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Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

The obstacle and wave pattern are stationary, so $\omega =\Omega =0$. Withthe dispersion relation becomesHence $\cos \varphi >0$, so $-\pi /2<\varphi <\pi /2$, and

Differentiating the dispersion relation with respect to the wavevector giveswhere the zero-frequency relation
$\alpha /\sqrt{\kappa }=V\cos \varphi$ was used. Its first component is always negative, so the ray angle $\psi$ lies inMoreover,

Put $q=|\tan \varphi |$. The magnitude of the slope relative to the negative $x_1$-axis isIts derivative isso its global maximum occurs at $q=1/\sqrt{2}$ and equalsTherefore all group-velocity rays lie in the stationary square-root-dispersion wakea wedge of semi-angle ${\tan }^{-1}(2^{-3/2})$ extending in the negative $x_1$-direction.

A function $f:{\mathbb{R}}^m\to {\mathbb{R}}^r$ is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^r$ such thatThe map $L$, which is unique, is the derivative $f^{\prime }(x)$.

Expanding in a matrix increment $H$ givesTherefore $p$ is differentiable everywhere and

The inverse function theorem states that if $f$ is continuously differentiable near $a$ and $f^{\prime }(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivativeIn the normalized case $f(0)=0$ and $f^{\prime }(0)=I$, continuity of $f^{\prime }$ gives a closed ball ${\overline{B}}_r(0)$ on whichFor small $y$, define $T_y(x)=x-f(x)+y$. The mean value inequality makes $T_y$ a contraction mapping with constant $q$. If $\Vert y\Vert \le (1-q)r$, thenso $T_y$ maps the ball into itself. The contraction mapping theorem gives a unique fixed point $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,so $\Vert x_y-x_z\Vert \le \Vert y-z\Vert /(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.

For the polynomial map in the question,The derivative is invertible, so the inverse function theorem supplies some $ϵ>0$ and a continuously differentiable $q:D_{ϵ}(I)\to {\mathcal{M}}_n$ such thatThere can be no inverse on all of ${\mathcal{M}}_n$, becauseso $p$ is not injective.

The one-dimensional Time-dependent Schrödinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

At a positive equilibrium, division by $N$ givesSince $b>d$, this has the positive solution

Write the vector field as $F(N)=N(b{e}^{-aN}-d)$. Its derivative isAt $N=N^{\ast }$, the first two terms cancel and $b{e}^{-a{N}^{\ast }}=d$, soThe one-dimensional linear stability analysis therefore shows that $N^{\ast }$ is locally asymptotically stable.

Define the smooth mapAt $(M,v)$ its derivative in the direction $(A,w)$ isThe restriction to variations $(0,w)$ is the surjective linear map $M:{\mathbb{R}}^n\to {\mathbb{R}}^m$, since $M$ has full row rank. Thus $0$ is a regular value of $\Phi$. The regular level set theorem now shows thatis an embedded submanifold of $X_{m,n}\times {\mathbb{R}}^n$, of dimension $mn+n-m$.

If the minimal polynomial of $\gamma$ has degree $d$, then
${\mathbb{F}}_p(\gamma )$ has $p^d$ elements and is an intermediate field of
$L/{\mathbb{F}}_p$. The tower law for field extensions givesso $d\mid n$.

The multiplicative group of a finite field is cyclic. Choose a generator
$\gamma$ of $L^{\times }$, of order $p^n-1$. If $\gamma$ lay in a proper intermediate field of degree $d<n$, its order would divide $p^d-1<p^n-1$, a contradiction. Thus
${\mathbb{F}}_p(\gamma )=L$ and its minimal polynomial has degree $n$.

For arbitrary $r\ge 1$, let $E$ be the splitting field of
$X^{p^r}-X$ over ${\mathbb{F}}_p$. Its roots form a field: the Frobenius endomorphism shows they are closed under addition, subtraction, multiplication, and inversion. The derivative is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose minimal polynomial over ${\mathbb{F}}_p$ has degree $r$, proving that an irreducible polynomial of every positive degree exists.

