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On the parabolic domain with the nonpositive real axis removed, first apply the principal square root, then scale by $\pi /2$, and then apply the map of part (b). This gives the stated composite $C(z)$ and maps bijectively onto the slit unit disc.

There is no genuine branch cut in the composite. The function ${\tan }^2(\pi s/4)$ is even in $s$, so changing the local square root has no effect. Near zero its Taylor series starts asgiving a holomorphic extension with $C(0)=0$. There are no poles within the parabolic domain: the possible squared locations are $z=(2+4k{)}^2$, all on the excluded real portion $x\ge 4$.

For $z=-r^2$ with $r\ge 0$, the continuation givesAs $r$ ranges from zero to infinity, these values run once through $(-1,0]$. They fill precisely the slit omitted by the previous mapping, and no point away from the negative real axis has an image in that slit. Therefore the extended map is a bijection onto the whole unit disc:It is holomorphic throughout, so this bijection is the desired conformal map from a parabolic domain to the unit disc; in particular $C^{\prime }(0)={\pi }^2/16\ne 0$.

At the origin, the Taylor series $\sin z=z+O(z^3)$ and $1-\cos z=z^2/2+O(z^4)$ giveThus $f_1$ has a pole of order three and $f_2$ a pole of order four at zero. For the third function, the Laurent serieshas infinitely many nonzero negative-power terms, so zero is an essential singularity of $f_3$.

To classify the point at infinity on the Riemann sphere, use the coordinate $w=1/z$. Thenso $f_3$ has a simple pole at infinity.

For the other two functions, an important qualification is necessary. The poles of $f_1$ at nonzero integer multiples of $\pi$, and those of $f_2$ at nonzero integer multiples of $2\pi$, approach infinity. Equivalently, their images in the $w$ coordinate accumulate at zero. Hence infinity is a non-isolated singularity of $f_1$ and $f_2$. Neither function is holomorphic on any punctured neighbourhood of infinity, so the classification of isolated singularities into removable singularities, poles and essential singularities does not apply there.

Expand the second stage by the Taylor theorem, writing $f=f(t_n,y_n)$ and $D_{y}f$ for the derivative in the state variable:The update therefore agrees with the exact solution's Taylor series:Thus the one-step local truncation error is $O(h^3)$, and the method has order at least two under the usual smoothness and Lipschitz continuity assumptions.

For $f=\lambda y$, the two-stage Runge-Kutta method givesOn the real axis, $R(z)=\frac{1}{2}(z+1{)}^2+\frac{1}{2}\ge \frac{1}{2}$, so $|R(z)|\le 1$ is equivalent to $(z+1{)}^2\le 1$. HenceThe decay convention gives $(-2,0)$. Since negative real values such as $z=-3$ lie outside the linear stability domain, the method is not A-stable. Its order is exactly two, since $R(z)$ lacks the cubic term of $e^z$.

Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.

Put ${\tilde{\xi }}_t={\varphi }_t({\xi }_t)$ and $p_t={\varphi }_t^{\prime }({\xi }_t)$. The given conformal change of half-plane capacity makes the image mapping-out equationIndeed its capacity speed is $2p_t^2$, and the stated Loewner local growth property identifies the indicated single driver. For $t<T$, differentiate the conjugacy identityat a fixed original point. The Chordal Loewner equation then gives the conformal conjugacy derivative for the chordal Loewner equationThe change of variable $z=g_t(w)$ covers the whole upper half-plane. The conformal automorphism of the upper half-plane ${\varphi }_t$ is a real Möbius transformation. Its pole is the image of the still-unswallowed boundary point $-1$, so it is holomorphic in a neighbourhood of ${\xi }_t$ before $T$.

To evaluate the apparent singularity, write $h=z-{\xi }_t$, $p={\varphi }_t^{\prime }({\xi }_t)>0$ and $q={\varphi }_t^{\prime \prime }({\xi }_t)$. The Taylor series isThus the first term in the derivative is $2p/h-q+O(h)$, while the second is $2p/h+2q+O(h)$. Their poles cancel and their constant terms differ by $-3q$. Therefore

Fix $R>|z|$. The Cauchy integral formula and the uniformly convergent geometric series on $|\zeta |=R$ giveDefiningthe remainder after degree $N$ has modulus at most $M_R(|z|/R{)}^{N+1}/(1-|z|/R)$, which tends to zero uniformly on each smaller disk. Differentiating the integral formula at zero identifies $c_j=f^{(j)}(0)/j!$, independently of $R$. Since $f$ is an entire function, $R$ can be chosen larger than any given $|z|$, proving the global Taylor series.

For the requested radius $r$, the same coefficient integral gives the Cauchy estimateTo characterize equality in the Cauchy coefficient estimate, parametrize its circle:If $M_r=0$, the identity theorem gives $f=0$. Otherwise equality in the modulus bound requires the continuous integrand to have both constant modulus $M_r$ and constant complex argument. To see this directly, rotate its average to the positive real axis: its real part is bounded above by $M_r$, and equality of its average with $M_r$ forces that real part to equal $M_r$ everywhere. Thus $f(r{e}^{i\theta })=c_n{r}^n{e}^{in\theta }$ on the circle. The identity theorem extends that equality to the plane. Conversely every such monomial attains the bound. Thereforeincluding the zero function and, when $n=0$, arbitrary constant functions.

