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The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.

Use rationalization of a square-root difference to avoid subtracting two large, nearly equal quantities:Since $1/n\to 0$ and the square root is continuous at $1$, the denominator tends to $2$. The reciprocal limit law established above is applicable because this limit is nonzero. HenceThe limit is one half; rationalization exposes the finite difference hidden by cancellation.

Put $A=(Gm/h^3)F(hk)$. The two amplitude equations have coefficient matrixIts determinant gives the dispersion relationWriting $q=A/{\Omega }^2\ge 0$, its squared growth rates areand the four growth rates are both square roots of each displayed value. DefineFor $0<q<q_{-}$, the discriminant is positive, the sum of the squared rates is $q-1<0$, and their product is $2q(3-q)>0$; both squared rates are negative. These modes have purely imaginary $s$. For $q_{-}<q<q_{+}$, the squared rates form a nonreal complex conjugate pair; their square roots include roots with positive real part. This is overstability, meaning growing oscillations. For $q\ge q_{+}$, at least the plus squared rate is positive because $q>1$, so a real growing root exists. Consequently every $q>q_{-}$ is exponentially unstable.

Take the maximizing mode $hk=\pi$, so $q=(Gm/(h^3{\Omega }^2))F(\pi )$. It is unstable wheneverThis is also the threshold for the onset of exponential growth among all lattice modes, since $0\le F\le F(\pi )$. At equality the maximizing mode has repeated imaginary roots and can have algebraic growth; it is a marginal collision of oscillatory modes, not a strict-decay assertion. The $k=0$ mode has $F=0$, an epicyclic motion sector and a zero-frequency epicyclic guiding center sector with possible secular shear drift. These neutral or algebraic behaviors are distinct from the strict exponential instability established by the requested inequality.

Before the lifetime, $X>0$, so apply the Itô formula to $g(x)=\sqrt{x}$ after stopping inside compact subintervals of $(0,\infty )$. The derivatives are $g^{\prime }(x)=1/(2\sqrt{x})$ and $g^{\prime \prime }(x)=-1/(4x^{3/2})$. Since the quadratic variation of $X$ satisfies $d[X{]}_t=X_{t}dt$, we getThe negative drift is the second-order correction in the Itô formula. This is a Lamperti transform up to a constant scale: the square root transformation makes the noise coefficient constant. Its drift is singular at zero, which is why this stochastic differential equation is stated only before the boundary lifetime.

Use ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ and retain terms linear in the displacement and perturbation potentials. Evaluation at $z=\eta$ may then be replaced by evaluation at $z=0$: the correction to a perturbation is second order. The kinematic boundary conditions becomeThe linearized pressure continuity isThe decaying normal modes give $-k{A}_1=ik(U_1-c)B$ and $k{A}_2=ik(U_2-c)B$. Thus $A_1=-i(U_1-c)B$, $A_2=i(U_2-c)B$. Substitution into the dynamic condition yields the Kelvin-Helmholtz dispersion relation with gravity,Put $\rho ={\rho }_1+{\rho }_2$ and $\overline{U}=({\rho }_1{U}_1+{\rho }_2{U}_2)/\rho$. Completing the square givesFor $k>0$, the temporal growth rate is $k{c}_i$. A growing complex conjugate branch therefore exists precisely when the radicand of the square root is negative:At equality the two phase velocities coincide and there is no exponential growth. Below it the roots describe real internal gravity waves. The denser lower fluid gives stable density stratification, but the tangential shear can overcome it. The assumption $U_1\ne U_2$ is needed for the displayed threshold; equal streams have no such shear instability.

For a square-free integer $\delta \equiv 2$ or $3(\mathrm{mod}4)$, the ring of integers of the quadratic field is ${\mathcal{O}}_{\mathbb{Q}(\sqrt{\delta })}=\mathbb{Z}[\sqrt{\delta }]$. Square-freeness is the usual convention underlying the first request: congruence alone would not suffice, for example $\delta =27$ gives the same field as $\delta =3$.

Now take the stipulated square-free $\delta \equiv 3(\mathrm{mod}4)$ and put $r=\sqrt{2}$, $s=\sqrt{\delta }$. For any integral $\alpha =a+br+cs+drs$, relative traces to the three quadratic subfields show $2a,2b,2c,2d\in \mathbb{Z}$. Those subfields all have integral bases $1$ and their square root, including $\mathbb{Q}(\sqrt{2\delta })$. Write $\alpha =(A+Br+Cs+Drs)/2$ with integer coefficients.

