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Odd terms vanish and direct evaluation gives $L(p)=(8/45)c_4$, so the rule is exact through degree $3$. With $n=3$, the Peano kernel isIts nonnegativity and $\int K=L(x^4)/4!=1/135$ give $|L(f)|\le \Vert f^{(4)}{\Vert }_{\infty }/135$, so $N=135$. Equality holds for a quartic with fourth derivative of constant sign.

Expanding the derivative limit along real and imaginary increments gives $u_x=v_y$ and $u_y=-v_x$. Since $\mathrm{Re}(z{e}^z)=e^x(x\cos y-y\sin y)$, the required function is $f(z)=z{e}^z$, and $f(0)=0$. Two such functions differ by a holomorphic function with zero real part; the open mapping theorem makes it constant, and the value at zero makes that constant zero.

Away from $x=y$ the Newton kernel is harmonic, while integrating its normal derivative over a small sphere gives one, proving ${\nabla }^2\hat{G}=\delta$. Green’s second identity is ${\int }_V(u{\nabla }^{2}v-v{\nabla }^{2}u)={\int }_{\partial V}(u{\partial }_{n}v-v{\partial }_{n}u)$. For the half-space, images give $G(x,y)=-(4\pi |x-y|{)}^{-1}+(4\pi |x-y^{\ast }|{)}^{-1}$. Substitution in the boundary formula yields the Poisson kernel $z/[2\pi ((x_1-y_1{)}^2+(x_2-y_2{)}^2+z^2{)}^{3/2}]$; polar integration against the Gaussian gives exactly the stated one-dimensional integral.

Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.

Write $q=1-p$, $\mu =q-p$, $\sigma =2\sqrt{pq}$, and ${\varphi }_S(x)={\prod }_{j\in S}(x_j-\mu )/\sigma$. The functions ${\varphi }_S$ form the p-biased product measure orthonormal basis, so the Fourier expansion is $f={\sum }_S{\hat{f}}_p(S){\varphi }_S$. The normalized discrete derivative of a Boolean function satisfiesApplying Parseval identity and then exchanging two finite sums gives

The noise operator on the Boolean hypercube acts diagonally on the same basis: $T_{\rho }{\varphi }_S={\rho }^{|S|}{\varphi }_S$. Hence the noise stability isIts derivative isTaking the right-hand value at $\rho =0$ leaves exactly the linear Fourier weight ${\sum }_i{\hat{f}}_p(\{i\}{)}^2$, while taking the left-hand value at $\rho =1$ gives $\mathbf{I}(f)$.

Fix $a>m$ and then $x>a$. By the assumptions on the derivative ${\psi }^{\prime }$, there is a unique ${\theta }_x>0$ with ${\psi }^{\prime }({\theta }_x)=x$. Apply exponential tilting to each summand:Under the product tilted law, the variables remain independent and identically distributed random variables and have mean $x$. Hence the strong law of large numbers implies that, for every $0<\delta <x-a$,Changing measure on this event givesThereforeFirst let $\delta \downarrow 0$ and then $x\downarrow a$. The continuity of a convex function gives the lower bound $-{\psi }^{\ast }(a)$. The endpoint $a=m$ follows by letting $a\downarrow m$, while for $a<m$ the strong law of large numbers makes the probability tend to one. This proves the required lower bound.

Differentiate a germ of a continuation of $F$ to obtain a germ of the continuation of $f$. This defines a surjective cover $R_{\tilde{F}}\to R_F$. Two primitive germs above the same derivative germ differ by an additive period. Translation by periods gives all deck transformations and acts transitively on each fibre, so the cover is regular.

Since $V$ is simply connected, $g$ has a primitive $G_0$. On connected $U\cap V$, $G_0-F$ has zero derivative and is constant. Adjust that constant to obtain $G=F$ on the overlap, hence the required continuation.

