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Use the Fourier transform convention $\hat{m}(t)=\int e^{-itx}dm(x)$. The complex measure is finite because ${\sum }_n|f(n)|n^{-\sigma }<\infty$. The Möbius divisor-sum identity, proved in the preceding solution, allows multiplication of absolutely convergent Dirichlet series and givesHere $\zeta$ is the Riemann zeta function. We prove the estimates in the intended range $1<\sigma \le 2$. A uniform estimate of the printed form for all $\sigma >1$ would be false: at $t=0$, $F(\sigma )\to 1$ as $\sigma \to \infty$, while its proposed upper bound tends to zero. A version valid for every $\sigma >1$ replaces $(\sigma -1{)}^{-3/4}$ by $1+(\sigma -1{)}^{-3/4}$.

For completeness, unique prime factorization and absolute convergence give $\zeta (s)={\prod }_p(1-p^{-s}{)}^{-1}$ andas the logarithm defined by that Euler product. The elementary nonnegative trigonometric polynomialtherefore proves the three-four-one inequality for Euler productsWe also need elementary uniform bounds for derivatives of the Riemann zeta function. For an integer $H\ge 1$, summation by parts givesThis continues the function to $\Re s>0$ with just the pole at one. For $1\le \sigma \le 2$, $|t|\ge 1$, choose $H=\lceil 2+|t|\rceil$. Differentiating $j$ times, the finite Dirichlet polynomial is bounded by ${\sum }_{n\le H}(\log n{)}^j/n{\ll }_j{\log }^{j+1}(2+|t|)$. Derivatives of the second term have the same bound, since $|s-1|\ge 1$. In the last term, differentiated integrands are bounded by $u^{-\sigma -1}(\log u{)}^j$, whose integral is $O_j(H^{-\sigma }{\log }^j(2H))$; multiplication by $|s|\ll H$ is harmless. It follows thatAs $\zeta (\sigma )\le 1+(\sigma -1{)}^{-1}$, the three-four-one inequality for Euler products gives, for $|t|\ge 1$,To handle bounded $t$, first note that $\zeta$ cannot vanish at $1+i{t}_0$ for $t_0\ne 0$: a zero of order $m\ge 1$ would make the three-four-one product $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound one. At $s=1$, the reciprocal $F$ extends holomorphically with a simple zero, since the pole of $\zeta$ has residue one. Thus $F$ and all its derivatives are bounded on the compact rectangle $1\le \sigma \le 2$, $|t|\le 1$. This proves the requested reciprocal zeta bounds near the line one:

For the derivative estimate, repeatedly differentiate $F=1/\zeta$. Every term of $F^{(k)}$ is a constant timesThe preceding derivative bounds and reciprocal estimate bound this by $O_k((\sigma -1{)}^{-3(k+1)/4}{\log }^{3k+1}(2+|t|))$. Compact-height bounds supply the remaining range. Since ${\partial }_t^{k}F(\sigma +it)=i^k{F}^{(k)}(\sigma +it)$, this has exactly the requested dependence, with an explicit logarithmic exponent linear in $k$ for $k\ge 1$.

Here is the unsmoothing a logarithmically weighted sum argument with the normalization repaired. For real $v$,Indeed the Fourier transform of $u{e}^{-\sigma u}{1}_{u\ge 0}$ is $(\sigma +it{)}^{-2}$, so Fourier inversion gives the identity, also at $v=0$ by continuity. In a vertical contour integral the coefficient is $1/(2\pi i)$ and the differential is $ds=idt$. The printed real-$t$ integral has $dt$ with $1/(2\pi i)$ and is missing this factor of $i$.