Its relative field norm to $\mathbb{Q}(s)$ isThus $AC$ is even, and the constant numerator is divisible by four. Since $\delta \equiv 3(\mathrm{mod}4)$, these conditions force $A,C$ both even: one odd makes the numerator odd, and both odd contradicts $AC$ even. They then force $B,D$ to have the same integer parity. Every algebraic integer consequently belongs to the lattice generated byConversely the last generator $w$ satisfies $w^2=(1+\delta )/2+\sqrt{\delta }$ and the monic integer equationAll four generators are integral and linearly independent over $\mathbb{Q}$, so this lattice is exactly ${\mathcal{O}}_L$, proving this integral basis of an odd biquadratic field containing square root of two without an unproved index assumption.

The cubic equation is $x_0{x}_2^2-x_1^3+x_0^2{x}_1=0$. On $x_0\ne 0$, put $u=x_1/x_0$, $v=x_2/x_0$, giving $v^2=u^3-u$. The projection is $u$, so its function-field extension is generated by a square root of $u^3-u$. This is not a square in $k(u)$, since its zeros at $0,1,-1$ have odd order. In characteristic different from two the extension is separable of degree two.

The only apparent basepoint is $P_{\infty }=(0:0:1)$. In the chart $x_2=1$, put $s=x_0$, $t=x_1$. The equation is $s-t^3+s^{2}t=0$, so $t$ is a local parameter and $s=t^3/(1+st)$. The ratio $s/t=t^2/(1+st)$ is regular and the projection extends there as $(s/t:1)$, with value $(0:1)$. Thus the displayed rational projection does define a morphism after this unique extension; it is not literally evaluable by the raw pair $(x_0,x_1)$ at that point.

At a finite point with $v\ne 0$, the derivative with respect to $v$ is $2v\ne 0$, so $u-u(P)$ is a parameter and $e_P=1$. At $v=0$, the roots $u=0,\pm 1$ are simple (also in characteristic three), and $v$ is a parameter with $u-u(P)$ of order two. At infinity the target parameter $s/t$ likewise has order two. ThereforeThe affine gradient and the nonzero $\partial f/\partial x_0$ at infinity also verify nonsingularity for every characteristic other than two. The ramification description is on geometric points; it descends to these stated rational points over $k$.

Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

A reduced positive definite binary quadratic form $f=[a,b,c]=a{x}^2+bxy+c{y}^2$ satisfieswith $b\ge 0$ when $|b|=a$ or $a=c$. Its discriminant of a binary quadratic form is $d=b^2-4ac<0$. The first displayed inequality already gives $|b|\le a$, while $c\ge a$ givesTaking the positive square root yieldsMoreover, $d=b^2-4ac\equiv b^2\equiv b(\mathrm{mod}2)$, so $b\equiv d(\mathrm{mod}2)$. For fixed negative $d$, the displayed bounds leave only finitely many possible integers $a$ and $b$, and then $c=(b^2-d)/(4a)$ is uniquely determined. Hence only finitely many reduced forms have discriminant $d$.

For $d=-15$, the bound gives $a\le \sqrt{5}$, so $a\in \{1,2\}$. The modular congruence condition makes $b$ odd. If $a=1$, then $b=\pm 1$ and $c=4$; the boundary convention $|b|=a\Rightarrow b\ge 0$ retains $[1,1,4]$. If $a=2$, then $b=\pm 1$ and $c=2$; the boundary convention $a=c\Rightarrow b\ge 0$ retains $[2,1,2]$. Thus the complete list is

Differentiate the squared Euclidean norm and use the skew-symmetric matrix identity $A^T=-A$:Thus $\Vert \mathbf{y}(t){\Vert }_2^2$ is constant, and continuity of the nonnegative square root gives

Put $Y=y^2$. On the inner supersonic branch, the equation from part (d) isIts asymptotic expansion as $x\to 0^{+}$ beginsTaking the positive square root givesSince $x=r/r_s$ and $y=u/c_s$,The leading term is the free-fall speed $\sqrt{2GM/r}$; the logarithmic term is the first pressure correction to the inner Isothermal Bondi accretion flow.

For any $x,y\in \mathbb{R}$, rationalizing the difference of square roots givessince $f(x),f(y)\ge 1$. Given $\epsilon >0$, use uniform continuity of $f$ with tolerance $2\epsilon$ to obtain $\delta >0$. Whenever $|x-y|<\delta$, the displayed estimate yields $|g(x)-g(y)|<\epsilon$. Thus $g$ has uniform continuity on $\mathbb{R}$. The square root function is in fact Lipschitz continuous with constant $1/2$ on $[1,\infty )$.

For $0\le x\le 7$, we have $2\le 2+x\le 9$. Taking the nonnegative square root givesThe derivative issoIn particular, the mean value theorem makes $G$ a contraction mapping on this interval.

The tail $(a_n{)}_{n\ge 1}$ is a monotone sequence bounded below by zero. The bounded monotone sequence theorem therefore gives $a_n\to L\ge 0$. Taking the limit of a sequence in $a_n^2\ge c$ yields $L^2\ge c>0$, so $L>0$. The reciprocal rule from part (a) permits passage to the limit of a sequence in the recurrence relation:Multiplication by $2L$ givesThis proves existence of the positive square root without assuming that existence when constructing the sequence.

Choose the first orthonormal basis vector to be ${\mathbf{e}}_1=\mathbf{a}$. If $\mathbf{b}$ is not parallel to $\mathbf{a}$, choose ${\mathbf{e}}_2$ along $\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$; otherwise choose any unit vector orthogonal to $\mathbf{a}$. Complete to an orthonormal basis with ${\mathbf{e}}_3$. The matrix with rows ${\mathbf{e}}_i^T$ is an orthogonal matrix, and expressing the three unit vectors in this basis givesThe last two components of ${\mathbf{c}}^{\prime }$ have polar coordinates of the displayed form; we may choose $w\ge 0$. If they vanish, $\theta$ is arbitrary.

An orthogonal transformation preserves dot products. Taking dot products with ${\mathbf{a}}^{\prime }$, and then using the unit Euclidean norms, givesThe remaining dot product givesAn orthogonal matrix has determinant $\pm 1$, so it preserves the absolute scalar triple product. In the new coordinates this scalar triple product is $yw\sin \theta$. ConsequentlyTaking the nonnegative square root provesNo division by $y$ or $w$ was used, so the proof includes parallel vectors. Equivalently, the expression under the square root is the Gram determinant of the three unit vectors.

Put $r=\sqrt{x^2+y^2}$. The definitions of the parabolic polar coordinates give $u^2=x+r$ and $v^2=-x+r$. Subtracting and multiplying these equations yieldsSince $u,v,y\ge 0$, taking the nonnegative square root givesThe Jacobian matrix has columns $(u,v)$ and $(-v,u)$. Its Jacobian determinant is thereforeHere the displayed fraction denotes the Jacobian determinant, as in the paper. The change of variables formula consequently gives $dxdy=(u^2+v^2)dudv$. The Jacobian determinant is positive except at the origin, and the definitions supply the unique nonnegative inverse throughout the region $y\ge 0$.

The hypothesis implies $v_1,v_2\ne 0$, since the zero vector is a scalar multiple of every vector. SetSince $v_1$ is not a scalar multiple of $v_2$, $w\ne 0$. Positive definiteness and bilinearity of the real inner product giveMultiplying by $\Vert v_2{\Vert }^2>0$ and taking square roots proves the strict Cauchy-Schwarz inequalitySince every real number $s$ satisfies $s\le |s|$, this proves, in particular,The argument derives the required inequality directly from the inner product axioms.

Rolle's theorem states that a continuous function $g:[u,v]\to \mathbb{R}$ with $u<v$, differentiable on $(u,v)$, and satisfying $g(u)=g(v)$ has $g^{\prime }(c)=0$ for some $c\in (u,v)$.

Fix $x>0$ and set $r=(\sqrt{1+x}-1)/x$. The functionis continuous and differentiable as required, and $g(0)=g(x)=0$. By Rolle's theorem, some $c\in (0,x)$ satisfiesSince $c>0$, this gives $r<1/2$. Multiplying by $x>0$ and adding one proves the strict square root bound

With the change of variables $z=y^2$, multiplication of the differential equation by $2y$ gives $z^{\prime }+2xz=2x$. The integrating factor is $e^{x^2}$, soThe initial condition gives $C=3$. Since $y$ starts negative and $z\ge 1$, continuity selects the negative square root throughout:This never reaches the singular value $y=0$.

For the cardioid, $r^{\prime }=-a\sin \theta$. The expression under the square root simplifies toSince $0\le \theta \le {\theta }_2\le \pi$, the cosine is nonnegative. ThereforeThe upper half of the cardioid starts at $(2a,0)$, passes through $(0,a)$, bends to negative $x$, and ends at the cusp $(0,0)$ when ${\theta }_2=\pi$. Its full upper-half arc length is $4a$; reflecting gives $8a$ for the full cardioid.

The fundamental theorem of calculus givesThis decreases from $2a$ to zero, so the arc length increases fastest at ${\theta }_2=0$, interpreted as the right-hand rate at the initial endpoint. If only positive endpoints are allowed, the rate has supremum $2a$ as ${\theta }_2\downarrow 0$, rather than an attained interior maximum.