One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

For each nonnegative integer $n$, putBecause $f$ is an entire function, every derivative $f^{(n)}$ is continuous and $Z_n$ is closed. No $f^{(n)}$ is identically zero: if $f^{(n)}\equiv 0$ for some $n$, repeated integration shows that $f$ is a polynomial. By the identity theorem, the zero set of the nonzero holomorphic function $f^{(n)}$ has empty interior. Thus each $Z_n$ is nowhere dense.

The complex plane is complete, so the Baire category theorem givesChoose $z_0$ outside this union. Then $f^{(n)}(z_0)\ne 0$ for every $n$, and hence every coefficient $f^{(n)}(z_0)/n!$ in the Taylor series about $z_0$ is nonzero. Such a point $z_0$ therefore exists.

A Laurent series about $a$ is an expansionthat converges to $f$ on an annulus $r<|z-a|<R$.

The partial fraction decomposition isFor $0<|z|<1$, expand both denominators by the geometric series:This is actually a Taylor series at zero, since the apparent puncture at zero contains no singularity.

For $1<|z|<2$, the $z+2$ term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:Hence

Taking $r\to 0$ and using $\sinh (r\sqrt{s})/r\to \sqrt{s}$ givesSet $x=a\sqrt{s}$. The Taylor series of the denominator isConsequentlyThe singular part at $s=0$ is thereforeUsing long-time asymptotics from a Laplace transform and part (a), the centre temperature has late-time behaviour

The function $f$ is real analytic at $y$ if there is a neighborhood of $y$ on which its multivariable Taylor seriesconverges to $f(x)$. Here $\alpha$ is a multi-index; equivalently, $f$ agrees locally with a convergent real power series centred at $y$.

Taking the natural logarithm of the finite Euler product and using the Taylor seriesgivesThe double series converges absolutely, and the Mertens second theorem therefore makes the right sideforExponentiating provesThis is the Mertens third theorem.

Let $S(t)=a\sinh t-t$. For $a>1$, $S^{\prime }(t)=a\cosh t-1>0$ on the integration interval, so its minimum is the endpoint $t=0$. The Taylor series isOn the contributing scale $t=O({\nu }^{-1})$, the higher terms are negligible. The endpoint form of Laplace's method, or the Watson lemma, therefore gives

The three fixed-$a$ estimates below cease to be uniform as the minimum of the phase approaches the endpoint. For the cubic endpoint-to-saddle transition, writeThe Taylor series of the hyperbolic sine gives, for bounded $s$ and fixed $l$,The cubic term controls the tail, so localization of the Laplace integral gives the uniform leading formulaHere $I$ is the cubic Laplace transition integral, equal to $\pi \mathrm{Hi}$ in terms of the Scorer Hi function.

To recover the endpoint regime, let $l\to +\infty$. Scale $s=v/(2^{1/3}l)$ in $I$; the cubic term becomes negligible and $I(-2^{1/3}l)\sim (2^{1/3}l{)}^{-1}$. Thereforewhich agrees with part (i) in the overlap $1\ll l\ll {\nu }^{2/3}$. At $l=0$, the Gamma integral gives $I(0)=3^{-2/3}\Gamma (1/3)$, recovering part (iii).

For $l\to -\infty$, set $M=-2^{1/3}l>0$. The exponent $Ms-s^3/3$ has its maximum at $s=\sqrt{M}$, with second derivative $-2\sqrt{M}$. Thus Laplace's method givesand the transition formula becomesFor $a=1-\eta$ with $\eta \downarrow 0$, part (ii) hasPutting $\eta =(-l){\nu }^{-2/3}$ reproduces both the exponential and its prefactor. For relative agreement of these leading exponentials, one may use the overlap $1\ll -l\ll {\nu }^{4/15}$, which makes $\nu {\eta }^{5/2}\to 0$. Thus the same transition integral connects all three regimes.

Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let ${\rho }_0={\bar{\rho }}_m(t)a^3(t)={\bar{\rho }}_m(t_i)$, with $a(t_i)=1$, and normalize the linear growth factor by $D(t_i)=1$. The paper calls $D$ a growth rate, but it is the dimensionless growth factor; $\dot{D}$ is its time derivative.

