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If $m\ge 1$ is an integer and $\mathbb{E}[|X{|}^m]<\infty$, the characteristic function has continuous derivatives through order $m$, withThus finite absolute moments justify recovering moments from derivatives.^[ref]

For a short justification of differentiation under the integral sign, the difference quotient for the $(k-1)$st derivative has integrand $(iX{)}^{k-1}{e}^{itX}(e^{ihX}-1)/h$. Its absolute value is at most $|X{|}^k$, since $|e^{iu}-1|\le |u|$ for real $u$. Lower absolute moments are finite because $|X{|}^k\le 1+|X{|}^m$. Apply the dominated convergence theorem successively, then use the same integrable bound to obtain continuity of each derivative.

For the Bernoulli distribution with parameter $1/3$, direct differentiation gives ${\phi }_B^{\prime }(0)=i/3$ and ${\phi }_B^{\prime \prime }(0)=-1/3$, agreeing with $\mathbb{E}[B]=\mathbb{E}[B^2]=1/3$. Its variance is therefore $1/3-(1/3{)}^2=2/9$. Existence and continuity of a characteristic function alone do not authorize these moment calculations: the standard Cauchy distribution, whose expected value is undefined, still has a characteristic function. The stated finite absolute moment hypothesis is a sufficient condition for the derivative formula, not an assumption built into the definition of a characteristic function.

An orthogonal change of coordinates relates two orthonormal bases of a real inner product space. With the new basis vectors as columns of an orthogonal matrix $Q$, old and new coordinate columns obey $x=Qy$. Then $x^{T}x=y^{T}y$, so Euclidean norm, Euclidean distance and angles are preserved. A quadratic form $x^{T}Ax$ becomes $y^T(Q^{T}AQ)y$. For a symmetric matrix, orthogonal diagonalization chooses $Q$ so that the transformed quadratic form is a sum of eigenvalues times squared coordinates. This explains the principal-axis descriptions of a quadric surface.

For a nondegenerate $d$-dimensional simplex $S$ with vertices $v_0,\dots ,v_d$ and an affine function $f$, the integral equals the volume times the value at the centroid $c=(v_0+\cdots +v_d)/(d+1)$. To prove this, write $x={\sum }_i{\lambda }_i(x)v_i$ in a barycentric coordinate system. On the standard simplex, symmetry permuting the vertices gives equal integrals of all ${\lambda }_i$; since ${\sum }_i{\lambda }_i=1$, each is $\mathrm{vol}(S)/(d+1)$. An invertible affine map transfers this result to any simplex. Affineness gives $f(x)={\sum }_i{\lambda }_i(x)f(v_i)$, so linearity of integration proves the formula. For a tetrahedron the factor is one quarter. The rule is exact for affine integrands; a nonlinear integrand generally needs additional moments.

Consider two equal balls of mass $m$, four light rods each of length $L$, a fixed upper pivot and a light sliding lower collar. Each upper and lower rod makes angle $\theta$ with the spindle. A spring of spring constant $k$ is uncompressed at $\theta =0$ and is compressed by collar motion. Its compression is $\delta =2L(1-\cos \theta )$. Collar balance gives $2T_l\cos \theta =k\delta$, while vertical ball balance gives $(T_u-T_l)\cos \theta =mg$. Radial Newton's second law gives $(T_u+T_l)\sin \theta =m{\omega }^{2}L\sin \theta$. Eliminating the rod tensions proves the displayed relation for nonzero $\theta$.

For total ball mass $m$, arm length $l$ and imposed constant shaft speed $\Omega$, the reduced effective potential is $V_{\mathrm{eff}}/(m{l}^2)=-{\Omega }^2{\sin }^2\theta /2-{\omega }_0^2\cos \theta$. Its equilibrium equation is $\sin \theta ({\omega }_0^2-{\Omega }^2\cos \theta )=0$. The downward state is stable for $|\Omega |<{\omega }_0$ and loses linear stability above it; nonzero stable openings then satisfy the displayed relation. Extending the angle to signed values gives a supercritical pitchfork bifurcation. Squared angular frequencies of small oscillations are ${\omega }_0^2-{\Omega }^2$ at the downward state and ${\Omega }^2-{\omega }_0^4/{\Omega }^2$ at the open states. At threshold the quadratic term vanishes but the quartic potential is positive, so zero linear frequency does not imply finite-amplitude free motion.

