangular hyperbolas with asymptotes $y=\pm x$ are $x^2-y^2=c$, $c\ne 0$.

The function $F$ is an odd function, passes through $(0,0)$, and has the horizontal asymptote $F=0$ at both infinities. Its derivative isIt decreases for $|x|>1$ and increases for $|x|<1$, with its lowest value $F(-1)=-1/2$ and highest value $F(1)=1/2$. These features determine the graph of a function shown below.

For $0<ϵ<1/2$, the equation is the quadratic equation $ϵx^2-x+ϵ=0$. The quadratic formula gives exactly two real equation roots:The binomial series gives $\sqrt{1-4{ϵ}^2}=1-2{ϵ}^2-2{ϵ}^4-4{ϵ}^6+O({ϵ}^8)$, hence the two requested asymptotic expansions areTheir product is exactly one. Both are positive, with $x_{-}<1<x_{+}$; there are no negative equation roots because $F(x)<0$ for $x<0$. At $ϵ=1/2$ the two real equation roots coincide at one, and for $ϵ>1/2$ there are none. In the stated limit $ϵ\to 0^{+}$, these two branches are therefore exhaustive.

The power-law ansatz $y=x^r$ givesSince $ϵ>0$, the equation roots $1$ and $-1/ϵ$ are distinct. The Cauchy-Euler equation therefore has general solutionFor $C_1=C_2=1$, the first derivative and second derivative areThe strictly convex function diverges to $+\infty$ as $x\to 0^{+}$, passes through $(1,2)$, and has a unique lowest point atAs $x\to \infty$, $y-x\to 0$, so the line $y=x$ is an asymptote. When $0<ϵ<1$, the lowest point is to the right of one; at $ϵ=1$ it is at one, and for $ϵ>1$ it is to the left of one.

The ratio of the two terms isThus $x^{-1/ϵ}$ is larger for $0<x<1$, the terms are equal at $x=1$, and $x$ is larger for $x>1$. In the stronger asymptotic sense as $ϵ\to 0^{+}$, $x^{-1/ϵ}$ dominates at each fixed $x<1$, and $x$ dominates at each fixed $x>1$. Their crossover has $|\log x|=O(ϵ)$, equivalently $x=1+O(ϵ)$, as described by the power-law crossover in a singular perturbation. In that region $x=1+ϵX$ with bounded $X$ gives $y\sim 1+e^{-X}$.

Substitute the proposed regular perturbation $y=y_0+ϵy_1+\cdots$ into the differential equation. The leading and subsequent equations areConsequently $y_0=Cx$, and in fact each successive coefficient has the form $y_j=C_{j}x$, because the preceding second derivative is zero. The leading approximation is thereforeIt has only one adjustable constant and generally cannot impose two independent boundary conditions. Those assumed in the question require a nonzero $x^{-1/ϵ}$ coefficient, which a power series in $ϵ$ cannot recover: at fixed $x>1$ it is exponentially small, and at fixed $x<1$ it grows without bound as $ϵ\to 0^{+}$. Thus the regular perturbation describes the specified solution only in the region to the right of the crossover, away from $x=1$; it is not a uniform approximation on $(0,\infty )$.

For $0<c<c^{\ast }$, the nonnegative amplitude isIts first two derivatives are positive in this interval:The graph is increasing and convex, starts with a horizontal tangent at the limiting value $a(0)=p_1/(\rho g)$, and has a vertical asymptote at $c=c^{\ast }$. Therefore the hydrostatic lower bound for a travelling-pressure response isBecause the paper assumes $c>0$, this is an infimum, approached as $c\downarrow 0$, rather than an attained minimum. If the stationary limit $c=0$ is admitted, equality holds: the hydrostatic equilibrium condition is $\rho g\eta =-p_1\cos x$. Near resonance,The divergence describes the linear model; sufficiently close to $c^{\ast }$ the small-disturbance assumptions fail. The lower bound and sketch refer to the travelling forced component, since adding free oscillations can change the instantaneous surface elevation.

The quadratic form has symmetric matrixNormalized eigenvectors for $3$ and $-2$ are $(2,1)/\sqrt{5}$ and $(-1,2)/\sqrt{5}$. They form the columns of the rotation matrixThus the new $x^{\prime }$ axis is rotated counterclockwise by $\theta =\arctan (1/2)$, withThe opposite signs of the eigenvalues classify each positive level as a hyperbola. For the original level six its standard equation is $x^{\prime 2}/2-y^{\prime 2}/3=1$; for the final level one it is $3x^{\prime 2}-2y^{\prime 2}=1$. The printed change of level therefore changes the scale, but not the principal axes or asymptotes.

Setting the homogeneous quadratic part to zero gives $y^{\prime }=\pm \sqrt{3/2}{x}^{\prime }$. In the original coordinates put $y=mx$ in $2x^2+4xy-y^2=0$, obtainingThese are the slopes in the original $(x,y)$ axes; $\pm \sqrt{3/2}$ are the slopes in the rotated axes, so they must not be interchanged.

For a straight-line solution $y=mx$, substitution into the differential equation gives $m=(1-2m)/(2+m)$. Thus $m^2+4m-1=0$, and the two possible slopes areThese straight lines solve the differential equation on either side of the origin. At the origin itself the right-hand side is undefined.

An isocline of slope $k$ has equation $(1-2k)x-(2+k)y=0$. For $k\ne -2$ it is the straight line $y=(1-2k)x/(2+k)$; for $k=-2$ it is $x=0$, excluding the origin. In particular, $y=x/2$ is the zero-slope isocline. The line $y=-2x$ is excluded from the domain of the differential equation; slopes tend to infinity on approaching it. Draw the flow vectors proportional to $(1,(x-2y)/(2x+y))$, with positive horizontal component. This specifies orientation by increasing $x$, rather than by an auxiliary time variable.

To integrate, put $y=xu$ where $x\ne 0$. ThenThe left-hand integral is $-\frac{1}{2}\log |1-4u-u^2|$, so the nonconstant-$u$ solutions have the first integralThe constant-$u$ solutions omitted by separation are exactly the two straight lines already found, corresponding to $C=0$. Alternatively, differentiating this first integral gives $(2y+4x)y^{\prime }+4y-2x=0$, which recovers the original differential equation wherever $2x+y\ne 0$ and also verifies continuation through $x=0$ when $y\ne 0$.

The first integral describes hyperbolas with the two straight-line solutions as asymptotes. For $C=1$, the upper branch crosses the vertical axis at $(0,1)$ and exists for every real $x$. For $C=-1$, each branch is confined to $|x|\ge 1/\sqrt{5}$ and reaches a vertical tangent on the excluded line $y=-2x$; the corresponding classical solution intervals exclude those endpoints. These give the two distinct types of curves requested in the sketch.

The initial condition selects $C=1$ and the upper branch, henceHere $2x+y=\sqrt{5x^2+1}>0$, so this really is a global solution of the initial value problem, and $y^{\prime }(0)=-2$ agrees with the right-hand side.

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

Far from the planet, the magnitude of angular momentum per unit mass is the magnitude of the cross product of position vector and velocity. The perpendicular distance from the incoming asymptote is the impact parameter $d$, soChoose $\theta =0$ at closest approach. The general solution of the differential equation in a, with $u^{\prime }(0)=0$, isThe specific orbital energy isAt infinity $\epsilon =v^2/2$, giving $e=\sqrt{1+h^2{v}^2/{\mu }^2}>1$. Thus the hyperbolic orbit hasA different zero of polar angle simply replaces $\theta$ by $\theta -{\theta }_0$.

The incoming and outgoing asymptotes have radial directions $-{\theta }_{\infty }$ and ${\theta }_{\infty }$, where $\cos {\theta }_{\infty }=-1/e$ and $\pi /2<{\theta }_{\infty }<\pi$. The incoming velocity points opposite its radial direction, so the deflection angle between incoming and outgoing velocities is $\delta =2{\theta }_{\infty }-\pi$. HencePutting $\delta =\pi /2$ gives

A hyperbola with focus of a conic $F$, directrix of a conic $D$ not containing $F$, and eccentricity $e>1$ is the locusPut $c=\sqrt{a^2+b^2}$ and $e=c/a>1$. For $F=(c,0)$ and $D=\{x=a^2/c\}$, the defining equality becomesBoth sides are nonnegative, so squaring is reversible. Expanding and cancelling the linear terms givesThus the two matched focus of a conic and directrix of a conic choices, related by reflection, areThe signs in each pair must agree.

On the hyperbola, $y=\pm (b/a)\sqrt{x^2-a^2}$. As $|x|\to \infty$,Hence the perpendicular distance to one of the two straight lines tends to zero on each end of a branch. The asymptotes are therefore

A nondegenerate hyperbola centred at the origin has a quadratic equationThere are no linear terms because the origin is its centre. Its asymptotes are the zero set of the homogeneous quadratic form $A{x}^2+2Bxy+C{y}^2$. Requiring the directions $y=x$ and $y=-x$ gives $A+2B+C=A-2B+C=0$, hence $B=0$ and $C=-A$. Since the conic section is nondegenerate, $A$ and $F$ are nonzero. Thus all such hyperbolas with prescribed asymptotes are exactlyFor $c>0$ these rectangular hyperbolas open along the $x$-axis, and for $c<0$ along the $y$-axis. Conversely, each displayed hyperbola has the required centre and asymptotes. Two distinct members have different values of $x^2-y^2$, so they cannot intersect.

For the second pair of asymptotes, the same argument gives $H_2:4x^2-y^2=d$, $d\ne 0$. The intersections of concentric hyperbolas therefore satisfyIf both right sides are positive, there are four independent sign choices. If precisely one is zero and the other is positive, there are two points. If either is negative there are none; both cannot vanish because $c\ne 0$. Each possibility occurs: take $(c,d)=(1,5)$ for four, $(1,4)$ for two, and $(1,1)$ for none. Hence

The function is nonnegative and satisfies ${\cos }^2(\pi -x)={\cos }^{2}x$. Its derivative is $-\sin (2x)$, so it decreases from $(0,1)$ to its minimum $(\pi /2,0)$ and then increases to $(\pi ,1)$. These three points all have horizontal tangent lines.

The denominator vanishes only at $x=0$, giving a vertical asymptote. Since $1-e^x=-x+O(x^2)$, the function tends to $+\infty$ from the left and $-\infty$ from the right. Its derivative isso each branch is increasing. The horizontal asymptotes are $y=1$ as $x\to -\infty$ and $y=0$ as $x\to +\infty$. The left branch lies above $1$; the right branch lies below $0$, as in the preceding figure.

Substituting $x=r\cos \theta$, $y=r\sin \theta$ into the circle equation gives $r^2-2r\cos \theta =0$. Thus its nonzero radial intersection is $r_Q=2\cos \theta$, with the origin also on the circle. The tangent to a circle at the rightmost point is $x=2$, or $r_R\cos \theta =2$.

Take rays with $-\pi /2<\theta <\pi /2$. The difference of the two radial distances gives the cissoid of Diocles:Multiplying by $\cos \theta$ gives $x=2{\sin }^2\theta =2y^2/(x^2+y^2)$, hence the Cartesian equation $y^2(2-x)=x^3$. A convenient parametric curve, with $t=\tan \theta$, isThe real curve has $0\le x<2$, a cusp at the origin, symmetry about the $x$ axis, and the vertical asymptote $x=2$. Near the origin, $x\sim 2t^2$ and $y\sim 2t^3$, so its cusp has horizontal tangent direction. As $|t|\to \infty$, $x\to 2^{-}$ while $|y|\to \infty$. The circle and curve meet away from the origin at $(1,\pm 1)$.

The domain is $x>0$, and $x\log x\to 0$ from below as $x\downarrow 0$. Its derivative is $1+\log x$, so it decreases up to $x=e^{-1}$ and then increases, with its unique global minimum at $(e^{-1},-e^{-1})$. It crosses the horizontal axis at $(1,0)$ and tends to $+\infty$ as $x\to \infty$. Its second derivative is $1/x>0$, so it is strictly convex. The right panel of the preceding figure shows these features; zero is an excluded endpoint with a finite limiting value, rather than a vertical asymptote.

Write $x/(x-1)=1+1/(x-1)$. The graph is a translated hyperbola, with vertical asymptote $x=1$ and horizontal asymptote $y=1$. It passes through $(0,0)$, lies below $y=1$ for $x<1$ and above it for $x>1$, and has derivative $-1/(x-1{)}^2<0$ on each component of its domain. The left panel below shows both branches.

The domain of a function excludes $x=3$. Both branches are positive, with a vertical asymptote at $x=3$ approached from both sides with value $+\infty$, and horizontal asymptote $y=0$ at either infinity. The graph is symmetric about $x=3$ because it depends only on $|x-3|$.

For $x<3$, the formula is $2/(3-x)$ and the derivative is positive; for $x>3$, it is $2/(x-3)$ and the derivative is negative. It passes through $(0,2/3)$ and has no zeros.