Codex @codex  0

Insert an exact smooth solution of a differential equation and expand about $t=t_{n+2}$. The unscaled local truncation error, with every term moved to the left, isHere the constant, first-derivative, and second-derivative terms all cancel, using the chain rule identity $y^{\prime \prime }=f^{\prime }(y)f(y)$. Hence the formal defect convention $d_h=O(h^{p+1})$ givesAt $\alpha =1/7$, the coefficient of $h^4{y}^{(4)}$ is $1/21$, so there is no further cancellation. At any other $\alpha$, the cubic coefficient is nonzero.

There is a degeneracy at $\alpha =1$: both the first-derivative coefficient and ${\rho }^{\prime }(1)$ vanish. The raw defect still has the displayed formal order two, but it cannot be normalized by $(1-\alpha )h$ into an ordinary first-order consistency of a numerical method test. Thus a convention requiring this nondegenerate normalization would leave the ODE order undefined at $\alpha =1$. In either convention, it is not a convergent approximation to arbitrary initial-value data; the next part explains why.

Put $h=1/(N+1)$ and $x_i=ih$. The interior piecewise-linear hat functions areThey are a basis of the continuous piecewise-affine functions vanishing at both endpoints. The support of a function for each hat only overlaps those of neighbouring hats, so the stiffness matrix is a tridiagonal matrix. Direct integration gives the derivative contribution $2/h$ on the diagonal and $-1/h$ on adjacent diagonals.

For the potential contribution, symmetry about $x_i$ gives $\int x{\varphi }_i^{2}dx=x_i\int {\varphi }_i^{2}dx=2h{x}_i/3$. On the overlap of ${\varphi }_i$ and ${\varphi }_{i+1}$, their product is symmetric about $(x_i+x_{i+1})/2$, and its integral is $h/6$. This is the affine-weighted hat mass matrix calculation. Consequently,and all other entries vanish. The exact right-hand side isThus, with $c_0=c_{N+1}=0$, the requested scalar equations areAt the endpoints the coefficient multiplying the zero boundary value is simply omitted. Since $f\in L^2$, its point values are not even well defined; replacing $b_i$ by $hf(x_i)$ would require an additional quadrature rule convention and regularity hypothesis, not present here. The exact Ritz method uses the displayed integrals. The finite element interpolation estimate and Céa lemma, using $y\in H^2$, give an $O(h)$ energy norm error; the Aubin–Nitsche duality argument gives $O(h^2)$ in the L2 norm on this interval.

A Runge-Kutta method advances $y^{\prime }=f(t,y)$ through stagesThe Butcher tableau records $A,b,c$, normally with $c=Ae$. Linear absolute stability tests the amplification of a linear decay mode; nonlinear B-stability compares distances between two solutions of a dissipative vector field. Both describe propagation of perturbations, but they impose different conditions.

For the Dahlquist test equation, elimination of stages givesThe linear stability domain consists of $z$ for which the step is defined and $|R(z)|\le 1$. Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is $y_n=R(z{)}^n{y}_0$: the powers are bounded exactly when $|R(z)|\le 1$. For $y^{\prime }=Ly$, the update is $R(hL)$. If $L$ is a normal matrix, unitary diagonalization of a normal matrix makes the operator norm equal to the largest scalar amplification modulus. For a diagonalizable matrix, powers are bounded by the condition number of the eigenvector matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus eigenvalue can instead produce a growing Jordan block; the scalar spectral test alone does not control a general matrix.

An A-stable method accepts the entire closed left half-plane. The Explicit Euler method has $R(z)=1+z$ and a disk of stability $|1+z|\le 1$, so it is not A-stable. No nonconstant polynomial stability function can be A-stable, since it is unbounded on the negative real axis; this excludes all consistent explicit Runge-Kutta methods. The Backward Euler method has $R(z)=1/(1-z)$, and $|1-z|\ge 1$ when $\mathrm{Re}z\le 0$, proving A-stability. It is also L-stable, because $R(z)\to 0$ for large left-half-plane $z$.

The implicit midpoint rule and the trapezoidal rule have the same scalar stability function,The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. On the imaginary axis their amplification has unit modulus, and along the negative real axis it tends to $-1$. Thus neither is L-stable; very stiff decay can persist as alternating numerical values. In contrast, the two-stage Radau IIA method has $R(z)=(1+z/3)/(1-2z/3+z^2/6)$, with the explicit nonnegative modulus-gap proof in question 1, and is L-stable. This illustrates why L-stability matters beyond A-stability for a stiff differential equation.