Set $a_m=\mu (m)(\log m{)}^k$, taking $(\log 1{)}^0=1$ when $k=0$. Its Dirichlet series is $(-1{)}^k{F}^{(k)}(s)$. The logarithmically smoothed Perron formula, justified by absolute convergence, givesFor $x\ge 3$ choose $\sigma =1+1/\log x$. The integral of ${\log }^{3k+1}(2+|t|)/|\sigma +it{|}^2$ is finite, with a constant depending only on $k$, and $x^{\sigma }=ex$. Thus, with $\beta =3(k+1)/4$,For $0<h\le 1$, subtraction gives the exact finite-difference identityThe last sum is $O_k((x{h}^2+h)(\log x{)}^k)$, since there are $O(xh+1)$ terms. ConsequentlyFor $k\ge 3$, choose $h=(\log x{)}^{(\beta -k)/2}$ to get exponent $(\beta +k)/2=7k/8+3/8$. For $0\le k<3$, the trivial bound $O_k(x(\log x{)}^k)$ is already no larger than that estimate. The extra $(\log x{)}^k$ term is absorbed for fixed $k$. HenceThis also yields the cancellation needed for the Prime number theorem. Choose a fixed $k>3$, discard $m\le \sqrt{x}$ at cost $O(\sqrt{x})$ for the unweighted sum, and use partial summation with $(\log m{)}^{-k}$ on the rest. The displayed bound gives $M(x)=O_k(x(\log x{)}^{-k/8+3/8})=o(x)$; the preceding solution then gives $\psi (x)\sim x$.

By the tip correspondence in the conjugacy,The time-dependent Itô formula is valid after localization away from the pole and the terminal time $S$. The map's time variation is of finite variation, so there is no extra stochastic cross-variation term. Using the supplied conjugacy derivative,ConsequentlyThe cancellation is special to $\kappa =6$: for a general SLE parameter the drift would be $(\kappa /2-3){\Phi }_t^{\prime \prime }({\xi }_t)dt$. This is the transformed SLE driving function, and explains the special Locality property of SLE at six.

The conformal automorphism of the upper half-plane with the indicated three boundary values isIts Möbius transformation matrix has determinant one, so it preserves $\mathbb{H}$. The inverse is ${\Phi }^{-1}(w)=w/(1-w)$. In particularon the extended real boundary. Therefore the image trace's first contact with the latter ray corresponds to the original time $S$. The map sends the original target at infinity to the new target $1$; it does not preserve the target of the chordal curve.

We use the image half-plane capacity as its time parameter:Since ${\Phi }_t$ is a conformal automorphism of the upper half-plane analytic at ${\xi }_t$, its real derivative there is strictly positive. Hence $u$ is continuous and strictly increasing before $S$. The original time $t$ and image time $u$ must be kept distinct throughout the comparison.

The PDF gives an inverse-gamma distribution with shape $k$ and scale $\theta$, and a gamma distribution prior with shape $\alpha$ and rate $\lambda$. Write $\mu (\theta )=\theta /(k-1)$. The supplied conditional moments, together with $\mathbb{E}\theta =\alpha /\lambda$ and $\mathbb{E}{\theta }^2=\alpha (\alpha +1)/{\lambda }^2$, give the three structural quantities of the Bühlmann model:Here $v$ is the variance of hypothetical means and $s^2$ the expected process variance. The law of total covariance and conditional independence of different years implyAll these quantities are finite because $k>2$.

For any coefficients $c_i$, minimizing the mean squared error over the intercept first gives $c_0=m(1-{\sum }_i{c}_i)$. Let $Y=\mu (\theta )-m$ and $W_i=X_i-m$. Expanding the remaining mean squared error givesIts normal equations are $s^2{c}_i+v{\sum }_j{c}_j=v$ for every $i$. Since $s^2>0$, subtracting any two equations shows that all $c_i$ agree. The quadratic part is strictly positive for every nonzero coefficient change, so the solution is the unique global minimizer. Solving givesA credibility estimate combines individual experience with the collective expected value, using an experience weight $Z\in [0,1]$. In this equal-exposure setting, the Bühlmann credibility premium isThus the computed coefficients give exactly a credibility estimate, with a fixed credibility factor determined by known parameters and the number of observed years. This is Bühlmann credibility for inverse-gamma observations.

