Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.

Choose a continuous branch of a multivalued function of the complex logarithm on a simply connected region avoiding zero. ThenAlternatively $u_r=-e^{-\theta }\sin (\ln r)/r=v_{\theta }/r$ and $v_r=e^{-\theta }\cos (\ln r)/r=-u_{\theta }/r$, verifying the Cauchy-Riemann equations. The holomorphy is local to a chosen branch of a multivalued function: increasing $\theta$ by $2\pi$ multiplies $g$ by $e^{-2\pi }$, so this is not a single-valued holomorphic function on the whole punctured plane.

The chain rule in polar coordinates gives $u_r=u_x\cos \theta +u_y\sin \theta$ and $u_{\theta }=r(-u_x\sin \theta +u_y\cos \theta )$, with analogous expressions for $v$. Substituting the Cauchy-Riemann equations givesThe restriction excludes the coordinate singularity at the origin.

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

Since $\gamma =1-\delta /r_2$, increasing hunting effort moves leftward on these diagrams. The first stability threshold corresponds toAbove this removal rate, a sufficiently small grey population can be eliminated while reds recover. Until the second threshold, however, the system is bistable: an initially large grey population can still win, so the history and initial abundances matter. The second threshold isAbove it, any positive red population recovers and the grey population tends to zero, regardless of its initial abundance in this deterministic model. This threshold is smaller than $r_2$, so competition from surviving reds can make eradication possible with less removal than would be necessary to suppress grey growth without competitors. If $\delta >r_2$, part (a)(iii) supplies the simpler competitor-independent extinction bound. These are model predictions about long-term dynamics; neither finite-time extinction nor uncertainty in ecological parameters is encoded in the equations.

Write $D=AB-1>0$. The physically nonnegative branches, omitting $S_1$, areOn both bifurcation diagrams, draw $S_2$ solid for $\gamma <B$ and dashed for $\gamma >B$; draw $S_3$ dashed for $\gamma <1/A$ and solid for $\gamma >1/A$. The coexistence branch $S_4$ is dashed throughout its interior because it is a saddle point. It connects $S_3$ at $\gamma =1/A$ to $S_2$ at $\gamma =B$, through two transcritical bifurcations when the branches are continued mathematically beyond the nonnegative quadrant.

Under $A>1$, $B<1$, $AB>1$, the thresholds are ordered $0<1/A<B<1$. As $\gamma$ decreases from one, $S_2$ first becomes stable atFor $1/A<\gamma <B$, both boundary equilibria are stable, and the positive coexistence point $S_4$ is a saddle point. Its stable manifold is the separatrix between their basins of attraction. For every fixed $u_1(0)>0$, a sufficiently small positive $u_2(0)$ lies in the red-survival basin. To justify the quantifier beyond a neighborhood of $S_2$, start on the red-only axis: its logistic equation approaches $S_2$. Continuous dependence over the finite time needed to enter a stable neighborhood shows that sufficiently small positive initial grey populations do so as well.

At the second threshold,$S_4$ meets $S_3$. For $0<\gamma <1/A$, $S_3$ is a saddle point, its stable manifold is the grey-only axis, and there is no positive coexistence equilibrium. Boundedness and the same Bendixson-Dulac theorem argument rule out an interior cycle. Consequently every initial state with $u_1(0)>0$ and $u_2(0)\ge 0$ approaches $S_2=(1,0)$. This is the removal regime that eliminates grey squirrels regardless of their initial abundance. At threshold equality linearization is nonhyperbolic; the strict parameter intervals above avoid inferring stability from a zero eigenvalue.

The non-axis nullclines are $u_1+A{u}_2=1$ and $B{u}_1+u_2=1$. For $A>1$, $B<1$, the first has intercepts $(1,0),(0,1/A)$ and the second has $(1/B,0),(0,1)$. Their axis intercept order rules out a positive intersection. Below the first line $u_1$ increases and above it decreases; below the second $u_2$ increases and above it decreases. Thus $S_2$ is a saddle point and $S_3=(0,1)$ is stable.

For strictly positive initial populations, both coordinates stay positive and are bounded above by their individual logistic equations. There are no interior periodic orbits: the Dulac function $1/(u_1{u}_2)$ makes the divergence of the multiplied vector field $-1/u_2-\rho /u_1<0$, so the Bendixson-Dulac theorem applies. The stable manifold of $S_2$ is the invariant red-only axis, and $S_1$ is repelling. The phase portrait, or the Poincaré-Bendixson theorem with these boundary directions, therefore givesThe qualification $u_2(0)>0$ matters: a grey population identically absent cannot drive extinction. In dimensional variables, positive grey presence leads to $N_1\to 0$, $N_2\to K_2$ in this no-hunting case.