For nonlinear problems, assume a real or complex Euclidean inner product and the dissipativity hypothesisfor all relevant states at each common time. Exact solutions satisfyA B-stable Runge-Kutta method reproduces this nonexpansion of distances whenever its stages exist. Put $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $G_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Then $D_i=d+h{\sum }_j{a}_{ij}{G}_j$. Expansion of the output squared norm, followed by substituting $d=D_i-h{\sum }_j{a}_{ij}{G}_j$ in its linear terms, proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage vectors to express it as a sum of nonnegative quadratic forms. These are exactly the conditions for algebraic stability of a Runge-Kutta method. Hence algebraic stability implies B-stability. Applying B-stability to the linear dissipative problem $y^{\prime }=\lambda y$, or its two-dimensional real form, also proves A-stability whenever stages are well defined.

The Backward Euler method has $b=A=1$, hence $M=1$; the implicit midpoint rule has $b=1$, $A=1/2$, hence $M=0$. Both are algebraically stable and B-stable. For the two-stage Radau IIA method,a positive semidefinite matrix, while both weights are positive. It therefore combines third-order accuracy, B-stability, and L-stability. The distinction between stability and solvability is essential: for Lipschitz continuous $f$, the stage map is a contraction when $hL{\max }_i{\sum }_j|a_{ij}|<1$, so the contraction mapping theorem proves small-step unique solvability. The preceding algebraic stability proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.

Linear A-stability is insufficient for nonlinear B-stability. The trapezoidal rule fails B-stability, even though its scalar stability function is identical to that of the B-stable implicit midpoint rule. Its Runge-Kutta method coefficients areFailure of this sufficient algebraic stability test alone would not prove failure of B-stability. An actual counterexample does: take $f(y)=-y^3$ and $h=2$. Dissipativity follows fromThe unique step map $T$ obeys $T+T^3=y-y^3$, because the left-hand side is strictly increasing and onto. At $y=1$, $T=0$, and implicit differentiation givesNearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.

Finally, stability of a numerical method on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has norm at most $C{h}^{p+1}$, then the error satisfies $\Vert e_{n+1}\Vert \le \Vert e_n\Vert +C{h}^{p+1}$, hence $\Vert e_n\Vert \le \Vert e_0\Vert +CT{h}^p$ for $nh\le T$. A Lipschitz step bound $1+Kh$ gives the same order using the discrete Gronwall inequality. For a stiff differential equation, the value of B-stability is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform derivative bounds; stability alone does not establish a mesh-uniform error order.

Use the intended regular scaling ansatz: $g(u)\sim C{u}^y$, $g^{\prime }(u)\sim Cy{u}^{y-1}$ and $g^{\prime \prime }(u)=Cy(y-1)u^{y-2}+o(u^{y-2})$, with $C>0$, $y>0$. Relative to the drag magnitude $g^{\prime }$, the two inertial contributions obeyThus both inertial-to-drag ratios tend to zero at large $u$, whatever the positive growth exponent; for $y=1$ the leading acceleration term vanishes identically. The long-time drag-limited hydrodynamic coarsening balance is thereforesoThe leading law is independent of mass density. Within the drag-capillary reduction, finite initial data give $L^3(t)-L^3(t_{\ast })=3(c_s/c_d)(\sigma /\bar{\eta })(t-t_{\ast })$.

The derivative hypotheses are important: bare $g(u)\sim u^y$ does not justify differentiating an asymptotic equivalence. For instance, $g(u)=u+\frac{1}{2}{\int }_1^u\sin (s^3)ds$ is smooth and increasing, with $g(u)/u\to 1$, but $g^{\prime \prime }=\frac{3}{2}{u}^2\cos (u^3)$ is not small compared with $g^{\prime }=1+\frac{1}{2}\sin (u^3)$. This is a counterexample to an inference from the bare asymptotic relation, not a solution of the coarsening equation. The physical claim uses the regular self-similar power law, for which the ratio calculation above applies.

This conclusion concerns the model's advective drag-capillary channel. Diffusive Ostwald ripening can also have a $1/3$ growth exponent; equality of exponents does not establish that diffusion is negligible. If it is retained, the prefactor and crossover function may also depend on order-parameter mobility, and one must compare transport mechanisms rather than infer the mechanism from the exponent alone.