For $a>b>0$ with $a,b\ne 1$, the solution of $(D_t+a)(D_t+b)y=e^{-|t|}$ that decays at both infinities isIts Fourier transform is $2/[(1+k^2)(a+ik)(b+ik)]$. The lower half-plane contains only the pole $-i$, while the upper half-plane contains $i,ia,ib$. Inverting by the residue theorem gives the two expressions. Direct substitution verifies the equation separately on each side, and $y,y^{\prime }$ match at zero, so there is no spurious impulsive source. The negative-time response is nonzero because the forcing already acts before zero, even though the Green function is causal. Coinciding rates $a=1$ or $b=1$ require limits with polynomial times exponential terms.

For $x^{\prime \prime }+2\gamma x^{\prime }+({\gamma }^2+{\mu }^2)x=g(t)$ with $\gamma ,\mu >0$, zero initial data give $x(t)={\mu }^{-1}{\int }_0^t{e}^{-\gamma (t-s)}\sin (\mu (t-s))g(s)ds$. If $|g(s)|\le C{e}^{-as}$ for $a>0$, the causal Green function of a damped oscillator givesBoth bounds tend to zero. A decaying homogeneous transient can be added for arbitrary finite initial data. The coincident decay rates require the extra polynomial factor and cannot be handled by dividing by zero in the first expression.

For $\beta >0$, $0<\gamma <{\omega }_0$ and zero prehistory, the equation $y^{\prime \prime }+2\gamma y^{\prime }+{\omega }_0^{2}y=F{e}^{-\beta t}H(t)$ hasOne obtains it by convolution with the causal Green function of a damped oscillator. Alternatively, a particular exponential has amplitude $F/[(\beta -\gamma {)}^2+{\Omega }^2]$; the two homogeneous terms enforce $y(0)=y^{\prime }(0)=0$. These continuity conditions avoid adding impulsive forcing at switch-on. The forcing may have either sign; its Fourier transform is $-iF/(\omega -i\beta )$ with the negative-exponential transform convention.

With inverse Fourier transform $G(t)=\int G(\omega )e^{-i\omega t}d\omega /(2\pi )$, a tempered retarded response vanishes for $t<0$ and is analytic in the upper frequency half-plane with suitable large-frequency growth bounds. Poles strictly below the real axis describe decaying modes; upper-half-plane poles obstruct a bounded retarded real-axis Fourier response. However causality alone does not imply stability: a retarded exponentially growing Green function can exist without an ordinary Fourier transform. Its Laplace transform inversion contour must pass above the growth poles in the frequency plane. Moving the contour to the real axis changes the boundary condition and can introduce a pre-response.

Suppose $q(t)=A/(t-c)+r(t)$, where $r$ is continuous, the pole $c$ is fixed, and differentiable endpoints obey $a(x)<c<b(x)$ locally. ThenApply the chain rule and the fundamental theorem of calculus to getThe same boundary expression as in the Leibniz integral rule thus applies to this explicitly specified Cauchy principal value. If $A\ne 0$, the two ordinary one-sided improper integrals diverge, so their sum is not an ordinary integral. A finite boundary derivative alone does not make that divergent integral meaningful.

Among curves of fixed arc length $L>2a$ joining $(-a,0)$ to $(a,0)$, the hanging catenary globally minimizes $\int yds$. Parametrize a competitor by arc length $s\in [-L/2,L/2]$, so $x^{\prime 2}+y^{\prime 2}=1$. Integration by parts and the fixed horizontal endpoint separation give, for every $b>0$,The inequality is the Cauchy-Schwarz inequality. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$. Integration gives the catenaryThe last equation has a unique positive $b$, since $\sinh q/q$ increases strictly for $q>0$. The equality curve is a single-valued graph below the chord, so the same global bound proves the result under those additional restrictions.

Let $\mathbf{p}\ne 0$ and choose an auxiliary vector $\mathbf{q}$ with $\mathbf{p}\times \mathbf{q}\ne 0$. Its component perpendicular to $\mathbf{p}$ is ${\mathbf{q}}_{\perp }=\mathbf{q}-(\mathbf{p}\cdot \mathbf{q})\mathbf{p}/|\mathbf{p}{|}^2$. Then all solutions of the Cartesian equation of a plane $\mathbf{x}\cdot \mathbf{p}=k$ areThe two free directions are nonzero, perpendicular to each other and perpendicular to $\mathbf{p}$, so they span its two-dimensional orthogonal complement. This proves completeness as well as direct substitution. If $\mathbf{p}\cdot \mathbf{q}=0$, then ${\mathbf{q}}_{\perp }=\mathbf{q}$. If the dot product is nonzero, another complete form is $\mathbf{x}=\alpha \mathbf{p}+[k-\alpha |\mathbf{p}{|}^2]\mathbf{q}/(\mathbf{p}\cdot \mathbf{q})+\beta (\mathbf{p}\times \mathbf{q})$.