Abbreviate $A={\alpha }_{12}$, $B={\alpha }_{21}$. The coordinate factors immediately give the three boundary equilibria $S_1,S_2,S_3$. The Jacobian matrix isAt $S_1$, the eigenvalues are $1,\rho \gamma >0$, so it is a repelling equilibrium. At the other two points, triangularity givesTherefore $S_2$ is asymptotically stable for $\gamma <B$ and a saddle point for $\gamma >B$; $S_3$ is asymptotically stable for $\gamma >1/A$ and a saddle point for $\gamma <1/A$. Equality gives a zero eigenvalue, so linearization alone is inconclusive there.

For later use, the coexistence point is physically positive exactly when $B<\gamma <1/A$ if $AB<1$, or $1/A<\gamma <B$ if $AB>1$. Its stability is as supplied in the question. If $AB=1$ the displayed formula for $S_4$ does not apply: when also $\gamma =B=1/A$, the coincident non-axis nullclines give a whole segment of equilibria.

The nonnegative quadrant is invariant because both right-hand sides have the corresponding population as a factor. There $u_2^{\prime }\le \rho \gamma u_2$. The differential inequality, or the Gronwall inequality, givesBiologically $\gamma <0$ means $\delta >r_2$: per-capita removal exceeds even the grey squirrels' maximum uncrowded growth rate. Competition can only strengthen the decline.

Since $d{N}_i/dt=K_i{r}_{1}d{u}_i/d\tau$, division of the equations by $K_i{r}_1$ givesThus the dimensionless parameters areEach is a ratio of quantities with matching units. In particular $\gamma$ can be negative even though every original model parameter is positive.

The terms $r_i{N}_i$ give intrinsic per-capita population growth. The terms $-r_i{N}_i^2/K_i$ model intraspecific crowding in a logistic equation, while $-r_i{b}_{ij}{N}_i{N}_j/K_i$ model interspecific competition. Here $K_i$ is the single-species carrying capacity, and $b_{ij}$ measures the competitive effect of one unit of species $j$ relative to one unit of species $i$ in that species' crowding term. The last term $-\delta N_2$ is per-capita removal of grey squirrels: constant effort, rather than a constant number removed per unit time.

The intrinsic rates $r_1,r_2$ and removal rate $\delta$ have units of inverse time. The capacities $K_1,K_2$ have the same density units as the respective $N_i$; with both densities expressed in the same units, $b_{12},b_{21}$ are dimensionless. The two equations form a Competitive Lotka-Volterra model with density regulation and harvesting.

When $0\le v,v^{\prime }\ll c$, expand the denominator:The leading term is ordinary Galilean velocity addition. When both $v,v^{\prime }\to c$ from below, the exact formula instead gives $\boxed{v^{\prime \prime }\to c}$, rather than $2c$. Indeed, writing $a=v/c$, $b=v^{\prime }/c$, we have $1-v^{\prime \prime }/c=(1-a)(1-b)/(1+ab)>0$ whenever $a,b<1$.

Using $v/c=\tanh \theta$ and the hyperbolic tangent addition identity,Consequently the relativistic velocity addition rule is

For the collinear boosts specified in the question, multiplication of the two $2$-by-$2$ blocks and the hyperbolic addition formulas giveThus the resulting rapidity is $\boxed{{\theta }^{\prime \prime }=\theta +{\theta }^{\prime }}$. Collinearity matters: boosts in different spatial directions do not in general combine into a pure boost of this form.

For increasing parameter $s$ along a future-directed timelike curve, the elapsed proper time isor $\int \sqrt{1-|\mathbf{v}{|}^2/c^2}dt$ when coordinate time is the parameter. A Lorentz transformation satisfies $L^T\eta L=\eta$, so replacing $\dot{x}$ by $L\dot{\tilde{x}}$ leaves the expression under the square root unchanged. Hence the integral is invariant. A monotone change of path parameter also leaves it unchanged by the ordinary substitution rule.

Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and coordinates $x^0=ct$. At a point, a smooth path is timelike if its tangent $\dot{x}$ is a timelike vector, here meaning ${\eta }_{ab}{\dot{x}}^a{\dot{x}}^b>0$. It is lightlike if its tangent is a null vector: the same expression is zero and the tangent is nonzero. In these coordinates, a future-directed timelike tangent has spatial speed less than $c$; a lightlike tangent has speed $c$.

Take a minimum spanning tree $T$, for example one constructed by Kruskal's algorithm. Fix $v$ and let $e$ be a longest edge of the tree's unique $xv$ path. Deleting $e$ partitions $T$ into two components separating $x$ and $v$. Any alternative $xv$ path crosses this graph cut. If all of its edges were shorter than $e$, a crossing edge $f$ would let us replace $e$ by $f$, producing a spanning tree of smaller total length. This contradicts minimality of $T$.

Therefore every competing path has an edge at least as long as $e$, and the tree path solves the bottleneck shortest path problem. The same minimum spanning tree works simultaneously for every $v$; this is the minimax path property of a minimum spanning tree.