For a wave travelling in $+x$,At $x=0$, chooseIt is parallel to $-\mathbf{B}$, so the magnetic force vanishes, while its derivative supplies $q\mathbf{E}/(\gamma m)$. Constant speed then givesThe circular radius is

For a confined system, $\mathcal{G}(t)$ stays bounded, so its long-time averaged derivative is zero. The preceding identities give the virial theorem

With $x=\ell \sin \theta$ and $y=-a{t}^2/2-\ell \cos \theta$, discard a total derivative from the kinetic energy to obtainThus $\ddot{\theta }+(g-a)\sin \theta /\ell =0$. If $a=g$, then $\ddot{\theta }=0$: the pendulum is weightless in the lift and rotates uniformly (or remains fixed).

The third Kramers-Moyal expansion coefficient is $F=\frac{1}{6}{\sum }_r{r}^{3}W=(a-b)/6$. At a first minimum the third derivative has no fixed sign, so the maximum-principle argument fails; finite truncations beyond second order need not preserve positivity.

Set ${\delta }_R=e^{-S}u$ and neglect $S^{\prime \prime }$, $(S^{\prime }{)}^2$, and slow variation relative to $k^2$. Cancelling the first derivative gives $S^{\prime }=\alpha k/(2\mathcal{H})$, hence $S={\int }^{\tau }\alpha k/(2\mathcal{H})d{\tau }^{\prime }$, and $u^{\prime \prime }+(c^2{k}^2/3)u\simeq 0$. Thus acoustic oscillations are exponentially damped, unlike pressureless matter perturbations, which have a growing mode.

An algebraic $x$ is separable when its minimal polynomial has distinct roots. In characteristic zero its derivative cannot vanish. In characteristic $p$, repeatedly write an inseparable minimal polynomial as a polynomial in $X^p$; after finitely many steps the minimal polynomial of $x^{p^n}$ has nonzero derivative and is separable.

An elliptic function is meromorphic and $\Lambda$-periodic. The argument principle on opposite sides of a period parallelogram cancels, so zeros and poles have equal total multiplicity. The Weierstrass elliptic functionis elliptic. Its half-period values $e_i$ are distinct, and comparing poles and leading coefficients proves ${\wp }^{\prime 2}=4{\prod }_i(\wp -e_i)$. The half-period addition formula givesDifferentiate and use the same formula at $z={\omega }_1/4$ to identify $(e_1-e_2)(e_1-e_3)=(\wp ({\omega }_1/4)-e_1{)}^2$, yielding the required derivative ratio.

$R_{\mu \nu \alpha \beta }=-R_{\nu \mu \alpha \beta }=-R_{\mu \nu \beta \alpha }=R_{\alpha \beta \mu \nu }$ and $R_{\mu [\nu \alpha \beta ]}=0$. In normal coordinates at a point the connection vanishes, so the expression reduces to derivatives of the metric and these symmetries are transparent; tensoriality extends them to all coordinates.

The Lie derivative formulation is ${\mathcal{L}}_{V}g=0$. Since $[{\mathcal{L}}_V,{\mathcal{L}}_W]={\mathcal{L}}_{[V,W]}$, two Killing fields give ${\mathcal{L}}_{[V,W]}g=0$, so their commutator is Killing.

For the standard symplectic form, $D_f={\sum }_a(f_{p_a}{\partial }_{q^a}-f_{q^a}{\partial }_{p_a})$; a transformation is canonical when it preserves the symplectic form, equivalently Poisson brackets. The Jacobi identity gives $[D_f,D_g]=D_{\{f,g\}}$ with the stated convention. For $Z^A=e^{-s{D}_f}{z}^A$, differentiate $\{Z^A,Z^B\}$ in $s$; Jacobi makes the derivative the Lie transport of the bracket, preserving the canonical constant brackets at $s=0$. Hence every map in the one-parameter flow is canonical.

For a local parametrization $\varphi :U\subset {\mathbb{R}}^k\to X$, the map $(q,w)\mapsto (\varphi (q),d{\varphi }_{q}w)$ parametrizes $TX$ locally. Its inverse uses the base chart and derivative, and its rank is $2k$. These charts make $TX$ a $2k$-manifold.