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A subset $X\subset {\mathbb{R}}^N$ is a smooth $k$-manifold if every $x\in X$ has a neighborhood parametrized by a smooth map $\varphi :U\subset {\mathbb{R}}^k\to X$ that is a homeomorphism onto its image and has derivative of rank $k$. DefineIf $\psi$ is another parametrization, the transition maps satisfyThe transition derivative is invertible, so the two images coincide. Thus the tangent space is independent of the parametrization.

A vector field $V=V_1{\partial }_x+\varphi {\partial }_u$ is a Lie symmetry whenon the equation manifold $\Delta =0$. With total derivative $D_x$, the prolongation coefficients obeyFor $V_1=f(x)$, direct iteration givesThus $\alpha =\beta =3$.

For $u^{\prime \prime \prime }=u^{-3}$, invariance requires ${\varphi }_3+3u^{-4}\varphi =0$ after substituting $u^{\prime \prime \prime }=u^{-3}$. Equating coefficients of the independent jet variables givesand the remaining terms force $\varphi =(3a/4)u$ and $f=ax+b$. Hence the Lie algebra is spanned by

Let $F$ be the splitting field over ${\mathbb{F}}_p$ of $X^q-X$. Its derivative is $-1$, so it has $q$ distinct roots. The roots are closed under addition, subtraction, multiplication, and inversion, using $(x+y{)}^q=x^q+y^q$ and $(xy{)}^q=x^q{y}^q$. They therefore form a field with $q$ elements.

Every field with $q$ elements has multiplicative group of order $q-1$, so every element satisfies $x^q=x$; it is therefore a splitting field of the same polynomial and is unique up to isomorphism. If an irreducible factor over ${\mathbb{F}}_p$ has a root $\alpha \in {\mathbb{F}}_q$, its degree is $[{\mathbb{F}}_p(\alpha ):{\mathbb{F}}_p]$, which divides $d$ by the tower law. In particular it is at most $d$.

Taking Fourier transforms givesThusThe multiplier and all its polynomially weighted derivatives have at most polynomial growth, so this defines a tempered distribution. In fact it maps $H^s$ continuously into $H^{s+4}$. Since the multiplier never vanishes, any homogeneous tempered solution has zero Fourier transform; uniqueness follows.

A compactly supported smooth function belongs to $H^s({\mathbb{R}}^n)$ for every real $s$. Part (a) therefore puts $u$ in $H^{s+4}$ for every $s$. Taking $s$ arbitrarily large and applying the Sobolev embedding theorem to every derivative shows that $u\in C^{\infty }({\mathbb{R}}^n)$.

Orthogonality and $\Vert \tilde{\varphi }(x)\Vert =\Vert x\Vert$ giveso ${\psi }_{\epsilon }$ maps each boundary sphere to itself. AlsoOn compact $A$, this derivative is uniformly close to the identity; for sufficiently small $\epsilon$ its determinant is positive and it is locally invertible. The map is properly homotopic to the identity and has degree one, so the local diffeomorphism has one sheet and is a global diffeomorphism $A\to A$.

At a first-order transition the Gibbs free energy is continuous but a first derivative such as entropy or volume jumps; there is latent heat. At a second-order transition the first derivatives are continuous while a second derivative, such as heat capacity or susceptibility, is discontinuous or divergent, and there is no latent heat.

At the critical point, $(\partial p/\partial v{)}_T=({\partial }^{2}p/\partial v^2{)}_T=0$. The first logarithmic derivative givesAt a stationary point the second-derivative condition reduces toCombining the equations gives $v_c=2b$, and substitution then gives

An allowable parametrisation is a smooth homeomorphism from an open subset of ${\mathbb{R}}^2$ onto an open subset of $S$, with derivative of rank two. Away from the axis, the rotation orbit has nonzero tangent. A transverse curve supplied by the submanifold theorem, followed by the rotation action, giveswith $|v|<\pi$ and $(f^{\prime },g^{\prime })\ne (0,0)$.