No. The Darboux theorem says that every derivative has the intermediate-value property. Choose $ϵ<1/3$. In a deleted neighborhood of $a$, the assumed limit puts $\varphi (x)$ within $ϵ$ of $\varphi (a)+1$. Darboux's theorem applied between $a$ and any such $x$ would require values near $\varphi (a)+1/2$, which that neighborhood excludes. This contradiction shows that $\varphi$ cannot be a derivative.

The probability generating function isTermwise differentiation givesThe coefficients are nonnegative, and $\mathbb{P}(X\ge 2)>0$, so both derivatives are positive and nondecreasing on $(0,1]$.

The derivative of a distribution is defined by integration by parts. Therefore

The measure, Gaussian factor, $r^{-1}$, coordinates, and derivatives all transform covariantly under rotations, so $T_{ijkl}$ is an isotropic tensor integral. Symmetry gives $\beta =\gamma$.

Contracting $i=j$ and using ${\nabla }^2(1/r)=0$ away from the origin givesNext contract $i=k$ and $j=l$. Sinceone obtainsSolving yields

Using the product rule for directional derivatives,Thus $f=|u{|}^2/2-P$ and $g=-h$. If $\nabla \cdot u=0$ and $w=\nabla \times u$, then $\nabla \cdot w=0$. Integrating the two directional derivatives as divergences gives only boundary fluxes, which vanish because $u\cdot n=w\cdot n=0$. This is helicity conservation by tangent boundary conditions, so $dH/dt=0$.

For the stated field, direct use of the cylindrical curl gives $u\cdot w=-2a{\rho }^4{\sin }^{2}z$. Therefore

Assume Newton's second law, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summingcancels internal pairs and gives $dP/dt=F:={\sum }_i{F}_i$. Taking moments about fixed $a$ cancels the internal torques pairwise and gives

The angular momentum about the centre of mass result follows similarly: differentiating ${\sum }_i(r_i-R)\times m_i({\dot{r}}_i-\dot{R})$ introduces no extra term because both total relative position weighted by mass and total relative momentum vanish. Thus its derivative is the external torque about $R$.

Finally, if every mass is $m$ and $F_i=-k{\dot{r}}_i$, then about a fixed pointTherefore

A subset $S\subset {\mathbb{R}}^3$ is a smooth surface if every point has a neighbourhood in $S$ parametrized by a map $X:U\to S$, where $U\subset {\mathbb{R}}^2$ is open, $X$ is a homeomorphism onto that neighbourhood, $X$ is smooth, and $DX$ has rank two everywhere. These are the embedded surface parametrization conditions.

For the given set use local angular intervals in the parametrizationIt is locally one-to-one and has tangent vectorsTheir cross product has magnitudeso the derivative has rank two. The angular charts cover $\Sigma$, proving that it is a smooth surface.

The area between heights $a_0$ and $b_0$ is thereforeBy hypothesis this equals $2\pi r(b_0-a_0)$ for every subinterval. Since the integrand is continuous,and squaring gives

If $0<\varphi (t)<r$, then ${\varphi }^{\prime }$ never vanishes, so its sign $\sigma \in \{1,-1\}$ is constant. The last equation givesand henceThus $\sqrt{r^2-\varphi (t{)}^2}=-\sigma (t-t_0)$ for a constant $t_0$, andThe graph lies on a circle of radius $r$, exactly as described by constant strip-area density of a surface of revolution.

The functionis continuous, satisfies the equation away from $\tilde{t}=y_0$, and has unequal one-sided derivatives there.

It is unique in the stated class. While a solution is positive its derivative is $-1$, so continuity forces it to follow $y_0-t$ until it first reaches zero at $t=y_0$. It cannot subsequently make a positive excursion: on any connected interval $(a,b)$ where $y>0$ and $y(a)=0$, integration of $y^{\prime }=-1$ gives $y(t)=-(t-a)<0$. A negative excursion similarly contradicts $y^{\prime }=1$. Hence the solution remains zero. This is the sign-decay differential equation.

For any $h$,If the normal equation $A^{T}A{c}^{\ast }=A^{T}y$ holds, the last two terms reduce to $\Vert Ah{\Vert }^2\ge 0$, so $c^{\ast }$ minimizes. Conversely, at a minimizer the directional derivative in every $h$ vanishes, forcing $A^T(A{c}^{\ast }-y)=0$.

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation