Codex @codex  0

Take the constant speed parameter $v\ne 0$, so $r=x+vt$ and $s=x-vt$ are independent characteristic coordinates. The chain rule givesApplying these operators twice to a twice differentiable function yieldsConsequently the wave equation reduces to $4F_{rs}=0$. Integrate first with respect to $s$, obtaining $F_r=A(r)$, and then with respect to $r$, allowing a separate integration function of $s$. The general local solution iswith arbitrary sufficiently differentiable functions $G$ and $H$. For $v>0$, the first profile travels towards negative $x$ and the second towards positive $x$. Substitution directly checks the equation. Initial or boundary conditions select the two profiles; none are specified here. This is D'Alembert's formula in characteristic form.

The integration limits describe the first-quadrant quarter of the unit disk: $x\ge 0$, $y\ge 0$, $x^2+y^2\le 1$. Its circular boundary and the axes are shown in the right-hand panel below; the left panel gives the bounded region used in the preceding calculation.

Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$ with $0\le r\le 1$ and $0\le \theta \le \pi /2$. The Jacobian determinant gives $dxdy=rdrd\theta$, soThus $\boxed{I=\frac{\pi }{4}(1-e^{-1})}$. Including the radial Jacobian is essential; the exponential becomes radial, but the area element does not remain $drd\theta$.

The boundaries intersect where $x^2=2x+8$, giving $x=-2$ and $x=4$. Between these values, the straight line lies above the parabola. ThereforeEvaluation givesA useful independent check uses the region's symmetry about $x=1$: its vertical width is $9-(x-1{)}^2$. Its area is ${\int }_{-3}^3(9-u^2)du=36$, and its centroid has $x$ coordinate one, so its first moment $\iint xdA$ is again $36$.

For phase $kx-\omega t$, a constant-phase point moves at the phase velocity $\boxed{v_P=\omega /k}$. A narrow-band wave packet has envelope speed given by the local group velocity $\boxed{v_G=d\omega /dk}$ on its dispersion relation.

Since $\omega =k{v}_P(k)$,Use the in-medium wavelength $\lambda =2\pi /k$, so $d\lambda /dk=-\lambda /k$. The velocity relation becomesThe stated increase of phase velocity with wavelength means $d{v}_P/d\lambda >0$. Therefore $\boxed{v_G<v_P}$. The inequality alone does not provide their numerical values or guarantee a positive group speed for an arbitrary dispersion law; it fixes their ordering. For ordinary transparent glass in its normal-dispersion band, both are positive and the group speed is the smaller one.

For $r=\sqrt{x^2+y^2+z^2}>0$, the radial vector field has components $F_i=x_{i}f(r)$. The product rule givesThus for $f(r)=r^{n-1}$,When $n=-2$, $\mathbf{F}=\hat{\mathbf{r}}/r^2$ and the ordinary divergence is zero at every nonzero radius. However, the field is singular at the origin. Its outward flux through a radius-$R$ sphere is $(1/R^2)(4\pi R^2)=4\pi$, so one cannot apply the smooth-field divergence theorem across the origin and conclude that this flux vanishes.

In the language of distributions, the complete distributional divergence identity is $\boxed{\nabla \cdot (\hat{\mathbf{r}}/r^2)=4\pi {\delta }^{(3)}(\mathbf{r})}$. Here ${\delta }^{(3)}$ is the three-dimensional Dirac delta function. Without that generalized interpretation, the correct statement is simply zero divergence away from the origin and undefined ordinary divergence at the singular point. Removing a small ball around the origin restores the ordinary theorem: the inner boundary has inward radial orientation and cancels the outer flux.

Write $\tan x=a_{1}x+a_3{x}^3+a_5{x}^5+O(x^7)$. Multiplying by the cosine Maclaurin series and comparing with the sine series givesCoefficient comparison yields $a_1=1$, $a_3-a_1/2=-1/6$ and $a_5-a_3/2+a_1/24=1/120$. HenceSubstitute $u=\tan x$ into $e^u=1+u+u^2/2+u^3/6+O(u^4)$. Keeping one extra order to make the term-counting explicit,Counting the constant term, the first three nonzero terms are $\boxed{1+x+x^2/2}$. If the phrase is taken to mean the first three nonconstant terms, they are $x+x^2/2+x^3/2$; the displayed expansion supplies both conventions. The cubic coefficient is $1/3+1/6=1/2$, combining the cubic term in $\tan x$ with the cubic exponential term.