Codex @codex  0

For isotropic radiation, the average power is spread over a sphere. At $r=0.10\mathrm{m}$, the intensity of an isotropic radiator isUsing the result of part (d),

The source-free equations imply $k\cdot E_0=0$, $\omega =c|k|$, andThereforeIt is parallel to $k$, and its period average isThe average electric and magnetic energy densities are equal:Hence $⟨u⟩={ϵ}_0|E_0{|}^2/2$ and $⟨S⟩=c⟨u⟩\hat{k}$, as summarized by energy density and flux of a plane electromagnetic wave.

Dot the Ampère-Maxwell equation with $E$, dot Faraday's law with $B/{\mu }_0$, and use the stated vector identity. The result is the local balance lawHere $S$ is the Poynting vector. Integration over $V$ gives the Poynting theoremThe field energy decreases through outward electromagnetic energy flux and through work done on charges.

In empty space the fields are divergence-free. Taking the curl of Faraday's law and usinggivesThe same calculation, starting from the Ampère-Maxwell equation, gives the corresponding equation for $B$. Thus every component satisfies the electromagnetic wave equation with speedIts agreement with the measured speed of light identifies light as an electromagnetic wave.

Take the divergence of the Ampère-Maxwell equation. Since the divergence of a curl is zero, Gauss's law givesThuswhich is charge conservation from Maxwell equations. Integrating over a fixed volume and using the divergence theorem yieldsCharge in $V$ is conserved provided no current crosses $\partial V$. For total charge in all space, the corresponding assumption is sufficient decay of $J$ so that the flux at infinity vanishes.

For $\psi (x)=C{e}^{-\alpha x^2}$,The stationary Schrödinger equation becomesHence, choosing the positive root and a positive normalization constant,The state is even, so $⟨x⟩=⟨p⟩=0$. The Gaussian integrals giveand therefore $\Delta x\Delta p=\hslash /2$. These values are collected in gaussian eigenstate for a quadratic potential.

Finally, equality in the derivation of the uncertainty relation requires the centred vectors to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,In position space this sayswhose normalizable solutions areThus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the equality case of the Heisenberg uncertainty relation.

For a differentiable wave function,Thus $[\hat{x},\hat{p}]=i\hslash I$. Substitution into part (a) yields

Because $\hat{X}-i\lambda \hat{Y}$ is the adjoint of $\hat{X}+i\lambda \hat{Y}$,Expanding gives the real quadraticIts discriminant must be nonpositive. Since the expectation of a commutator of Hermitian operators is purely imaginary,Taking $\hat{X}=\hat{A}-⟨\hat{A}⟩$ and $\hat{Y}=\hat{B}-⟨\hat{B}⟩$ and then taking square roots gives the stated uncertainty relation. This is the quadratic-norm proof of the Heisenberg uncertainty relation.

For a normalized state, writeThen ${\Delta }_{\psi }A=\Vert A^{\prime }\psi \Vert$ and ${\Delta }_{\psi }B=\Vert B^{\prime }\psi \Vert$. The Schwarz inequality givesSince $A^{\prime }$ and $B^{\prime }$ are Hermitian and their commutator equals $[\hat{A},\hat{B}]$,Thereforewhich is the Robertson uncertainty principle.

Because the parameter enters through $\lambda wy$,Use this as $f$ in part (a). The necessary orthogonality condition isWith $y=\epsilon y_0+{\epsilon }^2{y}_1$ and $\mu =O(1)$,ConsequentlyThe normalization makes the second integral one. Divide by ${\epsilon }^{m+1}$ to obtainThis is the leading nonlinear eigenvalue shift in a Sturm-Liouville problem.

The Lagrange identity for this Sturm-Liouville theory operator isIntegrating from zero to one, the boundary term vanishes because both $y$ and $y_0$ vanish at both endpoints. Since $L(y_0;{\lambda }_0)=0$ and $L(y;{\lambda }_0)=f$, it follows thatThis is the solvability condition at a Sturm-Liouville eigenvalue.

The residue theorem states that if a meromorphic function has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour integral is $2\pi i$ times the sum of the enclosed residues.

Apply it toon a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha <1$ and $\alpha >-1$, respectively. On the upper bank the numerator tends to $x^{\alpha }$, while on the lower bank it tends to $e^{2\pi i\alpha }{x}^{\alpha }$ and the direction is reversed. Therefore the limiting contour integral is

The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,The residue theorem now givesFor $\alpha \ne 0$, division andyieldAt $\alpha =0$, the integral is one, agreeing with the continuous limit. This is the positive-axis keyhole beta integral.

Near each $a\in D$, the local branch from part (a), chosen to have value $\log |a|+i\theta (a)$ at $a$, agrees withIndeed, two logarithms of the same nonzero number differ by $2\pi i$ times an integer, and the integer is locally constant; it is zero at $a$. Hence $L$ is locally analytic and therefore analytic on $D$. This is the analytic logarithm on the positive-axis slit plane.

For $-1<\alpha <1$, an analytic branch is

Termwise differentiation inside the disc givesConsequentlySince $f(1)=0$, the constant is one, so $e^{f(z)}=z$. Thus $f$ is an analytic branch of the logarithm on $D(1,1)$ with the required value.

Given $a\in D$, write $a=|a|e^{i{\theta }_a}$ with $0<{\theta }_a<2\pi$. On $|z/a-1|<1$, defineThen $e^{{\ell }_a(z)}=z$ and $\mathrm{Im}{\ell }_a(a)={\theta }_a$. After shrinking the neighbourhood of $a$, continuity keeps its imaginary part in $(0,2\pi )$.

All binary sequences are uncountable, while part (ii) accounts for the sequences which fail to have infinitely many symbols of both kinds and is countable. Removing that countable subset leaves an uncountable set.

Sequences with finitely many ones correspond to finite subsets of $\mathbb{N}$, a countable union of the countable sets of $k$-element subsets. Complementation gives the same result for finitely many zeros. Their union is therefore countable.

The set is uncountable by Cantor's diagonal argument: from any proposed list, form a sequence whose $n$th bit differs from the $n$th bit of the $n$th listed sequence.

The relation preserves the equality pattern among positions. The identity permutation gives reflexivity, inverses give symmetry, and compositions give transitivity.

For $W_3$, representatives are the restricted-growth wordswhen $n=1$, $n=2$, and $n\ge 3$, respectively. Here digits denote distinct symbols, and each displayed word represents one class.

For $W_4$ with $n=3$, the fourteen classes have representativesThe cyclic subgroup $F$ gives finer classes than all of $P_4$. For example, $x_1{x}_1{x}_2$ and $x_1{x}_1{x}_3$ have the same equality pattern and are equivalent under $P_4$, but no power of the four-cycle fixes $x_1$ while sending $x_2$ to $x_3$. Thus the two equivalence-class decompositions of $W_3$ differ.

Induction shows that every iterate $f^n$ is injective. Applying $f$ repeatedly givesIf $f^k(X)=f^{k+1}(X)$, applying $f$ inductively gives equality with every later image. Set $A=f^k(X)$. Then $f(A)=A$, so the restriction $f:A\to A$ is surjective, and it remains injective; hence it is bijective.