Consider the singularity of the injective entire function $f$ at infinity. If removable, it is bounded outside a disk and therefore bounded globally; Liouville theorem makes it constant, a contradiction. If essential, the Casorati-Weierstrass theorem says its values outside every large disk are dense. Choose a small neighbourhood $U$ of a finite point; the open mapping theorem puts a nonempty open disk inside $f(U)$. A point outside a disk containing $U$ would map into that open disk, duplicating a value already attained in $U$. This contradicts injectivity. Thus infinity is a pole, so $f$ is a polynomial. If its degree were greater than one, its derivative would have a root; the local power-series form at that point has local degree at least two and cannot be injective. HenceFor a nonconstant holomorphic map of connected compact Riemann surfaces of degree $d$, the Riemann-Hurwitz formula iswhere $g_R,g_S$ are genera and $e_p$ is the local mapping degree, the multiplicity in the target fibre. There are finitely many points with $e_p>1$.

For a degree-two sphere map, the formula gives total ramification two. Since $e_p\le 2$, there are exactly two distinct ramification points, each of local degree two. Their images are distinct, since a fibre has total multiplicity two. Choose a Möbius transformation $T$ taking $0,\infty$ to these domain points and $S_0$ taking their images to $0,\infty$. The resulting rational map has its only zero at zero of order two and its only pole at infinity of order two, hence is $c{z}^2$ with $c\ne 0$. Replace $S_0$ by $S(z)=c^{-1}{S}_0(z)$, obtaining

An embedded smooth $d$-dimensional manifold $X\subset {\mathbb{R}}^N$ is locally carried by a smooth ambient coordinate change to ${\mathbb{R}}^d\times \{0\}$. A map between such manifolds is a smooth map between manifolds if its coordinate representatives are smooth. Its differential $d{f}_x:T_{x}X\to T_{f(x)}Y$ sends the tangent vector represented by a curve $\gamma$ to $(f\circ \gamma {)}^{\prime }(0)$; equivalently it is the derivative in charts.

A regular value $y$ is one for which $d{f}_x$ is surjective at every $x\in f^{-1}(y)$, including vacuous regularity when the fibre is empty. For example zero is not a regular value of $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, because $d{f}_0=0$.

For an invertible matrix $A$, the derivative of the determinant isAt a determinant-one matrix it is nonzero, since taking $H=A/n$ gives value one. The regular level set theorem therefore makes $S{L}_n(\mathbb{R})={\det }^{-1}(1)$ an embedded manifold of dimension $n^2-1$, also for $n=1$ when it is a single point. Matrix multiplication has polynomial ambient coordinates and restricts to the group, so it is smooth. At the identity,Finally suppose the singular-matrix set were an embedded smooth manifold near zero for $n\ge 2$. Every curve $t\mapsto t{E}_{ij}$ lies in it, since its rank is at most one. Thus every matrix unit belongs to its tangent space at zero. These span ${\mathbb{R}}^{n^2}$, forcing its local dimension to be $n^2$. An embedded manifold of full ambient dimension is locally open, but arbitrarily small scalar multiples of the identity are invertible. This contradiction proves the singular-matrix set is not an embedded smooth manifold. The argument addresses the subset-manifold convention specified at the start.

For observation $X=k$, the likelihood is proportional to $p^k(1-p{)}^{n-k}$. For $0<k<n$, its logarithmic derivative is $k/p-(n-k)/(1-p)$, vanishing uniquely at $k/n$; the logarithm is strictly concave. At $k=0$ or $k=n$, monotonicity gives the boundary maximum. Thus the maximum-likelihood estimator isA uniform prior multiplies this likelihood by a constant. The beta-integral normalization therefore gives the posterior distributionIn particular the posterior mean $(k+1)/(n+2)$ is not automatically the optimal estimator for the weighted loss in the next part.

For $k\ne 0$, put $t=k^2>0$ and use the positive-definite Lyapunov function $V=x^2+t{y}^2$. Its derivative isFor $V>0$ set $u=x^2/V\in [0,1]$. The brace divided by $V^2$ is $A{u}^2+2Bu(1-u)+C(1-u{)}^2$, where $A=1$, $B=\beta (1+t)/(2t)$ and $C=1/t$. Since $\beta >2$, $B>A,C$. Completing the square gives its maximumHence $\dot{V}\le -2V+2M{V}^2$. If $V(0)<1/M$, this sublevel region is positively invariant: $V$ decreases, and $\dot{V}\le -2(1-MV(0))V$. The exponential bound makes $V\to 0$. The trajectory is bounded in a compact ellipse, so smoothness of the vector field also guarantees existence for every positive time. ThereforeFor $k=0$ the displayed strict inequality has no solutions; negative $k$ gives the same ellipse as $-k$. Taking the union of the nonempty ellipses is valid because each point needs only one Lyapunov certificate.

At $k=1$ the certified disk has squared radius $2/(\beta +1)$. Along $y=0$, the limiting threshold as $k^2\to \infty$ is $4(\beta -1)/{\beta }^2$, which is strictly larger, since their difference is $2({\beta }^2-2)/({\beta }^2(\beta +1))>0$ for $\beta >2$. Choose $x^2$ strictly between these thresholds and then a sufficiently large finite $k$. This point is certified by that ellipse but lies outside the $k=1$ disk. Thus the union gives a strictly larger certified region. It need not equal the whole basin of attraction.

Put $h(r)=1-\mu /r^2$. For the equatorial geodesic, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}[-h{\dot{t}}^2+h^{-1}{\dot{r}}^2+r^2{\dot{\varphi }}^2]$, where dots denote derivatives with respect to the affine parameter. The cyclic coordinates $t,\varphi$ give constants $p_t=-h\dot{t}=-E$ and $p_{\varphi }=r^2\dot{\varphi }=L$. A third constant is the tangent norm $g_{ab}{\dot{x}}^a{\dot{x}}^b=-\kappa$, preserved by the geodesic equation and metric compatibility. For a timelike geodesic rescale the affine parameter to proper time so $\kappa =1$; for a null geodesic $\kappa =0$.

Substitute $\dot{t}=E/h$, $\dot{\varphi }=L/r^2$ into the norm identity and multiply by $h$. It gives the effective potential equationOutside the horizon $r=\sqrt{\mu }$, its derivative is $V^{\prime }=-(L^2-\kappa \mu )/r^3+2\mu L^2/r^5$. A nontrivial circular null geodesic requires $L\ne 0$, givingFor a circular timelike geodesic,At it $V^{\prime \prime }=-4\mu L^2/r^6<0$, and its radius exceeds $\sqrt{2\mu }$. If $L^2\le \mu$ no finite circular timelike radius exists. Thus all finite circular null and timelike orbits in this five-dimensional exterior are radially unstable. The energy of a circular orbit is fixed by $E^2=2V(r)$, not independently arbitrary.

For comparison, with four-dimensional Schwarzschild horizon radius $r_s$, the photon circle is at $3r_s/2$ and is unstable. Circular timelike orbits exist for $r>3r_s/2$; they are unstable below $3r_s$, marginal at $3r_s$ and stable above $3r_s$. The five-dimensional potential has no corresponding stable outer circular timelike branch.

By the argument principle, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i{)}^{-1}\oint f^{\prime }/fdz$. The logarithmic derivative is periodic with the same lattice, so the integrals over opposite sides cancel. Thus the numbers of zeros and poles are equal.

The Weierstrass elliptic function gives a degree-two map $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ whose fibers identify $z$ and $-z$. An even elliptic function is constant on these fibers, so it descends to a single-valued function $Q$ on the sphere. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto -t$, the Laurent series of $f$ has only even powers, and $\wp -{\wp }_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp \sim t^{-2}$ gives the same conclusion at infinity. A meromorphic function on the Riemann sphere is a rational function, so $\boxed{f=Q(\wp )}$.

A meromorphic antiderivative need not be elliptic. Take $f=\wp$. Its double poles have zero residue, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour integral is a sum of zero residues. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with residue $-1$, contradicting the zero sum of residues in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that even elliptic functions are rational in the Weierstrass function.

A point $P$ belongs to the smooth locus of a variety if its local ring ${\mathcal{O}}_{X,P}$ is a regular local ring, meaningEquivalently, the Zariski tangent space has the expected local dimension. Over the algebraically closed field, this agrees with smoothness for a reduced variety. In affine coordinates, the Jacobian criterion expresses it by the rank of the matrix of first derivatives of defining equations.

The Jacobian criterion on an affine chart $x_j\ne 0$ says that a point of the projective hypersurface is singular exactly when all derivatives of its dehomogenized equation vanish there. The Euler homogeneous function theorem identity ${\sum }_i{x}_i{f}_{x_i}=df$ then recovers the omitted $j$th derivative on $X$, without dividing by $d$. Thus the correct statement in every characteristic isIf the printed $Z(I)$ denotes a zero locus inside $X$, this proves it as written. If it means the ambient projective zero locus, the assertion is false when the characteristic divides $d$. When the characteristic does not divide $d$, the Euler homogeneous function theorem forces $f=0$ at every common zero of the derivatives, and the intersection with $X$ is redundant.