The point assigned all-zero coordinates in a chosen Cartesian coordinate system. Choosing an origin identifies position displacements with vectors; a bare affine space has no distinguished origin.

For an axisymmetric magnetic vector potential, the conserved angular canonical momentum is $p_{\theta }=\gamma m{\rho }^2\dot{\theta }+q\rho A_{\theta }$. A particle arriving parallel to a solenoid axis from a field-free region has $p_{\theta }=0$. In a uniform field, $A_{\theta }=B\rho /2$ and $\dot{\theta }=-qB/(2\gamma m)$ wherever $\rho \ne 0$. The mechanical angular momentum changes through the fringe while the canonical quantity remains constant. Static magnetic fields also conserve $\gamma m{c}^2$; the azimuthal kick therefore reduces axial kinetic motion rather than increasing total speed.

On a common invariant domain for the operator products, let the position operator and momentum operator obey $[\hat{x},\hat{p}]=i\hslash I$. Then $\hat{x}\hat{p}=\hat{p}\hat{x}+i\hslash I$, so both displayed expressions equal $(\hat{p}\hat{x}{)}^2+i\hslash \hat{p}\hat{x}$. In position representation both act on a smooth function $f$ as $-{\hslash }^2(x^2{f}^{\prime \prime }+2x{f}^{\prime })$. This particular equality does not imply that arbitrary classical reorderings remain equivalent in quantum mechanics.

Putting $M=m$ in the diatomic-chain dispersion relation with equal springs gives ${\omega }_{-}=2\sqrt{K/m}|\sin (kd/2)|$ and ${\omega }_{+}=2\sqrt{K/m}|\cos (kd/2)|$ in the reduced zone $|k|\le \pi /(2d)$. The branches touch at its edge. If the sites are actually identical, the true primitive unit cell has length $d$; the apparent optical phonon branch is the outer part of the monatomic-chain dispersion relation folded by an unnecessarily doubled cell. A gap does not appear merely from redrawing the zone.

Measure $\theta \in [-\pi /2,\pi /2]$ from the vertically downward direction on the lower half of a cylinder of radius $R$, and let $z$ be axial distance. From rest at an edge, conservation of energy gives speed $\sqrt{2gR\cos \theta }$. The travel time is $T=(2gR{)}^{-1/2}\int \sqrt{R^2+z^{\prime 2}}/\sqrt{\cos \theta }d\theta$. Since $z$ is a cyclic coordinate, its conjugate quantity $z^{\prime }/[\sqrt{\cos \theta }\sqrt{R^2+z^{\prime 2}}]=K$ is constant, giving the displayed formula. For finite positive axial separation, $0<K<1$ is fixed uniquely by integrating $z^{\prime }$ between the edges: the separation grows strictly from zero to infinity as $K$ increases from zero to one. The time integrand is strictly convex in $z^{\prime }$, so its supporting-line inequality proves global minimality for the fixed endpoints. As the starting edge is approached, $z^{\prime }\to 0$: the initial direction is directly down the curved surface, with no axial component. This differs from the constant-pitch shortest geodesic on the unrolled cylinder.

For a real scalar lattice-periodic potential, let $|n⟩$ be a nondegenerate zone-centre periodic eigenstate with energy $E_n$. Conjugation by $e^{i\mathbf{k}\cdot \mathbf{r}}$ gives $H(\mathbf{k})=H(0)+\hslash \mathbf{k}\cdot \mathbf{p}/m+{\hslash }^2{k}^2/(2m)$. The zone-centre state can be chosen real, so $⟨n|\mathbf{p}|n⟩=0$. Second-order nondegenerate perturbation theory then givesTime-reversal symmetry makes this isolated spinless band even in $\mathbf{k}$. In a direction $\mathbf{e}$, the curvature givesA degeneracy requires a matrix treatment before extracting curvatures; a nonlocal potential or spin-orbit terms can also alter the simple momentum vertex.