For a ruled parametrisation, regularity is exactly

Rotate and translate along the axis so the specified ruling isHere $s\ne 0$ because the line is not parallel to the axis, and $c\ne 0$: if $c=0$, rotating the horizontal tangent line makes the ruled parametrisation singular at its closest point. Rotating $L$ givesa one-sheet hyperboloid. Rotation invariance puts this whole surface in $\Sigma$. Connectedness and the fact that a complete embedded hyperboloid cannot be a proper subset of another connected embedded surface force equality. Rescaling radial and axial coordinates gives a diffeomorphism with $x^2+y^2=1+z^2$.

Write $E=-{\hslash }^2{\kappa }^2/(2m)$. An even bound state is proportional to $\cos (kx)$ inside the well and to $e^{-\kappa |x|}$ outside, whereContinuity of the logarithmic derivative at $a$ givesFor sufficiently small $a$, $ka$ lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.

As $a\downarrow 0$, $\kappa \sim k^{2}a\to m/{\hslash }^2$. HenceThe wells converge distributionally to $V_0(x)=-\delta (x)$, whose normalized even bound state isIts derivative jump is precisely the one imposed by the delta potential.

Apply the residue theorem to $f(z)/(z-w{)}^{n+1}$; its residue at $w$ is $f^{(n)}(w)/n!$, which proves Cauchy's derivative formula.

For a compact subdisc choose a slightly larger contour still inside $D(a,R)$. Cauchy's derivative formula bounds the supremum of $g_k^{(n)}-g^{(n)}$ on the smaller disc by a fixed constant times the supremum of $g_k-g$ on the contour. Local uniform convergence therefore passes to every derivative.

On $|z-a|=\epsilon$, $g$ has a positive minimum modulus. For large $k$, $|g_k-g|<|g|$ there. Rouché's theorem then says $g_k$ and $g$ have the same number of zeros inside, namely $m$, counting multiplicity.

Here $L_3=-i\hslash (x_1{\partial }_{x_2}-x_2{\partial }_{x_1})$. The radial factor and $x_3^n$ are annihilated by the angular derivative, whileThe product rule therefore givesReplacing $i$ by $-i$ gives eigenvalue $-m\hslash$ for $(x_1-i{x}_2{)}^m{x}_3^{n}f(r)$.

In a parallel-plate rheometer, a point at radius $r$ on the upper plate moves at speed $\Omega r$. The thin-gap approximation therefore gives the local shear rateDefine the rim valueAn annulus of radius $r$ and width $dr$ has area $2\pi rdr$; its tangential force is $\tau (r)2\pi rdr$, and its moment arm is $r$. The measured torque is consequently

For a generalized Newtonian fluid, $\tau =\eta (\dot{\gamma })\dot{\gamma }$. Changing the integration variable from $r$ to $\dot{\gamma }$ givesThus the requested kernel isMultiplying by ${\dot{\gamma }}_R^3$ and taking a derivative with respect to ${\dot{\gamma }}_R$ turns the upper-limit contribution of the integral into the rim viscosity:ThereforeA measured torque curve and its slope hence determine $\eta$ over the range of imposed rim shear rates.

The experimental law impliesSubstitution givesThe graph is a decreasing rectangular hyperbola with a positive high-rate plateau $2/(\pi \alpha R^3)$ and a $1/\dot{\gamma }$ divergence at the origin. Equivalently,so the inferred material is a Bingham plastic with yield stress $3T_c/(2\pi R^3)$.

A critical point of a smooth map is a point where its derivative is not surjective; its image is a critical value. A regular value is a target value all of whose preimages have surjective derivative.

For $f(X)=X^{T}AX$ with $A=A^T$,Thus $X$ is critical exactly when $AX=0$, and every critical point has value zero. Hence zero is the only possible critical value; it is a critical value precisely when $A$ is singular. If $A$ is singular, every point of its kernel is critical, so there can be infinitely many critical points; if $A$ is invertible, only the origin is critical.