For the final Bayesian calculation, the likelihood function in $\theta$ is proportional to ${\theta }^{nk}\exp (-\theta {\sum }_i{x}_i^{-1})$. Multiplication by the gamma distribution prior gives the Gamma prior for an inverse-gamma scale updateUnder quadratic loss, the conditional risk of an estimate $d$ is $\mathrm{Var}(\mu (\theta )\mid x)+(d-\mathbb{E}[\mu (\theta )\mid x]{)}^2$. Its minimizer is therefore the posterior mean:This is not a fixed-weight arithmetic-mean credibility estimate. For $n=2$, the observations $(1,3)$ and $(2,2)$ have the same sample mean, but reciprocal sums $4/3$ and $1$, hence different posterior means. For any $n>2$, append the same positive observations to these two samples. For $n=1$, the estimate is $Cx/(1+\lambda x)$, with $C=(\alpha +k)/(k-1)$; its second derivative is $-2C\lambda /(1+\lambda x{)}^3\ne 0$, so it is not affine either. Symmetry in the observations would force equal slopes in any affine representation, so these examples also exclude a general fixed-coefficient linear representation.

One can express the posterior mean as a broader data-dependent blend. With $A={\sum }_i{x}_i^{-1}$ and harmonic mean $H=n/A$, it isThis weights a harmonic-mean-based experience estimate, and the weight itself depends on the observations. It is not the credibility estimate based on $\overline{X}$ and the fixed Bühlmann credibility factor derived above.

Assume $n\ge 1$ and $\Vert p{\Vert }_{[-1,1]}\le 1$. The given Bernstein inequality for algebraic polynomials shows that $q=p^{\prime }$ satisfies the nodal hypotheses of (a). Consequently, in the two endpoint regions $\cos (\pi /(2n))\le |x|\le 1$, $|p^{\prime }(x)|\le |T_n^{\prime }(x)|$.

For $0<\theta <\pi$, the finite geometric identitybounds its absolute value by $n$. It follows that $|T_n^{\prime }(\cos \theta )|\le n^2$, and the endpoint limit gives $T_n^{\prime }(1)=n^2$. In the central interval, the Bernstein inequality for algebraic polynomials givessince $\sin t\ge 2t/\pi$ on $[0,\pi /2]$. Combining the regions and rescaling proves the Markov inequality for polynomial derivatives:The Chebyshev polynomial itself attains equality, so this constant is sharp. Constant polynomials have zero derivative.

Put $s_j=t_{i+j}$ for $0\le j\le k$, with the knots in increasing order. The explicit formula for a divided difference at distinct nodes givesEach truncated power function is a piecewise polynomial function of degree $k-1$ and is $C^{k-2}$ for $k\ge 2$. Therefore their sum has the same global smoothness. The jump in the $(k-1)$st derivative at $s_j$ equalswhich is nonzero. Thus each $s_j$ really is a knot, and generally there is no additional smoothness.

If $t<s_0$, the values being differenced are the values of the degree-$k-1$ polynomial $(s-t{)}^{k-1}$, so their $k$th divided difference vanishes. If $t>s_k$, every sampled truncated power function is zero. To check that both support endpoints occur, for $s_0<t<s_1$ subtract the vanished full polynomial divided difference to obtainFor $s_{k-1}<t<s_k$, only the last term survives and is positive. The nonnegativity from the recurrence in (b) actually gives positivity throughout the interior: for each $s_0<t<s_k$, at least one lower-order B-spline covering $t$ has a positive recurrence coefficient, and induction starts with positive order-one indicators. Hence $\mathrm{supp}{M}_i=[t_i,t_{i+k}]$ and $M_i\in C^{k-2}$ for $k\ge 2$, with degree $k-1$ on its knot intervals. For $k=1$, the B-spline is the step function $1_{[s_0,s_1)}(t)/(s_1-s_0)$, so no continuity is asserted.

The unit-integral normalization of a B-spline also follows directly: integrating $k(s_j-t{)}_{+}^{k-1}$ over $s_0\le t\le s_k$ gives $(s_j-s_0{)}^k$. The $k$th divided difference of this monic degree-$k$ polynomial is one. Thus $\int M_i(t)dt=1$, explaining its normalization.

For real $x$, the function $F(x)=x^2{e}^{-x}$ has derivative $F^{\prime }(x)=x(2-x)e^{-x}$. It decreases from infinity to zero on the negative half-line, increases from zero to $4e^{-2}$ on $(0,2)$, and then decreases to zero. Hence sufficiently small positive $ϵ$ gives exactly three real roots: two near zero and one tending to positive infinity.

Set $\delta =\sqrt{ϵ}$ and write either small root as $x=s\delta +A{\delta }^2+O({\delta }^3)$, with $s=\pm 1$. Expanding $x^2{e}^{-x}$ gives ${\delta }^2+s(2A-1){\delta }^3+O({\delta }^4)$. Matching its value to ${\delta }^2$ fixes $A=1/2$. Thus the two asymptotic expansions areFor the large root, put $L=\log (1/ϵ)$. Taking logarithms gives $x-2\log x=L$, so $x/L\to 1$. Writing $x=L+r$ then gives $r=2\log L+2\log (1+r/L)$. Iteration yieldsIts two asymptotic expansion terms are of different logarithmic orders; a regular power series in $ϵ$ cannot describe it.

As an exact check, the real branches of Lambert W give $x_{-}=-2W_0(\sqrt{ϵ}/2)$, $x_{+}=-2W_0(-\sqrt{ϵ}/2)$ and $x_{\mathrm{large}}=-2W_{-1}(-\sqrt{ϵ}/2)$. If complex roots are intended instead, there are infinitely many, given by $-2W_j(\pm \sqrt{ϵ}/2)$. For each fixed nonprincipal branch their two logarithmic terms follow from $W_j(z)=L_j-\log L_j+O(\log |L_j|/|L_j|)$, where $L_j=\log z+2\pi ij$ and logarithms are continued on that branch. The three real roots are the ones selected above.

For fixed $0\le \alpha <1$, the phase derivative is $-1+\alpha \cos x\le -(1-\alpha )<0$ on the integration interval. Its maximum is at $x=0$, with phase zero and derivative $-(1-\alpha )$. The Laplace endpoint estimate, or the substitution $x=t/[\lambda (1-\alpha )]$, givesThis fixed-parameter formula is not uniform as $\alpha \uparrow 1$; its diverging coefficient signals the distinguished limit derived in the parent solution.

One convenient version of Watson's lemma is the following. Suppose $G(t)$ is integrable away from zero and has, as $t\downarrow 0$, an asymptotic expansion $G(t)\sim {\sum }_{j\ge 0}{c}_j{t}^{\beta +j-1}$, where $\beta >0$. On a finite interval, or on an infinite interval with suitable exponential growth control,To prove a finite truncation, split the integral at a fixed small $\delta >0$. The part away from zero is exponentially small. On $(0,\delta )$ subtract the first $N$ local terms and bound the remainder by $C{t}^{\beta +N-1}$; its integral is $O({\lambda }^{-\beta -N})$. Extending each polynomial integral to infinity changes it only exponentially, and the substitution $u=\lambda t$ produces the Gamma integral. A little-o local remainder gives the corresponding little-o asymptotic remainder. This is a proof by localization and remainder bounds, rather than merely formal integration of a series.

For the monotone-phase integral, assume the derivatives needed below exist and are bounded near $a$, and the rest of the integral is controlled. Put $t=\varphi (a)-\varphi (x)$. Since ${\varphi }^{\prime }(a)<0$, this is a regular endpoint coordinate andDifferentiating with $dx/dt=-1/{\varphi }^{\prime }$ gives $G(0)=-g(a)/{\varphi }^{\prime }(a)$ and $G^{\prime }(0)=g^{\prime }(a)/{\varphi }^{\prime }(a{)}^2-g(a){\varphi }^{\prime \prime }(a)/{\varphi }^{\prime }(a{)}^3$. Watson's lemma therefore gives the two-term monotone-endpoint Laplace expansionAn $O(e^{\lambda \varphi (a)}{\lambda }^{-3})$ remainder follows, for example, from a sufficiently smooth transformed amplitude with bounded second derivative near zero. The displayed second coefficient can vanish for special amplitudes without changing the expansion.

The unheaded transition request concerns a cubic endpoint-to-saddle transition. Near $x=0$ its exponent has the expansionAt the cubic endpoint $x$ has size ${\lambda }^{-1/3}$. Requiring the linear term to contribute on the same scale gives $1-\alpha =O({\lambda }^{-2/3})$. Thus the distinguished limit and its uniform leading integral areFor fixed $\nu$, the substitution $x={\lambda }^{-1/3}t$ makes the omitted exponent terms $O({\lambda }^{-2/3})$ on bounded $t$ intervals. Cubic decay controls the remaining tail, while the original interval away from zero is exponentially negligible. The estimate is uniform for $\nu$ in compact sets. In the printed range $\alpha \le 1$, $\nu \ge 0$; the same transition integral also makes sense for negative fixed $\nu$.

At $\nu =0$, the Gamma integral gives $h(0)=6^{1/3}\Gamma (1/3)/3$, recovering the cubic endpoint regime. For $\nu \to +\infty$, scale $t=u/\nu$ and expand $e^{-u^3/(6{\nu }^3)}$ to obtain $h(\nu )={\nu }^{-1}-{\nu }^{-4}+O({\nu }^{-7})$. Its leading term gives ${\lambda }^{-1/3}/\nu =1/[\lambda (1-\alpha )]$, recovering the ordinary endpoint regime in their overlap. Both matching limits are therefore checked explicitly.

Work above the plane, with $e^{-i\omega t}$ time dependence and $0<\theta <\pi$ away from grazing incidence. Put $a=k\cos \theta$ and $b=k\sin \theta$. The incident field and its chosen flat Dirichlet reflection cancel in value at the plane, while their normal derivative is $-2ib{e}^{iax}$. Thus the diffracted correction has zero trace on the negative half-line and prescribed normal derivative $2ib{e}^{iax}$ on the positive half-line.

There is an incompatibility in the stated edge data: a normal derivative that is identically zero on the Neumann half-line cannot diverge as that half-line approaches the junction. The inverse-square-root normal-derivative singularity belongs to the Dirichlet side, $x\to 0^{-}$; on $x>0$ the total normal derivative remains zero. The following solution satisfies the mixed conditions and bounded-field edge behavior, with that corrected side.

Take the transform and its inverse asLet $U_{+}$ be the transform of the diffracted boundary value, and $V_{-}$ the transform of its unknown normal derivative on the negative half-line. The outgoing transformed field is $U_{+}(\alpha )e^{i\beta (\alpha )y}$, whereContinue $k$ into the upper half-plane. Choose branches so that $\beta (0)=k$, outgoing propagating waves have positive vertical wavenumber, and evanescent waves decay. The branch point $-k$ lies below $C$ and belongs to ${\beta }_{+}$; $k$ lies above $C$ and belongs to ${\beta }_{-}$. Since ${\int }_0^{\infty }{e}^{i(\alpha +a)x}dx=i/(\alpha +a)$, the boundary transform equation isThe forcing splits by subtracting its numerator at the incident pole:The pole in $H_{-}$ is removable; $H_{-}$ is analytic below $C$, while $H_{+}$ is analytic above it. Hence $i{\beta }_{+}{U}_{+}-H_{+}=V_{-}/{\beta }_{-}+H_{-}$ is entire. The bounded mixed-boundary edge solution has $U_{+}=O({\alpha }^{-3/2})$ and $V_{-}=O({\alpha }^{-1/2})$, making both sides decay at infinity. These orders can also be obtained locally from the leading wedge solution $r^{1/2}\cos (\phi /2)$, which is Neumann at $\phi =0$ and Dirichlet at $\phi =\pi$. Liouville's theorem sets the entire remainder to zero. ThereforeThe solution of mixed Dirichlet-Neumann half-plane diffraction is consequentlyFor $k=k_0+i\epsilon$, take $C$ from left to right inside $-\epsilon <\mathrm{Im}\alpha <\epsilon$ and above the incident pole $-a$. Cuts from $-k$ run into the lower half-plane and cuts from $k$ into the upper half-plane, without crossing $C$. For example choose $\mathrm{Im}\alpha =c$ with $\max (-\epsilon ,-\epsilon \cos \theta )<c<\epsilon$. Then let $\epsilon \downarrow 0$, keeping the contour above $-a$ and $-k$ and below $k$. Any subsequent deformation for asymptotic evaluation must preserve these bypasses or include the residues of crossed poles. This prescription fixes the reflected-wave contributions and the radiation condition; a principal-value integral alone would not do so.

As a boundary check, $i\beta U_{+}=-2b{\beta }_{-}/[\sqrt{k+a}(\alpha +a)]$. On the positive half-line the lower-contour residue gives $2ib{e}^{iax}$, cancelling the incident-plus-reflected derivative. The boundary trace vanishes on the negative half-line by upper-half-plane analyticity. Finally $U_{+}=O({\alpha }^{-3/2})$ gives a square-root boundary value and hence a bounded field at the junction; the singular derivative occurs on its Dirichlet side.

Illuminate a surface window of length $D$, and define ${\hat{h}}_D(Q)={\int }_{-D/2}^{D/2}h(x)e^{-iQx}dx$. In the far field put $(x,z)=r(\sin {\theta }_s,\cos {\theta }_s)$, with $|{\theta }_s|<\pi /2$ away from grazing. The spectral phase $\xi \sin {\theta }_s+\beta (\xi )\cos {\theta }_s$ has stationary point ${\xi }_s=k\sin {\theta }_s$, value $k$, and second derivative $-1/(k{\cos }^2{\theta }_s)$. Stationary phase therefore givesLet $C_h(s)=⟨h(x+s)h(x)⟩$ and $S_h(Q)=\int C_h(s)e^{-iQs}ds$. Statistical stationarity gives exactlyIf $C_h$ is integrable, division by $D$ and dominated convergence yield $⟨|{\hat{h}}_D(Q){|}^2⟩/D\to S_h(Q)$. Thus the leading rough-surface far-field intensity per illuminated length isThis uses the field-intensity convention $I=|\psi {|}^2$; if $\psi$ is acoustic pressure, multiply by $1/(2\rho c)$ for physical far-field energy flux. The variance normalization is ${\sigma }^2=C_h(0)=(2\pi {)}^{-1}\int S_h(Q)dQ$. Consequently the diffuse intensity is of order ${\sigma }^2$ and samples the surface spectrum at the tangential momentum transfer. Its value is not determined by the r.m.s. height alone.

The finite window must first be in its far field, for example $r\gg k{D}^2$, before interpreting the large-window intensity density. An infinitely long stationary surface has infinite illuminated power and no finite total far-field intensity at a fixed point. For nonintegrable correlations the analogous result is a spectral measure, possibly with atoms. Zero mean height makes the linear rough mean field zero, but does not make its mean intensity zero; the flat specular reflection remains separate.

In first Born order the unknown potential enters linearly. Let $M_i$ multiply by the known incident field, $\mathcal{G}$ be the outgoing Green integral over the known support $D$, and $\mathcal{R}$ restrict the resulting field to the measurement set. With $y$ the measured scattered data,The formal least-squares, minimum-norm least-squares solution is $V=A^{†}y$, or $V=(A^{\ast }A{)}^{-1}{A}^{\ast }y$ on a domain where that inverse is meaningful. The refractive index is thenchoosing the physical branch approaching one outside the scatterer. At weak contrast $n-1\simeq V/(2k^2)$.

A noisy stable reconstruction uses Tikhonov regularization:A smoothness penalty $\alpha \Vert LV{\Vert }^2$ may be substituted when appropriate, giving $A^{\ast }A+\alpha L^{\ast }L$ in the normal equation for a linear inverse problem, under the corresponding coercivity assumptions. The parameter balances noise amplification and smoothing bias.