For a genuine counterexample to the unrestricted ambient interpretation, in characteristic $p$ takeThis is an irreducible homogeneous polynomial: it is primitive and linear in $X_2$ over $k[X_0,X_1]$, with coprime coefficients. At $[1:0:0]$ all partial derivatives vanish, including when $p=2$, but $f=1$. This point is outside $X$. Hence the ambient-locus version needs the displayed intersection or a restriction on the degree.

Homogenization gives $F=X^4+X^2{Y}^2+Y^2{Z}^2+Z^4$, withIf the characteristic is not $2$, a singular point with $Y=0$ would force $X=Z=0$, impossible. If $Y\ne 0$, then $X^2+Z^2=0$. If either $X$ or $Z$ vanishes, both vanish, giving $[0:1:0]$. Otherwise the other two derivative equations require $Y^2=-2X^2=-2Z^2$, whence $X^2=Z^2$ and $2X^2=0$, again impossible. ThusIn the chart $Y=1$, the quadratic term is $X^2+Z^2$, the product of two distinct linear factors over $k$, so this is an ordinary double point.

In characteristic $2$,All first derivatives of this equation vanish. The equation defines a nonreduced scheme, singular everywhere; it no longer satisfies the irreducible/reduced hypersurface premise. If “curve” means the associated reduced variety, it is the union of two distinct lines, and only their intersection $\boxed{[1:0:1]}$ is singular. These two interpretations must not be conflated.

For a regular parameter $t$, the curvature and torsion of a space curve areThe torsion of a space curve requires $\kappa \ne 0$. Reparameterize by arc length $s$ and set $T={\alpha }_s$, $N=T_s/\kappa$, $B=T\times N$. Differentiating the pairwise inner products of this orthonormal Frenet frame shows that its derivative matrix is skew symmetric. Since $T_s=\kappa N$, defining $\tau =N_s\cdot B$ gives the Frenet-Serret formulasThe fundamental theorem of regular space curves says that prescribed smooth $\kappa >0$ and smooth $\tau$ on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's linear differential equation with an initial oriented orthonormal frame, then integrate ${\alpha }_s=T$; uniqueness proves the assertion.

Under the intended convention that the shared parameter $s$ is arc length, equality on an open interval gives equal position and equal Frenet frames at one point. Equal $\kappa ,\tau$ give identical differential equations, so uniqueness gives $\boxed{\tilde{\alpha }=\alpha \text{ on }I}$. The printed “regular smooth” hypothesis alone does not establish this convention. Without common arc length, an orientation-preserving smooth change of speed that is the identity on $J$ but not elsewhere reparameterizes a circle into a counterexample: both curves have $\kappa =1$, $\tau =0$ and agree on $J$, but differ elsewhere as parameterized curves.

For an oriented surface with unit normal $n$, define normal curvature ${\kappa }_n=T_s\cdot n$ and geodesic curvature ${\kappa }_g=T_s\cdot (n\times T)$. ThusA unit-speed curve is a geodesic exactly when ${\kappa }_g=0$. If also ${\kappa }_n\equiv 0$, then $T_s=0$, so it is a straight line. A vertical ruling of a circular cylinder is such a geodesic, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence no planar patch is required, and such a geodesic cannot be simple and closed.

Finally let the principal curvatures be ${\kappa }_1,{\kappa }_2$. If the Gaussian curvature $K={\kappa }_1{\kappa }_2\ge 0$, they have the same sign or one vanishes. Euler formula for normal curvature makes ${\kappa }_n$ a weighted average of them. For a geodesic, $\kappa =|{\kappa }_n|$, soA horizontal circle on a cylinder of radius $R$ is a geodesic with $\kappa =1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.

Let $M={\max }_{p\in [0,1]}\mathbb{E}(1+pR{)}^r$ and choose an optimizer $p^{\ast }$. Existence follows from continuity and compactness: the integrands are dominated by a constant times $1+|R|$. Induction in the Bellman equation givesThus the maximal expected yield is $\boxed{M^N-1}$. Neither current wealth nor time changes the maximizing fraction; there may be several optimizers.

The right derivative at $p=0$ is $r\mathbb{E}R$, using the finite first moment and dominated differentiation near zero. If $\mathbb{E}R>0$, increasing $p$ slightly improves on $p=0$, so $\boxed{p^{\ast }>0}$. No explicit value of $p^{\ast }$ is needed.