Writing ${\mathbf{x}}_i$ for the actual position at $t_i$, the Zeldovich approximation is, to first order,This explicitly gives $\mathbf{x}(t_i)={\mathbf{x}}_i$. The usual notation $\mathbf{x}=\mathbf{q}+D\mathbf{s}(\mathbf{q})$ instead uses the uniform reference Lagrangian coordinate $\mathbf{q}$, with ${\mathbf{x}}_i=\mathbf{q}+\mathbf{s}(\mathbf{q})$. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine $\mathbf{s}$, mass conservation between the initial and current positions givesDemanding $\delta (t)=D(t){\delta }_i$ gives $\nabla \cdot \mathbf{s}=-{\delta }_i$. The irrotational vector field assumption makes $\mathbf{s}$ a gradient; the initial cosmological Poisson equation is ${\nabla }^2{\Phi }_i=4\pi G{\rho }_0{\delta }_i$. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,The evolution of $D$ follows from the linearized peculiar-motion equation $\ddot{\mathbf{x}}+2H\dot{\mathbf{x}}=-\nabla \Phi /a^2$. Since cosmological Poisson equation gives $\Phi (t)=(D/a){\Phi }_i$ to this order, substitution yields the linear growth equationThus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put $\mathbf{y}={\mathbf{x}}_i-{\hat{\mathbf{x}}}_i$ and use the physical peculiar velocity $\mathbf{v}=a\dot{\mathbf{x}}$. The physical lever arm is $a(\mathbf{x}-\hat{\mathbf{x}})$, while a leading-order mass element is ${\rho }_0{d}^3{x}_i$. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass givesTerms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because ${\int }_{V_L}\mathbf{y}{d}^3{x}_i=0$ at this order.

Apply the divergence theorem componentwise:The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element $d\mathbf{S}$,The original PDF prints a different coefficient, $-\dot{D}/(4\pi G{\bar{\rho }}_m{a}^2)$. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical $V_L$ centered on its barycentre, the outward normal is parallel to $\mathbf{y}$, so $\mathbf{y}\times d\mathbf{S}=0$ pointwise. HenceFor an equipotential surface, ${\Phi }_i={\Phi }_{\ast }$ on ${\Sigma }_L$, andThereforeThe second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:The constant has no force; $\mathbf{g}$ has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensorwe obtainThe second equality swaps the indices $k,l$ and uses the symmetry of both tensors. Define the initial acceleration tidal tensor byThen the formula in the question has exactly the stated sign:If one defines $T$ as the positive Hessian matrix of ${\Phi }_i$ instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's $I$ is a mass second-moment tensor; the mechanical inertia tensor is $\mathrm{tr}(I)1-I$.

Only the anisotropic parts contribute: a multiple of the identity produces zero contraction with the Levi-Civita symbol. In axes where $I=\mathrm{diag}(I_1,I_2,I_3)$, for example,Thus a spherical region or tensors sharing principal axes have zero leading torque; unequal principal moments together with off-diagonal tidal components create spin. The Taylor series truncation assumes that higher spatial derivatives of the tidal field are sufficiently small across the region.

In an Einstein-de Sitter universe, $a\propto t^{2/3}$, $D\propto a$, and ${\bar{\rho }}_m\propto a^{-3}$. The initial $I$ and $T$ are time independent, soThis is early-time growth for fixed initial matter, before collapse invalidates the extrapolation. It does not predict indefinite linear spin growth for a virialized halo.

Two reasons for only approximate agreement with simulations are that nonlinear collapse and shell crossing change the trajectories and the tidal field, invalidating the first-order displacement and fixed leading tidal tensor; and that real halos undergo dark-matter halo mergers, anisotropic accretion, and exchange of material and angular momentum, so the halo identified at a later time need not be the same isolated collection of initial particles. These processes can change both the magnitude and the direction of the spin.

Watson lemma states that ifand the Laplace integral has suitable growth control away from zero, then

The Taylor series of the amplitude isTermwise application of Watson lemma therefore gives the asymptotic expansionIn particular, the leading asymptotic approximation is $I(x)\sim 2x^{-3}$.

Put $w=z-1$. ThenThusThis is actually a Taylor series, because $z=1$ is a regular point. The point $z=0$ is a simple pole. At infinity,extends analytically to $w=0$, so infinity is a removable singularity of $f$, with extended value zero.

For the real integral, first writeLetWith $z=e^{i\theta }$,The denominator has roots $1/2$ and $2$, so only $1/2$ lies inside the contour. Its residue is $1/(3i)$, and the residue theorem givesConsequently

SetThenThe Taylor series at the endpoint isand henceThe Watson lemma, together withtherefore yields

Put $g=1/f$. The hypothesis $|f(z)|\to \infty$ gives $g(z)\to 0$ as $z\to a$, so defining $g(a)=0$ makes $g$ continuous on $U$ and analytic there by the stated assumption. Its zero at $a$ has some finite order $m\ge 1$, and henceThereforehas a pole of order $m$: its Laurent series has $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$.

Now let $f$ be entire and tend to infinity at infinity. The functiontends to infinity as $z\to 0$, so the preceding argument says that $h$ has a pole at zero. If the Taylor series of $f$ is $f(w)={\sum }_{n\ge 0}{a}_n{w}^n$, thenA pole has only finitely many negative powers, so $a_n=0$ for all sufficiently large $n$. Thus $f$ is a polynomial.

HereFor $0<h\le 1$, the Lipschitz bound on $f$ givesso part (ii) applies with a constant independent of sufficiently small $h$.

The exact solution has the Taylor expansion>Meanwhile,so one numerical step from $y$ isThe local error is therefore $O(h^3)$. Taking $p=2$ in part (ii) proves the required second-order global-error bound. This method is a two-stage Runge-Kutta method.