The data geometry matters: knowing a scattered field on a measurement surface does not automatically mean knowing it throughout the support. A single incident wave and fixed-frequency far field sample only a shifted sphere of the potential's Fourier transform, and in general do not determine an arbitrary three-dimensional potential uniquely. Tikhonov regularization stabilizes a selected minimum-norm least-squares solution; it cannot create missing information or guarantee uniqueness from inadequate data. If the total field and its derivatives were known throughout $D$, the first-Born equation could instead be inverted formally as $V\simeq -(\Delta +k^2){\psi }_s/{\psi }_i$ where ${\psi }_i\ne 0$, but differentiation still amplifies measurement noise.

Write $f(u)=a{u}^2/(b^2+u^2)$, with $b>0$. Besides zero, the fixed points satisfy $u^2-au+b^2=0$. If $a^2>4b^2$, they areThe derivative is $f^{\prime }(u)=2a{b}^{2}u/(b^2+u^2{)}^2$. At a positive fixed point, $b^2+u^2=au$, so $f^{\prime }(u)=2b^2/(au)=2(1-u/a)$. Thus zero is attracting, $u_{-}$ is repelling because $f^{\prime }(u_{-})>1$, and $u_{+}$ is attracting because $0<f^{\prime }(u_{+})<1$.

Moreover,For $0<u_0<u_{-}$ the iterates strictly decrease and stay positive. Their limit is a fixed point below $u_{-}$, hence zero. For $u_{-}<u_0<u_{+}$ the iterates increase towards $u_{+}$; for $u_0>u_{+}$ they decrease towards $u_{+}$. Monotonicity of $f$ prevents crossing the limiting fixed point. $u_{-}$ is the extinction threshold. The cobweb plot illustrates these alternatives.

For $a^2<4b^2$ there are no positive fixed points, and $f(u)<u$ for every $u>0$, so all populations tend to zero. At $a^2=4b^2$, the positive fixed point $a/2$ is semistable: initial populations below it tend to zero, and those above it decrease towards it. This is a saddle-node bifurcation in the discrete dynamics.

Work geometrically over an algebraic closure; the degree and ramification do not change under this base extension. The coordinate squaring map lands in $W$ because its defining equation pulls back to that of $V$. Above a generic target point with all coordinates nonzero there are $2^3$ square-root choices, identified in pairs by simultaneous sign change. HenceThe effective sign-change group is $C_2\times C_2$. At a point with all coordinates nonzero its stabilizer is trivial, and the squaring map is locally étale, since characteristic is not two. Thus $e_p=1$ there. If one coordinate is zero, a sign change of that coordinate fixes the point, giving stabilizer order two. More explicitly, if $X_0=0$, use $t=X_0/X_2$ as a local parameter; the curve equation has nonzero derivative in the other coordinate, so this is valid, and the corresponding target parameter is $t^2$. ThereforeTwo coordinates cannot vanish on either curve. For each vanishing coordinate, the other-coordinate ratio solves $z^8=-1$, giving eight distinct points; there are 24 ramification points. As a check, the smooth plane-curve genera are $21$ and $3$, and Riemann-Hurwitz formula gives $2\cdot 21-2=4(2\cdot 3-2)+24$.

The Theorema Egregium states that Gaussian curvature is determined by the first fundamental form, and therefore preserved by local isometries. For a local surface parametrization $r$, write $g_{ij}=r_i\cdot r_j$, choose a unit normal $N$, and set $b_{ij}=r_{ij}\cdot N$. Decomposing the second derivatives into tangent and normal parts givesCommute the third partial derivatives of $r$ and take tangent components. The terms involving derivatives of $b$ are normal; the tangent terms give the Gauss equationwhere $R$ is computed from derivatives and products of the displayed Christoffel symbols, using $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. For example the tangent term from differentiating $b_{ij}N$ is $-b_{ij}{b}_k{}^{\ell }{r}_{\ell }$, which produces exactly the difference of the two products on the right. ConsequentlyThe last expression depends only on $g$ and its derivatives. Pulling back $g$ under a local isometry therefore preserves $K$, proving the theorem.